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I

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Active LearningIn

Chemistry Education

"ALICE"

Copyright 1997, A.J.Girondi 505 Latshmere Drive Harrisburg, PA 17109 [email protected]/Athens/Oracle/2041

Welcome to ALICE! You are about tobegin a course which will probably beunlike anything you have experiencedthus far in your formal education. ALICEis a unique program in which you willstudy chemistry in small groups. Yourgroup will be able to progress faster orslower than other groups (to a limitedextent). ALICE does not make use of aregular textbook. Instead, you will begiven printed pages which you will placeinto your notebook. These pages includereading material, laboratory activities, andworksheets. By the end of the schoolyear, you will be leaving the course withyour own completed copy of ALICE.

Your instructor’s role will not be typical.There will not be many long lectures. Theinstructor will not be the center of attentionin ALICE. You will be! As a student inALICE, you will be given a great deal ofindependence. A typical class period willhave several lab groups working on avariety of different activities, while othergroups will be completing worksheets orreading some text material. Others may

be talking to the instructor or witnessing ademonstration. Not everyone will bedoing the same thing at the same time!ALICE is an example of active learningwhich means that the student is almostalways actively involved in what is goingon. This program is constantly beingupdated and revised based on commentsand suggestions from students like you!

Welcome

ALICEis taught

here

Even your reading assignments will keepyou active! You will be answeringquestions or solving problems constantlyas you read. No more going to the end ofthe chapter to find these things.Hopefully, the use of active learningtechniques will make your study ofchemistry one that you will findinteresting, challenging, and enjoyable.On most days you will find that the classtime seems to pass very quickly.

Of course, with more independ-ence comes more responsibility. You willbe expected to use class time properly,complete your work promptly, and cometo class prepared. Good class attendanceis a must. The responsibility for yourlearning is yours! But, your instructor,“the guide by the side,” will offer anyassistance you may need. Good luck!

ACTIVE LEARNING IN CHEMISTRY EDUCATION“ALICE”

Copyright 1997, A.J. Girondi, Ph.D.

ACTIVE LEARNING IN CHEMISTRY EDUCATION

TABLE OF CONTENTSChapter

1. Laboratory Techniques and Concepts

2. Handling and Presenting Data

3. Physical States of Matter

4. The Gas Laws

5. Atoms and Molecules

6. Chemical Equations - Part 1

7. Chemical Equations - Part 2

8. The Mole Concept - Part 1



11. Periodic Classification of the Elements

12. Modern View of Atomic Structure - Part 1

13. Modern View of Atomic Structure - Part 2

14. Chemical Bonding - Part 1

15. Chemical Bonding - Part 2

16. Solutions - Part 1

17. Solutions - Part 2

18. Rates of Chemical Reactions

19. The Energy of Chemical Processes

20. Chemical Equilibrium

21. Acids and Bases - Behavior in Water

22. Acid – Base Reactions

23. Oxidation and Reduction - Part 1

24. Oxidation and Reduction - Part 2

25. Introduction to Organic Compounds - Part 1



28. Nuclear Chemistry - Part 1

29. Nuclear Chemistry - Part 2

Additional Materials: APPENDICES: A, B, C, D, E, F; Reference Notebook;

ALICE Activities - Materials & Equipment List

Active Learning In Chemistry EducationMaterial and Equipment List for Laboratory Activities

CHAPTER 1

ACTIVITY 1.3 Measuring Lengths

Meter stick or metric ruler

ACTIVITY 1.4 Uncertainty In The Measurement Of Temperature

Small beaker (100 or 150 mL); glass thermometer;

ACTIVITY 1.5 Uncertainty In Measuring Volumes Of a Liquid

50 mL graduated cylinder; small beaker of water

ACTIVITY 1.6 Use Of The Electronic Balance

Electronic balance; weighing papers; rock salt;

ACTIVITY 1.7 Uncertainty In The Measurement Of Masses

Electronic balance; small beaker (100 or 150 mL)

ACTIVITY 1.8 Qualitative and Quantitative Observations

Candle; cardboard; matches or lighter

ACTIVITY 1.9 Use of the Laboratory Burner

Laboratory burner; lighter

ACTIVITY 1.10 Working with Glass Tubing

Glass tubing; high-temperature burner or lab burner with flame spreader (wing top); steel file

ACTIVITY 1.11 Heating Liquids in Tubes - The Hot Water Bath

250 or 400 mL beaker; ring stand with ring and wire gauze; burner; dextrose solution; Benedict's solution

ACTIVITY 1.12 Decanting Liquids

Solutions A and B (any combination which will form a precipitate will do); 100 or 150 mL beaker;stirring rod; 10 mL graduated cylinder

ACTIVITY 1.13 How to Properly Filter Liquids

Funnel; filter paper; small beaker containing precipitate from Activity 1.12; ringstand with small ring; stirring rod; balance

Materials List - Page 1

CHAPTER 2

ACTIVITY 2.13 Collecting and Graphing Experimental Data

Two 250 mL beakers; sodium chloride; stirring rod; two thermometers; eight ice cubes; timer or watch with second hand

ACTIVITY 2.14 Cooling Rates of Evaporating Liquids

Thermometer; cotton or paper toweling; rubber band; timer or watch with second hand; string or paper clip; ethyl alcohol; ring stand with utility clamp

CHAPTER 3

ACTIVITY 3.2 Identifying Metals By Physical Properties

Pieces of Metals "A" and "B"; 50 mL graduated cylinder; balance;Handbook of Chemistry and Physics

ACTIVITY 3.3 Comparing the Densities of Liquids

10 mL graduated cylinder; bottles of ethyl alcohol, oil, and vinegar; balance

ACTIVITY 3.5 Energy Changes Which Accompany Phase Changes

250 mL beaker; burner; stopwatch or watch with second hand; test tube containing paradichlorobenzene (PDB); thermometer; ringstand with ring and wire gauze

ACTIVITY 3.7 Physical and Chemical Changes

Four test tubes (three 100 mm tubes and one 150 mm tube); test tube rack; 0.1 M solutions of AgNO3, K2CrO4, Pb(NO3)2, NaI, NaOH; 3 M solutions of HCl and H2 SO4; solid NaOH pellets; thermometer; mossy zinc metal

ACTIVITY 3.9 Determining the Thickness of Aluminum Foil

One piece of regular aluminum foil and one piece of heavy duty aluminum foil (each 10 cm X 10 cm); metric ruler; balance

ACTIVITY 3.12 (Optional) The Barge Problem

100 mL beaker; 400 mL beaker; pieces of lead metal; grease pencil or marker or label tape

ACTIVITY 3.12 Sublimation (Teacher Demonstration)

Beaker; watch glass; ice cube; iodine crystals; burner; ringstand with ring and wire gauze


CHAPTER 4

ACTIVITY 4.2 Measuring the Effect of Pressure on Gas Volume

Boyle's law apparatus; 5 identical textbooks;

ACTIVITY 4.4 Measuring the Effect of Temperature on Gas Volume

Charles' law apparatus (glass tube containing mercury bead); metric ruler; large baker or flask;burner; ringstand with ring and wire gauze

ACTIVITY 4.9 Measuring the Effect of Temperature on Pressure

Special manometer arrangement (see Figure 4.6)

ACTIVITY 4.10 Comparing Equilibrium Vapor Pressures

Special manometer arrangement (as used in Activity 4.9); acetone; ethyl alcohol;isopropyl alcohol

ACTIVITY 4.12 A Teacher Demonstration of the Pressure–Temperature Relationship

Computer with special syringe and thermal probe attachment; Data Logger or equivalent software

CHAPTER 5

ACTIVITY 5.2 Properties of Iron In a Compound and In a Mixture

Mixture of iron filings and sulfur in a small container; small container of FeSO4

CHAPTER 6

ACTIVITY 6.2 An Experiment Involving the Heating of Copper

Copper turnings; crucible; ringstand, ring, and triangle; burner; balance

ACTIVITY 6.5 The Direct Combination Reaction of Copper and Sulfur

150 mm test tube; powdered sulfur; granular copper; small balloon; burner; balance;ringstand with utility clamp

ACTIVITY 6.8 Observing the Decomposition Reaction of H2O2

150 mm test tube; 6% H2O2 Solution; MnO2 powder; wood splint; matches or burner


CHAPTER 7

ACTIVITY 7.2 Determination of the Activities of Selected Metals

10 mL graduated cylinder; six 150 mm test tubes; test tube rack; three small pieces each of copper and zinc; 0.2 M solutions of KNO3, AgNO3, Pb(NO3)2

ACTIVITY 7.3 "Activities" of the Halogens

Five or six 100 mm test tubes; test tube rack; one or two long-stemmed pipets; small corks to fit the test tubes; dropper bottles containing the following: TCE (trichloroethylene), fresh chlorine water, 0.1 M NaCl, 0.1 M NaBr, 0.1 M NaI; a waste container for used TTE

ACTIVITY 7.5 Observation of Some Double Replacement Reactions

Dropper bottles containing 0.1 M BaCl2, 0.1 M K2 SO4, 0.1 M CuCl2, 1% (NH4)2S, 0.25 M NaI,

0.25 M Pb(NO3)2; a spot plate

ACTIVITY 7.9 Predicting the Formation of Precipitates

Dropping plate; 0.1 M Fe(NO3)3; 0.1 M CuCl2; 0.5 M KOH

ACTIVITY 7.10 (Optional) The Reaction Between NaOH and Ni(NO3)2

Nine 150 mm with the same internal diameter; test tube rack; grease pencil or marker; 5 mL and 10 mL volumetric pipets; 0.1 M Ni(NO3)2; 0.1 M NaOH; phenolphthalein solution;corks to fit the test tubes

CHAPTER 8

ACTIVITY 8.1 Determining the Mass of Extremely Small Objects

Vial of rice; vial of lead shots; vial of staples; balance; empty vial

ACTIVITY 8.11 Percentage of Oxygen in Potassium Chlorate

Crucible and cover; triangle; ringstand with ring; KClO3; MnO2; balance; burner

CHAPTER 9

ACTIVITY 9.2 Determination of the Gram-Atomic Mass (GAM) of Silver

Crucible with cover; ringstand with ring; triangle; crucible tongs; burner; balance; Ag2O

ACTIVITY 9.4 Balancing an Equation by Experiment

250 mL beaker; CuCl2; two nails; sandpaper; forceps or crucible tongs; balance; 1.0 M HCl


ACTIVITY 9.6 The "Silver Tree" Reaction

Two beakers (100 or 150 mL); balance; copper wire; silver nitrate crystals; oven; the instructor will provide a container for the product (silver metal)

CHAPTER 10

ACTIVITY 10.2 Determination of the Formula of an Oxide of Tin

Evaporating dish with watch glass cover; ringstand with ring and wire screen (gauze); granulated tin (30-mesh); balance; 8 M HNO3 (nitric acid); burner; stirring rod

ACTIVITY 10.5 Determination of the Molar Volume of a Gas

Large beaker (600 mL or larger); magnesium metal ribbon; balance (accurate to 0.001 g); scissors;fine copper wire; ringstand with clamp; gas-measuring tube; 6 M HCl (hydrochloric acid); rubber stopper with 1 or 2 holes to fit gas tube; barometer

CHAPTER 11

ACTIVITY 11.3 Metals, Nonmetals, & Metalloids – Comparing Properties

Glass vials containing samples of aluminum, copper and tin, sulfur, and silicon; Handbook of Chemistry and Physics; electrical conductivity testing apparatus

ACTIVITY 11.5 Properties of Alkaline-Earth Element Compounds

Four 150 mm test tubes; 0.1 M Mg(NO3)2; 0.1 M Ca(NO3)2; 0.1 M Na2CO3; 0.1 M Sr(NO3)2;0.1 M Ba(NO3)2

ACTIVITY 11.6 The Properties of Halogen Compounds

0.1 M NaCl; 0.1 M AgNO3; 0.1 M NaBr; 0.1 M NaI; 6.0 M NH4OH; solution of an "unknown" halogen; four 100 mm test tubes with corks to fit

ACTIVITY 11.8 Making, Collecting, and Studying Hydrogen Gas

Pneumatic trough with shelf; 125 or 250 mL Erlenmeyer flask with one-hole stopper to fit;piece of glass tubing to fit stopper hole; rubber tubing; three or four 150 mm test tubes;3 M HCl; mossy zinc metal; wooden splints; matches or burner

ACTIVITY 11.10 Reactions of Sodium and Potassium with Water (Teacher Demonstration)

Sodium metal; potassium metal; phenolphthalein solution; large beakers with watch glass covers


CHAPTER 12

ACTIVITY 12.4 The Emission Spectra of Elements

Burner; nichrome or platinum wire with loop; dropping plate; 0.1 M solutions containing strontium, lithium, barium, calcium, copper, and sodium ions; 3.0 M solution of hydrochloric acid; high voltage apparatus; gas discharge tubes containing different gases; hand-held spectroscopes

ACTIVITY 12.10 (Optional) The Colors of Transition Metal Chemistry

96-well microplate; plastic micropipets; toothpicks; 0.1 M solutions of KNO3, Ca(NO3)2, NH4VO3, Cr(NO3)2, Mn(NO3)2, Co(NO3)2, Fe(NO3)3, Ni(NO3)2, Cu(NO3)2, Zn(NO3)2; 6.0 M ammonia (NH3); 1 M potassium thiocyanate, KSCN; 6 M hydrochloric acid (HCl); one sheet of white paper

CHAPTER 13

ACTIVITY 13.8 Creating a Probability Density Plot

Crayon; target (page 13-21)

CHAPTER 14

The are no laboratory activities in Chapter 14.

CHAPTER 15

ACTIVITY 15.5 Comparing Properties of Ionic and Covalent Compounds

Sodium chloride; naphthalene; table sugar; metal file; metal can lid; ringstand with ring;burner; 100 mL beaker; stirring rod; electrical conductivity tester

ACTIVITY 15.7 Determining the Formula of a Hydrate

Porcelain crucible with cover; barium chloride hydrate crystals; balance; ringstand with ring;triangle; burner

CHAPTER 16

ACTIVITY 16.3 A Study of Molecular Models

Molecular models of Cl2; HCl; BBr3; AlF3; BeI2; H2O; CF4; NH3; BHBr2; CF2Br2

ACTIVITY 16.4 Testing the Miscibility of Liquids

Large test tube (25 x 200 mm) with stopper or cork that fits; 2-butanol; colored water; ethyl alcohol; bottle used as waste container for these liquids


ACTIVITY 16.6 Testing the Polarities of Solutes and Solvents

Four 150 mm test tubes with stoppers or corks to fit; marker or grease pencil; table sugar; small piece of styrofoam; small iodine crystal; toluene

ACTIVITY 16.8 Determining the Solubility of Sodium Chloride

Saturated salt solution; 25 mL graduated cylinder; thermometer; evaporating dish with watch glass cover; ringstand with ring and wire screen (gauze); burner;

ACTIVITY 16.9 Making a Saturated Solution (A Teacher Demonstration)

Large stoppered test tubes containing saturated solution of sodium acetate; 250 mL beaker; ringstand with ring and wire screen (gauze); small crystals of sodium acetate; burner

ACTIVITY 16.10 Developing a Solubility Curve for NH4Cl

Large beaker (400 or 600 mL); ringstand with ring and wire screen (gauze); burner; NH4Cl;thermometer

CHAPTER 17

ACTIVITY 17.2 Making and Using a 0.10 Molar Solution

100 mL volumetric flask; Pb(NO3)2; two 150 mL beakers; balance; K2CrO4; 50 mL graduated cylinder; ringstand with ring and wire screen (gauze); triangle; filter paper; stirring rod; beaker tongs; funnel; oven

ACTIVITY 17.5 The Freezing and Boiling Points of Salt Water

Saturated salt solution; thermometer; small beaker (100 or 150 mL); 3 or 4 crushed ice cubes;stirring rod

ACTIVITY 17.6 The Relationship Between Boiling Point and Pressure (A Teacher Demonstration)

Vacuum pump; pressure probe; temperature probe; small beaker of water; bell jar fitted with 2-holed stopper; computer; appropriate software (such as Data Logger)

ACTIVITY 17.10 A Laser-Aided Look at a Colloidal Suspension (A Teacher Demonstration)

Small bottles containing a suspension, a true solution, and a colloidal suspension; laser


CHAPTER 18

ACTIVITY 18.1

3 M HCl; two 150 mm test tubes; mossy zinc metal; piece of Mg metal ribbon (2 cm); wood splint;matches or lighter or burner

ACTIVITY 18.5 Comparing the Rates of Two Reactions

Dropper bottle of 0.01 M KMnO4; two 100 or 150 mL beakers; FeSO4; oxalic acid; stirring rod

ACTIVITY 18.7 Solution Concentration and Reaction Rate

Solution 1:Solution 2:50 mL flask or beaker; stopwatch or watch with second hand

ACTIVITY 18.8 The Effect of Temperature on Reaction Rate

Solution 1:Solution 2:600 or 1000 mL beaker; ice cubes; burner; ringstand with ring and wire sreen (gauze); two large (25 x 200 mm) test tubes;

ACTIVITY 18.9 The Effect of Surface Area on Reaction Rate

Steel wool; burner; iron filings or powdered iron; lycopodium powder; dust explosion apparatus

ACTIVITY 18.10 The Effect of a Catalyst on Reaction Rate

6% H2O2 solution; four 150 mm test tubes; MnO2; Fe2O3; dirt; granular aluminum

ACTIVITY 18.11 Catalysts in Matches and in Cigarette Tobacco (Teacher Demonstration)

Book of safety matches; two sugar cubes; ash from cigarette tobacco; watch glass

CHAPTER 19

ACTIVITY 19.4 Determination of the Specific Heats of Three Metals

400 mL beaker; ringstand with ring and wire sreen (gauze); burner; two cylinders each of iron, copper, and aluminum metal; balance; 3 foam cups; 100 mL graduated cylinder; crucible tongs;thermometer; stirring rod; cardboard lid for foam cups

ACTIVITY 19.9 How Much Heat is Required to Melt Ice?

Thermometer; foam cup; cardboard lid for cup; 100 mL graduated cylinder; 2 or 3 ice cubes;balance

ACTIVITY 19.10 Measuring Enthalpy Changes

150 mm test tube; thermometer; forceps; pellets of NaOH; NH4NO3; 1 M HCl; 1 M NaOH


ACTIVITY 19.11 Heat of Condensation of Water

Foam cup with cardboard lid; 125 mL Erlenmeyer flask; one-hole stopper to fit flask; U-shaped glass bend; burner; 50 mL graduated cylinder; thermometer; ringstand with ring and wire sreen (gauze); balance

ACTIVITY 19.13 Superheated Steam (Teacher Demonstration)

1000 mL Erlenmeyer flask; 2 large burners; ringstand with ring, clamp, and wire sreen (gauze); coiled copper tubing

CHAPTER 20

ACTIVITY 20.4 LeChatelier's Principle and Changes in Concentration

0.1 M Fe(NO3)3; 0.2 M Fe(NO3)3; 0.1 M KSCN; 100 or 150 mL beaker; four 150 mm test tubes;6 M NaOH; 0.1 M AgNO3

ACTIVITY 20.9 Testing LeChatelier's Principle With Cobalt Ions

Two 50 mL Erlenmeyer flasks or beakers; CoCl2•6H2O; 6 M HCl; hotplate or burner; ice bath; filter paper; crucible tongs

ACTIVITY E.2

0.1M Solutions of: NaHCO3; HCl; BaCl2; CuSO4; KNO3; Cu(NO3)2 (add blue food coloring to the colorless solutions so that they look like the copper solutions); dropping plate;red and blue litmus paper

CHAPTER 21

ACTIVITY 21.4 Comparing the Conductivity of Strong and Weak Acids

0.1 M HCl (hydrochloric acid); 0.1 M HC2H3O2 (acetic acid); 0.1 M H8C6O7 (citric acid); four 100 or 150 mL beakers (one for each acid and one for water); conductivity tester

ACTIVITY 21.7 Comparing the COnductivity of Strong and Weak Bases

0.1 M NaOH; 0.1 M NH3; three 100 or 150 mL beakers (one is for water); conductivity tester

CHAPTER 22

ACTIVITY 22.2 The Production of a Salt by an Acid-Base Reaction

Evaporating dish and watch glass; 1 M HCl; 1 M NaOH; burner; ringstand with ring and wire sreen (gauze);


ACTIVITY 22.5 Acid–Base Indicator Solutions

Dropper bottle containing a solution with pH < 3.1; dropper bottle containing a solution with pH > 4.4; dropping plate; dropper bottle of methyl orange solution; dropper bottle containing a solution with pH < 6.0; dropper bottle containing a solution with pH > 8.0; dropper bottle of bromthymol blue solution; dropper bottle containing a solution with pH < 8.2; dropper bottle containing a solution with pH > 10.0; dropper bottle of phenolphthalein solution; dropper bottle of white vinegar; dropper bottle of household ammonia solution; dropper bottle of laboratory detergent solution; dropper bottle of colorless soft drink; dropper bottle of tap water; pH meter

ACTIVITY 22.7 A Titration of Vinegar

Two 50 mL burets; ringstand with double buret holder; two small funnels; 0.50 M NaOH; whilte vinegar; three 125 mL flasks; dropper bottle of phenolphthalein solution; magnetic stirrer (if available)

CHAPTER 23

ACTIVITY 23.6 Observing Redox Reactions

0.2 M CuSO4; powdered zinc metal; 125 mL Erlenmeyer flask; 3.0 M HCl; 2 cm piece Mg ribbon;small piece copper metal; 0.1 M HCl

CHAPTER 24

ACTIVITY 24.1 Redox Reactions in a Petri Dish

Two petri dishes; four iron nails; zinc wire; copper wire; agar-agar powder; 0.1 M K3Fe(CN)6;phenolphthalein solution; stirring rod; 250 mL beaker; ringstand with ring and wire screen;degreasing solvent such as 1,1,1-trichloroethane; burner; balance; pliers; forceps or crucible tongs

ACTIVITY 24.3 Construction and Testing of Voltaic Cells

Voltaic cell apparatus (porcelain cup; glass cup or beaker; voltmeter; two wire leads with alligator clips; strip of copper; strip of zinc; strip of lead; fine sandpaper or steel wool; solution of CuSO4 (0.1 M to 0.5 M will do); solution of ZnSO4 (0.1 M to 0.5 M will do); solution of Pb(NO3)2 (0.1 M to 0.5 M will do)

ACTIVITY 24.5 Batteries

1.5 volt cell and 9 volt battery each cut in half to expose the interiors

ACTIVITY 24.7 The Electrolysis of Water

Direct current power supply with adjustable voltage control; electrolysis of water equipment (such as Hoffman apparatus); teacher will put some Na2SO4 into the water in the apparatus

ACTIVITY 24.8 Electroplating

A coin such as a nickel or quarter; 1 M HCl; two wire leads with alligator clips; 100 or 150 mL beaker; copper strip; solution of copper sulfate in sulfuric acid; power supply or battery


NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 1

LABORATORYTECHNIQUES

ANDCONCEPTS

1-1 © 1997, A.J. Girondi

NOTICE OF RIGHTS

All rights reserved. No part of this document may be reproduced or transmitted in any form by any means,electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author.

Copies of this document may be made free of charge for use in public or nonprofit private educationalinstitutions provided that permission is obtained from the author . Please indicate the name and addressof the institution where use is anticipated.

© 1997 A.J. Girondi, Ph.D.505 Latshmere DriveHarrisburg, PA 17109

[email protected]

Website: www.geocities.com/Athens/Oracle/2041

1-2 © 1997, A.J. Girondi

SECTION 1.1 An Introduction To "ALICE"

Welcome to ALICE (Active Learning in Chemistry Education). In this course you will learn thebasic principles of chemistry. It will help you to acquire some of the knowledge and skills which areneeded to understand many of the scientific and technological issues of our times. In addition toproviding you with a good background in chemistry, ALICE will help you to develop some logical methodsof solving problems. Unlike more traditional chemistry courses, ALICE is a program of study in which youwill be very independent. The teacher will not be the center of attention in your classroom. You will be!Well, you and your fellow students will be. You will find that the teacher will be not be lecturing to yourclass very often. Most of the time you will be working with other students in small groups. The teacher'srole will be that of a guide who will offer help and assistance to the small groups. He/she will provide youwith the laboratory materials you need, and from time to time he/she will perform various demonstrationsfor your group or your class.

There is no formal textbook for ALICE. Instead, you will be given special printed materials, onechapter at a time. These materials will be kept in your notebook which will eventually grow into a textbookby the end of the year. However, it will be a textbook that you will help to write. ALICE involves "activelearning." That means that you will have an active role in almost everything that happens in the classroom.As you read the materials, you will be responding to questions or solving problems. Most of the pagesyou read will require one or more responses from you. The correct answers to questions and problemscan be found at the end of each chapter. You will frequently see sentences like the following: Themeasurement of the amount of mass in a given volume of a substance is known as {35} . The{35} which appears as a subscript before the blank is the method used to identify the response in theanswer key at the end of the chapter. The laboratory activities are built right into the reading materials.The data that you collect and the conclusions that you draw will all become a part of your book. You will,however, be expected to assume a lot of responsibility. You should budget your time wisely, come toclass prepared, be a contributing worker in your group, and stay on task when you are in class.

You should do your own work, even though you will be encouraged to cooperate with others inyour group. The kind of small group work that you will experience in ALICE is often called "cooperativelearning." That simply means that you will help others to learn, and that they will help you. Discussionswith your lab partner, classmates, and teacher will give you more insight into the concepts beinginvestigated. Hopefully, this style of learning will make the class more interesting to you, and will actuallyhelp you to learn better than if you were in a more passive and traditional role.

The grade you earn is always determined by you. Quantity will not be the decisive factor inobtaining a grade, but quality will be very important. Your grade will be your responsibility. Each chapterwill be checked by your teacher who will further explain the details of how you will be evaluated.

This may be the first time you will experience this form of freedom in the classroom. You can learnas much, if not more, than you would in a conventional course. You have much more responsibility foryour own learning. The old saying, "you get as much out of something as you put into it," applies here.Learning can be accomplished by doing, seeing, and listening. Learning is your responsibility. It alwayshas been, but now your responsibility has increased. Some of the equipment you will be using may beexpensive. Careful handling of this equipment is expected, and you must obtain an understanding ofhow to use it properly. Your teacher will be available to help you with this. Some of this equipment ispictured on the previous page. Learn the name and spelling of each piece illustrated.

Practice safety in the lab. Safety in the laboratory is a primary concern for everyone. Always andimmediately report any accident, injury, or equipment breakage to your teacher. Always wear safetyglasses and an apron when working in the laboratory. You can be sure that no childishness will betolerated in the lab.

Write careful responses. Your written work should be complete with proper grammar, punctuation,and correct spelling. In addition, please be neat. No response has any value if it can't be read!

1-3 © 1997, A.J. Girondi

beakerburner

dropper

volumetric flasks

clay triangle

crucible with lid

pipet

iron ring

triangular file

test tube holderwith test tube

funnel

wing tip

graduated cylinder

crucibletongs

utilityclamp

watch glass

Florenceflask

thistle tube

glass bend

wide-mouth bottle

evaporating dish

buret

double buret holder

ring stand

test tube brushes pneumatic

trough

Erlenmeyer flask

Figure 1.1 Names of Laboratory Equipment

1-4 © 1997, A.J. Girondi

Some of the most important skills you will be developing in this course are those of observing andmeasuring. Observation and measurement allow you to collect data and then to interpret that data in orderto figure out some of the "whys" and "hows" around you. Experimentation followed by observation andmeasurement will allow you to answer many of these questions. Some such questions might include thefollowing. "Why do we put baking soda in most baked goods?" "What makes diamonds such hardsubstances?" "Why are some things in our environment so harmful to plant and animal tissue?" You useyour senses to collect data everyday. You see, taste, smell, feel, and hear details from your surroundings.Your senses are essential to the collection of data in your immediate environment. In the study ofchemistry or any other science, however, you will find that your senses alone will fall short of what isneeded. Your senses are valuable, but limited in scientific study. For example, through your sense ofsight, you may get the impression that one object is longer than another. However, it is extremely difficultto get more than an impression without the instruments necessary to obtain more accuratemeasurements. In the following exercises, you will be investigating the usefulness and accuracy of variousmeasuring instruments. You will first estimate the mass, length, volume, and temperature of severalobjects. Then, you will check your estimates using various laboratory instruments. Laboratory measuringinstruments have their limits, just as your senses have their limits. One of your tasks, in addition to learninghow to use various measuring instruments properly, will be to determine the limits of accuracy of theinstruments. Follow the directions carefully, recording your data in the space provided.

SECTION 1.2 Uncertainty in the Measurement of Length

Distances are normally measured with a meter stick or ruler. The limit of accuracy of a meter stickis indicated by how "close" you can read the length on the meter stick's scale (that is, how well you canestimate the fractions of degrees between the marks). On the portion of the meter stick shown in Figure1.2, the distance between the closest marks is 0.1 cm. The dotted line to the right of the meter stick is at alength of 6.65 cm. The last decimal place, the 0.01 place, in the measurement is estimated.

You can usually estimate to one decimal place beyond the closest marks on any measuringdevice. On this ruler, the closest marks are 0.1 cm apart, so you can estimate to the hundredths column,0.01 cm. However, on this ruler the smallest marks are so close that it is all we can do just to determine thatthe dotted line is between two of them - about half way. Therefore, your best estimate of the position ofthe dotted line is 6.65 centimeters. We can say that the measuring instrument is readable to ±0.05 cm.(The ±0.05 cm means that your measurement may be off by as much as 0.05 cm above or below its truevalue. This value is called the uncertainty of the instrument.)

There are two additional dotted lines in Figure 1.2.

1. What distance is represented by dotted line a?{1}______________ cm

2. What distance is represented by dotted line b? {2}______________ cm

(The centimeter scale below has been enlarged for ease of reading.)

4 5 6 7a b 6.65

FIGURE 1.2 Measuring With a Metric Rule

The scale shown in Figure 1.3a represents hypothetical units called "gorks." The closest lines on thescale each represent 1 gork. The length of the dark line above the scale is 27.5 gorks. Since we can readone decimal place beyond the closest lines, we can read this scale out to the tenths column. The lineappears to be about half-way between 27 gorks and 28 gorks. The length of the line appears to be about0.5 gork away from each of the two closest lines. Therefore, the uncertainty is ±0.5 gorks.

1-5 © 1997, A.J. Girondi

4 8 12 16 20 24 28 32 36 40 44 48

Figure 1.3a The "Gork" Scale

The line above the scale in Figure 1.3b appears to extend right up to the 22 gork mark. Should weexpress its length, then, as 22 gorks? Well, we noted above that the closest marks on this scale represent1 gork. We recall that we can usually read a scale to one decimal place beyond the closest marks. We can,therefore, read this scale to the tenths column. We can actually express the length of this line as 22.0gorks. The lines on this scale are so close that about all we can do is tell whether a measurement lies on aline or mid-way between two lines. The closest lines represent 1 gork each, so the uncertainty here isagain ±0.5 gork. So, the length of the line above Figure 1.3b is 22 gorks ±0.5.

4 8 12 16 20 24 28 32 36 40 44 48

Figure 1.3b The "Gork" Scale

The scale in Figure 1.4 below is calibrated in units called "flings." The closest lines on this scale eachrepresent 0.5 fling. The length of the dark line above the scale is determined to be between 16.5 and17.0 flings. We, therefore, read its length as 16.75 flings. The value of 16.75 is 0.25 fling away from thetwo closest lines. Since all we really know is that the length is somewhere between these two lines, theuncertainty can be expressed as ±0.25 fling.

2 4 6 8 10 12 14 16 18 20 22 24

Figure 1.4 The "Fling" Scale

Problem 1. What is the length of the line above the "slurp" scale in Figure 1.5 below? _____________

How much uncertainty is there in this measurement? ______________

2 4 6 8 10 12

Figure 1.5 The "Slurp" Scale

Problem 2. What is the length of the line above the "klump" scale in Figure 1.6?_________________

How much uncertainty is there in this measurement?_______________

10 20 30 40 50 60 70 80 90 100 110

Figure 1.6 The "Klump" Scale

1-6 © 1997, A.J. Girondi

Problem 3. What is the length of the line above the "glip" scale in Figure 1.7?___________________

How much uncertainty is there in this measurement?_______________

1 2 3 4

Figure 1.7 The "Glip" Scale

Problem 4. What is the length of the short line above the "zorch" scale in Figure 1.8? _____________

What is the uncertainty of this measurement? ____________ What is the length of the long line above

Figure 1.8?__________ What is the uncertainty of this measurement? ____________

1 2 3 4 5 6 7 8 9 10

Figure 1.8 The "Zorch" Scale

ACTIVITY 1.3 Measuring Lengths

Estimate the length and width of this page in centimeters (cm) without using a measuring instrument.

1. What is your estimate of its length? __________ cm

2. What is your estimate of its width? __________ cm

Get a ruler or meter stick and measure the length and width of this page.

1. What is the length (in cm) between the closest marks on your measuring device? ________cm

2. What decimal place is your measuring device accurate to? ___ cm

3. What is the length of this page? cm

4. What is the width of this page? cm

5. How much uncertainty (in cm) is there in your measurements? ±__________cm

(The answers to the above questions will depend on the type of ruler you are using.)

ACTIVITY 1.4 Uncertainty In The Measurement Of Temperature

Next, you will be trying your skills at estimating temperature and learning how to read athermometer properly. This will not be as easy as it may seem. In science, temperature is usuallymeasured and recorded in degrees Celsius, oC. You are probably familiar with the Fahrenheit scale. TheFahrenheit scale is most commonly used in daily weather reports. You should be able to make a ballparkestimate in oC knowing that room temperature, usually about 72oF, is equal to 25oC.

1-7 © 1997, A.J. Girondi

Fill a small beaker about halfway with water. Now obtain a thermometer with a range of about 110 to 120oC. Use the same procedure used in determining the uncertainty of the meter stick to determine theuncertainty of the thermometer.

How many degrees are represented between the closest two marks on the thermometer's

scale? {3} oC

To what column can the thermometer reading be estimated? {4}_____________________

Hold the thermometer in the water in the small beaker. Allow it to remain immersed until the temperaturestabilizes. Be sure to read the temperature to one decimal place beyond the smallest divisions marked onthe thermometer.

What is the temperature of the water? ___________oC

What is the uncertainty in the temperature value you obtained? {5} ±____________oC

Return the thermometer to the lab shelf, and save the beaker of water for the next section.

ACTIVITY 1.5 Uncertainty In Measuring Volumes Of A Liquid

The volume of a liquid is normally measured in liters (L) or in milliliters (mL) in the chemistrylaboratory. If you are not familiar with the quantity of a liquid that makes a liter or milliliter, you may havesome trouble estimating volumes. But don't worry, because by the end of this course, you will probablybe an expert. Almost every lab activity you will be doing throughout this course involves measuringvolumes.

Obtain a 50 mL graduated cylinder. Look at the cylinder and then at the water in the beaker usedin the previous section. What is your estimate in mL of the volume of water in the beaker? ________ mL

Before measuring the actual volume of theliquid, you must learn how to read a graduatedcylinder properly. Many liquids in a graduated cylinderhave cohesive properties. This means the particles ofthe liquid have a tendency to stick to each other.There is also an adhesive property of most liquids toglass, meaning unlike particles (liquid and glass) stickto each other. These two properties (adhesion andcohesion) combine to cause a liquid to "climb" thewalls of a graduated cylinder and form a bend or dipon the surface of the liquid. This dip is called themeniscus. When measuring the volume of a liquid, aninaccurate measure will result if the meniscus is nottaken into account. When reading a volume from agraduated cylinder, you should place your eye evenwith the bottom of the meniscus. Read the scalewhere the bottom of the meniscus rests. StudyFigure 1.3 carefully. The volume of the liquid shownthere is 55 mL.

Figure 1.3 Reading a Meniscus

20

40

Graduated cylinders come in a variety of sizes. Each size has a different degree of uncertainty.When measuring small volumes, a small graduated cylinder should be used. Large graduates are used formeasuring large volumes. Let's compare the degree of uncertainty of various sizes of graduated cylinders.

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Problem 5. Apply the procedure used previously to determine the uncertainty of the ruler andthermometer. Obtain three different sizes of graduates (10 mL, 50 mL, and 100 mL). Examine eachclosely and record your data in the chart below. Do not forget the units! In each case assume that thelevel of liquid is located either at one of the closest marks or is halfway between two of them.

Size of Graduated Volume represented Amount of uncertainty Cylinder by closest two marks (±)

a. 10 mL mL ± mL

b. 50 mL mL ± mL

c. 100 mL mL ± mL

Now pour the liquid from your beaker (from section 1.5) into the appropriate graduated cylinder (accordingto size).

What is the measured volume of the liquid? _____________ mL (Be sure to express the volume to onedecimal place beyond the smallest divisions.)

ACTIVITY 1.6 Use Of The Electronic Balance

In chemistry, mass is generally measured in grams (g). There are various types of laboratorybalances used in chemistry classrooms. The balances used in high school laboratories today may beeither manual or electronic. We will assume that your lab is equipped with electronic balances.

Balances are delicate and expensive pieces of equipment and should always be handledcarefully. Locate an electronic balance in your laboratory. Note the positions of the "On/Off" control andthe "Tare/Rezero" control. The purpose of the on/off control is obvious. After they have been turned on,electronic balances should be left on for the remainder of the class period. DO NOT turn them on and offduring each use. The "Tare" control is used to cancel out the mass of any object which is resting on thebalance pan. It is also use to "rezero" the balance. Let's now play with the balance a bit, to become familiarwith its use and controls.

Turn the balance off if it is already on. Now turn the balance on and notice what happens on thedisplay. The balance goes through a self-check each time it is turned on during which time the electroniccircuits are each tested to be sure they are working properly. The balance cannot be used until the self-check is completed. This does not take very long. When the self-check is complete, the display shouldread zero. If it does not, press the "Tare/Rezero" control. The display should now read zero. Note thatthe balance at this point can measure out to either two or three decimal places (depending on the balanceyou are using). This represents a high degree of sensitivity, so that even small breezes of wind may causefluctuations. Some balances will display masses to three decimal places up to a certain mass, and thenthey will switch to two decimal places above that mass. Place an object on the pan and note the display.Most of the work you will be doing in chemistry will require masses measured to two decimal places. Thereis no scale to read on electronic measuring devices. You simply read a digital display. Therefore,uncertainty in not estimated in the same way it is when reading a scale. The last digit in any measurementis always uncertain. So, if the display reads 45.85 g, then the last 5 (in the hundreds column) is uncertain.This means that the minimum uncertainty is ±0.01 since the uncertain digit is in the hundreds column.However, the uncertainty could be greater than this, and you would have to check the specifications ofthe instrument to determine that.

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Dry chemicals are never placed directly on the balance pan. Instead they are placed on either apiece of weighing paper or in a small cup or container. Note that a box of weighing paper is always foundclose to the balances in the lab. Obtain a piece of weighing paper and put in on the balance pan. Note themass of the paper. We are now going to measure out about 5 grams of rock salt onto the weighing paper.However, we do not want to include the mass of the paper. So, before we begin measuring out the salt,press the "Tare/Re-zero" and note what happens. The mass of the paper has been eliminated, so that thebalance now reads 0.000 with the paper on the pan. Therefore, any mass we read on the display will bethat of the salt. Obtain a bottle of rock salt from the materials shelf, and use a spoon or scupula to addabout 5 grams of salt to the paper. Read the display.

What is the mass of salt on the pan? ________________ g

Be very careful if you decide to remove some substance from the balance pan. DO NOT press on thebalance pan with the spoon! (And, of course, never handle chemicals with your hands.) Remove thepaper and salt from the balance pan and return the salt to the container from which you obtained it. Do notreuse weighing paper. Discard it after use.

Sometimes you may need to measure out more of a substance than a piece of weighing papercan safety hold. In that case it is recommended that you use an object such as a small paper or plastic cupin place of weighing paper. Some chemicals may react with paper or plastic, in which case glass containerssuch as beakers may be used. The "Tare/Rezero" control is used to eliminate the mass of the cup orcontainer in that same way as was done with weighing paper.

Note that the "Tare" control is also the "Rezero" control. If you attempt to use the balance andnote that the display does not read zero even though nothing is on the pan, simply press the"Tare/Rezero" control. BE SURE to leave the balance and the surrounding area clean after you haveused it. DO NOT leave chemical substances on or around the balance! Note that the pan can be lifted offfor easy cleaning underneath. Use a damp paper towel for cleaning.

Tare/Rezero ON/OFF

45.85 g

Figure 1.9An Electronic Balance

1-10 © 1997, A.J. Girondi

ACTIVITY 1.7 Uncertainty In The Measurement Of Masses

Now it's time to make an estimate of mass and then to check that estimate using the laboratorybalance. Clean and dry the small beaker used in section 1.5. Place the beaker on the balance anddetermine its mass in grams.

What is the measured mass of the beaker? ______________ g

What is the minimum amount of uncertainty for the balance you are using? ± ___________g

What was the unit used on the numerical answers above?

Note: After you have finished a laboratory activity, it is your responsibility to clean all of the equipment and to return it to its proper place. You may also find it necessary to wipe up the lab table.

ACTIVITY 1.8 Qualitative and Quantitative Observations

Most of the information gathered in chemistry requires some type of direct observation. Some ofthese observations rely solely on our senses. For example, observing that a particular solution is yellowrelies solely on our sense of sight. But, to determine that there are 15 mL of this yellow solution relies onour sense of sight aided by a measuring instrument, the graduated cylinder in this case. Measuring toolshelp us to extend our senses by allowing us to collect more accurate information. The two observationsmade above are two rather different types of observations. In general, they are called qualitative andquantitative observations. Qualitative observations are those that describe, in general, how somethinglooks, feels, tastes, and/or smells. Quantitative observations involve numerical measurements.

Example: I have a test tube in my hand containing 5 mL of a solution that is green in color. The factthat I have identified a measurement (5 mL) would indicate a quantitative observation. By identifying thecolor of the solution, I have made a qualitative statement.

In science, it is necessary to be able to make accurate observations. Most people need toimprove their observation skills. So let's do this. Obtain a candle, a square to set the candle on, and amatch or lighter. Light the candle and stand it on the cardboard. Some students are capable of making anextraordinary number of observations of a lit and unlit candle. You will be asked to make only 10observations - 5 qualitative and 5 quantitative. Use as many of the laboratory measuring devicesintroduced earlier as possible to make your quantitative observations. Be certain that you are using theinstruments properly, that you always write the appropriate units after any numbers, and that you carry yourmeasurements out to as many decimal places as possible. Your observations may be of both a lit and anunlit candle. (10 total observations) Write your observations in the table which follows in completesentences.

Qualitative Observations:

1.____________________________________________________________________________

2.____________________________________________________________________________

3.____________________________________________________________________________

4. ____________________________________________________________________________ 5.____________________________________________________________________________

1-11 © 1997, A.J. Girondi

Quantitative Observations:

1.____________________________________________________________________________

2.____________________________________________________________________________ 3.____________________________________________________________________________

4.____________________________________________________________________________ 5.____________________________________________________________________________

ACTIVITY 1.9 Use of the Laboratory Burner

As you work in the chemistry laboratory, there are several procedures or techniques that arecommonly used. You will now be given an opportunity to practice some of these basic procedures. Onepiece of equipment you will frequently be using is a burner. Figure 1.10 illustrates the anatomy of a typicalburner. Your burner may be different from the one shown. Most burners have an adjustment for air flow -usually this involves turning the barrel of the burner. Some also have a gas flow adjustment - usuallycontrolled by a valve at the base of the burner.

Attach your burner to the gas outlet at your lab station using a two or three foot length of rubbertubing which is in good condition. Be sure that the connections are tight. Obtain a flame igniter or lighter.Be very careful if you have long hair. Hair will burn very quickly if it gets too close to the flame! It is a goodidea to tie long hair back. Turn the gas on so that the valve is completely open - this means that the handleof the valve should be located directly overtop of the valve itself. The flow of gas should be not becontrolled from the main valve, but rather it should be controlled by the valve on the burner itself. Themain valve should be completely open when in use. Step back from the burner and light the gas. If youcannot light it, do not allow gas to continue to flow. Turn it off and ask the teacher for assistance.

Barrel - rotate to adjust air flow

Gas Inlet

Valve - rotate to adjust gas flow

FIGURE 1.10 Anatomy of a Laboratory Burner

The burner flame should be blue in color and about 1 to 2 inches in height. If the flame is yellow,more air is needed (turn the barrel counterclockwise). If the flame keeps blowing out, less air is needed(turn the barrel clockwise). If your flame is too long or too short, adjust the gas flow by turning the valve atthe base. Ask your teacher to help you if you are uneasy with the process. The hottest part of the flame isat the tip of the inverted light blue cone within the flame. The tip of this cone should touch the bottom of

1-12 © 1997, A.J. Girondi

the object being heated for maximum efficiency. Before moving on, be sure you can light and adjust yourlaboratory burner properly and with confidence.

Safety is always a concern, and it will be brought to your attention many times. The use of safetyglasses to protect your eyes may well be the most important piece of equipment you will ever use in achemistry lab. They must be worn at all times in the lab. Wear them during this next activity.

A larger high-temperature burner is also available in your lab. These burners are used for bendingglass tubing and for specific experiments where high temperatures or heating large volumes of liquids arerequired. You will be told when high-temperature burners should be used. DO NOT use them unless youare directed to do so. You will be using high-temperature burners during activity 1.10.

ACTIVITY 1.10 Working with Glass Tubing

Occasionally, you will be doing laboratory activities that require the manipulation of glass tubing.At this point, your teacher will demonstrate to you the proper way to cut, bend, fire-polish, and seal glasstubing using a large high–temperature burner or a regular burner with a "wing top" attachment. Theteacher will demonstrate these techniques in front of the entire class. Caution! Hot glass cools veryslowly, so allow it to cool completely before touching it. Water may cause hot glass to shatter - even Pyrex!Never put a liquid into hot glass - even Pyrex. After the teacher demonstration, you will be given specifictasks to perform with the glass tubing. Each student will be required to:

1. File and break two pieces of glass tubing to a proper length (15 to 20 cm in length). Use those piecesof tubing to:

2. Make a 90-degree glass bend fire-polished at each end.

3. Make a stirring rod (about 15 to 20 cm long) sealed at both ends.

Check with your teacher after you have completed the tasks, and show him/her your glass bend andstirring rod.

ACTIVITY 1.11 Heating Liquids in Tubes - The Hot Water Bath

If you wish to heat a liquid in a test tube, you should never put the test tube containing the liquiddirectly into the burner flame. If placed directly into the flame, the solution may "bump" and shoot out ofthe tube, possibly in someone's face. It's safer to heat the tube in a hot water bath as shown in Figure1.11. Let's try an activity using this technique. To set up the water bath you will need a 250 mL or 400 mLbeaker about 1/2 full of tap water. Put this on a wire gauze supported by an iron ring attached to a ringstand. A test tube containing the sample to be heated can then be placed into the water. Light the burnerunder the water bath. Begin heating the water. The tip of the light blue cone of the flame should touchthe bottom of the wire gauze. Next, obtain the dextrose solution and the Benedict's solution from thematerials shelf. Measure out 5 mL of dextrose solution and place it in a large test tube.

Qualitatively describe the dextrose solution._____________________________________________

Quantitatively describe the dextrose solution.____________________________________________

Qualitatively describe the Benedict's solution.____________________________________________

1-13 © 1997, A.J. Girondi

Next, add 3 drops of Benedict's solution to the dextrose solution in the test tube. Immerse the test tube

in the water bath. When it gets hot enough, the solution in the tube will change.

Describe any changes that occur as you heat it. ___________________________________________

Was your observation above more QUALitative or QUANTitative?______________________________

Allow your apparatus to cool, and then clean up the lab area.

The tip of the light blue conein the flame should touch

the bottom

Figure 1.11 Hot Water Bath

stirring rod

Figure 1.12 Decanting

ACTIVITY 1.12 Decanting Liquids

It will frequently be necessary for you to separate a solid from a liquid. There are several ways inwhich to do this. The simplest way is to let the solid settle out and then pour off the liquid - leaving thesolid in the container. This technique is called decanting. To avoid splashing, the liquid is slowly poureddown a stirring rod. The technique is shown in Figure 1.12. It takes a bit of practice to do this properly, sotry it. Get solutions A and B from the materials shelf. Mix 7 mL of solution A with 7 mL of solution B in asmall beaker. Describe what happens using qualitative terms. ________________________________

______________________________________________________________________________

The solid that is formed is called a precipitate . Decant the liquid from the precipitate following theprocedure shown in Figure 1.12. The liquid that you pour off is called the decantate. Save the beakercontaining the precipitate for Activity 1.13.

ACTIVITY 1.13 How to Properly Filter Liquids

When a solution is decanted, the precipitate left behind is quite wet. In many experiments it will benecessary to determine the mass of the precipitate. In these cases, it is not enough to decant the liquidoff of the precipitate. Instead, the solid is filtered through fine paper. In the filtration technique, the liquidsoaks through filter paper while the solid becomes trapped. The filter paper containing the precipitate canthen be dried and weighed. Follow the procedure below to determine the mass of your precipitate.

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1. Obtain a piece of filter paper and measure its mass using a balance. Record the mass of the paper inTable 1.1. Fold the filter paper as shown in Figure 1.9 below.

2. Pull one layer of the folded paper away from the rest forming a cone. Place the filter paper cone in afunnel and wet it with distilled water. Press the paper carefully with your thumbs so that it fits snugly in thefunnel. Put the funnel into a small iron ring attached to a ring stand. (If you can't find a small enough ring tohold your funnel, you can put the funnel into a pipestem triangle which can then be placed on a largerring.) Note that the tip of the funnel should touch the upper inside wall of the beaker to prevent splashing.

3. Pour the precipitate and any liquid still remaining in the beaker into the filter paper cone. If all the soliddoes not come out of the beaker, rinse the beaker with extra water from your wash bottle, and then pourthis extra water and the remainder of the precipitate into the filter paper cone.

4. Use a magic marker to put your initials on a 100 mL beaker. Then, when filtering is complete, put thefilter paper cone containing the precipitate into the beaker.

5. Allow the filter paper and the precipitate in the beaker to dry overnight in an oven. After that time,remove the dry paper cone from the beaker, and determine the mass of the filter paper and precipitate.Record all of your data in Table 1.1. Discard the paper and precipitate.

2

1

3

4

Press moistened paper against funnel with thumbs to seal.

Figure 1.9 Folding Filter Paper and Filtering Apparatus

Filtrate

TABLE 1.1 Determining the Mass of a Precipitate

Mass of Filter Paper g

Mass Paper + Precip g

Mass of Precipitate g

1-15 © 1997, A.J. Girondi

Write two qualitative observations about the activity above.

(1)____________________________________________________________________________

(2)____________________________________________________________________________

Write two quantitative observations about the activity above.

(1)____________________________________________________________________________

(2)____________________________________________________________________________

SECTION 1.14 Chemistry In Our Daily Lives

From the historical point of view, the development of the science of chemistry is very fascinating.In one sense, chemistry can trace its origins to prehistoric times when humans first discovered fire andhow it could be used to cause chemical changes. (But is was not a science then!) About 30,000 years agopeople were using paints and pigments for a variety of uses. About 8,000 to 10,000 years ago peoplewere producing copper and tin metal by treating ores and minerals with fire. About 3,000 to 4,000 yearsago people were producing bronze, iron, and steel for use as weapons and utensils. They also hadbecome fascinated with such metals as gold and silver, which they regarded as solid forms of sunlight andmoonlight. A certain group of individuals became so interested in metals that they attempted to convertthe more common metals (such as iron, copper, and tin) into gold and silver. These people becameknown as ALCHEMISTS , a term from which the modern word chemistry comes. For 1,500 yearsalchemists tried in vain to find a way to produce gold from other things. In their attempts, they developedmany methods and techniques to separate and purify chemical substances in nature.

The science of chemistry started approximately 4 0 0 years ago when people motivated bycuriosity became interested in studying the properties and behavior of natural substances. They learnedthe differences between solids, liquids, and gases. They discovered the existence of many elements andcompounds. Through laboratory experimentation, it became apparent that elements and compoundswere composed of atoms and molecules. It also became clear that there were groups of elements thatacted in a similar fashion, and they came to be known as "chemical families." A large amount of chemicalinformation accumulated during the period from 1700 to 1900 as scientists learned to recognize that therewere a limited number of chemical elements (about 90) of which the "stuff" of nature was composed.

During the 20th century we have learned much more about chemical substances and the lawsgoverning their behavior. We have created new elements in our laboratories that do not exist in nature.We have even learned enough that we know how to change certain metals into gold, though the processwould be very difficult and expensive. The dream of the alchemists turned out to be within our reach!

The historical view of the development of chemistry provides a further reason to study theknowledge that we have gained over the past 10,000 years. We are privileged to inherit 100 centuries ofaccumulated knowledge. One of the most important reasons that we should learn as much as we canabout chemistry is because of the role it plays in our lives. Almost every aspect of our lives involveschemistry to some extent. The food we eat contains chemical substances added to it to help prevent itfrom spoiling. The farmer uses chemicals to protect food crops from pests and weeds. The farmer usessynthetic chemicals produced in a laboratory to fertilize his land. All aspects of modern agriculture rely tosome degree on chemicals and chemical knowledge.

Our knowledge of chemistry has allowed us to develop drugs to treat and prevent sickness and tohelp make life more comfortable. Research is currently progressing toward the development of new andbetter drugs for treating diseases and health problems that are a serious threat to the well-being of us all.

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The clothing that most of us wear is made largely from synthetic fibers that have been developed inchemistry laboratories. Many materials are now made from plastics which have helped to raise our standardof living by making inexpensive products available to a larger number of people. The household productsused for cleaning and washing are another example of better living made possible through chemistry.These products are being improved continuously by the chemical industry as we learn new and betterways to produce them.

We all use energy, and in our country each person uses much more than most people in otherparts of the world. As energy supplies are being consumed, we will need to find ways to make coal, ourmost plentiful energy resource, usable without polluting our environment. Recovering more oil fromdepleted well and from oil shale is yet another challenge for chemists. From these examples, you can seethat chemistry is an important part of your life. The more you know about it, the more responsible you willbe as a citizen when you and others encounter the problems of a highly technical society.

SECTION 1.15 Learning Outcomes

Review the learning outcomes listed below. Check off each one which you feel you havemastered. Go back and review any material which you still feel uncertain about. When you have checkedoff all of the outcomes, you will be prepared for the exam on Chapter 1.

_____1. Identify the proper names of some common pieces of laboratory equipment.

_____2. Set up and use a hot water bath.

_____3. Distinguish between qualitative and quantitative observations.

_____4. Read laboratory measuring instruments properly, recognizing the amount of uncertainty in the measurements.

_____5. Properly decant and filter solutions.

_____6. List several reasons illustrating the important role of chemistry in our daily lives.

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SECTION 1.16 Answers to Questions and Problems

Questions:

{1} 5.25 cm; {2} 5.85 cm; {3} oC; {4} tenths; {5} 0.5oC

Problems:

1. 7 slurps, ±1 slurp2. 75 klumps, ±5 klumps3. 1.45 glips, ±0.05 glip4. 4.5 zorches, ±0.5 zorch; 5.0 zorches, ±0.5 zorch5. a. 0.1 mL; 0.05 mL (as long as you can read a level between the closest lines. If you can't the

uncertainty would be 0.1 mL.)b. 1 mL on most 50 mL graduates; 0.5 mLc. 1 mL; 0.5 mL (as long as you can read a level between the closest lines. If you can't the uncertainty would be 1 mL.)

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NAME_______________________________________ PER_____________ DATE DUE______________


"ALICE"

CHAPTER 2

HANDLINGAND

PRESENTINGDATA

2-1 © 1997, A.J. Girondi

NOTICE OF RIGHTS


Copies of this document may be made free of charge for use in public or nonprofit private educational institutionsprovided that permission is obtained from the author . Please indicate the name and address of the institutionwhere use is anticipated.


[email protected]


2-2 © 1997, A.J. Girondi

SECTION 2.1 Use of Scientific Notation

Much of this chemistry course will deal with experimentation. In chapter 1 you tried many of the morecommonly used laboratory techniques. Being skilled with the laboratory equipment will help you to carry outexperiments more efficiently. Every experiment you perform has a purpose. You are usually trying to answer aquestion or to discover some relationship that may exist between variables. The conclusions you draw fromexperiments often depend on how you organize your data. This is a very important part of chemistry. As a result,we are devoting this entire chapter to the handling of data.

When you work with chemistry concepts, you will often be using extremely large or extremely smallnumbers. For example, the speed of light is 30,000,000,000 cm per second. The mass of the Earth is6,000,000,000,000,000,000,000,000 kg. The mass of an electron is very, very small. In fact it is0.000,000,000,000,000,000,000,000,000,091 g. The wavelength of yellow light is about 0.000059 cm.Numbers expressed in this way are awkward and have little meaning to us. They cannot be quickly comprehendedat first sight. Therefore, a more convenient way to express such numbers is in a form called "exponentialnotation."

In exponential notation, numbers are expressed as multiples or powers of ten. Exponential numbers are aform of what is called "scientific notation." The following examples illustrate how common numbers may beexpressed in exponential notation.

10 = 1 X 101 0.01 = 1 X 10-2 123 = 1.23 X 102

0.1 = 10-1 100 = 1 X 102 1000 = 1 X 103

0.001 = 1 X 10-3 10,000 = 1 X 104 0.0001 = 1 X 10-4

When writing in scientific notation, there should be only one digit (not a zero) to the left of the decimal point.

The exponent is determined by the number of digits the decimal is moved to get it to the proper position.For example, 1,300 can also be written as 1.3 X 103 - the decimal moves 3 places to the left so the exponent onthe ten is a 3. If we would have had to move the decimal point three places to the right, then the exponent on the10 would have been a negative three (-3). That is, 0.0013 = 1.3 X 10-3. Notice also that the trailing or leadingzeros in the original numbers are no longer used.

Problem 1. Complete the following by writing the numbers in either exponential form or regular form.

a. 190,000 _______________ b. 1.986 X 105 _______________

c. 528 _______________ d. 1.986 X 108 _______________

e. 4,400,000 _______________ f. 1.75 X 107 _______________

g. 9700 _______________ h. 3.33 X 104 _______________

i. 49 _______________ j. 2.5 X 101 _______________

k. 0.056 _______________ l. 9.28 X 10-8 _______________

m. 0.113 _______________ n. 1.411 X 10-3 _______________

o. 0.00000035 _______________ p. 6.275 X 10-6 _______________

q. 0.00077 ______________ r. 2.79 X 10-4 _______________

s. 0.00000512 ______________ t. 1.3 X 10-1 _______________

2-3 © 1997, A.J. Girondi

NOTE! NOTE! NOTE! Check with your teacher at this point to see if you should read and complete the workon uncertainty in measurement in Appendix A. If so, do that now and then return to section 2.2, and finish thischapter.

To Appendix A -----> ??????

SECTION 2.2 Calculations Involving Scientific Notation

Using exponential numbers can save you a lot of time when doing calculations. For example, if you wereasked to multiply a number such as 1,400,000 by 3,200, the amount of time spent carrying decimal places wouldbe considerable. Assuming that you did not have a calculator available, it would take time to complete such acalculation. You can solve the problem quite easily using exponential notation. The problem would take this form:

(1.40 X 106) X (3.20 X 103) = 4.48 X 109

How was this answer obtained? First multiply the regular numbers by one another, then add the exponents to getthe correct final exponent in the answer. To divide using exponential numbers, you must change the sign of theexponent in the denominator and add it to the exponent in the numerator.

Example 1: 8.0 X 106

2.0 X 103 = 8.0 X 10+6(-3)

2.0 = 4.0 X 103

Example 2: 9.0 X 104

4.5 X 10-7 = 2.0 X 10+4(+7) = 2.0 X 1011

You should also try to do these problems using a scientific calculator. The exponent key on most scientificcalculators is designated as the "EXP" key or the "EE" key. In order to enter an exponential number into acalculator the exponent key is used as follows. To enter 8.6 X 106, you should enter 8.6 then press the exponentkey then enter the exponent. That would be: 8.6 [EE] 6. Notice that you would NOT press the multiplication keyand you would NOT enter the ten. On most calculators, the exponent is shown at the extreme right of the displayand usually without the ten showing (we assume it is there).

Many students get confused about how to solve more complex problems when using a calculator. Let'stry an example:

(5.6 X 103 )(2.4 X 102 )

(5.8 X 105)(3.6 X 103 ) = 6.4 X 10-4

To solve this problem on a calculator without using any subtotals, simply enter the first number in the numerator(5.6 X 103), then divide by (5.8 X 105). Now, multiply by (2.4 X 102), and then divide by (3.6 X 103). Calculators arefussy about the way you enter your data. It is best to alternately multiply and divide such as was done above.

Here is another example:

45

(32) (88) = ??

If you enter 45 in the calculator, and then divide by the product of 32 and 88 you will get the wrong answer. Youcan solve this problem correctly in one of two ways. You can enter 45 in the calculator, then divide by 32, thenmultiply by 1, and finally divide by 88. This is the alternating pattern mentioned above. Or, you can enter 45 intothe calculator and then divide by the quantity (32 X 88). This method requires the use of the parentheses foundon scientific calculators. Try it both ways. You should get 0.016 (rounded off).

2-4 © 1997, A.J. Girondi

Problem 2. Practice doing calculations such as those below. You should give each answer in scientific notation.You may use a calculator to solve these problems, but try doing several of them using only a pencil - just to be sureyou can.

Sample Problem: (7.0 X 1027)(4.0 X 10-10) = 28 X 1027-10 = 28 X 1017 = 2.8 X 1018

a. (4.0 X 1020)(3.0 X 1015) =

b. (9.0 X 1017)(4.0 X 10-4) =

c. (8.0 X 1022)(8.0 X 1022) =

d. (5.0 X 1012)(5.0 X 10-12) =

e. (4.0 X 1015) ÷ (2.0 X 1020) =

f. (3.0 X 10-12) ÷ (3.0 X 10-14) =

g. (7.0 X 106) ÷ (3.5 X 10-10) =

h. (8.2 X 10-20) ÷ (4.1 X 10-20) =

i. (2.88 X 1023) ÷ (2.0 X 1013) =

j. (3.0 X 1017 )(4.0 X 10-4 )

(6 X 10 -5 ) =

k. (7.0 X 10-10 )(7.0 X 10 -4 )

(7 X 10-14 ) =

l. (4.0 X 102 )(4.0 X 106 )

(1.6 X 106 ) =

m. (3.2 X 101)(6.4 X 101)

(1.6 X 103 ) =

n. (5.0 X 105)(6.0 X 104 )

(3.0 X 107 )(1.0 X 10 -6 ) =

o. (6.4 X 104)(7.0 X 103 )

(3.2 X 10-4 )(3.5 X 10-3 ) =

SECTION 2.3 "GuEstimating"

Because it is easy to make an error when entering a number into a calculator, you should estimate (or"guEstimate") your answer first. In this way, you will be able to quickly recognize any unreasonable answerobtained with a calculator. For example, suppose we wanted to multiply 6821 by 499. By rounding the numbersto 7000 and 500 and putting them in exponential form, we get (7 X 103)(5 X 102) = ? It is easy to guesstimate theanswer as 35 X 105 or 3.5 X 106. The actual answer obtained with a calculator is 3,403,679. Our guesstimateindicates that the answer obtained is, indeed, reasonable. (It is about 3.4 X 106.)

2-5 © 1997, A.J. Girondi

Problem 3. To practice this procedure, solve the problems below. Round off the numbers first, then convert toexponential form where necessary, then perform the operation and obtain your "guesstimate." (Don't worry aboutrounding the GuEstimate properly.) Do all this without the use of a calculator. Finally, use a calculator to obtain theanswer and compare it to your guesstimate.

"GuEstimate" Calculated Answer

a. (4,851,300 m)(6,200,111 m) ____________________ ______________________

b. 80,109 mm2 ÷ 411 mm ____________________ ______________________

c. (0.0000500 cm)(49,850 cm) ____________________ ______________________

d. (30,011 km)(299 km) ____________________ ______________________

e. (400,102 m)(6.0 X 1023 m) ÷ 310 m ____________________ ______________________

SECTION 2.4 Metric Abbreviations and Equivalents

Scientists normally use the metric system to define physical and chemical quantities. In so doing,abbreviations are used to eliminate the need to write the words defining a unit. The following are the properabbreviations for frequently used units of measurement in chemistry. You should study and remember theseunits and abbreviations.

Problem 4. After each measure, indicate whether each is a measure of mass, volume, or length. Note whichletters are CAPITALIZED, and which are lowercase.

a. gram (g) ____________ b. milliliter (mL) ____________ c. meter (m) ____________

d. milligrams (mg) ____________ e. millimeter (mm) ___________ f. liter (L) ___________

g. centimeter (cm) ____________ h. kilogram (kg) ____________

i. cubic centimeter (cm3) ____________

You may frequently have to change one metric unit into another. The metric equivalents listed below arecommonly used in chemistry and are needed to make conversions. You will be expected to know them.

1 kg = 1000 g; 1 m = 100 cm; 1 L = 1000 mL; 1 g = 1000 mg; 1 m = 1000 mm; 1 cm = 10 mm

Problem 5. Table 2.1 contains masses of several objects each given in a particular unit of measure. Completethe table by converting the masses to the units needed to fill in the blanks. Any very small or very large numberscan be expressed in exponential form. 1 mg = 0.001 g 1 kg = 1000 g

2-6 © 1997, A.J. Girondi

Table 2.1 Mass Conversions

Object g mg kg

small beaker _____78.96_____ _______________ _______________

pencil _______________ _____6783_____ _______________

large flask _______________ _______________ ____ 0.378______

SECTION 2.5 Celsius Versus Fahrenheit Temperatures

Another important property that we frequently measure in science is temperature. You have already useda Celsius thermometer in Chapter 1. The temperature scale that is commonly used in our everyday lives is theFahrenheit scale. Because of the way this scale was originally devised, the temperatures 0oF and 100oF are notequal to the freezing and boiling points of water. Scientists more commonly use the Celsius scale. This scale wasdevised by the Swedish astronomer, Anders Celsius, in the early 1700's. The reference point of 0oC is thetemperature at which water freezes, and the 100oC reference point is the temperature at which water boils (bothunder standard pressure of 1 atmosphere). If room temperature is taken to be 77oF, this is then equivalent to25oC. There are formulas available to change a temperature in one scale to an equivalent temperature in theother. However, since Fahrenheit temperatures are rarely ever used in chemistry, we will not do such conversionshere.

SECTION 2.6 Rules for Handling Units During Calculations

Notice that all of the problems which you have completed so far have units as a part of the answer. InChapter 1, it was explained that numbers without units are meaningless. Therefore, we must be able to handleboth numbers and their units with efficiency. In fact, your ability to work with units in a chemistry problem will be animportant aid to your success in chemistry. There are two rules that you must use when units are involved inproblems. They are:

1. Only quantities with the same units can be added or subtracted. You cannot add or subtract apples and oranges.

2. When quantities are multiplied or divided, their units are also multiplied or divided.

To illustrate this, imagine trying to add 50.0 mm Hg to 3.50 atm. Both mm Hg and Atm are units of pressure. Weknow that the mathematical sum is 53.5, but what are the units? Are the units "mm Hg" (millimeters of mercury) or"atm" (atmospheres)? The answer of 53.5 has no meaning because the units on each number are not the same.Therefore, can we add the pressures in the forms given? No! However, if you know that 1 atm = 760 mm Hg, youcan convert 3.50 atm to mm Hg and add the results to 50.0 mm Hg. This is done for you below.

3.50 atm X 760 mm Hg

1 atm = 2660 mm Hg

Now use this information to solve problem 6.

2-7 © 1997, A.J. Girondi

Problem 6. 50 mm Hg + 2660 mm Hg = ___________ mm Hg

Suppose we wanted to convert 4.00 hours to seconds. One method of setting up the problem is shown below.

4.00 hr X 60 min

1 hr X

60 sec1 min

= 14,400 sec

Notice that all of the units except seconds canceled out.

SECTION 2.7 Dimensional Analysis - A Powerful Way To Solve Problems

Units are a part of all problems. If the units cancel out correctly leaving you only with the units you want inthe answer, you know you have set up the problem correctly. This powerful method of setting up problems isknown as "dimensional analysis" or "unit analysis." Because of the way the set-up of the problem looks on paper,it is sometimes referred to as a "fencepost."

You are ready to try some sample problems involving units and their cancellation. After you have someexperience with these types of problems, you will realize that you are capable of working many types of problemswithout memorizing formulas. Dimensional analysis is not always the shortest way to solve a problem. However, asproblems get more and more complex, dimensional analysis gets more and more powerful because it can guideyou through the logic required, and it allows you to set up large parts of the problem at once, rather than doing acomplicated problem in small parts - one at a time. In addition, you will find that you can solve problems withoutknowing anything about the meanings of the terms involved.

Work the following problems. The problems have been partially set up for you. Cancel out units and placethe correct units on your answers. Never express answers as fractions!

Problem 7. If 1 tree = 10 branches, and 15 nests = 15 eggs, and 1 baby bird = 1 egg, and 2 trees = 1 yard, and5 branches = 5 nests, how many baby birds are there in 1 yard behind a house? (Start with the information given,1 yard, and solve for baby birds.)

1 yard X 2 trees1 yard

X 10 branches

1 tree X

5 nests5 branches

X 15 eggs15 nests

X 1 baby bird

1 egg

= ________ baby birds(Notice how all units cancel except baby birds.)

Problem 8. Two warts = 1 querk, 3 querks = 1 gag, 5 gags = 6 nerfs, and 4 nerfs = 5 wigs. How many warts arethere in 1 wig? Don't panic! You are looking for warts, and you are given wigs. So, we start the fencepost with 1wig and set up the units so that they cancel and leave you with warts. Finish the set-up below and calculate theanswer.

1 wig X nerfs wigs

... = ______ warts

2-8 © 1997, A.J. Girondi

SECTION 2.8 Use of Complex Units of Measure

A word that we use a lot with units is per. We say miles per gallon (mi/gal) or miles per hour (mi/hr) or gramsper mole (g/mol). These are common units, and they are used in the following problems. Solve these problemsmaking certain that you show all of your work, including units.

Problem 9. 1.00 case of apples costs $16.00. What is the cost per dozen if a case contains 14.0 dozen apples.(You want to end with units of $/doz.)

1.00 case14 dozen

X = $ 1 dozen

Problem 10. A car travels 300.0 miles on 11.0 gallons of gas. How many miles is the car able to travel when 143gallons of gas are used? (Be sure to use dimensional analysis.)

143 gal X

= ________ miles

The term "per" can be denoted by a single line. For example, miles per gallon can be written as mi/gal. The termalso gives a clue as to the mathematical process that is involved. Is the process addition, subtraction,

multiplication, or division? {1}___________________________

SECTION 2.9 Practice Problems Involving Units of Measure

The following problems are designed for you to practice canceling out units as they fit into the problems.Note that singular and plural units that are otherwise the same are considered identical and can cancel each other.For example, apple can cancel apples.

Problem 11. Indicate for each problem, what units are left after all canceling has been done.

Examples:

L X mLL

= mL sec2 X feetsec

= (sec)(feet)

a. mole X g

mole =

b. hrs X miles

hr =

c. feetsec

X secmin

X minhr

=

d. g

mole X

moleL

=

2-9 © 1997, A.J. Girondi

e. moles X mole

g =

f. moles X g

mole =

g. g X mole

g X

gmole

=

h. mole X mgL

X L

mole X

atmmg

=

i. cm3

m X

mdm

X dmcm

X 1

hec =

j. g

mole X

moleg

=

If you can handle numbers and units, you should be able to do problems such as the following example. Ifa person can run 100. yards in 10.5 seconds, how fast is he/she running in miles per hour? (1760 yd = 1 mi). Firstset up the unit analysis for solving the problem:

ydsec

X miyd

X secmin

X minhr

= mihr

Notice how all units cancel except those we wanted to keep in the answer. Substitute the proper numbers andsolve the problem in the space below. See if you get the answer given.

yd sec

X mi yd

X sec min

X min hr

= 19.5 mi 1.00 hr

Problem 12. Make the following conversions using dimensional analysis. Show your work neatly.

a. Change 400.0 ounces to its comparable figure in tons. (16 ounces = 1 pound, and 2000 pounds = 1 ton.)

b. Calculate the number of seconds in 1.000 week.

c. Convert 2.00 miles to fathoms. (1 fathom = 6 feet; 1 mile = 5280 feet)

2-10 © 1997, A.J. Girondi

d. Change 400. cubic feet per 1.00 second to quarts per minute. (0.265 gallons = 0.0353 ft3; 4 quarts = 1 gallon)

400. ft3

1.00 sec X

SECTION 2.10 Practice With Dimensional Analysis

Problem 13. Use the information below to solve the problems by unit analysis (dimensional analysis). Someproblems have been started for you.

1 sack = 7 bips; 4 tolls = 3 smacks; 12 tolls = 1 lardo; 5 smacks = 1 bip; 8 lardos = 7 fleas

a. Calculate the number of smacks in 1.00 lardo.

b. Calculate the number of lardos in 1.00 bip.

c. How many sacks are in 1.00 smack?

d. 12 bips equal how many fleas?

2-11 © 1997, A.J. Girondi

e. How many tolls are equivalent to 49 fleas?

Problem 14. Since you will be using the metric system exclusively in chemistry class, you should be able toconvert one metric unit into another. The metric equivalents presented earlier in this chapter are given againbelow for your convenience. Solve the problems below using "fenceposting" to set up and solve the problems,and be sure to SHOW YOUR WORK.

1 kg = 1000 g; 1 m = 100 cm; 1 L = 1000 mL; 1 g = 1000 mg; 1 m = 1000 mm; 1 cm = 10 mm

a. Convert 1.4 kg to g.

b. Convert 896 mL to L

c. Convert 6785 mg to kg

d. Convert 0.458 m to mm

2-12 © 1997, A.J. Girondi

e. Try this one now! Convert 4.5 m3 to cm3. Hmmmm. This could be tough because you were not given anyrelationship between cubic meters (m3) and cubic centimeters (cm3). Therefore, you will need to derive such arelationship before you can calculate the answer. Do it this way. A scale model of a cubic meter is drawn in thespace below. Since all sides of a cube are equal in length, each side has a length of 1 meter. Now label eachside's length in centimeters. (You will recall that the relationship between meters and centimeters is 1 m = 100cm.) O.K. Now, since the volume of a cube is s3 where s is the length of a side, you can calculate the volume ofthe cube in cubic centimeters. Do that in the space below the box.

Now, complete the following: 1 m3 = ___________________ cm3. You can check your answer by using it in part fbelow.

f. Using the relationship you just calculated, you can now convert 4.5 m3 to cm3. Show the set-up and answerbelow.

SECTION 2.11 Practice With Algebra Skills

Chemistry uses a lot of mathematics. But, do not be alarmed! We are going to give you examples of howto work through the problems, and if you get stuck, you can get plenty of help from your teacher. One of the funmath forms is algebra. The problems below introduce you to some basic algebra problems. These are the types ofmath computations you will use most often in this course.

The first type of algebraic equation you will be using is:ab

= cd

This equation can be rearranged to solve for a as follows:

a = cbd

2-13 © 1997, A.J. Girondi

The equation to solve for b would be:

b = a dc

Problem 15. Use your algebra skills to solve the following.

a. Write the equation you would use to solve for c.

b. Write the equation you would use to solve for d.

Another algebraic form you will use is ax + b = cx + d. To solve for x in this equation, begin by moving all xterms to the left side of the equation: ax - cx = d - b. Factor out the x term: x(a - c) = d - b. Then solve for x bydividing both sides of the equation by the quantity a - c:

x(a-c)a -c

= d-ba-c

which yields x = d-ba-c

c. Now follow the steps shown above and solve this next problem. Show all of the steps in solving for x.

5x + 2 = 3x + 8

Problem 16. Solve the following problems for x. Show work as well as the answer. Since measurements arenot involved here, do not be concerned about rounding.

a. 8x

= 43

b. 12x

3 = 24

c. 2x + 5

30 =

4 - 2x12

d. 2x - 3 = 1 - 5x

2-14 © 1997, A.J. Girondi

e. 40(x + 4) = 88

f.

16 x1

= 50.0

2

g.

4x + 16

2 = 12

SECTION 2.12 Graphing Skills and Terms

You will be collecting data from time to time that will be more easily interpreted in the form of a graph.There is a correct way to graph data. A well-designed graph should tell the observer a lot about the data collected.It also includes a descriptive title that describes what the graph is all about. A complete key is a good explanationthat allows viewers to know what data they are looking at. The sample graph below is a plot of data collected from alaboratory investigation. Note that each set of data has been graphed with a symbol. This allows each set of datato be easily distinguished from the other. Answer the following questions related to the graph.

What would a good title be for this graph? {2}___________________________________________________

___________________________________________________________________________________

What is the unit used on the vertical, or Y, axis? {3}_______________________________________________

What is the unit used on the horizontal, or X, axis? {4}_____________________________________________

All properly constructed graphs should have equally spaced numbers on both axes. (It is not necessary, however,

to use the same spacing on both axes.) What is the value assigned to each square on the X axis?

{5}_________________ What is the value assigned to each square on the Y axis? {6}____________________

The normal practice to follow when constructing a graph is to place the independent variable on the X axis and the

dependent variable on the Y axis. The independent variable is controlled by the experimenter and the dependent

variable responds accordingly. What variable is being controlled in the graph below? {7}____________________

What is the dependent variable? {8}_______________________

2-15 © 1997, A.J. Girondi

3

2

1

0 5 10 15 20 25 30 35 40

* *

*

x x * x x x * x x x

Volume of Sodium Carbonate Used (mL)

Rate of PrecipitateFormation (g/min)

* = situation 1o = situation 2x = situation 3

Figure 2.1 A Sample Graph

ACTIVITY 2.13 Collecting And Graphing Experimental Data

Before you are given some practice at graphing, it's important to identify two major terms that relate to anyscience investigation. They are variable and control. A variable can be defined as something that changes -usually a property that is being measured. Unlike a variable, a control does not change and thus, gives ussomething with which to compare a variable. To illustrate a variable and a control, get two 250 mL beakers and put100 mL of tap water in each. Place about 30 g of sodium chloride (table salt) in one beaker and stir until most or allof the salt is dissolved. (Note that the quantity is about 30 g. This means it does not have to be exact.) Use amagic marker to label this beaker as the variable. The beaker containing the pure water is the control.

Use two thermometers to measure the initial temperature of the contents of each beaker. In Table 2.2record these original temperatures at time = 0 sec. Now place four ice cubes in each beaker, start timing and startstirring. Measure and record the temperature of the contents of each beaker at 30-second intervals (continuestirring). Record all data in Table 2.2.

What was one thing that was different between your variable and control beakers? {9}___________________

____________________________________________________________________________________

What was the purpose of the control in this experiment?___________________________________________

____________________________________________________________________________________

2-16 © 1997, A.J. Girondi

Table 2.2An Experiment Using Variable and Control Factors

Time (sec) Temp. Water Temp. Water(variable) (control)

0 __________________________________

30 __________________________________

60 __________________________________

90 __________________________________

120 __________________________________

150 __________________________________

180 __________________________________

Keep in mind that the line you draw on a graph does not necessarily have to actually pass through each data point.The best line may actually pass between the points, depending upon the amount of error in your data. It is usuallybetter to draw a smoothly flowing line with curves, rather than a choppy one obtained by trying to connect thepoints. See the example below.

Figure 2.2 A Smooth Curve

2-17 © 1997, A.J. Girondi

Now graph your data on the blank grid below. Follow the procedures outlined in section 2.13 for

preparing graphs. Remember, a well-designed graph has:

1. a descriptive title

2. independent variable on the X axis; dependent variable on the Y axis

3. both axes labeled correctly, including units

4. a key, if more than one line is plotted on the same graph

5. equally spaced increments on both axes.

2-18 © 1997, A.J. Girondi

ACTIVITY 2.14 Cooling Rates Of Evaporating Liquids

To reinforce your skills in data collecting and graphing, complete this lab activity. Collect the necessarydata and graph the results.

1. Obtain a thermometer and wrap somecotton (or paper toweling) around thebulb at the end so that the wrapping isabout as thick as the thermometer itself.Use a rubber band to secure thewrapping to the bulb.

2. Attach a buret clamp to a ring stand,and use a piece of string or a paper clip tohang the thermometer from the clamp.(See Figure 2.3 below.)

3. Read the temperature of thethermometer and record this as thetemperature at time zero in Table 2.3.

4. Obtain a bottle of ethyl alcohol whichis at room temperature.

5. Briefly dip the wadded end of thethermometer into the alcohol. Beginmeasuring time as soon as thethermometer is removed from thealcohol.

6. Read the thermometer every 30seconds and record the temperature inTable 2.3.

7. Continue this process for five minutesor until you obtain two temperaturereadings which are equal.

Table 2.3Time Versus Temperature

Time (sec) Temp. (oC)

0

Prepare a graph of temperature versus time on the grid on the next page. Remember to place theindependent variable on the horizontal (X) axis.

Figure 2.3

2-19 © 1997, A.J. Girondi

What is the title of your graph? _____________________________________________________________

What is the dependent variable in this experiment? {10}___________________

What is the label for the vertical (Y) axis? {11}_______________________

What is the label for the horizontal (X) axis? {12}_________________________

What is the interval you used on the Y axis? __________X axis?_________

Referring to your data, what data did you collect that would be considered quantitative? {13}_________________

Make a general statement about how the temperature changed with respect to time: ______________________

____________________________________________________________________________________

2-20 © 1997, A.J. Girondi


This chapter was designed to lay the groundwork for future chapters. Review Chapter 2 (and Appendix Aif applicable) carefully and when you are satisfied that you understand the learning outcomes listed below, checkoff each one. Arrange to take the quiz or exam on chapter 2.

_____1. Express numbers in scientific notation and back to ordinary form.

_____2. Multiply and divide exponential numbers.

_____3. Use metric units for mass, length, and volume.

_____4. Use dimensional (unit) analysis to solve problems.

_____5. Perform simple algebraic computations.

_____6. Distinguish between variables and controls in an experiment.

_____7. Properly construct and interpret a graph.

You should check the remaining learning outcomes only if you were directed to complete the work in Appendix A.

_____8. Distinguish between those digits in a measurement which are significant and those which are not.

_____9. Express answers to calculations using the correct number of significant digits (also called significant figures).

2-21 © 1997, A.J. Girondi


Questions:

{1} division; {2} Rate of Precipitate Formation With Respect to Volume of Sodium Carbonate Used;{3} g/min; {4} mL; {5} 5 mL; {6} 1 g/min; {7} volume; {8} rate; {9} salt is variable & control used for comparison;{10} temperature; {11} temperature (oC); {12} Time (sec); {13} temperature and time

Problems:

1. a. 1.9 X 105; b. 198,600; c. 5.28 X 102; d. 198,600,000; e. 4.4 X 106; f. 17,500,000; g. 9.7 X 103;h. 33,300; i. 4.9 X 101; j. 25; k. 5.6 X 10-2; l. 0.0000000928; m. 1.13 X 10-1; n. 0.001411;o. 3.5 X 10-7; p. 0.000006275; q. 7.7 X 10-4; r. 0.000279; s. 5.12 X 10-6; t. 0.13

2. a. 1.2 X 1036; b. 3.6 X 1014; c. 6.4 X 1045; d. 2.5 X 101; e. 2.0 X 10-5; f. 1.0 X 102; g. 2.0 X 1016;

h. 2.0 X 100; i. 1.4 X 1010; j. 2 X 1018; k. 7 X 100; l. 1.0 X 103; m. 1.3 X 100; n. 1.0 X 109; o. 4.0 X 1014

3. a. GuEstimate: (5 X 106)(6 X 106) = 3 X 1013 m2; Calculated answer: 3.0079 X 1013 m2

b. GuEstimate: (8 X 104 mm) ÷ (4 X 102 mm) = 2 X 102 mm; Calculated answer: 195 mm or 1.95 X 102 mmc. GuEstimate: (5 X 10-5 cm)(5 X 104 cm) = 2.5 X 100 cm2; Calculated answer: 2.49 cm2

d. GuEstimate: (3 X 104 km)(3 X 102 km) = 9 X 106 km2; Calculated answer: 8.97 X 106 km2

e. GuEstimate: (4 X 105 m)(6 X 1023 m) ÷ (3.00 X 102 m) = 8 X 1026 m; Calculated answer: 7.7 X 1026

4. a. mass; b. volume; c. length; d. mass; e. length; f. volume; g. length; h. mass; i. volume

5. small beaker: 78.96 g, 78,960 mg, 0.07896 kgpencil: 6.783 g, 6783 mg, 0.006783 kglarge flask: 378 g, 378,000 mg, 0.378 kg

6. 2710 mm Hg

7. 20 baby birds

8. 4 warts

9. $1.14/1 doz

10. 3900 mi.

11. a. g; b. miles; c. feet/hr; d. g/L; e. moles2/g; f. g; g. g; h. atm; i. cm2/hec; j. 1

12. a. 0.01250 ton; b. 6.048 X 105 sec; c. 1760 fathoms; d. 7.21 X 105 qts/min

13. a. 9.00 smacks; b. 0.556 lardos; c. 0.0286 sacks; d. 5.8 fleas; e. 672 tolls (670 rounded)

14. a. 1400 g; b. 0.896 L; c. 6.785 X 10-3; d. 458 mm; e. 1 X 106 cm3; f. 4.5 X 106 cm3

15. a. c = ad/b; b. d = bc/a; c. x = 3

16. a. x = 6; b. x = 6; c. x = 0.7; d. x = 0.57; e. x = -1.8; f. x = 1.56 g; g. x = 2

2-22 © 1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


“ALICE”

CHAPTER 3

PHYSICALSTATES

OFMATTER

3-1 © 1997, A.J. Girondi

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3-2 © 1997, A.J. Girondi

SECTION 3.1 Physical Properties of Matter

You have already seen that chemistry is a science that deals with the composition of substancesand the changes in composition that they may undergo. By simply observing a chemical substance, youcannot always determine the composition of that substance. You can usually distinguish one chemicalsubstance from another by experimentation. Sometimes the experiments you perform will be qualitativeactivities, and at other times they will be mainly quantitative.

Chemical substances are usually distinguished by their appearance, taste, odor, feel, and othersimilar properties. Such characteristics are called physical properties. We can recognize specificsubstances by their physical properties. Just as you can recognize your friends by their physicalappearance, you can also recognize chemical compounds by their physical appearance and properties.To obtain a better understanding of physical properties, we will attempt to distinguish between twodifferent metals on the basis of their physical properties.

ACTIVITY 3.2 Identifying Metals By Physical Properties

Get a sample of metal A and a sample of metal B from the materials shelf. Metals A and B look quitesimilar. It would be difficult for the untrained eye to determine the identities of these metals based on theirappearance alone. But, there are ways of identifying these two metals based on their physical propertieslisted in Table 3.1.

If you were to refer to a reference book, such as the Handbook of Chemistry and Physics, youwould probably have a difficult time trying to determine the identity of metal A and metal B. There aresimply too many metals with the same or similar properties because the physical properties that you havedescribed are very general.

Would you categorize the first four observations called for in Table 3.1 as quantitative or qualitative?

{1}_____________________ Why?{2}_______________________________________

There are several physical properties of substances that are more quantitative in nature. Manysubstances are more easily identified by quantitative means. For example, every pure substance (elementor compound) has a specific temperature and pressure at which it boils. If we were to melt metals A and B,we could identify them by comparing their measured melting points with the melting points of metalsfound in a reference book.

Another physical property that is a good quantitative measure is the density of a substance.Density is the mass of a given volume of a substance. You would probably agree that a block of cement isheavier than a block of styrofoam of the same volume. As a result, we would say that the cement is moredense than styrofoam. We will try to identify metals A and B by calculating the density of each.

density =

mass (grams)

volume (mL)

Density can be measured in a very simple way. Themathematical formula for density is given at right. Themeasurements you will need to determine density are themasses of your metals and the volumes occupied by themetals. To determine the densities of metals A and B use thefollowing procedure. .

Procedure:

1. At this time, make the first four observations of metals A and B listed in Table 3.1. Record yourobservations in the table. Following the remaining steps to determine the densities of metals A and B.

3-3 © 1997, A.J. Girondi

2. Obtain a 50 mL graduated cylinder. Add 40.0 mL of water to the graduate. Measure the mass of thegraduate containing 40.0 mL of water. Record this mass on a piece of paper.

3. Add as many pieces of metal A as needed to the water in the graduate to make the total volumebetween 47.0 and 50.0 mL. All of the pieces of metal must be totally submerged. Calculate the volume ofthe metal in the graduate by subtracting the original volume of water (40.0 mL) from the final volumereading. Record this result as the volume of metal A in Table 3.1.

4. Weigh the graduate and content (water + metal). From this final mass subtract the original mass of thegraduate + water. Record this difference as the mass of metal A in Table 3.1.

5. Dry the pieces of metal A with a towel and return them to the materials shelf.

6. Repeat steps 1 through 4 using metal B.

Table 3.1Observed Properties of Metals A and B

Physical Property Metal A Metal B

1. Physical State ________________ ________________(solid, liquid, gas)

2. Color ________________ ________________

3. Luster ________________ ________________(shiny, dull, etc.)

4. Texture ________________ ________________(rough, smooth)

Table 3.2Densities of Some

Common Metals

Metal Density (g/mL)

lead 11.34zinc 7.14mercury 13.59tin 5.75nickel 8.90platinum 21.45aluminum 2.70

Table 3.2 lists various common metals and their densitiesincluding metals A and B. Calculate the density of each metal in thespace provided below. Be sure to include units with your answers.Enter the results into Table 3.1. Compare your calculated densitiesof metals A and B to the densities of the metals in Table 3.2. Find themetals in Table 3.2 whose densities most closely match those formetals A and B. Enter the identities of metals A and B in Table 3.3.

Using the Handbook of Chemistry and Physics or whateverother reference resource is available to you, look up metals A and Band list any other properties that are given. A copy of the handbookmay be available in your lab area and/or in the school library. Enterthe additional properties of these metals into Table 3.3.

3-4 © 1997, A.J. Girondi

Metal A: Show density calculations in space below.

Metal B: Show density calculations in space below.

Table 3.3Identities and Densities of Metals A and B

Metal A. Mass: _____________ g; Volume: _____________ mL

Density: ____________ g/mL; Identity: _____________________________

Metal B. Mass: _____________ g; Volume: _____________ mL

Density: ____________ g/mL; Identity: _____________________________

ACTIVITY 3.3 Comparing the Densities of Liquids

Liquids also have densities, but they are determined in a slightly different way. Use the procedurebelow to determine the density of liquids.

1. Measure the mass of a clean, dry 10 mL graduated cylinder to the nearest 0.01 g. Record the mass inTable 3.4.

2. Obtain a bottle of alcohol, oil, or vinegar from the materials shelf. Carefully measure exactly 10.0 mL ofone of the liquids into the graduate. It is extremely important for you to read the meniscus correctly in thisactivity. Use a dropper to adjust the volume to exactly 10.0 mL, if needed.

3. Measure the mass of the liquid to the nearest 0.01 g, and record this mass in Table 3.4. Discard theliquid into the sink after use.

4. Repeat steps 1 through 3 for the other liquids. When finished with the oil, wash the graduate using asmall test tube brush and some detergent. Rinse well with water.

5. Calculate the mass of each liquid using the data collected. Calculate the density of each liquid bydividing its mass by its volume (10.0 mL). Be sure that each measurement or calculation in your table hasa unit with it.

6. Wash and return all glassware.

3-5 © 1997, A.J. Girondi

Table 3.4C a l c u l a t e d D e n s i t i e s o f S e l e c t e d L i q u i d s

Alcohol Oil Vinegar

Mass of graduate + liquid ________ ________ ________

Mass of graduate ________ ________ ________

Mass of 10 mL of liquid ________ ________ ________

Density of liquid (g/mL) ________ ________ ________

Volume of liquid used _10 mL__ _10 mL__ _10 mL__

Use the data in Table 3.4 to answer the following questions.

1. Which liquid appears to be most dense? _____________________________________________

2. Which liquid appears to be least dense? _____________________________________________

3. If oil is added to water, 2 layers form. One layer is water, while the other is oil. Which substance wouldyou expect to find on the bottom and which on top? (The density of water is about 1.00 g/mL) Explain:

_____________________________________________________________________________

_____________________________________________________________________________

Put about 2 mL of oil into a test tube. Carefully add about 2 mL of water. Observe the behavior of the

water as you add it. Which liquid is on top?_________________ Was your prediction correct?_______

(Use detergent and water again to clean the oily glass.) Discard the liquids into the sink.

4. Specific gravity is the ratio of the density of a liquid compared tothe density of water (1.00 g/mL). Since it is a ratio, specific gravityhas no units. For example, suppose the density of a liquid is 1.80g/mL. The specific gravity of that liquid is calculated as shown atright.

S.G. = 1.80 g/mL1.00 g/mL

= 1.8

Notice that the only difference between the density of a liquid and its specific gravity is thatdensity has units and specific gravity does not. What is the specific gravity of the oil that youused?__________

SECTION 3.4 Phase Change Equations

Phase changes are examples of changes in physical properties. Phase changes can bedescribed in the form of equations. Most solids when heat sufficiently undergo a change in state from asolid to a liquid phase. Most of us are familiar with the process called melting. This phase change can bewritten as an equation:

water(s) ----------> water(l)∆

The ∆ symbol means “heat added” while the subscripts (s) and (l) refer to solid and liquid, respectively.

3-6 © 1997, A.J. Girondi

The reverse process can be written as: water(l) ----> water(s) and is called freezing. Using these twoexamples, write a similar equation to show the phase change involved in boiling. Use the subscript (g) torepresent a gas or vapor. Use any appropriate symbols as well.

{3}_______________________________________________

Condensation is the opposite of boiling. Write an equation to show the phase change involved incondensation.

{4}_______________________________________________

ACTIVITY 3.5 Energy Changes Which Accompany Phase Changes

In this activity you will be observing a phase change firsthand. You are going to obtain data anddraw what is known as a cooling curve. Remember to always wear safety glasses and an apron.

1. Fill a 250 mL beaker about 1/3 full of tap water and bring the water to a boil. Obtain a stopwatch, ifpossible, or have available a wrist watch or classroom clock with a second hand.

2. Obtain a test tube which already contains a small amount (about 2 cm) of paradichlorobenzene (PDB)from the materials shelf. Keep the tube of PDB away from any flame! Your lab should have goodventilation during this activity.

3. Turn off the heat under the beaker of water and place the tube of PDB into the hot water. Allow it toremain there until the solid is completely melted. At that point, place a thermometer directly into themelted PDB. Do not remove the tube of PDB from the hot water until the temperature of the melted PDBis between 65oC and 80oC. (The higher the better within that range if you have time.)

Do not try to use a flame around or under the melted PDB. It is very flammable!

4. When the appropriate temperature has been reached, remove the tube of PDB from the hot water andplace it into a test tube rack on your lab table. Start timing immediately, and record the temperature of thecooling PDB every 30 seconds.

5. Record the temperature as precisely as possible to the nearest 0.5oC, and use the thermometer tocarefully stir the PDB 10 seconds before each reading is taken. You will need a partner to help keep time,read the thermometer, and record the temperatures. Enter all data into Table 3.5.

6. After you have obtained 3 or 4 consecutive readings which are within 1 degree of each other, you canstop if your class is close to ending. However, if you have time you should continue taking readings tocomplete as much of the data table as possible. If the PDB gets too solid to stir, discontinue that. Stopthe readings if the temperature reaches 40oC.

7. When you have collected all needed data, reheat the water in your beaker and put the tube of PDB init. Remelt the PDB so that you can remove the thermometer. Remove the tube from the hot water andallow it to cool. DO NOT remove the PDB from the test tube. Return all equipment to the materials shelf.

3-7 © 1997, A.J. Girondi

Table 3.5Cooling Temperatures

Time (sec) Temp(oC)

____0____ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

Time (sec) Temp(oC)

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

_________ _________

Prepare a graph of your data on the grid below. Record temperatures on the Y axis and time on

the X axis. What was the dependent variable in this activity?{5}____________________ What was the

independent variable?{6}______________________. Compare your graph to the graph of a normal

cooling curve for a pure substance (element or compound) shown in Figure 3.1. Notice how the graph

has been divided into five sections. The data you collected reflects only a part of the overall cooling curve

for PDB. An interesting thing to notice on the cooling curve is that there are two plateaus where

temperature remains constant. Each plateau signifies a phase change (phase changes occur at constant

temperatures). In section B-C of the graph in Figure 3.1, condensation is occurring. Section D-E

represents freezing.

3-8 © 1997, A.J. Girondi

Temp (oC)

A

B C

D E

F

Time (sec) ----->

vapor vapor + liquid liquid liquid + solid solid

Figure 3.1 Cooling Curve of a Pure Substance

If we were to pretend that the curve in Figure 3.1 is a cooling curve for PDB, what letter on the curve mostclosely approximates the area in which you collected data in this activity?{7}__________

Before you go any further, draw a circle around that area of the graph in Figure 3.1 above which bestrepresents the range in which you collected your cooling curve data. Seek help if you need it.

SECTION 3.6 Kinetic and Potential Energy Changes During Phase Changes

You may ask, “Why is the temperature constant while a phase change is occurring?” In order tounderstand the answer, you must be aware that temperature is a measure of kinetic energy - the energydue to the motion of the particles. When particles lose kinetic energy, their temperature decreases (sincetemperature is a measure of kinetic energy). However, when particles move farther apart or closertogether, they gain or lose potential energy which is energy of position. Particles move closer togetherwhen they lose potential energy, and farther apart when they gain potential energy. Potential energychanges do not involve temperature changes.

When molecules have a lot of kinetic energy, their motion keeps them from being drawn closertogether by attractive forces that exist between them. However, if the particles lose kinetic energy, therewill come a point at which the attractive forces will pull them closer together. This involves a loss ofpotential energy which occurs at a constant temperature. The result will be a phase change likecondensation or freezing.

When substances are heated, the molecules move faster and gain kinetic energy as temperatureincreases. Eventually, they may gain enough kinetic energy to overcome some of the attractive forcesbetween them and move farther apart. When they move apart they are gaining potential energy, and theydo this at a constant temperature. This results in a phase change such as melting or vaporization. Thereare plateaus on a warming curve just as there are on a cooling curve. On such a curve, particles are movingfaster when temperature is rising. They are moving farther apart when the temperature is constant duringthe heating process. Pure liquids boil at constant temperatures (at constant pressure) because the

3-9 © 1997, A.J. Girondi

particles are moving farther apart and potential energy is being stored as the liquid becomes a vapor.

Refer back to Figure 3.1 as you answer the following questions. In which sections of the cooling

curve is the temperature changing?{8}____________________________ According to the curve, does

the temperature change or remain constant during a phase change?{9}__________________________

In which section(s) of the curve is the substance changing temperature and getting

cooler?{10}__________________ In which section(s) of the curve are the particles moving

slower?{11}_____________________ In which section(s) of the curve are the particles moving at a

constant rate?{12}______________________ In which section(s) are particles moving closer

together?{13}_______________________

The freezing point of a pure substance (element or compound) is the temperature at which it

changes from the liquid to the solid phase. On the graph, the freezing point (FP) of that substance is the

temperature at which the plateau occurs. Look at the graph you plotted in Activity 3.5. What is your

estimate of the freezing point of the PDB? {14}_______________

Steam at 100oC contains more heat and can, therefore, cause a more serious burn than boiling

water at the same temperature. Explain this in terms of the kinetic and potential energy content of the

steam and of the boiling water. {15}_____________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

Figure 3.2 is a graph showing the warming curve of water. It is the opposite of a cooling curve.Divide the graph into sections. Label with a pencil which phase or phases (solid, liquid, vapor) are presentin each section of the graph, and label those areas of the curve where phase changes (melting andvaporization) are occurring.

According to Figure 3.2, what is the melting point of ice?{16}__________ In the same way, what is

your estimate of water’s boiling point based on Figure 3.2?{17}_____________ How do the melting and

freezing points compare? {18}_________________________________________________________

One rather interesting point about phase changes is that even though phase changes occur, youalways have the same chemical substance you began with. The solid, liquid, and gaseous phases of wateror PDB are easily converted back and forth by simply adding or removing heat energy. (See Figure 3.3)

3-10 © 1997, A.J. Girondi

-25

0

25

50

75

100

125

Time (sec) ------->

oC

Figure 3.2 Warming Curve of Water

LIQUID WATER

LIQUID WATER

WATER VAPOR

ICE

add energy

remove energyremove energy

add energy

Figure 3.3 Energy and Phase Changes

Every pure substance melts and boils at aspecific temperature. Table 3.5 contains a list ofsome selected pure substances. As can be seen,each substance has its own characteristic meltingand boiling point. As a result, some substancesexist as gases at room temperature while others aresolids or liquids. Use the melting and boiling pointinformation provided to determine which of thesubstances listed are solids, which are liquids, andwhich are gases at room temperature (25oC) andnormal pressure.

Table 3.5Melting and Boiling Points

Substance M.P. (oC) B.P. (oC)

Neon -248.7 -245.8Water 0 100.0Acetone -95.0 56.6Sodium 97.5 899.0Ethyl Alcohol -117.3 78.5Copper 1083.0 2582.0

3-11 © 1997, A.J. Girondi

The solids are: {19}________________________________________________________________

The liquids are: {20}________________________________________________________________

The gases are : {21}________________________________________________________________

ACTIVITY 3.7 Physical and Chemical Changes

Now that we have learned the fine points about phase changes, we are going to look at two otherkinds of changes - physical changes and chemical changes. A physical change refers to a change in theappearance of a substance. Like a phase change, when a physical change occurs, you also end up withthe same substance you began with - it just looks different. If you tear a sheet of paper into small pieces,you have brought about a physical change. You still have paper, which is chemically unchanged. Is

tearing a piece of paper also a phase change?{22}_________Why or why not?{ 23}__________________

______________________________________________________________________________

When ice is changed to liquid water and then to steam, is water undergoing a physical change?________

Why or why not? {24}_______________________________________________________________

A somewhat different type of change is called a chemical change. After a chemical changeoccurs, you no longer have the same chemical substance. A new and different chemical substance isformed. If you were to set fire to paper, would you still have paper? Obviously not! Chemical changesresult in the formation of substances with new chemical compositions. There are some key signs that youcan look for to determine whether or not a chemical change has occurred. For example, if a gas is given offor if a precipitate (solid) is formed during a change, you can be fairly certain that a chemical change hasoccurred. Often (but not always) a color change can indicate a chemical change. A temperature changemay or may not indicate that a chemical change has occurred. Color changes and temperature changessometimes occur during physical changes, so they are not clear evidence of chemical changes.

In this activity you will witness some physical and chemical changes. You will need 4 test tubes, atest tube rack, and the chemicals labeled “ALICE 3.7” on the materials shelf. Follow the directions below,and record all of your observations in Table 3.6. In Table 3.6, you must place check marks in the columnthat best describes what happens during each change. You may check more than one column for eachchange as needed. Pay close attention to whether or not a solid substance (precipitate) forms. It willeither lie on the bottom of the tube, or will remain suspended and make the liquid contents cloudy. Clearsolutions do not contain precipitates. (Do not confuse the terms clear and colorless. Clear liquids may ormay not be colorless.)

Chemical Change 1: Rinse a small test tube with distilled water. Add about 20 drops of AgNO3 (solutionA) to the tube. Add about 10 drops of K2CrO4 solution (solution B) to the same tube. Record anyobservations in Table 3.6 by placing check marks in the proper boxes.

Chemical Change 2: Add about 10 drops of Pb(NO3)2 (solution C) to a small clean, rinsed test tube. Addabout 20 drops of NaI (solution D) to the same tube. Shake the contents gently. Record observations.

Chemical Reaction 3: Add about 2 mL of HCl (solution E) to a clean 150 mm (medium) test tube. Place asmall piece of zinc metal into the tube. Record observations.

3-12 © 1997, A.J. Girondi

Chemical Reaction 4: Add about 2 mL of H2SO4 (solution F) to a 150 mm test tube. Note the temperatureof the solution by feeling the outside of the tube with your hand. Next, add about 4 mL of NaOH solution(solution G). Caution: keep these solutions off skin and out of eyes! Feel the tube again. Record.

Table 3.6Evidence of Chemical Changes

Color Change Gas Given Off Precipitate Forms Temp. Change (yes or no) (yes or no) (yes or no) (yes or no)

Change 1:

Change 2:

Change 3:

Change 4:

In chemical change 4, an energy change was the only easily noticeable evidence of a chemicalchange. However, energy changes by themselves are not enough evidence to indicate for sure that achemical change has occurred, because many physical changes are also accompanied by energychanges. To observe a physical change in which there is an energy change, place about 2 mL of water ina clean test tube. Note the general temperature of the water by pressing the tube on the inside of yourwrist. Using forceps, place 2 or 3 pellets of solid sodium hydroxide (NaOH) into the tube of water. Shakethe tube gently for 30 seconds or so, and press the bottom of the tube on the inside of your wrist. Do notlet the solution or the NaOH pellets contact you! Note the relative temperature of the contents of thetube. NaOH does not react with water, it merely dissolves. Dissolving is a physical change. What change

in temperature did you observe?{25}____________________________________________________Discard the contents of the tube into the sink and run the water. Rinse the tube with water.

Your experience from this activity should tell you that if you note a color change, a gas beingproduced, a temperature change, or the formation of a precipitate, you should at least suspect that achemical change has occurred. This is particularly true if more than one of these events has occurred,because color and temperature changes themselves are not conclusive evidence.

Identify each of the following changes as either physical (P)or chemical (C).

a. getting a flat tire {26}___________

b. a campfire burning {27}___________

c. seeing your “breath” when you exhale on a cold day {28}___________

d. digesting food {29}___________

e. rain changing to ice {30}___________

f. drying the laundry {31}___________

g. a bomb explosion {32}___________

3-13 © 1997, A.J. Girondi

SECTION 3.8 Density Problems

Earlier in this chapter, you worked with the concept of density as a physical property of matter. Dothe problems below. The units used to express density are usually g/mL or g/cm3. The g/mL unit is usedmore frequently for liquids, while g/cm3 is used more often for solids. This is really just a matter ofpreference, since 1 mL and 1 cm3 represent exactly the same volume. That is, 1 mL = 1 cm3. Usedimensional analysis wherever possible. Show all work neatly, even if only one or two steps are needed.

Problem 1. What is the density of gasoline if 350 mL has a mass of 245 grams?

Problem 2. A block of metal has the dimensions 5.0 cm X 7.0 cm X 20.0 cm, and its mass is 5.0 kg.What is the density of the metal in g/mL?

Problem 3. What is the density of hydrogen gas (in g/mL) if 100. liters have a mass of 8.93 X 10-3 kg?

Problem 4. What is the density of milk in g/mL if 1.0 quart has a mass of 1.0 kg? (1.06 qt. = 1.00 L)

ACTIVITY 3.9 Determining the Thickness of Aluminum Foil

In this activity you will use what you have learned about density to calculate the thickness ofaluminum foil. Now you can imagine how tough this would be to do using only a ruler. Aluminum foil isn’tvery thick! We will accomplish this task in an indirect way. The formulas you will use are already familiar toyou. The volume of a regular object can be found using the formula V = L x W x H, where L = length, W =width, and H = height. If the object is a piece of aluminum foil, we can alter the formula to V = L x W x T,where T = thickness. Since this activity involves finding the thickness, we can solve the formula for Twhich gives T = V/(L x W). From this formula you can see that in order to determine the thickness, youneed to know the length and width of the piece of foil, and you also need to know the volume of it.

3-14 © 1997, A.J. Girondi

How do you calculate the volume of a piece of aluminum foil? That’s where density becomesuseful. The density (d) of aluminum is 2.70 g/1 cm3. If you measure the mass of the foil and if you knowthe density of the metal, then you can use the formula at right below to find the volume:

density = g

cm3V = g X

cm3

g= cm3 V = g X

1

dSince: and since: then:

Use this formula to get the volume of the foil

Do you see how these two formulas can be used to derive this one?

T = V

L x W

Once you have the volume, you can substitute that value into theformula shown at right. Measure the length and width (in cm)of therectangular piece of foil, and substitute those values along with thevolume into the formula. When you solve the formula for T, the units willcancel and leave you with centimeters. You will have calculated thethickness using the mass and density!

Procedure: Cut or obtain 2 square pieces (about 10.0 cm x 10.0 cm) of aluminum foil - one regular andone heavy duty. Weigh each piece on the laboratory balance, and measure the length and width of eachpiece in centimeters. Enter the data into Table 3.7 below. Return the foil squares to the appropriatecontainers. Complete the calculations, showing all work neatly in the spaces provided.

Table 3.7Aluminum Foil Data

Regular or Heavy Duty Length (cm) Width (cm) Mass (g)

_________________ _________ _________ _________

_________________ _________ _________ _________

SHOW your calculations in the spaces below. Express results to two decimal places.

Step 1. Calculation of the volume of the foil.

a. regular foil

b. heavy duty foil

Step 2. Calculation of the thickness of the foil.

a. regular foil

b. heavy duty foil

3-15 © 1997, A.J. Girondi

Step 3. According to your calculations, determine how many times thicker the heavy duty foil is comparedto the regular foil. (Divide the thickness of the heavy duty foil by the thickness of the regular foil.)

Result: the heavy duty foil is _____________ times thicker.

SECTION 3.10 Accuracy, Precision, Percentage Error and Percentage Deviation

Let’s use your data from Activity 3.9 to determine your accuracy and your precision . Accuracyrefers to the closeness of your result to the actual or accepted value. Precision refers to the closeness ofyour results to each other (if you did an experiment more than once) or to that of other lab groups who didthe same experiment. It is possible to be precise without being accurate. In other words, you can get aresult that is close to what others got, while still being inaccurate. This can happen if your class is usingpoor measuring instruments or a bad technique. However, in order to be accurate you must also beprecise. In other words, if everyone in your class is getting a result which is close to the accepted value(accurate), then those results will also be close to each other (precise). Accuracy is often expressed aspercentage error, while precision is often expressed as percentage deviation .

Step 1. Calculation of Error. Obtain the accepted value for the thickness of the two kinds of aluminum foilfrom your instructor. Calculate your error and percent error as shown below. Quantities enclosed invertical bars such as: |O - A| refer to absolute value.

Error = |O - A| O = your observed value A = the accepted value

% error =

|O - A|

A X 100

Be careful when you use the formula above to calculate % error. It involves subtraction, division, andmultiplication. When using a calculator to solve such equations, you must do all your addition andsubtraction and then get a subtotal before doing the multiplication and/or division. If you follow the rulesfor significant digits, you will need to round twice. List your error results below.

% error for the regular foil: __________% % error for the heavy foil: __________%

In this experiment, you should be able to achieve less than 5% error. How did you do? ______________

______________________________________________________________________________

Step 2. Calculation of Deviation. Get thickness results from three other lab groups in your class. Enterthe data into Table 3.8. Calculate the average (mean) thickness for each foil. To calculate your deviation(precision) use the formulas below.

Deviation = |O - M| O = your observed value M = average (mean) value

% deviation =

|O - M|

M X 100

Note that the calculation of deviation also involves an absolute value. Use the same precautions whenusing a calculator as you did when calculating % error. List your deviation results below Table 3.8.

3-16 © 1997, A.J. Girondi

Table 3.8Lab Group Data

Thickness of Thickness ofGroup Regular Foil (cm) Heavy Foil (cm)

Your Group ____________ ___________

Lab Group 1 ____________ ___________

Lab Group 2 ____________ ___________

Lab Group 3 ____________ ___________

Average (Mean) ____________ ___________

% deviation for the regular foil: __________% % deviation for the heavy foil: __________%

Did you manage to achieve a small deviation (5% or less)? ___________________________________

If your group had a small error but a high deviation, that means that your small error was probablyjust due to luck, or maybe off-setting errors. If your group had a large error (over 5%) but a small deviation,that means that your high error was due to the measuring instruments or the procedure – not to humanerror. If your group had a small error and a small deviation, that means that the instruments and your labtechniques and skills were good (you were both accurate and precise)! If you got a large error and also alarge deviation, plan to pursue a career in something other than science, because the error was all your

fault! Comment on the performance of your group: ________________________________________

______________________________________________________________________________

______________________________________________________________________________

SECTION 3.11 Review Problems

Problem 5. A piece of nickel sheet metal has a width of 11.5 cm, a length of 4.66 cm, and a mass of58.18 grams. What is the thickness of this sheet of nickel metal? (Use any needed data from Table 3.2.)

3-17 © 1997, A.J. Girondi

Problem 6. A student experimentally tried to determine the thickness of the metal mentioned inproblem 5. Her observed (O) result was 0.126 cm. The results from three other students were 0.130 cm,0.114 cm, and 0.134 cm.

a. Calculate her % error. (Your answer to problem 5 can be used as the accepted (A) value.)

b. Calculate her % deviation. (You will need to calculate the mean, M, of her value and the three others.)

c. Based on her % error and % deviation, comment on these results assuming an acceptable value of 5%

for both error and deviation. _________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

Problem 7. The “Miracle Thaw” is a heavily advertised sheet of “space age” metal on which you canplaced frozen foods for rapid defrosting (according to the manufacturer). The Miracle Thaw is 45 cm long,25 cm wide, and 0.50 cm thick. It has a mass of about 1519 grams. What is the “space age” metal whichcomposes this “miracle?” Use data from Table 3.2. (Let the consumer beware.)

Problem 8. A cork cube weighs 500.0 grams and has a density of 0.025 g/cm3. Calculate the length ofone side of the cube in cm. Hint: the volume formula is V = s3 where s is the length of a side. You willneed to take a cube root in this problem. Use the appropriate key(s) on your calculator.

3-18 © 1997, A.J. Girondi

ACTIVITY 3.12 The Barge Problem

--> --> --> Teacher Demonstration <-- <-- < --

With the background you have gained thus far in this chapter, explain what would happen in thesituation described below. You must use the concept of density in your explanation. Once you haveformulated your explanation, you will carry out an experiment to illustrate what actually happens. Take thedensity of water as 1.00 g/mL , and keep in mind that , in order to float, a boat must displace a volume ofwater equal to its own mass.

A barge is hauling metal beams through the Panama canal. While the barge is in a closed lock inthe canal, all of the metal falls overboard into the water in the lock. Do you think the water level in the lockwill rise, fall, or remain unchanged when the metal falls in? ___________________________________

Your teacher will demonstrate an experiment to test your hypothesis. A 100 mL beaker will be

used as the barge, a 400 mL beaker about half-full of water for the lock, and pieces of lead metal for the

beams. Enough metal will be used to make a difference, but not to sink the barge! A piece of tape or a

grease pencil will be used to mark the water level before and after the metal spills. Describe the result of

the experiment:_________________________________________________________________

Do the results support your original hypothesis?____________ Whether your answer is yes or no,

explain this phenomenon: _________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

_____________________________________________________________________________

ACTIVITY 3.13 Sublimation

--> --> --> Teacher Demonstration <-- <-- < --

Because toxic fumes are involved, your teacher will conduct this demonstration under an exhaustfume hood in your lab area. You are going to witness a rather unusual type of phase change. It involvessolid iodine crystals which change directly to a vapor without going through the liquid state. This is knownas sublimation. You may be familiar with dry ice (which is solid CO2) which also undergoes sublimation.Care must be taken because iodine vapors are very dangerous and can cause serious stains and burns.

Look at Figure 3.4. Note that an ice cube is placed on top of a watch glass which is located on topof a beaker or evaporating dish containing some iodine crystals. As heat is applied, the iodine crystalsslowly turn into a deep violet-colored vapor which is much heavier than air. When the vapor comes intocontact with the cool underside of the watch glass it turns directly from the vapor phase back to the solidphase in a phase change known as deposition. Crystals may also form on the cool sides of the container.Do not allow iodine to come into contact with you or your clothing! On the line below, write a phasechange equation for the sublimation of solid iodine. Remember to use the symbol for added heat.

{33}_______________________________________

3-19 © 1997, A.J. Girondi

ice cube

watch glass

deposited crystals

iodine vapor

iodine crystals

Figure 3.4 The Sublimation of Iodine Crystals


Read through the learning outcomes below. Place a check mark in front of each outcome that youhave mastered. When complete, arrange to take the test on Chapter 3, and move on to Chapter 4.

_____1. Identify general physical properties of matter (color, phase, texture, luster, etc.)

_____2. Solve problems involving the density of a solid or a liquid.

_____3. Given proper information, determine the specific gravity of a liquid.

_____4. Label, interpret, and explain the parts of heating and cooling curves for pure substances.

_____5. Distinguish between physical changes, chemical changes, and phase changes.

_____6. Distinguish between accuracy and precision, and calculate percentage error and percentage deviation given the necessary information.

3-20 © 1997, A.J. Girondi


Questions:

{1} qual; {2} no measurement involved; {3} water(l) ---∆--> water(g); {4} water(g) ------> water(l) ;{5} temperature; {6} time; {7} area around D and up to E; {8} A-B, C-D, E-F; {9} remains constant{10} A-B, C-D, E-F; {11} A-B, C-D, E-F; {12} B-C, D-E; {13} B-C, D-E; {14} about 53oC;{15} Steam at 100oC has the same amount of kinetic energy but has more potential energy than water at100oC since it has gone through an additional phase change.{16} 0oC; {17} 100oC; {18} the same; {19} sodium, copper; {20} water, ethyl alcohol, acetone; {21} neon;{22} no; {23} no change in phase - still a solid; {24} yes, because it is still water in each phase{25} increase; {26} physical; {27} chemical; {28} physical; {29} chemical; {30} physical; {31} physical;

{32} chemical; {33} I2(s) ----∆----> I2(g)

Problems:

1. 0.70 g/mL

2. 7.1 g/mL

3. 8.93 X 10-5 g/mL

4. 1.1 g/mL

5. 0.122 cm

6. a. 3.28 %b. 0 %c. Since both error and deviation were under 5% in this type of activity, she has done well. The fact that there was 0% deviation indicates that the 3.28% error was probably due to the equipment or procedure.

7. aluminum

8. 27 cm.

3-21 © 1997, A.J. Girondi

SECTION 3.16 Student Notes

3-22 © 1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 4

THE GASLAWS

Boyle's LawCharles' Law

Gay-Lussac's LawThe Combined Gas Law

4-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


4-2 ©1997, A.J. Girondi

SECTION 4.1 Kinetic Theory and the Properties of Gases

We know from chapter 3 that matter normally exists in three forms - solid, liquid, and gas. While afourth state, plasma, does exist, it is not in our realm to be able to study this form of matter firsthand. Wecan study the other three, however. This chapter will deal with gases, exclusively. The other states ofmatter will be covered more extensively in other chapters.

If we attempt to define gas, we can say that it is a state of matter made up of particles that distributethemselves uniformly throughout the space in which the gas is confined. This differs from a liquid and asolid. A liquid takes the shape of the container that it is in. In addition, a liquid does not distribute itselfuniformly in its container. Figure 4.1 will help you to visualize this fact. A solid, of course, retains its ownshape regardless of the shape or size of the container into which it is placed.

GAS

Figure 4.1 The Three Phases of Matter

LIQUID SOLID

Gases have three basic characteristics. They are: (1) expansibility, (2) diffusion, and (3)compressibility. Expansibility is the ability of a gas to expand to the shape and volume of its containerwhether it is a small beaker, a room, or something larger. The opposite of expansibility is compressibility.A gas can be compressed to fill a volume many times smaller than its original volume. Diffusion is the abilityof a gas to move from a region of high concentration to a region on low concentration. If you were to opena bottle of perfume in a room, it would evaporate into a gas, and you would soon be able to smell the odorat all points in the room. The perfume vapor molecules have diffused through the air molecules to all partsof the room.

The properties of gases just described provide the basis for the kinetic-molecular theory. Thistheory gets its name from its close tie with the concept of kinetic energy. Any object that is in motion haskinetic energy. The faster something is moving, the more kinetic energy it has. The kinetic-moleculartheory helps to explain why gases behave the way they do.

The main assumptions of the kinetic theory are:

1. Gases consist of molecules in continuous, random motion. The molecules collide with each other andwith the walls of the container. The pressure exerted by the gas is caused by the collisions of the gas particles with the walls of the container.

2. Molecular "collisions" are elastic. This means that although there is a lot of "bouncing off" going on asthe molecules approach each other, there is no energy lost to friction when the gas molecules "collide" with each other.

4-3 ©1997, A.J. Girondi

3. Gas molecules in a container move at a variety of different speeds; but, overall, an increase intemperature increases the average speed (kinetic energy) of the molecules. This increase in the speed of the molecules results in more frequent and more forceful collisions between the molecules and thewalls of the container. Consequently, gases exert more pressure at higher temperatures.

How does the kinetic theory explain the gas property of diffusion? {1}___________________________

______________________________________________________________________________

______________________________________________________________________________

How does the kinetic theory explain the fact that gases exert pressure? {2}________________________

______________________________________________________________________________

______________________________________________________________________________

ACTIVITY 4.2 Measuring the Effect of Pressure on Gas Volume

The background you now have is sufficient to introduce you to the gas laws. There are several ofthese. The first one we will look at was named after Robert Boyle, the scientist who first discovered thatthere was a definite relationship between the pressure and the volume of a gas. Boyle's Law simplystates:

"At a constant temperature, if the pressure on a gas is increased, thevolume of the gas will decrease proportionately."

Figure 4.2 illustrates the laboratory setup forinvestigating the relationship between the pressure and volumeof a gas. Your teacher has a similar setup for you. Once you havegotten the equipment and materials, begin by setting the end ofthe piston on the 30 cm3 (or cc) mark. (To set the piston whereyou want it, air must be allowed to escape from the syringe. Thewire that is attached to the syringe is there for that purpose.Place the wire into the syringe before inserting the piston. Thiswill allow air to escape. Then, remove the wire while holding thepiston in place.) If you gently push down on the piston and thenlet up, the piston should return to its starting point. Record yourstarting point in Table 4.1. Once you have done this, add 1 "unit"of mass to the platform top (a unit of mass can be a textbook. Ifyou decide to use textbooks, be sure they are all the same).Record the new volume in Table 4.1. Continue adding "units" ofmass to the plunger by 500 g or 1 book increments until youhave 5 or 6 readings. Record each new volume in theappropriate space in Table 4.1.

Figure 4.2

What happens to the volume of the gas as you increase the mass (pressure)? {3}_____________________________

______________________________________________________________________________

Does your result above agree with the general statement given for Boyle's law in section 4.1? Explain.

{4}___________________________________________________________________________________________________________________

______________________________________________________________________________

4-4 ©1997, A.J. Girondi

It was mentioned that temperature must remain constant for Boyle's law to be valid. At what temperature

did you conduct your experiment? ___________oC. Do you think it remained constant? ____________

Table 4.1Effect of Pressure on Volume

Pressure Applied (books) Volume (cm3)

1 ___________

2 ___________

3 ___________

4 ___________

5 ___________

On the grid below, prepare a graph of the information in data table 4.1. Your plot should actuallybe a slightly curved line. Do not try to make it absolutely straight. Remember to follow the instructionsgiven in chapter 2 regarding the correct way to construct a graph.

4-5 ©1997, A.J. Girondi

Your plot is slightly curved. It would be even more curved if we had used a greater range of pressures.

You may wonder why your plot is not a straight line. When two variables are both changing at constant

rates, their resulting graph will be a straight line. However, in this activity, only one variable was changing at

a constant rate. Was it pressure or volume?{5}_________________. Was this the dependent or the

independent variable? {6}__________________________. You may wonder why both variables don't

change at constant rates. Well, as pressure continues to increase in equal amounts, the amount of

decrease in volume gets smaller and smaller. In other words, as pressure changes at a constant rate,

volume changes at a rate which is not constant. Why is this? Gas atoms or molecules – like all atoms and

molecules – have electrons. These electrons are negatively charged and repel each other when they get

too close. As gas particles get closer to each other, the repulsion between them gets stronger and it's

more difficult to force them closer together. In other words, each additional "book" of pressure, has less

effect on the volume of the gas.

SECTION 4.3 Solving Problems Involving Boyle's Law

Boyle's law can be expressed mathematically as: P1V1 = P2V2. This equation means that theproduct of the pressure (P) and the volume (V) of a gas remains constant, provided the temperature doesnot change. (If temperature were a variable, it would have to be included in the equation). So, eventhough P and V can change, the product, PV, maintains a constant value. When the product of twovariables remains constant, the variables are said to be inversely proportional, meaning that as oneincreases, the other decreases.

The sample problem below illustrates how Boyle's law works, mathematically. Study it carefully.

Sample problem: A quantity of air has a volume of 100. mL (V1) at 720. mm of pressure (P1). What willthe new volume (V2) be if the pressure is changed to 760. mm (P2)? Follow the steps below.

1. Write the formula P1V1 = P2V2.

2. Solve the equation for the unknown quantity (V2):

V2 =

P1V1P2

3. Substitute the values into the formula including units:

V2 =

(720. mm)(100. mL)760. mm

4. Do the math: V2 = 94.7 mL.

Solve the following problems. Show the set-up for the problem work, and be sure to include units on allmeasurements used in the problem and given in the answer. (The period appearing after some numbersindicates that the trailing zeros are significant figures.)

4-6 ©1997, A.J. Girondi

Problem 1. A gas has a measured volume of 100. mL under a pressure of 740. mm Hg. What would thevolume be under a pressure of 780. mm Hg at constant temperature?

Problem 2. A sample of gas is confined to a 100. mL flask under a pressure of 740. mm Hg. If this samegas were transferred to a 50.0 mL flask, what would the resulting pressure be?

Problem 3. You are given a gas that you measure under a pressure of 720. mm Hg. When the pressureis changed to 760. mm, the volume becomes 580. mL. What was the original volume of the gas?

Problem 4. At a certain original pressure, the volume of a given amount of air is 134 mL. If the pressureis changed to 1200. mm Hg, the volume is reduced to 45.0 mL. What was the original pressure?

Problem 5. A balloon has a volume of 800. mL when it is held at sea level where the pressure is 760.mm Hg. Calculate the volume of the same balloon after it has floated up to where the atmosphericpressure is only 720. mm Hg.

According to your calculations, does the balloon get larger or smaller? {7}________________________

4-7 ©1997, A.J. Girondi

ACTIVITY 4.4 Measuring the Effect of Temperature on Gas Volume

Gas volumes are also affected by temperature changes. The gas law that explains this is calledCharles' law. Charles' law states that:

"As the absolute temperature of a gas increases, its volume alsoincreases proportionately, when the pressure remains constant."

As we discuss Charles' law, we will explain the meaning of the words absolute and proportionately as theyare used above.

One way that we can see Charles' Law demonstrated in the lab is to use the piece of equipmentpictured in Figure 4.3. It consists of a piece of glass capillary tubing which contains a small bead of mercurymetal. One end of the tube is sealed, while the other is open. A volume of air is trapped in the tubebetween the sealed end and the mercury bead. As the volume of the trapped air changes, the mercurybead will move up or down the tube. The volume of trapped air can be calculated if you assume that thecontainer is a cylinder. The formula for the volume of a cylinder is V = π r2h , where π = 3.1416. Notice thatyou will need to know the internal radius of the capillary tube as well as the height (or length) of the aircolumn between the sealed end of the tube and the mercury bead. Measure the internal diameter of thetube in mm. Half of that is the internal radius, r. All measurements should be made in millimeters, whichwhen substituted in the formula above will give volumes in units of cubic millimeters (mm3).

h

Figure 4.3 Charles' law apparatus

Obtain a thermometer, a metric ruler, and a capillary tube with mercury bead from the materialsshelf. Measure the height of the air column between the closed end of the tube and the mercury bead inmm at room temperature. Record this height and temperature in Table 4.2. Next, place the tube andthermometer in a large beaker or flask of hot water at constant temperature (between 80oC and 90oC) sothat most of the trapped air column is immersed. When the mercury bead stops moving, read thetemperature. Without removing the tube from the hot water, measure the height of the air column in mmagain. Record the data. Finally, place the capillary tube and thermometer in an ice water bath which will beavailable in the lab. The mercury bead will move down the tube. When it has stopped moving, measurethe temperature and, without removing the tube from the ice water, measure the height of the air columnin mm using a ruler. Record all data in Table 4.2. Return all materials to the proper place. Using your data,calculate the volume of the trapped air in mm3 at each of the three temperatures.

Table 4.2Temperature Versus Volume

Temp (oC) Air Column Height (mm) Air Volume (mm3)

________ _________________ _____________

________ _________________ _____________

________ _________________ _____________

internal radius of tube = ________ mm

4-8 ©1997, A.J. Girondi

Using your data in Table 4.2, plot a graph of temperature (X axis) versus volume (Y axis) on the gridbelow. The scales on each axis do not necessarily have to start with zero. You may start each scale at anyconvenient value. After you have plotted your three points, draw the best straight line that you canthrough them. (We know from experiment, that this curve has to be a straight line, so use a ruler to drawthe best straight line you can through your points. Do not expect this line to actually intersect with each ofyour points.)

Temperature is a measure of the amount of kinetic energy that particles have. The more kinetic energy

that particles have, the faster they move and the harder they "collide" with the walls of their containers.

What effect does increased temperature have on the pressure exerted by a gas? {8}_________________

______________________________________________________________________________

The graphic representation in Figure 4.4 on the next page depicts the relationship between thevolume of gas and its temperature. Note that volume decreases as temperature decreases. Study thiscarefully.

Do your data seem to follow a similar trend?_________. What happens to the volume of a gas as thetemperature goes up and down?

As you examine the graph in Figure 4.4, note that eventually a temperature is reached where thevolume of the gas appears to go to zero. Actually, it is impossible for a real gas to have a volume of zerobecause no matter how close the molecules get to each other, they have volume themselves. In addition,repulsive forces between the molecules limit how close they can get to each other. These factors are notaccounted for in the formula for Charles' law. Scientists have invented the concept of an "ideal" gas whichis a gas that obeys the gas laws, including Charles' law, at all times and under all conditions.

4-9 ©1997, A.J. Girondi

300 mL

200 mL

100 mL

VOLUME

-273 Temperature in oC 0 +50

A gas with a volume of 273 mL at 0oC (273 K)

...should occupy zerovolume at -273oC (0 K)

coldestpossible

temperature

Figure 4.4 Volume-Temperature Relationship

While ideal gases do not really exist, the ideal does help us to understand the gas laws. It ispossible for an "ideal" gas to have a volume of zero. Note that this would be the case at -273oC. It isimpossible for any gas including an ideal gas to have a negative volume. Therefore, at -273oC a gas wouldhave the smallest volume possible, and, therefore, -273oC (or 0 K) must be the coldest possibletemperature which a gas - or anything else for that matter - can have. If colder temperatures were possible,then gases would have to have negative volumes! This temperature, -273oC, has come to be known asabsolute zero. It is the starting point for the Kelvin temperature scale. (-273oC = 0 K) Zero on the Kelvinscale (0 K) is absolute zero. (The degree symbol is not used when expressing Kelvin temperatures.) Atabsolute zero, the molecules have no more kinetic energy. Although we have been able to get very closeto this temperature in laboratories, it has never been reached.

SECTION 4.5 STP - What It Is and Why We Need It

Let's say that you have two one–liter containers. One contains gas A and the other contains gasB. Since a gas expands to fill its container, the volume of each gas is one liter. Which container holds thegreatest amount of gas – meaning the greatest number of gas molecules?

1.00 Liter

GAS A

1.00 Liter

GAS B

28.0oC & 700 mm Hg 26.0oC & 662 mm Hg

Which of these 1.00 Liter Containers Holds More Gas?

4-10 ©1997, A.J. Girondi

Just because the two containers have the same volume, they do not necessarily contain the samequantities of gas. After all, most of the volume occupied by a gas is empty space. A one–liter containercan be filled by many gas molecules or only a few. In order to determine which container holds the mostmolecules, you would need to know the temperature and pressure of each gas. These factors affect howmuch space exists between the molecules and, therefore, how much gas you can get into the container.Only if both of our gases (A and B) are at the same temperature and pressure would both one–litervolumes contain the same number of gas molecules. (This is known as Avogadro's hypothesis, and youwill learn more about it in Chapter 8.) If the two gases are not at the same temperature and pressure, thenwe will have to determine what the volumes of the gases would be if they were at the same temperatureand pressure. Only then could we determine which container holds the most molecules. When this isnecessary, the conditions of temperature and pressure which have been chosen for use are 0oC and 760mm Hg (millimeters of mercury) pressure. These conditions are known as standard temperature andpressure, or STP.

0.835 Liter

GAS A GAS B 0.795 Liter

Converting to STP RevealsThat There is More of Gas A

0oC & 760 mm Hg 0oC & 760 mm Hg

Why did we develop the concept of STP? {9}___________________________________________________________________

______________________________________________________________________________

The units of mm Hg (millimeters of mercury) for measuring pressure may seem somewhat mysticalto you. If you fill a test tube with water and put your thumb over it, and invert it into a shallow dish of water,you will notice that the water will not drain out. This is because atmospheric pressure is "pushing up" onthe water column. In fact, atmospheric pressure is great enough to support a column of water over 34 feethigh! As the atmospheric pressure changes from day to day, the height of the water column which it cansupport varies. Thus, we could use the height of the water column as a measure of pressure in millimetersof water. This device would be a barometer. But, who wants a barometer that is over 34 feet high?Mercury metal is about 13.5 times more dense than water, and can be used in place of water in abarometer.

760 mm

Figure 4.5Mercury Barometer

On an average day at sea level, the atmosphere supports acolumn of mercury 760 mm high. This is more manageable,and is why mercury barometers are found in labs and atweather bureaus. We can, therefore, measure atmosphericpressure by measuring the height of a column of mercurywhich it can support (units of "mm Hg"). We will use mm fromnow on, but "mm" really implies "mm Hg." 760 mm is alsoknown as 1 atmosphere Figure 4.5 (1 atm) of pressure. (760mm Hg = 1 atm)

Hg

4-11 ©1997, A.J. Girondi

We can summarize standard conditions as follows:

STP (Standard Temperature and Pressure) Means: pressure = 1 atm or 760 mm Hgtemperature = 0oC or 273 K

When solving gas law problems that involve temperature changes, it is always necessary to usethe absolute, or Kelvin scale. Why? Well, you see the zero on the Kelvin scale really means zero. At zeroon the Kelvin scale, molecules have lost all their kinetic energy. The zeros on the Fahrenheit and Celsiusscales do not really mean zero. At those temperatures, the molecules still contain quite a bit of kineticenergy, which is why 0oF and 0oC are not the lowest temperatures on those scales. For this reason, onlythe Kelvin scale can be used in gas law calculations. We know from Charles' law, that as the temperature ofa gas goes up, its volume goes up in proportion. In other words, if the temperature of a gas doubles, thenits volume also doubles.

Let's say that you have 1 L of a gas at 20oC. What will the volume be if you heat the gas to 40oC?You might think that the new volume would be 2 L. After all, doesn't the temperature double from 20oC to40oC? And shouldn't the volume, therefore, double from 1 L to 2 L? The answer is no, because a changefrom 20oC to 40oC is NOT a doubling of the temperature. Remember that 0oC does not really mean 0. Itwould be a doubling if 0oC really meant 0. The real zero point on the Celsius scale is -273oC. Is 40oC twiceas far from -273oC as is 20oC? The answer, of course, is no! However, if we repeat the problem usingKelvin temperatures, it will work. If you have 1 L of a gas at 20 K, and you heat it to 40 K, what will the newvolume be? It will be 2 L. 40 K is, indeed, twice as much as 20 K. Remember, you must ALWAYS useKelvin temperatures when solving gas law problems.

The conversion of temperatures in oC to the Kelvin scale is easy to do. Simply add 273o to theCelsius number. Follow the examples below.

oC + 273 = K25oC + 273 = 298 K

What is the volume of an "ideal" gas at a temperature of 0 K? {10}____________________________________

Problem 6. For additional practice, convert the temperatures below to Kelvin.

a. 56oC _________ b. -34oC __________ c. 12.3oC __________

Problem 7. Convert the temperatures below to oC.

a. 123 K _________ b. 35.6 K __________ c. 358 K ___________

SECTION 4.6 Solving Problems Involving Charles' Law

Like Boyle's law, Charles' law can be represented in mathematical form. One form is the following:

V1T1

= V2T2

(When the ratios of two variables are equal, the variables are directly proportional – as one increases, theother increases.)

By substituting into the formula, you can calculate the final volume from the change in temperature andthe original volume. Now let's look at a sample problem:

4-12 ©1997, A.J. Girondi

Sample problem: A quantity of air has a volume of 80.0 mL (V1) at a temperature of 20.0oC (T1). Whatwill the new volume (V2) be if the temperature is changed to 100.oC (T2), assuming the pressure remainsconstant?

Follow these steps to solve this problem:

1. You must ALWAYS use Kelvin temperatures when solving gas law problems!!! Make the changes!!

20.0oC + 273 = 293 K; 100.oC + 273 = 373 K

2. Write out the formula, and solve for the unknown variable. In this case, that would be V2.

V2 =

V1T2T1

3. Substitute correct values into the formula, and do the math.

V2 =

(80.0 mL)(373 K)293 K

= 102 mL

To gain practice with Charles' law, solve the problems below. Show set-up as well as the answers.Assume that pressure remains constant in all problems.

Problem 8. At 60.0oC, a gas has a volume of 600. mL. What is the volume of this gas at 10.0oC?

Problem 9. If 105 mL of oxygen at 25.0oC were heated until its volume expanded to 120. mL, whatwould its final temperature (in oC) be?

Problem 10. A quantity of hydrogen has a volume of 103 mL at a temperature of 20.0oC. To whattemperature (oC) would this gas need to be cooled in order to reduce the volume to 92.0 mL?

4-13 ©1997, A.J. Girondi

Problem 11. If 136 mL of nitrogen at 25.0oC is cooled to 0.00oC, what will the new volume be?

Problem 12. A helium balloon has a volume of 2.50 L at a temperature of 25.0oC. What temperature(oC) is needed to decrease the size of the balloon to 1,800. mL?

SECTION 4.7 The Effect of Pressure on the Temperature of a Gas

Gay-Lussac's law deals with temperature and pressure and has the following formula:

P1T1

= P2T2

What this formula says is that the ratio of pressure divided by temperature for a gas is constant, eventhough the temperatures and pressures may change. Gay Lussac's law assumes that the volume of thegas remains constant. The reason that this ratio remains constant can be explained using the kinetictheory. As the temperature increases, there are more collisions of the faster-moving gas molecules withthe walls of their container, and those collisions are more forceful since the molecules have more kineticenergy. As a result, as temperature goes up, so does the pressure being exerted by the gas. In otherwords, as T increases, P increases in proportion. Therefore, the ratio of P over T remains the same. Youshould use the same procedure for solving problems involving Gay-Lussac's law as you used for solvingCharles' law. Note the similarity between the equation for Gay-Lussac's Law and the equation for Charles'Law. How do you think a graph of the P vs. T relationship would compare to a graph of the V vs. Trelationship?

{11}__________________________________________________________________________________________________________________

Problem 13. A collapsible cylinder contains a gas at 765 mm Hg pressure. As external force causes thecylinder to collapse, the pressure reaches 988 mm Hg. The final temperature in the cylinder is 86.2oC.What was the original temperature of the gas in the cylinder before it collapsed? (Remember, always useKelvin temperatures!)

4-14 ©1997, A.J. Girondi

Problem 14. If an automobile tire contains air with a pressure of 26.0 psi (pounds per square inch) at atemperature of 20.0oC. After driving for several miles, the temperature of the air in the tire increases to48.0oC. Assuming volume remains constant, what is the new pressure of the air in the tire?

Problem 15. A steel cylinder contains a gas at 25.0oC and 2.50 atm pressure. The cylinder is designedto sustain a maximum internal pressure of 3.00 atm. How high can the temperature go before the cylinderwill explode?

Explain why a basketball seems to lose some firmness when you take it outside to shoot baskets on a cold

day. {12}____________________________________________________________________________________________________________

______________________________________________________________________________

SECTION 4.8 Solving Combined Gas Law Problems

The volume of a gas can change with changes in both pressure and temperature. By combiningCharles' law and Boyle's law, the following mathematical formula results:

P1V1T1

= P2V2

T2

This is called the combined gas law. The sample problem below illustrates the use of the combined gaslaw. Carefully study the four steps used to solve the problem. Many students like to eliminate step twoand go directly to step three. Don't do that. Complete step two before going to step three as you solvethe problems.

Sample problem: A volume of gas measured 900. mL at a temperature of 51.0oC with a pressure of700. mm. What volume will it occupy at a temperature of 27.0oC and a pressure of 760. mm? Follow thesesteps:

Step 1. Write the equation.

P1V1T1

= P2V2

T2

Step 2. Solve for the unknown variable.

V2 =

P1V1T2T1P2

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Step 3. Substitute numbers and units.

V2 =

(700. mm)(900. mL)(300. K)

(324 K)(760 . mm )

Step 4. Do the math. V2 = 768 mL

Try the problems below to gain practice with the combined gas law. Show all work as well as your answer.

Problem 16. In the laboratory, a volume of gas measures 600.mL at 22oC and 735 mm of pressure.What new volume will be established if the gas is placed under standard conditions (STP)?

Problem 17. Bromine vapor occupied a volume of 2.5 L at 575oC and 780.mm. In order to reduce thevolume to 2.3 L, the temperature is changed to 17oC. What must the pressure be changed to underthese conditions?

Problem 18. A gas has a volume of 237 mL when the pressure is 2.20 atm and the temperature is45.0oC. What volume will the gas occupy at 3.40 atm and 23.0oC?

Problem 19. A sample of gas if found to occupy a volume of 38.2 mL at 25.0oC and an unknownpressure. If this same gas occupies 37.5 mL at STP, determine what its unknown original pressure was inmm Hg.

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Problem 20. A gas occupies a volume of 250. mL at STP. To what Celsius temperature must thesystem be changed to increase the volume to 0.500 L, while the pressure is adjusted to 1.50 atm? (1.00atm = 760. mm)

Note that the three simpler gas laws can be derived from the combined gas law as follows. The combinedlaw is:

P1V1T1

= P2V2

T2

If temperature is not changing (is constant), then T1 and T2 are equal. If that's true, then T1 and T2

can be eliminated from the equation. Right? This would leave us with P1V1 = P2V2 which is Boyle's Law,the relationship between pressure and volume!

If pressure is not changing (is constant), then P1 and P2 are equal and can be eliminated from theequation for the combined laws. This would leave us with V1 / T1 = V2 / T2 which is Charles' Law, therelationship between temperature and volume!

If volume is not changing (is constant), then V1 and V2 are equal and can be eliminated from theequation for the combined laws. This would leave us with P1 / T1 = P2 / T2. This is Gay-Lussac's Law, therelationship between pressure and temperature!

(So, if you know the equation for the combined laws, you can always use it to derive the other laws.)

ACTIVITY 4.9 Measuring The Effect Of Temperature On Pressure

(This activity may be done by the class as a whole, rather than by lab groups. Ask your instructor.)

One way that we can measure gas pressures inthe lab is to use a piece of equipment known as amanometer. Figure 4.6 illustrates a manometer. Todemonstrate that temperature does have an effect onthe pressure exerted by a gas, obtain a clean, dry 125mL flask. Note the assembly drawn in Figure 4.6.Before inserting the stopper into the flask, open orremove the clamp attached to the tubing. Insert thestopper assembly into the flask so that it is tight. Handlethe flask with only your finger tips so that you do notwarm it. One tube should lead to the manometer andthe other should be open to the atmosphere. Tightenor replace the clamp to seal the flask. At this point thedifference in levels of liquid in the manometer should bezero since the pressure in the flask is the same asatmospheric pressure.

Table 4.3Temperature vs. Pressure

Condition Pressure of Air

Room Temp. ____0___ mm Hg

Warm Hands ________ mm Hg

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Glass Tubing

liquid

Ruler

difference in levels

Figure 4.6 A Manometer

rubber tubingclamp

atmospheric pressure

Before you actually start the investigation, predict whether the pressure in the flask will rise or fall when theair in the flask is warmed by your hands.

Prediction: _____________________________________________________________________

You are now ready to begin. Pick up the flask and wrap your hands around it. Note any change in thelevels of liquid in the manometer. When the levels have stopped moving, determine the difference in thelevels in mm. Enter the data into Table 4.3.

The liquid in the manometer is colored water, so the difference in levels that you measured isexpressed as mm H2O. Normally, pressures are measured in mm of mercury. To convert the difference inlevels from mm H2O to mm Hg, you must now divide mm H2O by 13.5. (Mercury is 13.5 times more densethan water.) Do this and enter the result into Table 4.3. Comment on how your prediction matches theresult. _________________________________________________________________________

Why does the air exert more pressure when it is warmed? {13} _________________________________

______________________________________________________________________________

ACTIVITY 4.10 Comparing Equilibrium Vapor Pressures

(This activity may be done by the class as a whole, rather than by lab groups. Ask the instructor.)

Manometers are used to measure the vapor pressure of liquids. When a liquid in a closedcontainer evaporates it forms a vapor. This vapor exerts a pressure. As the liquid continues to evaporate,the pressure increases. Some of the vapor molecules collide with the surface of the liquid in the containerand return to the liquid phase again. This is called condensation. At first, the evaporation process goes

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faster than the condensation, so that the amount of vapor in the container increases (as does thepressure). As the amount of vapor in the container increases, the rate of condensation will also increase.Finally, a point is reached in which the rates of evaporation and condensation are equal. At this point, theamount of vapor in the container will be constant, as will the pressure. (As indicated when the mercurylevels stop moving in Activity 4.9.) This condition is known as equilibrium. It is characterized by twoopposing forces going on at equal rates. When equilibrium has been reached, the pressure of the vaporin the container is constant and is known as the equilibrium vapor pressure.

Figure 4.7 Equilibrium Vapor Pressure in a Closed System

The equilibrium vapor pressure will remain constant provided that the temperature remainsconstant. If temperature changes, that will offset the equilibrium by changing evaporation andcondensation rates. Eventually, a new equilibrium will be established at the new temperature, and a newequilibrium vapor pressure will exist. In other words, a substance has a specific equilibrium vapor pressureat any given temperature. Different substances have different equilibrium vapor pressures.

A substance that has strong forces of attraction between its molecules is harder to vaporize andwill have a lower equilibrium vapor pressure (at a given temperature) than would a substance with weakerforces of attraction between its molecules. We conclude that the weaker the attraction between themolecules of a substance, the higher its equilibrium vapor pressure will be at any given temperature. Wecan use the apparatus in Figure 4.6 (which you used in the previous activity) to compare the equilibriumvapor pressures of various liquids. NOTE! All of the liquids used in this activity are flammable.

Procedure:

1. Obtain the manometer and remove the clamp from the hosing on the 125 mL Erlenmeyer flask; putabout 10 mL of acetone into the DRY flask and replace the stopper. (If the flask is not dry, rinse it with alittle acetone first. Discard the acetone used for the rinsing into the sink and run the water for a minute.)Clamp the hose shut (as shown in Figure 4.6). Pick up the flask and swirl the liquid for about 10 seconds tohasten the evaporation. Remove your hands from the flask. After two minutes, record the difference inthe levels of the liquid on the manometer (even if the levels are still moving). This difference is in mm ofwater. To convert to mm of mercury, divide the difference by 13.5. (Mercury is 13.5 times more densethan water.) Record this difference in levels in mm Hg in Table 4.4. Next, pick up the flask and wrap yourhands around it to warm the acetone. The bottom of the flask should press against one of your palms.You should see the water levels moving in the manometer. After two minutes, read the difference in thelevels (in mm). Again, convert to mm Hg and record the data in Table 4.4. Part of the difference in levelsis due to the vapor pressure exerted by the acetone. Another part is due to the expansion of air which isalso present in the flask. Even though the air adds to the total pressure, this procedure will allow us tocompare the vapor pressure of acetone to that of anther substance. Discard the acetone into the sink andrun some water.

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2. Rinse the flask with a little ethyl alcohol and discard it into the sink while running some water. Repeatstep 1 using about 10 mL of ethyl alcohol instead of acetone. Record the data in Table 4.4.

Table 4.4Equilibrium Vapor Pressures

Pressure of Vapor Pressure of VaporLiquid (Room Temp) (Warm Hands Temp)

Acetone __________mm Hg __________mm Hg

Ethyl Alcohol __________mm Hg __________mm Hg

According to your results, which liquid seems to have the weakest attractive forces between its

molecules?{14}_________________ Liquids which evaporate most readily have high vapor pressures.

Which liquid do you think will evaporate most readily? {15}____________________________Why do you think that

some liquids evaporate more easily than others?{16}____________________________________________________________

_____________________________________________________________________________________________________________________

SECTION 4.11 Review Problems To Solve

If you look back at the problems you have solved which make use of Boyle's law, Charles'law, Gay-Lussac's law, and the combined gas law you will notice that all of these problems include what wemight call the "original" conditions and the "new" conditions. For example, you might have been given thevolume of a gas at a certain temperature and pressure, and you may have been asked to find the "new"volume after the original temperature or pressure had changed.

There is yet another gas law, known as the ideal gas law which can be used to solve gas problemswhich involve only one set of conditions. For example, what is the volume of 2.00 grams of hydrogen gasat STP? Notice that only one set of conditions is given? You can not solve this problem using the lawsthat you have learned up to this point. In order to understand the ideal gas law, you first need to learnabout the mole concept which is presented in chapters 7, 8 and 9 of ALICE. After you are familiar with themole concept, we will learn about this final gas law.

You will find some review problems below. Read each problem to determine which gas law you willuse to solve it. In the space to the left of each problem, indicate which law you have chosen by writing Bfor Boyle's law, CH for Charles' law, G for Gay-Lussac's law, or COM for combined gas law. Then, solveeach problem showing all work.

Problem 21. _______ A basketball has a volume of 6.00 L when it is kept in a closet that is at 25.0oC.What does the volume of the ball become when it is taken outdoors on a chilly 10.0oC day? (Assumeconstant pressure.)

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Problem 22. ________ A 10.0 L sample of water vapor at 200.oC and 1.00 atm pressure is cooled to150.oC. The gas now occupies a volume of 7.00 L. Calculate the resulting pressure in the container.

Problem 23. ________ The volume of a gas at 1800. mm Hg pressure is 10.0 L. If the temperature iskept constant, what will be the volume of gas at standard atmospheric pressure, 760. mm Hg?

Problem 24. ________ The pressure gauge on a cylinder of oxygen gas at a temperature of 15.0oCreads 50.0 atm. What would the pressure gauge read if the temperature of the gas is raised to 38.0oC?

Problem 25. ________ A gas occupies a volume of 4.00 L at 470. mm and -70.0oC. What will itsvolume be at 963 mm and 122oC?

Problem 26. ________ At 755 mm Hg pressure, 1.00 cm3 of a gas has a mass of 0.0341 grams. Whatwill the density of this gas be (in grams per cm3) at a pressure of 905 mm Hg pressure? (Hint: the pressurechange will affect the volume, but not the mass.)

Density = __________ g/cm3

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ACTIVITY 4.12 A Teacher Demonstration of the Pressure–Temperature Relationship

The relationship between pressure and temperature can also be demonstrated (with the aid of acomputer) using a syringe connected to a temperature probe. Your teacher will demonstrate. Themechanical energy used to push the plunger of the syringe into the cylinder is transferred to the gasmolecules in the syringe where it is converted into kinetic energy. Watch the change in the temperatureof the gas in the syringe (shown on the screen of the computer display) as the gas in the syringe iscompressed.

What do you observe? {17}__________________________________________________________

______________________________________________________________________________


Understanding the gas laws is essential to your mastery of chemistry. Before you move on tochapter 5, review the learning outcomes listed below. Check the ones that you are certain you canhandle. Arrange to take the chapter quiz and move on to chapter 5.

_____1. Describe the basic principles of the kinetic-molecular theory.

_____2. Define pressure and how it is measured and expressed.

_____3. Convert between the Celsius and Kelvin temperature scales.

_____4. Explain what is meant by the absolute temperature scale.

_____5. Mathematically adjust temperature and pressure to standard conditions.

_____6. Qualitatively explain the effects of pressure and temperature changes on the volume of a gas.

_____7. Solve mathematical problems using Charles', Boyle's, Gay-Lussac's, and the combined gas laws.

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Questions:

{1} Constantly moving gas particles collide with air molecules causing mixing and the spread of the gasthrough air; {2} Constantly moving gas particles collide with the wall of their container; {3} decreases{4} yes - as pressure increases volume decreases; {5} pressure; {6} independent; {7} larger; {8} itincreases the pressure exerted; {9} so that volumes of gases can be compared under the sameconditions of T and P, which allows us to relate volumes of gases to actual amounts of gas; {10} zero; {11}they would look similar; {12} when temperature decreases so does pressure; {13} molecules collide withthe walls of the container more frequently and with greater force; {14} (the one that had the highest vaporpressure); {15} (the one that had the lowest vapor pressure); {16} liquids which evaporate more easilyhave weaker attractive forces between the molecules; {17} Temperature should increase as pressureincreases.

Problems:

1. 94.9 mL2. 1480 mm Hg3. 612 mL4. 403 mm Hg5. 844 mL6. a. 329 K; b. 239 K; c. 285.3 K7. a. -150oC; b. -237.4oC; c. 85oC8. 510 mL9. 68oC10. -11oC11. 125 mL12. -58.0oC13. 5.00oC (278 K)14. 28.5 psi15. 84.6oC (357.6 K) - rounds to 85oC (358 K)16. 537 mL rounds to 540 mL17. 290 mm18. 143 mL19. 814 mm Hg20. 546oC21. CH, 5.70 L22. COM, 1.28 atm23. B, 23.7 L24. G, 54.0 atm25. COM, 3.8 L26. B, 0.0409 g/cm3

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NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 5

ATOMSAND

MOLECULESAtomic Theory

Naming CompoundsWriting Formulas

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NOTICE OF RIGHTS




[email protected]


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SECTION 5.1 The Classification Of Matter

In order to work effectively with chemical concepts, it is important for you to learn the "language ofchemistry" and how certain words and symbols are used by chemists. In this chapter you will learn aboutthe meaning of terms such as element, compound, mixture, atom, molecule, ion, and polyatomic ion (orradical). You will also learn to identify chemical elements by chemical symbols. Finally, you will learn howto write chemical formulas and how to name chemical compounds. An idea that can be attributed to theancient Greeks is the concept of the atom. They developed the idea that all of the material substance inthe world was composed of fundamental building blocks that could not be divided into smaller parts. Theproperty of being indivisible into smaller parts led them to coin the word atom, which means "indivisible."Today, we realize that atoms can be broken up into smaller pieces, but the name atom is still used todescribe the fundamental unit of matter for chemists. Atoms have the special ability to combine togetherto form larger groups of atoms called molecules. Two types of molecules are possible - elements andcompounds. When a molecule is composed of only one kind of atom, it is referred to as an element. Forexample, the element gold is composed of millions of gold atoms connected together. A sample of themetal copper is made of copper atoms linked together. Atoms of some elements tend to combine in pairs.These paired elements are called diatomic molecules. Diatomic means "two atoms" in a molecule.Elements that exist as diatomic molecules include hydrogen and oxygen. In the pure form they are writtenas H2 and O2. You will learn about other diatomic gases in chapter 6.

The idea that substances in nature are composed of certain basic or fundamental elements thatcannot be further reduced to simpler substances is an ancient one dating back to early Greece. TheGreek elements were considered to be earth, air, fire, and water. Today, we realize that nature is muchmore complex than that. Of the first 92 elements, 90 are naturally-occurring ones of which all other matteris composed. Technitium (#43) amd promethium (#61) are the exceptions. We have also been able toproduce at least an additional 19 elements in the laboratory. These are called human-made elements.They are more complicated than the basic ninety-two and are very unstable, which means theydecompose readily into simpler elements. A listing of all of the elements is undoubtedly presentsomewhere in your chemistry classroom. This list is called the periodic table of elements. Each elementon the table is represented by a symbol.

In instances where molecules are composed of more than one kind of atom, this cluster of atomsis referred to as a compound. Examples of compounds include water (H2O), carbon dioxide (CO2), andtable salt (NaCl). Although these substances consist of more than one kind of atom, they are consideredto be pure substances. Elements are also considered to be pure substances.

A mixture is yet another way to describe combinations of elements and compounds. A mixture iscomposed of materials that have been placed together but which are not chemically combined. A mixturemay be composed of elements or compounds. The important thing to remember is that the substancesmaking up a mixture can always be separated by physical methods which means without using a chemicalreaction. Sand and salt can be made into a mixture by simply stirring these two substances together. Thesand and salt mixture can then be separated by adding water and filtering. The salt will dissolve in thewater and pass through the filter paper, while sand would not be able to pass through the filter paper.Dissolving is a physical change and filtering is a physical method of separation. The next activity will help toillustrate the difference between a mixture and a compound. Homogeneous materials are those whichhave the same composition throughout. Heterogeneous materials do not have the same compositionthroughout. Solutions such as salt water are homogeneous mixtures since they have the samecomposition throughout and can be separated by physical methods (such as boiling the water away). Themineral called marble is an example of a heterogeneous mixture since its composition can vary, evenwithin one small sample as well as from one location to another on earth.

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Matter

Pure SubstancesMixtures

Homogeneous Heterogeneous Elements Compounds

Solutions

Figure 5.1 The Classification of Matter

ACTIVITY 5.2 Properties Of Iron In A Compound And In A Mixture

Obtain the materials labeled 5.2 from the materials shelf. Container A contains iron filings andpowdered sulfur mixed together. Container B contains a chemical compound containing iron, sulfur, andoxygen (FeSO4). In the compound, iron is chemically bonded to the other elements.

1. Hold a magnet up to each container and move it around. One of the properties of elemental iron is thatit is attracted to a magnet.

2. Is the iron in container A attracted to the magnet?__________ Is the iron in container B attracted to

the magnet?__________ Based on your observations, do elements retain their original properties when

they form compounds with other elements? {1}__________ How does this activity support your

conclusion? {2}________________________________________________________________

______________________________________________________________________________

SECTION 5.3 Symbols Of The Elements

Each chemical element has a name assigned to it for the purpose of identifying it. The names forthe elements have been developed as each element has been discovered, and there are manyinteresting and colorful names. In order to avoid having to write a long name in describing an element, ithas become customary in chemistry to use a chemical symbol in place of the name. Most of the symbolsare derived from the names of the elements themselves.

Quite often, the symbol consists of two letters that are the first two letters in the name of theelement. As an example, the element calcium is denoted by the symbol Ca. Another example is argon,which is denoted by the symbol Ar. However, there are exceptions to this practice. The element arsenicis denoted by the symbol As so that it is not confused with argon. Another exception in chemical symbolsis the fact that some elements are denoted by only one letter. Thus, hydrogen is represented by H,oxygen by O, fluorine by F, and so on. In addition, the symbols for some of the elements are not related totheir more modern names but have come from older names for those elements that are no longer used.

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An example is sodium which has the symbol Na. This symbol is derived from the older name "natrium" thatwas originally given to sodium. Other examples of this are potassium, K; mercury, Hg; iron, Fe; tungsten,W; etc.

Table 5.1 contains the names and symbols for 38 of the more common elements. You areexpected to MEMORIZE the names of these elements and their symbols. (Spelling counts!) Manystudents misspell the name of the element which has the symbol F. Write its name - spelled properly - in

this space: _____________________ Remember, the symbol of this element is F - not Fl! And, it is notspelled flourine!

You will notice that the first letter in the symbol of any element is a capital letter, while the secondletter (if there is one) is in lowercase. You need to use care in writing the symbols. As an example, theelement cobalt has the symbol Co. If you were to write CO as the symbol for cobalt, you would actually bewriting the formula for a compound, carbon monoxide, which is very different from cobalt.

Table 5.1 Selected Elements And Their Symbols

Aluminum Al Lithium LiAntimony Sb Magnesium MgArsenic As Manganese Mn Barium Ba Mercury Hg Bismuth Bi Nickel NiBromine Br Nitrogen NCalcium Ca Oxygen OCarbon C Phosphorus PCesium Cs Platinum Pt Chlorine Cl Potassium KChromium Cr Silicon Si Cobalt Co Silver Ag Copper Cu Sodium Na Fluorine F Strontium Sr Gold Au Sulfur S Hydrogen H Tin SnIodine I Titanium Ti Iron Fe Tungsten W Lead Pb Zinc Zn

SECTION 5.4 Atomic Theory

Although the Greeks invented the idea of the atom about 2,500 years ago, their concept was notbased on experimental evidence gathered in a laboratory. Our modern atomic theory is the result of thework of several European scientists dating back to the 1600's. The most notable among these scientistsis John Dalton, an Englishman. Isaac Newton and Robert Boyle, two other Englishmen, had suggestedthe possibility of atoms through their work, but Dalton put the idea of the existence of atoms on a firmexperimental basis.

Dalton was able to show that when elements combined to form compounds, the masses of eachelement that went into making a compound always were present in definite ratios to each other. Thus,when he combined hydrogen with oxygen to form water, he always found that the ratio of the mass of

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hydrogen to that of oxygen was 1:8 (1 gram of H to 8 grams of O). This convinced him that the only way for

1 gram ofhydrogen

8 grams ofoxygen

9 grams ofwater

+this to make sense was if each element consisted of basic units, or atoms, that had definite masses.According to his reasoning, if basic units of matter did not exist, then1 gram of hydrogen should be able tocombine with any number of grams of oxygen.

John Dalton assumed that water consisted of one hydrogen atom and one oxygen atom.Furthermore, he reasoned that if the hydrogen atom had a mass of one mass unit, then the oxygen atommust have a mass of eight mass units.

+H O HO Really ? ? ?(He was incorrect, of course, in his assumption, because we now know that two atoms of hydrogencombine with one atom of oxygen to form a single molecule of water.) Water is H2O, not HO. Now if themass ratio is 1 to 8 ( H to O) in water, and if the atom ratio is 2 to 1 in water , what this means is that theoxygen atom must have a mass which is 16 times greater than that of the hydrogen atom – not 8 times asgreat (as Dalton had assumed).

Amedeo Avogadro (whom you will learn about later) recognized the problem in Dalton'sassumptions through his work with the gas laws and was able to show that certain elements, includinghydrogen and oxygen, actually existed in molecular (diatomic) form (H2 and O2). When water is formed,two molecules of hydrogen (which is four atoms of hydrogen) react with one molecule of oxygen (which istwo atoms of oxygen) to form two molecules of water:

2 H2 + O2 -----> 2 H2O

Nevertheless, Dalton's work represented the pioneering effort to experimentally establish the existenceof atoms. Dalton was also aware that two elements can produce two completely different compounds. Forexample, carbon can combine with oxygen to form either CO (carbon monoxide) or CO2 (carbon dioxide).He found that in CO, 16 parts by weight of oxygen always combined with 12 parts by weight of carbon. InCO2, he found that 32 parts by weight of oxygen always combined with 12 parts by weight of carbon. So,if he started with two equal weights of carbon and reacted them both with different quantities of oxygenhe always ended up with either a 12:16 ratio or a 12:32 ratio:

CO12 grams carbon

16 grams oxygen

CO212 grams carbon

32 grams oxygen

If you take enough of eachcompound such that the mass ofone element in them is the same

then the mass of the other element presentin the two compounds will be in a ratio ofsmall whole numbers (1:2 in this case)

Notice in the example above that for oxygen, 16 to 32 is a 1 to 2 ratio. This seemed to indicate thatelements could not combine in just any quantities. 12 grams of carbon have to combine with 16 grams ofoxygen or with 32 grams of oxygen - nothing in between! For example, he never found 12 parts by weightof carbon combining with 24 parts by weight of oxygen. Wonder why? Maybe, he reasoned, it's becauseelements don't come in just any quantity. Maybe they come in definite discrete amounts or particles. Thiswould explain the simple whole number ratio for oxygen (1:2 in this case). He called this regularity in theway two elements can combine his law of multiple proportions. This law together with the law of definitecomposition served as strong evidence suggesting the existence of atoms.

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This can all sounds somewhat confusing, so here is an analogy. Suppose you go into an icecream store and order a dish of chocolate and vanilla. When it comes you measure the mass of the icecream to make sure you are not being cheated (you just happen to have a balance with you). You find thatthe dish contains 100 grams of vanilla and 100 grams of chocolate. Since you like chocolate so much, yousend the dish back to the kitchen for more chocolate. This time when you get it back it contains 100 gramsof vanilla and 200 grams of chocolate. Hmmm. That's interesting. Exactly a 2:1 ratio between the originalamount of chocolate and the new amount. You repeat the order once more, and when you get the dishback it contains 100 grams of vanilla (starting to melt) and 300 grams of chocolate. Hey. Now thechocolate has varied in a 1:2:3 ratio - a ratio of small whole numbers. Whole numbers, mind you! Why?This leads you to believe that maybe the ice cream comes in discrete amounts - not just any amount. Afterall, the chocolate did not increase by fractional amounts. You ask the waiter if this might be true, and sureenough, the ice cream does come in discrete amounts! They are called scoops. Ah! The scoop theory ofice cream. Brilliant! See the analogy to Dalton's reasoning? Ice cream comes in scoops, while elementscome in the form of atoms!

Law of Multiple Proportions: Sometimes the same two elements can combine in differentproportions to form different compounds. (Example: CO and CO2) When they do this, ifyou hold the mass of one element in the compounds constant, the mass of the otherelement present will vary in a ratio of small whole numbers.

Dalton's atomic theory, first conceived in 1803, can be summarized by the following four statements:

1. An element is composed of extremely small particles called atoms.2. All atoms of a given element are identical to all other atoms of that element, but differ from atoms of

other elements.3. Atoms are indivisible and cannot be created or destroyed or changed into atoms of another element.4. Chemical changes take place when atoms of elements combine with each other in new ways.

Dalton's original theory was not entirely correct. For example, he thought that all atoms of a given elementhad exactly the same mass. Today, we know that atoms of an element do not all have the same mass. Wehave discovered the existence of isotopes, which are atoms of an element which have different masses.

As you have discovered, elements are represented by their chemical symbols. These symbolsare also used to describe chemical compounds that are formed when the elements react with each other.The use of these symbols to describe chemical compounds results in a chemical formula for thecompound. This formula contains the appropriate symbols and, in addition, also has small numbers writtenas subscripts (below the element) that indicate how many of each kind of atom are present in a molecule ofthe compound. The chemical formula for water is H2O. This formula means that in the water molecule,there are two hydrogen atoms and one oxygen atom. (When only one atom is involved, it is customary notto write a subscript "1" below the element.) In other words, when there is no number, it is understood thatthere is only one atom of that type present. Some compounds contain only two elements and these arecalled binary compounds. Other compounds contain more than two elements. As an example, theelements copper (Cu), sulfur (S), and oxygen (O) combine to form a compound with the chemical formulaCuSO4. In this compound for every one atom of copper, there is one atom of sulfur and four atoms ofoxygen.

Problem 1. Using the same idea, how many atoms of calcium are there in calcium chromate, CaCrO4?

Calcium?__________ Chromium?__________Oxygen?__________.

Problem 2. The formula for sodium acetate is NaC2H3O2. How many atoms of each element are present

in one sodium acetate molecule? Sodium?_________ Carbon?__________ Hydrogen?__________

Oxygen?__________.

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SECTION 5.5 Oxidation Numbers Of Elements and Polyatomic Ions

We are now at the point where you are ready to learn how to write the formulas for chemicalcompounds. In order to accomplish this task, we will be using what are known as oxidation numbers. Youwill not understand where these oxidation numbers come from until you study atomic structure in a laterchapter, but you will be able to use them, nonetheless. A list of some common oxidation numbers forselected elements can be found in Table 5.2 below. There are other oxidation numbers for theseelements besides those listed here, but this list will suit your purposes for now.

Table 5.2 Common Oxidation Numbers of Selected Elements(Note: Sometimes these elements can assume oxidation numbers other than those listed.)

Aluminum Al +3 Lithium Li +1Antimony Sb +3,+5 Magnesium Mg +2Arsenic As +3,+5 Manganese Mn +2,+4,+7Barium Ba +2 Mercury Hg +1,+2Bismuth Bi +3 Nickel Ni +2Boron B +3 Nitrogen N -3,+3,+5Bromine Br -1,+5 Oxygen O -2Calcium Ca +2 Phosphorus P +3,+5Carbon C +2,+4 Platinum Pt +2,+4Cesium Cs +1 Potassium K +1Chlorine Cl -1,+5,+7 Silicon Si +4Chromium Cr +2,+3,+6 Silver Ag +1Cobalt Co +2,+3 Sodium Na +1Copper Cu +1,+2 Strontium Sr +2Fluorine F -1 Sulfur S -2,+4,+6Gold Au +1,+3 Tin Sn +2,+4Hydrogen H +1 Titanium Ti +3,+4Iodine I -1,+5 Tungsten W +6Iron Fe +2,+3 Zinc Zn +2Lead Pb +2, +4

(A copy of Table 5.2 (oxidation numbers) which you will be permitted to use during tests and quizzes canbe found in the Reference Notebook which you were given as a part of ALICE.) You do NOT need tomemorize Table 5.2.

Common oxidation numbers for other elements can be found on some periodic tables and inother reference sources. Remember, you do not need to memorize oxidation numbers. Table 5.2contains the most commonly used oxidation numbers of the elements listed. It is possible that sometimesan element will exhibit an oxidation number which is not listed in the table.

There is yet another kind of fundamental unit present in some substances called an ion. An ion isa particle that carries an electric charge. In certain specific events that happen in chemistry, atoms ormolecules can end up with a positive (+) or negative (-) charge. These atoms or molecules then becomeions. Another name for ions which contain more than one atom is a polyatomic ion (they used to be calledradicals). Examples of ions include H1+, OH1-, NH41+, and SO42-. Ions like SO42- can also be written asSO4-2. It means the same thing. The ion has a charge of minus two.

There are oxidation numbers for polyatomic ions (see table 5.3). You ARE required to commit thislist (Table 5.3) to MEMORY including the names, the formulas, and the charges!

5-8 ©1997, A.J. Girondi

A table of polyatomic ions (like Table 5.3) can also be found in your ALICE Reference Notebook.You will be given a quiz to insure that you have learned the names, formulas, and charges of thepolyatomic ions. After that quiz, you will be allowed to refer to the list of polyatomic ions in your referencenotebook during future tests and quizzes. The reason for having you memorize them is to help you torecognize them as polyatomic ions when you see them. When you write polyatomic ions, you should writethe charge with the formula. For example, sulfate should be written as SO42-, while permanganate shouldbe written as MnO41-, etc. This list is not complete. There are many other polyatomic ions besides thoselisted in Table 5.3.

Table 5.3Common Oxidation Numbers of Selected Polyatomic Ions

Name Formula Charge

Ammonium NH41+ +1Acetate C2H3O21- -1 Chlorate ClO31- -1Perchlorate ClO41- -1 Cyanide CN1- -1 Hydrogen carbonate HCO31- -1 (or bicarbonate) Hydrogen sulfate HSO41- -1 Hydroxide OH1- -1 Nitrate NO31- -1 Nitrite NO21- -1 Permanganate MnO41- -1 Thiocyanate SCN1- -1Carbonate CO32- -2 Chromate CrO4 2- -2 Dichromate Cr2O72- -2 Sulfate SO42- -2 Sulfite SO32- -2Phosphate PO43- -3

SECTION 5.6 Writing Formulas for Chemical Compounds

The process of writing a chemical formula using oxidation numbers is really rather simple. The onerule that you must remember is that

"the sum of the oxidation numbers of the atoms in the formula of a compound must be zero."

For example, hydrogen's oxidation number is +1 and oxygen's is -2. Therefore, in order for the oxidationnumbers to add up to zero, we need two hydrogens. Two hydrogens = +2 and one oxygen = -2, so theformula for water is H2O. The subscript "2" to the right of the H indicates the presence of two hydrogenatoms. When a symbol is present without a subscript to its right, we assume that a subscript of "1" is there.We don't actually write the subscript if it is a one. Notice, too, that the element with the positive oxidationnumber is usually written first. Let's try more.

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Example 1: What is the formula for a compound of calcium and chlorine? Ca = +2 and Cl = -1. Therefore,in order for the oxidation numbers to add up to zero, these two elements must combine in a one to tworatio: CaCl2

Example 2: What is the formula for a compound of aluminum and oxygen? Al = +3 and O = -2. Therefore,in order for the oxidation numbers to add up to zero, these two elements must combine in a two to threeratio: Al2O3

One method used to write formulas involves the use of a lowest common multiple (LCM). Inexample 1 above, the lowest common multiple (disregard the signs) of the two oxidation numbers (+2 and-1) is 2. Now divide each oxidation number into the lowest common multiple (LCM) to determine thesubscript for that element in the formula. For Ca: 2/2 = 1; and for chlorine: 2/1 = 2. Therefore, the formula isCaCl2. In the case of example 2 above, the LCM of the oxidation numbers involved (+3 and -2) is 6. Foraluminum: 6/3 = 2; and for oxygen: 6/2 = 3. So, the formula for the compound is Al2O3. Notice how thesum adds up to zero! [2 Al = +3 X 2 = +6; 3 O = -2 X 3 = -6]. Then, (+6) + (-6) = 0. Practice now by doingproblem 3.

Problem 3. Using oxidation numbers from Tables 5.2 and 5.3, write correct formulas for compounds ofthe following substances. Keep in mind that if one element has only a positive oxidation number, youmust use a negative oxidation number for the other element. Some elements have more than oneoxidation number. So, when you see a symbol followed by a Roman numeral in parentheses in theproblems below, the Roman number equals the oxidation number which you should use for that element.Iron has two oxidation numbers: +2 and +3. Fe(II) refers to Fe2+.

Example: Manganese can be +2, +4, or +7. Mn(IV) = Mn4+. See? The oxidation number is the same asthe Roman numeral. Note, however, that the Roman numeral is not used in the formula. For example,when Mn(IV) and Cl combine, the compound's correct formula is MnCl4. It is incorrect to include theRoman numeral in the name. Therefore, Mn(IV)Cl4 is WRONG!

a. Ba and Cl _______________ i. Cu(I) and S _______________

b. Hg(I) and Br _______________ j. As(V) and O _______________

c. Ca and O _______________ k. C(IV) and O _______________

d. Hg(II) and Cl _______________ l. Sn(IV) and Br _______________

e. Al and S _______________ m. P(III) and I _______________

f. Ag and S _______________ n. As(V) and Cl _______________

g. N(III) and O _______________ o. Sn(II) and F _______________

h. Na and O _______________ p. Sr and Br _______________

This same method is used when the formula you are trying to write contains a polyatomic ion. Justkeep in mind that polyatomic ions stay together as a group and act as though they were a single atom witha single oxidation number. Whenever more than one polyatomic ion appears in a formula, it must beenclosed by parentheses with a subscript outside.

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Example 3: Write the formula for a compound of potassium and permanganate. K = +1 and MnO41- = -1.LCM = 1, so, the formula for potassium permanganate is KMnO4.

Example 4: Write the formula for a compound of calcium and chlorate. Ca = +2 and ClO31- = -1. LCM = 2 So, the formula for calcium chlorate is Ca(ClO3)2

Example 5: Write the formula for a compound of ammonium and sulfate. NH41+ = +1 and SO42- = -2.LCM = 2, so, the formula for ammonium sulfate is (NH4)2SO4 .

Problem 4. Write correct formulas for compounds of the following:

a. Mg and SO42- _______________ g. Al and SO32- _______________

b. Sn(II) and CrO42- _______________ h. Zn and CO32- _______________

c. Na and HCO31- _______________ i. Cu(II) and OH1- _______________

d. Fe(II) and OH1- _______________ j. Fe(III) and SO42- _______________

e. Pb(II) and PO43- _______________ k. Hg(I) and NO31- _______________

f. NH41+ and Cl1- _______________ l. NH41+ and Cr2O72- _______________

Check your answers for problem 4, and then try problem 5 below.

Problem 5. Write correct formulas for compounds of the following:

a. Mg and F _______________ f. Bi (III) and S ______________

b. Ba and ClO31- _______________ g. K and Cl ______________

c. N(V) and O _______________ h. H and S ______________

d. Ca and PO43- _______________ i. Cr(III) and C2H3O21- ______________

e. Al and OH1- _______________ j. S (IV) and O ______________

Earlier in this chapter you determined the number of atoms found in the formulas of compounds.Problem 6 will help you to improve this skill and to better understand the meaning of chemical formulas. Ineach of the following problems indicate the total number of atoms in each formula. For example, the totalnumber of atoms in the formula H2O is three. In a formula which includes parentheses, such as Ca(NO3)2,the subscript to the right of the parentheses multiplies everything inside the parentheses. The totalnumber of atoms in Ca(NO3)2 is nine (1 Ca + 2 N + 6 O = 9). The formula Na2SO4 has seven atoms.

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Problem 6. Determine the total number of atoms contained in each of the following formulas.

a. NaCl __________ f. Pb(C2H3O2)2 __________

b. AlPO4 __________ g. Sr(NO3)2 __________

c. Fe(ClO3)3 __________ h. Ba3(PO4)2 __________

d. Ag2SO3 __________ i. Al(HCO3)3 __________

e. Na2Cr2O7 __________ j. (NH4)2CO3 __________

The formulas in Problem 7 below belong to a group of compounds known as "hydrates." They arecompounds that have water molecules included in them. The water is tacked on to the end of the formulafollowing a raised dot. The raised dot does NOT mean multiplication, as it might in algebra; rather, it means"plus." So, a formula such as BaCl2 •2H2O includes 2 water molecules. The total number of atoms in thisformula is nine (1 Ba, 2 Cl, 4 H, and 2 O). For each of the following hydrates, indicate the total number ofatoms in each formula.

Problem 7. For each of the following hydrates, indicate the total number of atoms in each formula.

a. CuSO4 •5H2O __________ b. Na2CO3 •10H2O __________ c. CoCl2 •6H2O ___________

SECTION 5.7 Rules For Naming Compounds

There are three methods for naming chemical compounds. They are the:

1. prefix method 2. Latin name method 3. Roman numeral method

Although there are some exceptions, generally speaking, these methods are used as follows.

The prefix method is most often used to name compounds which contain only nonmetals . Nonmetals arefound to the right of the "staircase" on the periodic table.

The Latin name method is used for compounds containing certain metals including iron (Fe), copper (Cu),tin (Sn), or mercury (Hg).

The Roman numeral method is used to name all compounds which contain metals. Metals are found tothe left of the "staircase" on the periodic table (excluding hydrogen).

A. The Prefix Method

You should MEMORIZE the following prefixes:

1 = mon or mono 4 = tetr or tetra 7 = hept or hepta2 = di 5 = pent or penta 8 = oct or octa3 = tri 6 = hex or hexa

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In this method prefixes are used to indicate the number of atoms of each element present in theformula of a compound. For example, As2O5 is called diarsenic pentoxide. Note that pent is used insteadof penta in order to avoid the awkward sound of the double vowel. If only one atom of the first element inthe formula is present the use of "mono" is usually avoided; if there is only one atom of the secondelement, then the prefix "mono" is always used. For example, SO2 is called sulfur dioxide (rather thanmonosulfur dioxide); however, the molecule CO is called carbon monoxide. Note that if the compound isbinary (contains only two elements), the name of the second element is changed so that it always endswith the suffix "ide". (Examples: sulfur dioxide and carbon monoxide.) Finish spelling the name on this

binary compound: CO2 is carbon diox_______. {3}

Problem 8. Name the following nonmetallic compounds using the prefix method.

a. SO3 _________________________ e. N2O ________________________

b. As2O3 _________________________ f. SF6 ________________________

c. PBr5 _________________________ g. CCl4 ________________________

d. SeF2 _________________________ h. NO ________________________

Caution: Some students confuse the use of Roman numerals in names with the use of prefixes in names.Here is the difference. Roman numerals indicate the oxidation number of an element. Prefixes indicatethe number of atoms of an element represented in the formula. For example, iron (III) oxide is Fe2O3.The Roman numeral (III) in the name tells us that the oxidation number of iron in this compound is +3. Itdoes NOT mean that there are three iron atoms represented in the formula. As you can see, there areonly two iron atoms represented. The compound N2O4 is called dinitrogen tetroxide. The "di" before thenitrogen means that there are two nitrogen atoms represented in the formula. The prefix does NOTindicate the oxidation number of the nitrogen.

B. The Latin Name Method

You need to MEMORIZE the Latin names for the elements listed in Table 5.4.

Note that the Latin name of the lower oxidation state of each element ends in "ous", while thename of the higher oxidation state ends in "ic". (You should refer to Table 5.2 for the possible oxidationstates of elements.) For example, the compound CuCl2 contains copper in the +2 oxidation state[copper(II)]. So, the Latin name of this compound is cupric chloride. On the other hand, CuCl containscopper in the +1 oxidation state. Its name is cuprous chloride. Therefore, in order to use this method ofnaming, you must first determine the oxidation state of the metal and then choose the proper Latin name.Note that if the compound contains only two elements, the name of the second element is changed sothat it always ends with the suffix "ide".

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Table 5.4 Latin Names of Four Selected Elements

Element Symbol Latin Name

copper (I) Cu1+ cuprouscopper (II) Cu2+ cupric

iron (II) Fe2+ ferrousiron (III) Fe3+ ferric

mercury (I) Hg1+ mercurousmercury (II) Hg2+ mercuric

tin (II) Sn2+ stannoustin (IV) Sn4+ stannic

(These Latin names can also be found in your Reference Notebook.)

In problem 9 you will be asked to assign Latin names to compounds. You must first determine theoxidation number of the metal in each compound. To do this, you should first determine the oxidationnumber of the other element present. Note the examples below.

Example 1: Let's find the Latin name for SnS2. According to Table 5.2, tin (Sn) has oxidation numbers of+2 and +4. But which of these is being used in the compound SnS2? Well, note from Table 5.2 that sulfur(S) can have oxidation numbers of -2, +4, or +6. Since the tin has only positive oxidation numbers, wemust use the negative oxidation number of the sulfur which is -2. Now, if sulfur is -2 here, and since thereare two atoms of sulfur in the formula, the total oxidation number of the sulfur in SnS2 is -4. Since the totalof the oxidation numbers in the formula must equal zero, the oxidation number of the single tin atom in theformula must be +4. When tin has an oxidation number of +4 its Latin name (see Table 5.4) is stannic.Thus, SnS2 is called stannic sulfide.

Example 2: What is the Latin name for Cu2O? Since oxygen is -2, the total oxidation number for thecopper (Cu) must be +2. Hey, but wait. There are two atoms of copper represented in the formula.Therefore, each individual copper atom must have an oxidation number of +1. Therefore, checking Table5.4, Cu2O is called cuprous oxide.

Problem 9. Name the following using the Latin name method. Calculate the oxidation number of themetal being used, then name the compound.

Ox. No. of Metal Name

a. SnS ____________ _________________________________________

b. HgCl ____________ _________________________________________

c. FeO ____________ _________________________________________

d. CuS ____________ _________________________________________

e. HgF2 ____________ _________________________________________

f. SnCl2 ____________ _________________________________________

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g. Fe2S3 ____________ _________________________________________

h. CuI ____________ _________________________________________

C. The Roman Numeral Method

When naming metallic compounds using this method, first determine the possible oxidationstates of the metal. If the metal has more than one positive oxidation state, then you must use a Romannumeral in the name. If the metal has only one positive oxidation state, then you should not use a Romannumeral in the name.

For example, let's try to name FeCl3 according to this method. Checking a reference sheetcontaining oxidation states we find that iron can have states of either +2 or +3. Therefore, we must use aRoman numeral in the name. The oxidation state of iron in FeCl3 is +3. The name of this compound is iron(III) chloride. Note that when a Roman numeral is used, it is the same number as the oxidation state of themetal. That is, if the oxidation state of iron in a compound is +3, then the Roman numeral used is (III).

Let's try another one. Name FeO. In FeO, the oxidation state of iron is +2. Therefore, the nameof FeO is iron (II) oxide. Note that the Roman numeral is always enclosed in parentheses. Let's name thecompound: BaCl2. The metal barium has only one positive oxidation state which is +2. Therefore, noRoman numeral is needed and the name is simply barium chloride. Note that if the compound containsonly two elements, the name of the second element is changed so that it always ends with the suffix "ide".

Problem 10. Name the following using the Roman numeral method. Use Roman numerals in the nameonly if needed. Check your list of oxidation numbers to see if a metal has more than one positive oxidationstate. If it does, use a Roman numeral in the name.

a. MnO2 ___________________________________

b. KBr ___________________________________

c. CrCl3 ___________________________________

d. HgS ___________________________________

e. FeBr2 ___________________________________

f. CaF2 ___________________________________

g. CrO3 ___________________________________

h. CuO ___________________________________

i. Al2O3 ___________________________________

j. Co2O3 ___________________________________

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D. Compounds Containing Polyatomic Ions

Polyatomic ions are charged particles which consist of more than one atom. Formulas which beginwith one of these ions are named by first naming the polyatomic ion, then naming the element whichfollows it and changing the ending by using the suffix "ide.". Examples include NH4Cl (ammoniumchloride) and (NH4)2S (ammonium sulfide). Formulas which end with one of these polyatomic ions arenamed by naming the first element and then naming the polyatomic ion. Example: CaSO4, calciumsulfate. A Roman numeral is added to the name only if a metal is involved that has more than one positiveoxidation number. For example, since copper can be +1 or +2, CuSO4 is called copper (II) sulfate.However, since aluminum can only be +3, Al(NO3)3 is simply called aluminum nitrate. Some formulasconsist of two polyatomic ions. They are named simply by naming the first polyatomic ion followed by thename of the second one. NH4NO3 is ammonium nitrate; (NH4)2SO3 is ammonium sulfite.

Do not use the prefix method when naming compounds containing polyatomic ions.

Problem 11. Name the following compounds which contain polyatomic ions.

a. Ca(C2H3O2)2 _____________________________________

b. Ba(NO2)2 _____________________________________

c. Fe(OH)2 _____________________________________ (2 names)

_____________________________________

d. (NH4)2O _____________________________________

e. Ag2SO4 _____________________________________

f. KMnO4 _____________________________________

g. CuCO3 _____________________________________ (2 names)

_____________________________________

h. NaHSO4 _____________________________________

i. NH4C2H3O2 _____________________________________

j. (NH4)3PO4 _____________________________________

Note: Do not mix the methods for naming compounds. For example, if we name FeCl2 ferrous (II) chloride,that is wrong because we mixed the Latin name method with the Roman numeral method. Wait! There isone exception to never mixing the methods. In hydrates (described below) you will notice that the Romannumeral and prefix methods are both used in a name.

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The compounds known as hydrates, which you studied earlier in this chapter, are named as follows:

CuSO4 •5H2O is called copper (II) sulfate pentahydrate. The prefix "penta" means "five", so pentahydratemeans "five waters," which is added to the name of the compound.

BaCl2 •2H2O is called barium chloride dihydrate. The prefix "di" means "two," so dihydrate means "twowaters." (You should MEMORIZE the prefixes for 1 through 8 which were listed on a previous page.)

Problem 12. Perform the following tasks dealing with hydrates.

a. Name this hydrate: CoCl2 •6H2O _____________________________________

b. Finish writing this formula for sodium sulfate heptahydrate: Na2SO4 • ______H2O

SECTION 5.8 Writing Correct Formulas From Names

Problem 13. Write correct formulas for the compounds named below. Use the table of oxidationnumbers in this chapter or in your reference notebook. Remember that the sum of the oxidation numbersof all the atoms and/or polyatomic ions in a formula must add up to zero.

a. calcium nitrate __________________ n. ammonium fluoride _________________

b. strontium chloride __________________ o. iron (III) sulfide _________________

c. phosphorus triiodide __________________ p. sodium carbonate _________________

d. silver phosphate __________________ q. carbon tetrachloride _________________

e. dinitrogen pentoxide __________________ r. cobalt (II) chloride _________________

f. stannous chlorate __________________ s. ferrous phosphate _________________

g. chromium (III) oxide __________________ t. lead (IV) oxide _________________

h. ammonium chromate __________________ u. antimony (V) sulfide _________________

i. cesium sulfite __________________ v. sulfur hexabromide _________________

j. strontium fluoride __________________ w. silver oxide _________________

k. hydrogen iodide __________________ x. manganese (IV) chloride _____________

l. iron (III) chloride hexahydrate __________________________

m. calcium sulfate dihydrate __________________________

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SECTION 5.9 More Practice With Naming Compounds

Problem 14. Name the compounds below. Before deciding upon a name, first check to see if thecompound contains a metal. Use the prefix method for compounds that do not contain a metal; use theRoman numeral method for compounds that do contain a metal. (Metals are found to the left of the"staircase" on the periodic table.) Remember you should not mix the three methods of naming.

Formula Name a. CaCl2 _______________________________________________

b. Al2S3 _______________________________________________

c. Ba(OH)2 _______________________________________________

d. CaCO3 _______________________________________________

e. MgSO3 _______________________________________________

f. Pb3(PO4)2 _______________________________________________

g. As2O5 _______________________________________________

h. PBr3 _______________________________________________

i. KOH _______________________________________________

j. AsCl5 _______________________________________________

k. Ag2S _______________________________________________

l. SrCr2O7 _______________________________________________

m. CsHSO4 _______________________________________________

n. Co(NO2)2 _______________________________________________

o. SrSO4 _______________________________________________

p. NaHCO3 _______________________________________________

q. MgO _______________________________________________

r. CrF3 _______________________________________________

s. (NH4)2Cr2O7 _______________________________________________

t. Ni(ClO3)2 _______________________________________________

u. CBr4 _______________________________________________

v. Na3PO4 _______________________________________________

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w. N2O3 _______________________________________________

x. Pb(C2H3O2)2_______________________________________________

y. P2O5 _______________________________________________

z. KCN _______________________________________________

aa. LiNO3 _______________________________________________

bb. Na2CrO4 _______________________________________________

Problem 15. Give two names for each of the following. Put the Latin name in the left column and theRoman numeral name in the right column.

Latin Name Roman Numeral Name

a. HgBr _______________________________ ______________________________

b. Fe(OH)2 _______________________________ ______________________________

c. HgCl2 _______________________________ ______________________________

d. Fe2(SO4)3 _______________________________ ______________________________

e. Cu2S _______________________________ ______________________________

f. SnCrO4 _______________________________ ______________________________

g. Cu(OH)2 _______________________________ ______________________________

h. HgNO3 _______________________________ ______________________________

i. SnO2 _______________________________ ______________________________

SECTION 5.10 A Summary Of Things To Memorize

1. Symbols of selected elements (Table 5.1)

2. Names, symbols and charges of polyatomic ions (Table 5.3)

3. Selected Latin names of elements (Table 5.4)

4. Selected Greek prefixes for the numbers 1 through 8.

It will also be assumed that you can recognize and write Roman numerals for at least the first eightnumbers: I, II, III, IV, V, VI, VII, VIII

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You are now at the end of chapter 5. Check the learning outcomes below. When you feel thatyou have mastered all of them, arrange to take any test or quizzes on chapter 5, and then go to Chapter 6.

_____1. Distinguish between elements, compounds, heterogeneous mixtures, homogeneous mixtures, and pure substances.

_____2. Distinguish between atoms, molecules, and ions.

_____3. Write from memory the names (spelled correctly) and symbols of selected common elements.

_____4. Be able to identify the four important ideas which composed Dalton's Atomic Theory.

_____5. Write from memory the Latin names (spelled correctly) for the lower and higher oxidation states of copper, iron, mercury, and tin.

_____6. Write from memory the names (spelled correctly), formulas, and charges of the common polyatomic ions.

_____7. Name compounds using the Roman numeral method.

_____8. Name compounds using the Latin name method.

_____9. Name compounds using the prefix method.

_____10. Name compounds which are hydrates.

_____11. Calculate the number of atoms in the formulas of compounds (including hydrates).

_____12. Use oxidation numbers to write correct formulas for compounds given their names or the elements they contain.

_____13. Write from memory the common Greek prefixes.

_____14. Write from memory the names and symbols of common elements.

5-20 ©1997, A.J. Girondi


Questions:

{1} No; {2} Magnet is attracted to pure iron, but not attracted to iron when it is in a compound;{3} ide

Problems:

1. 1,1,42. 1,2,3,23. a. BaCl2; b. HgBr; c. CaO; d. HgCl2; e. Al2S3; f. Ag2S; g. N2O3; h. Na2O; i. Cu2S;

j. As2O5; k. CO2; l. SnBr4; m. PI3 ; n. AsCl5; o. SnF2; p. SrBr24. a. MgSO4; b. SnCrO4; c. NaHCO3 d. Fe(OH)2; e. Pb3(PO4)2; f. NH4Cl; g. Al2(SO3)3;

h. ZnCO3; i. Cu(OH)2; j. Fe2(SO4)3; k. HgNO3; l. (NH4)2Cr2O7

5. a. MgF2; b. Ba(ClO3)2; c. N2O5; d. Ca3(PO4)2; e. Al(OH)3; f. Bi2S3 g. KCl; h. H2S; i. Cr(C2H3O2)3j. SO2

6. a. 2; b. 6; c. 13; d. 6; e. 11; f. 15; g. 9; h. 13; i. 16; j. 147. a. 21; b. 36; c. 218. a. sulfur trioxide; b. diarsenic trioxide; c. phosphorus pentabromide; d. selenium difluoride;

e. dinitrogen monoxide; f. sulfur hexafluoride; g. carbon tetrachloride; h. nitrogen monoxide9. a. +2, stannous sulfide; b. +1, mercurous chloride; c. +2, ferrous oxide; d. +2, cupric sulfide

e. +2, mercuric fluoride; f. +2, stannous chloride; g. +3, ferric sulfide; h. +1, cuprous iodide10. a. manganese (IV) oxide; b. potassium bromide; c. chromium (III) chloride

d. mercury (II) sulfide; e. iron (II) bromide; f. calcium fluoride; g. chromium (VI) oxide;h. copper (II) oxide; i. aluminum oxide; j. cobalt (III) oxide

11. a. calcium acetate; b. barium nitrite; c. iron (II) hydroxide, ferrous hydroxide;d. ammonium oxide; e. silver sulfate; f. potassium permanganate;g. copper (II) carbonate, cupric carbonate; h. sodium hydrogen sulfate; i. ammonium acetate;j. ammonium phosphate

12. a. cobalt (II) chloride hexahydrate; b. Na2SO4 • 7 H2O13. a. Ca(NO3)2; b. SrCl2; c. PI3; d. Ag3PO4; e. N2O5; f. Sn(ClO3)2; g. Cr2O3; h. (NH4)2CrO4;

i. Cs2SO3; j. SrF2; k. HI; l. FeCl3 • 6 H2O; m. CaSO4 • 2 H2O; n. NH4F; o. Fe2S3; p. Na2CO3;q. CCl4; r. CoCl2; s. Fe3(PO4)2; t. PbO2; u. Sb2S5; v. SBr6; w. Ag2O; x. MnCl4

14. a. calcium chloride; b. aluminum sulfide; c. barium hydroxide; d. calcium carbonate;e. magnesium sulfite; f. lead (II) phosphate; g. diarsenic pentoxide; h. phosphorus tribromide;i. potassium hydroxide; j. arsenic pentachloride; k. silver sulfide; l. strontium dichromate;m. cesium hydrogen sulfate; n. cobalt (II) nitrite; o.strontium sulfate; p. sodium hydrogen carbonate;q. magnesium oxide; r. chromium (III) fluoride; s. ammonium dichromate; t. nickel chlorate;u. carbon tetrabromide; v. sodium phosphate; w. dinitrogen trioxide; x. lead (II) acetate;y. diphosphorus pentoxide; z. potassium cyanide; aa. lithium nitrate; bb. sodium chromate

15. a. mercurous bromide, mercury (I) bromideb. ferrous hydroxide, iron (II) hydroxidec. mercuric chloride, mercury (II) chlorided. ferric sulfate, iron (III) sulfatee. cuprous sulfide, copper (I) sulfidef. stannous chromate, tin (II) chromateg. cupric hydroxide, copper (II) hydroxideh. mercurous nitrate, mercury (I) nitratei. stannic oxide, tin (IV) oxide

5-21 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 6

CHEMICALEQUATIONS

(Part 1)

Direct Combinationand

Decomposition6-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


6-2 ©1997, A.J. Girondi

SECTION 6.1 The Meaning Of Chemical Equations

So far, you have learned about chemical elements, symbols, and formulas. The next thing youneed to learn is how these are used in describing chemical reactions. Elements combine to formcompounds, compounds can be broken down into elements, and compounds can react to form newcompounds. These processes are referred to as chemical reactions.

Chemists have developed a definite way to describe chemical reactions through the use of whatare called chemical equations. These are shorthand statements of what is occurring in a chemical reaction.These statements involve the use of appropriate chemical formulas to describe the elements andcompounds involved in the reaction.

In this chapter you will learn how chemical formulas are used to write chemical equations thatdescribe what happens when chemical reactions occur. There are certain basic types of chemicalreactions, such as direct combination, decomposition, single replacement, and double replacementreactions.

You will find that when these reactions do occur, the number of each type of atom for theelements and compounds involved does not change during the reaction; the atoms merely getrearranged. In a later chapter, you will learn how to classify reactions in a different way (depending onwhether they are "oxidation-reduction reactions" or not). There is much to learn, so we better get started.

The reason for having a shorthand method is very simple. It literally eliminates having to writechemical reactions in the longhand form. Referring back to the reaction in which water is formed fromhydrogen reacting with oxygen, this process could be described in words as follows:

Two molecules of hydrogen react with one molecule of oxygen to yield two molecules of water. It is muchsimpler to write the following statement, which means the same thing:

2 H2 + O2 ----> 2 H2O

This is a "balanced" chemical equation because the number of atoms on the left-hand side of the arrow isequal to the number of atoms on the right- hand side of the arrow. The arrow (which is read as "yields")indicates that the substances on the left change to the substances on the right during the reaction.

In the equation, you will note that a number (coefficient) appears before the hydrogen moleculesymbol and the water molecule symbol (the number 2, in both cases). This simply tells how many of eachkind of molecule is (are) involved in the reaction. As was the case when using subscripts to write chemicalformulas, 1's are seldom written. The number 1 is not written before the symbol for the oxygen molecule,but it is understood to be there. For example, Na1Cl1 is written as NaCl.

In general, the substances on the left-hand side of the equation are called the reactants, while the

substances on the right-hand side are called the products. In the reaction 2 H2 + O2 ----> 2 H2O, the

reactant(s) is/are {1}_________________. The product(s) is/are {2}___________________.

Chemical equations often include special subscripts and symbols which offer additionalinformation. The subscript, (aq), stands for "aqueous" which means the substance is dissolved in water;the subscripts (s), (l), and (g) are used to identify solids, liquids, and gases; when the Greek symbol delta,∆ , is placed over the arrow in the equation, it indicates that heat is being added.

H2CO3(aq) -------> H2O(l) + CO2(g)

CaCO3(s) -------> CaO(s) + CO2(g)∆

6-3 ©1997, A.J. Girondi

In the two equations above, the liquid (l) is {3}___________, the substance which is dissolved in water (aq)

is {4}____________, the gaseous substance (g) is {5}______________, and the solids (s) are

{6}________________________. The substance which is being heated is {7}____________.

ACTIVITY 6.2 An Experiment Involving the Heating of Copper

How can we tell when a chemical compound has been formed? By making measurements, we can"quantize" a chemical change in an experiment to show that a reaction actually has happened. Let's give ita try.

You will need to set up some lab equipment that resembles Figure 6.1. You will also need a

balance sensitive to 0.01 or 0.001 g. You are going to heat some copper metal in a crucible. You will

measure the mass of the contents of the crucible before and after heating. Do you think there will be a

change in the mass of the contents as a result of heating?______________ If you answered yes, do you

think the mass will increase or decrease?__________________ Explain:_______________________

_____________________________________________________________________________

1. Obtain enough copper turnings so that whenyou roll the turnings between your hands, youget a round ball about the size of a large marble.Put the copper into a crucible and determine themass of the crucible and contents to the nearest0.01 or 0.001 g (depending on the balanceavailable). Record this mass in Table 6.1.

2. Place the crucible on a pipestem triangle whichin turn is supported by a ring on a ring stand.Heat strongly for a about 10 minutes. The tip ofthe flame should touch the bottom of thecrucible. Allow it to cool. Reweigh the crucibleand contents together. Record the data.

crucible

triangle

iron ring

Figure 6.1 Heating a Crucible

Table 6.1Heating Copper Turnings

Mass of crucible + copper before heating __________ g

Mass of dish + contents after heating __________ g

Difference in mass of dish + contents __________ g(before and after heating)

6-4 ©1997, A.J. Girondi

3. Complete Table 6.1, if you have not already done so, by determining the mass of the crucible contentsafter heating.

Was there a change of mass for the substance in the crucible after heating?___________. If so, was

it a gain or a loss? ____________. Was your hypothesis correct?___________ How do you explain the

result?_________________________________________________________________________

SECTION 6.3 Equations For Direct Combination Reactions

Now that you know something about chemical equations, the next thing to learn is how chemicalreactions can be categorized as general types of reactions. The four general types that we will bestudying in this chapter include: direct combination, simple decomposition, single replacement, anddouble replacement.

A direct combination reaction is one in which two elements directly combine to form a compound.An example of this is the formation of sodium chloride (NaCl) from the reaction of sodium (Na) with chlorinegas (Cl2). You read in Chapter 5 about diatomic elements like H2 and O2. Chlorine - when it is notcombined with other elements - occurs in the diatomic form, Cl2.

2 Na (s) + Cl2(g) -----> 2 NaCl (s)

reactants product

The arrow means "yields"

Notice that there are equal numbers of each kind of atom on both sides of the equation. This has to betrue in order for this reaction to be called an equation. Another way to describe this requirement is to saythat the equation must be balanced. The word balanced implies that the atoms on the reactant side (left-hand) of the equation are in balance (that is, are equal in number) with the atoms on the product side(right-hand). The arrow which separates the reactants from the products is read as "yields." You shouldbe able to balance simple chemical equations merely by looking at what is reacting and what is formed todetermine what numbers (coefficients) are needed to give a balanced equation. This is called balancingequations by inspection.

Analyze the equation for the formation of NaCl, above. How many atoms of Na are shown on the reactant

side? {8}__________ How many atoms of Na are on the product side? {9}__________ Are they equal?

{10}__________ Are the numbers of Cl atoms equal on both sides? {11}_________________

The requirement that the equation be balanced illustrates a law of chemistry called the law ofconservation of matter. This law simply means that:

In a chemical reaction, matter is neither created nor destroyed.

In a balanced chemical equation, the amount of matter on the left side of the equation must equalthe amount of matter on the right side. This means that there must be equal numbers of atoms of eachelement on both sides of the equation.

6-5 ©1997, A.J. Girondi

Now let's look at direct combination reaction which you saw earlier in this chapter, the formation ofwater from hydrogen and oxygen. Suppose we write the equation as follows:

H2 + O2 -----> H2O

How many atoms (not molecules) of hydrogen are on the left side? {12}________ right side? {13}________

Are these figures equal? {14}__________ Is the hydrogen balanced? {15}_________

How many atoms of oxygen are on the left side? {16}_____ right side? {17}______ BUT, are these figures

equal? {18}_______ Is the oxygen balanced? {19}_________

Uh, oh. We cannot create or destroy any atoms in ordinary chemical reactions. The same number ofatoms of each element must appear on both sides of an equation. In order to "balance" the equation forthe formation of water, the coefficient "2" must be placed in front of the H2O. This balances the number ofoxygen atoms. Examine the equation below.

H2 + O2 -----> 2 H2O

(Note: a coefficient such as the 2 in front of the H2O in this equation, implies a multiplication operation. Itmeans two times everything in the formula to its right.

1. H2 represents one molecule of hydrogen consisting of two atoms of hydrogen. How many atoms (not

molecules) of H are on the left side of the equation?{20}______ Right side? {21}__________ Are these

figures equal? {22}______

2. How many atoms of oxygen are on the left side of the equation? {23}________ Right side?

{24}________ Are these figures equal? {25}_________

To finish balancing the equation for the formation of water, a coefficient "2" must be placed in front of theH2 on the left side. Write the balanced equation below.

{26} _____________________ -----> _____________________

In the equation above notice that since hydrogen and oxygen are diatomic elements, they are alwayswritten in diatomic form with the subscript "2" whenever they are not combined with other elements: H2

and O2. When they do combine with other elements, the subscript "2" is changed to whatever subscript isrequired to obtain the correct formula for the compound using oxidation numbers. Remember the sum ofthe oxidation numbers in the formula of a compound has to be zero, so: H= +1, O = -2, and H2O = 0.

Before going on you should MEMORIZE the complete list of seven diatomic elements. Whenthese elements occur uncombined with other elements, they occur as pairs of atoms and are written asfollows:

The Seven Diatomic Elements

Hydrogen H2, Nitrogen N2, Oxygen O2, Fluorine F2, Chlorine Cl2, Bromine Br2, and Iodine I2

To help you remember the seven diatomic elements, this phrase is offered: "Henry Needs Oxygen ForClimbing Brown Icebergs!" (Silly things like brown icebergs are easy to remember.) Note the first letter ineach word in the phrase. You might also note that the positions of the 7 diatomic elements on the periodic

6-6 ©1997, A.J. Girondi

table kind of form the number 7 (except for H). In addition, you should now MEMORIZE two otherpolyatomic elements which exist in special forms when they are uncombined with other elements. Theyare:

Two Polyatomic Elements: sulfur: S8 and phosphorus: P4

SECTION 6.4 Balancing Direct Combination Equations

When balancing an equation, only coefficients (the numbers in front of the formulas) may bechanged or added. You must never change or add any subscripts to the formulas - assuming the formulasare correct when you start.) For example, the following equation needs to be balanced.

H2 + Cl2 ----> HCl

The formula of the product is written correctly as HCl. To balance the equation, we need 2 hydrogenatoms and 2 chlorine atoms on the right side since there are two of each on the left side. It would beincorrect to do this by changing the formula to H2Cl2, because HCl is the correct formula for the product.

H2 + Cl2 ----> H2Cl2 <--- WRONG!!

Instead, we will change the coefficient in front of the HCl formula from 1 to 2 as follows:

H2 + Cl2 ----> 2 HCl <--- RIGHT

Note that there are now two atoms of H and two atoms of Cl represented on both sides of the equation.Let's try one more example.

Suppose we want to balance the equation representing the reaction between arsenic, As, andO2. The product will obviously be a compound of arsenic and oxygen.

As + O2 -----> As?O?

Use oxidation numbers to find the correct formula for the product.

Oxygen is diatomic when it is not combined with other elements.

Before trying to balance this equation, we must first write the correct formula for the product. We will useoxidation numbers of +5 for arsenic and -2 for oxygen. Therefore, the correct formula for the product mustbe As2O5. The unbalanced equation will look like this:

As + O2 -----> As2O5

When balancing equations by "inspection", it is often helpful to start with those elements which have thelargest subscripts. Since oxygen has a subscript of 5, let's start with it. Note that there are 5 oxygen atomson the right side of the arrow and only 2 on the left. We can balance the number of oxygen atoms byputting a 5 in front of the O2 and a 2 in front of the As2O5 as follows:

As + 5 O2 -----> 2 As2O5

The 2 in front of the As2O5 means 2 times everything in that formula. Therefore, we now have 10 oxygenatoms represented on both sides. Great. However, note that we have not yet balanced the As atoms. To

6-7 ©1997, A.J. Girondi

do that, we need to put a 4 in front of the As on the left side of the equation. The balanced equation isnow complete:

4 As + 5 O2 -----> 2 As2O5

The coefficient ratio in this balanced equation is 4 to 5 to 2. Note that this is a simple whole number ratiothat cannot be further reduced. This equation would also be balanced if the coefficient ratio were 8 to 10to 4. Be this would not be correct, since you should always use the lowest possible ratio of coefficients.

The equations below represent direct combination reactions. Notice that in direct combination reactions,two simpler substances are combining to form a single more complex product. The correct formulas forthe products have already been written for you.

Problem 1. Balance the following equations simply by adding coefficients in the blanks provided. If nocoefficient is added, "1" is understood to exist. In this problem, the formulas for all reactants and productshave already been correctly written for you. This will not be the case in problem 2.

a. _____As + _____Cl2 ----> _____AsCl3

b. _____K + _____Cl2 ----> _____KCl

c. _____Cu + _____O2 ----> _____Cu2O

d. _____SO3 + _____H2O ----> _____H2SO4

e. _____P4 + _____Br2 ----> _____PBr5

f. _____Fe + _____Cl2 ----> _____FeCl3

g. _____Zn + _____O2 ----> _____ZnO

h. _____Si + _____S8 ----> _____SiS2

i. _____Al + _____N2 ----> _____AlN

j. _____Al + _____O2 ----> _____Al2O3

ACTIVITY 6.5 The Direct Combination Reaction of Copper and Sulfur

The purpose of this activity is to use the direct combination reaction of copper and sulfur to testthe validity of the Law of Conservation of Matter. You are going to witness a reaction between coppermetal and sulfur. When you do, the product will be copper (II) sulfide, CuS. However, some of the sulfurmay also react with air and form sulfur dioxide gas, SO2. We don't want any of this gas to escape from thetest tube in which the reaction will take place, so we will fit a balloon over the mouth of the tube to trap thegas. If the Law of Conservation of Matter is true, then the reactants (Cu and S8) and the products (CuSand SO2) should have the same mass: total mass of the reactants = total mass of the products.

You will need a 150 mm test tube (the regular size), 1.00 gram of powdered sulfur, 2.00 grams ofgranular copper (do not use powdered copper!), a small balloon, a burner, and a balance. Some of thesethings are available on the materials shelf.

6-8 ©1997, A.J. Girondi

Cu + S8

balloon

1. Examine the copper granules and sulfur powder.

Describe each substance: __________________

______________________________________

______________________________________

______________________________________

______________________________________

2. Mix 2.00 g of the copper and 1.00 g of the sulfuron a piece of paper and then pour them into the testtube.

Figure 6.2 Test Tube With Balloon

3. Fix a balloon over the mouth of the tube, using a rubber band - if necessary - to hold the balloon tightly.Measure the mass of the test tube with its contents and the balloon. Enter all data in Table 6.2. Be sure towear safety glasses!

4. Use a utility clamp to mount the tube on a ring stand. The clamp should be positioned near the top ofthe test tube. Tilt the tube so that the mouth is pointed away from everybody around you, and rests atabout a 45 degree angle. Heat the mixture gently until the contents begin to glow. Then stop heatingimmediately by moving the burner away from the tube! Watch the contents glow!

5. Allow the tube to cool, and again find the mass of the tube with contents and balloon. Record theresult.

Table 6.2The Copper – Sulfur Reaction

mass of tube and contents before heating __________g

mass of tube and contents after heating __________g

Examine the contents of the tube after it has cooled. Do you think that a new chemical substance has

formed?_________. Explain:_______________________________________________________

6. The test tube should be thrown away, since the material formed is very difficult to remove. The balloonshould be returned for others to use.

If you have not created or destroyed any matter during this reaction, then the masses before and after thereaction should be the same. Experimental error is always a factor, so let's say that the masses before and

after should be close. Do you feel your results are close enough to support the law? _______________

Why?_________________________________________________________________________

6-9 ©1997, A.J. Girondi

SECTION 6.6 More Practice With Direct Combination Reactions

The direct combination equations which you have balanced thus far in this chapter only requiredthat you add coefficients. The products of the reactions were written for you. You did not have to predictwhat the products would be. You will have to predict the products when you try to balance the next set ofequations. In an earlier activity, you carried out a direct combination reaction by heating a ball of copperturnings in a crucible. If you recall, the copper seemed to gain mass after you heated it. Actually, the heatcaused the copper to react with oxygen in the air to form copper (II) oxide.

Cu + O2 -----> ???

To balance the equation above, we must first write the correct formula for the product using oxidationnumbers. We will use +2 for Cu, and -2 for O. Therefore, the correct formula for the product is CuO. Theequation then becomes:

Cu + O2 -----> CuO

However, we must now balance the equation. We do this by putting a coefficient of 2 in front of the Cu and also in front of the CuO:

2 Cu + O2 -----> 2 CuO

Note that just because the oxygen is written in diatomic form as O2 on the left side, this does not mean it isto be written in diatomic form on the right. It only has to be written as diatomic when it is not combined withother elements.

The products of these reactions will usually be fairly obvious to you. Reminder: whenever any ofthe seven diatomic elements occur uncombined with other elements in an equation, they are alwayswritten in diatomic form. Remember, too, that "Henry Needs Oxygen For Climbing Brown Icebergs." Inaddition, remember that two other elements exist in a special form when they are uncombined with otherelements. They are: phosphorus (P4) and sulfur (S8).

Problem 2. Use oxidation numbers to write the correct formulas for the products of the followingequations. Then, add the needed coefficients to balance the equation.

Here is a little help to guide you through the first one. Aluminum and fluorine will combine to forma compound: Al?F?. Using the oxidation numbers (Al = +3, F = –1), determine the correct formula for thiscompound and write it in the space provided below to the right of the arrow. Now put the neededcoefficients in front of each formula on the right and left of the arrow to balance the number of atoms ofeach element on both sides.

a. _____Al + _____F2 ----> ________________________________________

b. _____Na + _____Br2 ----> ________________________________________

c. _____Ca + _____O2 ----> ________________________________________

d. _____H2 + _____Cl2 ----> ________________________________________

e. _____Mg + _____S8 ----> ________________________________________

f. _____P4 + _____S8 ----> ___________ <----(use +3 as the oxidation number for P)

g. _____P4 + _____O2 ----> ___________ <----(use +5 as the oxidation number for P)

6-10 ©1997, A.J. Girondi

In problem 3, you will see "word equations." These are equations which use words in place of chemicalsymbols. You will have to write the correct reactant and product formulas before you balance theequations.

Problem 3. Complete and balance the following equations. First, write the correct formulas for eachreactant and product. For your convenience, the reactant which assumes a positive oxidation number iswritten first. The first one is done for you as an example. The oxidation numbers used are Al = +3 and O =–2. Thus, the correct formula for the product is Al2O3. Since oxygen is uncombined on the left side ittakes the form of a diatomic gas and is written as O2. The coefficients 4, 3, and 2 are then used to balancethe equation. Before completing the remaining equations, you should recall the seven diatomic elementsand the two polyatomic ones.

a. aluminum + oxygen ---> ?? 4 Al + 3 O2________ -----> _______2 Al2O3________

b. silicon (IV) + fluorine ----> ?? _____________________ -----> _____________________

c. carbon (II) + oxygen ---> ?? _____________________ -----> _____________________

d. phosphorus (III) + chlorine ----> ?? _____________________ -----> _____________________

e. iron (III) + sulfur ----> ?? _____________________ -----> _____________________

f. sulfur (VI) + bromine ----> ?? _____________________ -----> _____________________

g. silver + iodine ----> ?? _____________________ -----> _____________________

h. hydrogen + fluorine ----> ?? _____________________ -----> _____________________

SECTION 6.7 Balancing Equations Of Decomposition Reactions

Decomposition reactions are the opposite of direct combination reactions. They involve thebreaking down of compounds into simpler parts. An example is the decomposition of sodium chloride:2 NaCl ---> 2 Na + Cl2. You can recognize decomposition reactions by noting that there is usually only onereactant on the left side of the equation.

Problem 4. Balance the decomposition equations below by adding the needed coefficients. Allformulas for reactants and products have been correctly written for you.

a. _____Hg2CO3 ----> _____Hg + _____HgO + _____CO2

b. _____H2O ----> _____H2 + _____O2

c. _____AuCl3 ----> _____Au + _____Cl2

d. _____NCl3 ----> _____N2 + _____Cl2

e. _____(NH4)2Cr2O7 ----> _____Cr2O3 + _____N2 + _____H2O

f. _____CaCO3 ----> _____CaO + _____CO2

g. _____HgO ----> _____Hg + _____O2

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h. _____KClO3 ----> _____KCl + _____O2

i. _____H2SiO3 ----> _____SiO2 + _____H2O

j. _____MgCO3 ----> _____MgO + _____CO2

Decomposition equations are not as simple as direct combination equations. It is more difficult topredict what the products will be. Some categories of decomposition reactions are listed in Table 6.3.

Table 6.3Four Kinds of Decomposition Reactions

1. Metallic carbonates decompose into metallic oxides and carbon dioxide.

CaCO3 ----> CaO + CO2

K2CO3 ----> K2O + CO2

H2CO3 ----> H2O + CO2

2. Many metallic hydroxides decompose into metallic oxides and water.

2 KOH ----> K2O + H2O

Ca(OH)2 ----> CaO + H2O

2 Al(OH)3 ----> Al2O3 + 3 H2O

3. Metallic chlorates decompose into metallic chlorides and oxygen gas.

2 KClO3 ----> 2 KCl + 3 O2

Ba(ClO3)2 ----> BaCl2 + 3 O2

2 Al(ClO3)3 ----> 2 AlCl3 + 9 O2

4. Many binary (two-element) compounds decompose into pure elements.

2 HgO ----> 2 Hg + O2

2 H2O ----> 2 H2 + O2

2 NaCl ----> 2 Na + Cl2

Note: There are many other kinds of decomposition reactions in addition to those

shown above. A copy of this table can be found in your ALICE reference notebook.

Before trying the next problems, let's look at another example. Let's complete the equation whichrepresents the decomposition of aluminum chlorate: Al(ClO3)3. This compound looks similar to those ofcategory 3 in Table 6.3, which tells us that metallic chlorates decompose into metallic chlorides andoxygen gas. Thus, Al(ClO3)3 will decompose into aluminum chloride and oxygen. We must now write thecorrect formulas for these products. Al has an oxidation number of +3, while chlorine has an oxidationnumber of -1. Thus, the correct formula is AlCl3. Oxygen is diatomic and must be written as O2. Theequation becomes:

6-12 ©1997, A.J. Girondi

Al(ClO3)3 ----> AlCl3 + O2

formula correctly writtenusing oxidation numbers

a diatomic gas

Al = +3; Cl = -1substance being decomposed

Next, to balance this equation we need to add some coefficients as shown below:

2 Al(ClO3)3 ----> 2 AlCl3 + 9 O2 (It's done!)

Problem 5. Complete and balance the decomposition equations below. Refer to Table 6.3 forexamples that may help you. Use oxidation numbers to check the formulas of products. Here is somehelp with the first one. The first compound below is a chlorate (ClO31-) compound. Table 6.3 includessome chlorate (ClO31-) compounds. Note that when chlorate compounds decompose, one of theproducts is diatomic oxygen gas. The other product is a compound between chlorine and the metalpresent in the compound. In part a below, it will be a compound of sodium (Na) and chlorine (Cl). Useoxidation numbers to write the correct formula for this compound (Na = +1, Cl = –1). Write the correctformulas for these two compounds on the right side of the arrow below, and then add coefficients tobalance the equation.

a. _____NaClO3 -----> _______________________________________

b. _____FeO -----> _____Fe + ______

c. _____BaCO3 -----> _______________________________________

d. _____KCl -----> _______________________________________

e. _____Na2CO3 -----> _______________________________________

f. _____K2O -----> _______________________________________

g. _____NaBr -----> _______________________________________

h. _____CaCO3 -----> _______________________________________

i. _____Mg(ClO3)2 -----> _______________________________________

j. _____H2CO3 -----> _______________________________________

On tests and quizzes, you will be permitted to use your reference notebook which includes a copy ofTable 6.3 – Four Kinds of Decomposition Reactions

Problem 6. Complete and balance equations which represent the decomposition of the compoundsgiven below. Again use Table 6.3 as a guide. Use oxidation numbers when writing formulas forcompounds. To begin the first one, write the correct formula for strontium hydroxide in the space to theleft of the arrow (Sr = +2, OH = –1). Write the correct formulas for the products in the spaces on the right.Add coefficients to balance the equation.

a. strontium hydroxide _____________ ------> _____________+______________

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b. barium carbonate _____________ ------> _____________+______________

c. calcium chlorate _____________ ------> _____________+______________

d. silver oxide _____________ ------> _____________+______________

e. sodium carbonate _____________ ------> _____________+______________

ACTIVITY 6.8 Observing The Decomposition Reaction of H2O2

In this activity you will witness a simple decomposition reaction, and the role of a catalyst.

Procedure:

1. Obtain a 150 mm test tube a fill it about 1/4 full of 6% hydrogen peroxide solution, H2O2.

2. Obtain a small amount of manganese dioxide, MnO2. Don't do it yet, but when you put the MnO2 intothe H2O2 solution, the solution will begin to decompose into water and oxygen gas which you will seebubbling out of the solution:

Unbalanced equation: ? H2O2(aq) ----> ? H2O(l) + ? O2(g)

The MnO2 does not actually take part in the reaction, it simply speeds up the otherwise very slowdecomposition of H2O2. It is called a catalyst.

3. Before you put the MnO2 catalyst into the tube, have a glowing wood splint ready. (A glowing splint isprepared by burning a splint briefly, and then blowing it out.) Add the MnO2, tilt the tube to a 45 degreeangle and place your thumb loosely over the mouth of the tube. Allow the reaction to continue for about15 seconds. Then, remove your thumb and put the glowing splint down into the tube - but do not touchthe liquid with it. If oxygen is being given off, the splint should glow brighter or burst into flame. If it doesnot, try again.

Describe your results:_____________________________________________________________

Write the balanced equation for this decomposition reaction: {27}______________________________


This is the end of Chapter 6. Check the learning outcomes below. When you feel you havemastered them, take the test on chapter 6 and move on to Chapter 7. Chapter 7 will introduce you to twomore kinds of chemical equations.

_____1. Write the formulas of seven diatomic and two polyatomic elements from memory.

_____2. Identify the reactants and products in chemical equations.

_____3. Identify chemical equations as either direct combination or decomposition.

_____4. Correctly complete and balance direct combination and decomposition equations.

6-14 ©1997, A.J. Girondi


Questions:

{1} H2 and O2; {2} H2O; {3} H2O; {4} H2CO3; {5} CO2; {6} CaCO3 and CaO; {7} CaCO3; {8} 2; {9} 2;{10} yes; {11} yes; {12} 2; {13} 2; {14} yes; {15} yes; {16} 2; {17} 1; {18} No; {19} No; {20} 2; {21} 4;{22} No; {23} 2; {24} 2; {25} yes; {26} 2 H2 + O2 ---> 2 H2O; {27} 2 H2O2 ---> 2 H2O + O2

Problems:

1. a. 2,3,2; b. 2,1,2; c. 4,1,2; d. 1,1,1; e. 1,10,4; f. 2,3,2; g. 2,1,2; h. 4,1,4; i. 2,1,2; j. 4,3,2

2. a. 2 Al + 3 F2 ---> 2 AlF3

b. 2 Na + Br2 ---> 2 NaBr c. 2 Ca + O2 ---> 2 CaOd. H2 + Cl2 ---> 2 HCle. 8 Mg + S8 ---> 8 MgSf. 4 P4 + 3 S8 ---> 8 P2S3

g. P4 + 5 O2 ---> 2 P2O5

3. a. 4 Al + 3 O2 ---> 2 Al2O3

b. Si + 2 F2 ---> SiF4

c. 2 C + O2 ---> 2 COd. P4 + 6 Cl2 ---> 4 PCl3e. 16 Fe + 3 S8 ----> 8 Fe2S3

f. S8 + 24 Br2 ----> 8 SBr6g. 2 Ag + I2 ---> 2 AgI

h. H2 + F2 ---> 2 HF

4. a. 1,1,1,1; b. 2,2,1; c. 2,2,3; d. 2,1,3; e. 1,1,1,4; f. 1,1,1; g. 2,2,1; h. 2,2,3; i. 1,1,1; j. 1,1,1

5. a. 2 NaClO3 ----> 2 NaCl + 3 O2

b. 2 FeO ----> 2 Fe + O2

c. BaCO3 ----> BaO + CO2

d. 2 KCl ----> 2 K + Cl2e. Na2CO3 ----> Na2O + CO2

f. 2 K2O ----> 4 K + O2

g. 2 NaBr ---> 2 Na + Br2h. CaCO3 ----> CaO + CO2

i. Mg(ClO3)2 ----> MgCl2 + 3 O2

j. H2CO3 ----> H2O + CO2

6. a. Sr(OH)2 ----> SrO + H2Ob. BaCO3 ----> BaO + CO2

c. Ca(ClO3)2 ----> CaCl2 + 3 O2

d. 2 Ag2O ----> 4 Ag + O2

e. Na2CO3 ----> Na2O + CO2

6-15 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


“ALICE”

CHAPTER 7

CHEMICALEQUATIONS

(Part 2)

Single Replacementand

Double Replacement7-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


7-2 ©1997, A.J. Girondi

SECTION 7.1 Balancing Equations of Single Replacement Reactions

In a single replacement (or single displacement) reaction one element displaces (or replaces)another element that is present in a compound. You can recognize a single replacement equation rathereasily since the most common examples involve an element reacting with a compound. You can see thatin the examples below, that the element zinc replaces the hydrogen in the hydrogen iodide compound.In the first example, zinc has a positive oxidation number (+2), and it replaces the element with the positiveoxidation number in the compound (hydrogen, +1). In the second example, chlorine has a negativeoxidation number (-1) and replaces the bromine in the HBr, since bromine also has a negative oxidationnumber. Metals tend to replace metals in these equations, and nonmetals tend to replace nonmetals. Inthese equations hydrogen reacts like a metal. Therefore the zinc metal replaces the hydrogen "metal,"and the nonmetallic chlorine replaces the nonmetallic bromine. Remember, metals replace metals andnonmetals replace nonmetals.

Notice, that all of the formulas are written correctly (the sum of the oxidation numbers of the atomsin each formula add up to zero).

Zn + 2 HI --------> ZnI2 + H2

+1 –1 +2 –1

(Note that the zinc metal replaces the hydrogen "metal" in the compound.)

Cl2 + 2 HBr ----> 2 HCl + Br2

+1 –1 –1+1

(Note that the nonmetallic chlorine replaces the nonmetallic bromine in the compound.)

Notice how the diatomic elements in the examples above (H2 and Br2) are written in diatomic form whenthey occur uncombined with other elements. Before going on, be sure you have memorized the sevendiatomic elements: H2, N2, O2, F2, Cl2, Br2, and I2.

Which element does chlorine replace in the equation above? {1}_________ Why that element?

{2}______________________________________. Which element does zinc replace in the equation

above? {3}____________. Why that element? {4}__________________________________________

Single replacement reactions do not always occur. Just because you can write and balance anequation, doesn't mean that the reaction will happen. For example, the equation:

2 Ag + CuCl2 ----> 2 AgCl + Cu

is balanced and looks good. However, it is wrong! The reason it is wrong is that this reaction does notoccur. When you mix these reactants together, nothing happens. The equation should be:

2 Ag + CuCl2 ----> NR

where "NR" means "no reaction." You may ask, "How am I supposed to know whether a reaction happensor not?" Glad you asked! The "activity series" is a listing of elements that you can use to determinewhether or not a single replacement reaction will occur.

Examine the activity series in Table 7.1. In order for silver to replace copper, it would have to bemore "active" than copper. Silver would have to be listed above copper on the activity series in order to

7-3 ©1997, A.J. Girondi

replace it. Silver is not listed above copper, so the reaction cannot occur. Note that there are two columns(metals and nonmetals). The activity of metals can only be compared to other metals. Nonmetals can onlybe compared to other nonmetals. Although hydrogen is a nonmetal, it reacts like the metals and is,therefore, listed with the metals.

Table 7.1The Activity Series

Metals Nonmetals

lithium fluorinepotassium chlorinecalcium brominesodium iodinemagnesiumaluminumzincchromiumironnickeltinleadhydrogen Although hydrogen iscopper not a metal, it ismercury compared to metals insilver the activity series.platinumgold

(A copy of the activity series can also be found in the reference section of your ALICE materials.)

Now let's look at two types of single replacement reactions.

Example 1. Replacement of a metal in a compound by another metal:

Mg + CuCl2 ----> MgCl2 + Cu+2 +2-1 -1

Notice that one metal replaces a second metal in a compound. Since metals generally have positiveoxidation numbers, you can think of this type of reaction as having the elements with the positiveoxidation numbers switch places. Magnesium can replace copper because Mg is listed above Cu in theactivity series.

In the reaction below Al has an oxidation number of +3 when it combines with chlorine, while Cuhas an oxidation number of +2 when it combines with chlorine. The aluminum is going to replace thecopper in the CuCl2. Aluminum can replace copper because Al is listed above Cu in the activity series.When it does, the aluminum will then combine with the chlorine; however, the correct formula foraluminum chloride is AlCl3, whereas the formula for cupric chloride is CuCl2.

Given: Al + CuCl2 ---->

Predict products and write formulas correctly using oxidation numbers:

Al + CuCl2 ----> AlCl3 + Cu+2 +3-1 -1

7-4 ©1997, A.J. Girondi

Finally, balance the equation by adding coefficients:

2 Al + 3 CuCl2 ----> 2 AlCl3 + 3 Cu

Note that the balanced equation above reveals the same number of atoms of each element on both sides.

Don't try to balance the equation by changing the subscripts in the formulas! Some students would try tobalance the equation above by changing the formulas of the products as follows:

Wrong!---> 2 Al + 3 CuCl2 ----> Al2Cl6 + 3 Cu <--- Wrong!

This is wrong! It is balanced, but the subscripts in the empirical formulas which you have learned to writeshould reflect the simplest ratio possible. So Al2Cl6 should be AlCl3. Once the formulas of thecompounds in an equation are written correctly, don't change them! In other words, don't change anysubscripts.

Example 2. Replacement of a nonmetal in a compound by another nonmetal:

Cl2 + 2 NaBr ----> 2 NaCl + Br2+1 +1-1 -1

(Notice that the chlorine and bromine are diatomic when they are not combined with other elements.)

Chlorine can replace bromine in the equation above because chlorine is listed above bromine in theactivity series. We say that chlorine is "more active."

Sometimes this second type of replacement reaction involving nonmetals may not occur. Forexample:

I2 + KCl ----> NRIodine cannot replace the chlorine in the reaction above because iodine is not listed above chlorine in theactivity series. We say that iodine is "less active" than chlorine.

In order for one metal to replace another metal, it must be listed above the metal it is going toreplace. If this is not the case, no reaction (NR) will occur. The same test is applied when a nonmetal istrying to replace another nonmetal, but notice that nonmetals are listed separately in the series. Checkthe activity series to make sure that you understand the examples shown above.

Problem 1. Balance the equations below. (Note: in the first equation, water is written as HOH ratherthan H2O. This often makes it easier to understand what replaces what.)

a. _____Na + _____HOH ----> _____NaOH + _____H2

b. _____Ca + _____HCl ----> _____CaCl2 + _____H2

c. _____Fe + _____CuCl2 ----> _____FeCl3 + _____Cu

d. _____Zn + _____Fe2O3 ----> _____ZnO + _____Fe

e. _____Cl2 + _____AlBr3 ----> _____AlCl3 + _____Br2

f. _____Zn + _____FeCl3 ----> _____ZnCl2 + _____Fe

g. _____Cl2 + _____ NaI ----> _____NaCl + _____I2

h. _____K + _____NiCl2 ----> _____KCl + _____Ni

i. _____Ca + _____HCl ----> _____CaCl2 + _____H2

j. _____Li + _____BaSO4 ----> _____Li2SO4 + _____Ba

7-5 ©1997, A.J. Girondi

Problem 2. Complete and balance the single replacement reactions below. Be sure to answer with"NR" instead of completing the equation if there is no reaction. Remember, that Roman numerals inparentheses tell you which oxidation number to use (ex: Mn(II) = Mn2+). However, the Roman numeralsdo not appear in the formulas of the compounds. For example, do not write a formula such as Mn(II)Cl2.Simply write MnCl2. Before you attempt to balance the equation, make sure you have written the formulasfor the products correctly - using oxidation numbers.

a. _____Al + _____H2SO4 -----> _________________________________________

(Reminder: hydrogen reacts like a metal)

b. _____Cl2 + _____KI -----> _________________________________________

(Reminder: iodine is diatomic)

c. _____Fe(III) + _____Cu(NO3)2 -----> _________________________________________

d. _____Ag + _____Pb(C2H3O2)2 -----> _________________________________________

(Caution: check the activity series)

e. _____Br2 + _____NaF -----> _________________________________________

f. _____K + _____ CaSO4 -----> _________________________________________

g. _____Hg(I) + _____LiCl -----> _________________________________________

h. _____Fe(II) + _____CuCl2 -----> _________________________________________

i. _____I2 + _____AlBr3 -----> _________________________________________

j. _____Cl2 + _____AlBr3 -----> _________________________________________

ACTIVITY 7.2 Determination of the Activities of Selected Metals

In this activity you will gather data which will help you to see how the activity series was developed.You will mix various combinations of chemicals. Some combinations will result in single replacementreactions, while others will do nothing at all.

You need a 10 mL graduate; 6 test tubes (150 mm); a test tube rack; three small pieces of copper,three small pieces of zinc; 0.2M solutions of KNO3, AgNO3, and Pb(NO3)2. Caution: Pb(NO3)2 is toxic,and AgNO3 can cause temporary skin stains.

Note: You may occasionally see terms such as 0.1M or 0.2 M when referring to solutions. These terms refer to the concentrations ofthe solutions. At this point this information is primarily for your instructor. You will learn more about this concept in an upcoming chapter.

Procedure:

1. Clean 6 test tubes and rinse with distilled water. There is no need to dry them. Label the tubes 1through 6.

2. Place 5 mL of KNO3 into two of the tubes, 5 mL of AgNO3 solution into 2 of the tubes, and 5 mL ofPb(NO3)2 into 2 of the tubes.

7-6 ©1997, A.J. Girondi

3. Put a piece of each of the two metals into the two tubes containing KNO3, then repeat this for the tubescontaining the other two solutions. Make note of which tubes contain which reactants. Observe thecontents of the tubes occasionally for at least 10 minutes. If time is short, you can allow the tubes toremain in your lab drawer overnight.

A change in the appearance of the metal or of the color of the solution is evidence that a singlereplacement reaction has occurred. In some tubes, nothing will happen. Fill each block in the table belowwith either "R" (reaction) or "NR" (no reaction), depending on your observations. Decant the solutions -but not the solids - into the waste bottle provided for that purpose. Add some water and decant again.Dump the solids into a paper towel and throw it into the waste container.

Table 7.2 Single Replacement Reactions

Metal KNO3 AgNO3 Pb(NO3)2

Zn

Cu

Based on the number of reactions for each metal shown in Table 7.2, which is more active, copper or zinc?

{5}_______________. Based on your results which are summarized in Table 7.2, indicate by underlining

which metal in each of the following pairs is the more active:

K or Cu; K or Zn; Cu or Ag; Zn or Ag; Pb or Cu; Zn or Pb

Based on your conclusions above, arrange the following metals in order from most active to least active:

Cu, Zn, K, Ag, and Pb. {6} ______________________________________ Are your results supported

by the Activity Series found in Table 7.1? ____________ If iron metal had been included in this activity

between which two metals listed above would it have "fit" in terms of activity?{7}____________________

ACTIVITY 7.3 "Activities" Of The Halogens

The elements in Group 7A (as known as Group 17) are called the halogens. They includefluorine, chlorine, bromine, iodine, and {8}_______________. The word halogen means "salt-former."Salts are compounds which are combinations of metals and nonmetals. Their chemical property of formingsalts (such as NaCl and KBr), along with other chemical and physical properties which they have incommon, is the reason that these elements are placed in the same chemical group or family. Thehalogens have similar chemical reactions, but their chemical "activities" vary. A more "active" halogen willreplace a less active halogen in a compound. For example: F2(g) + 2 NaCl(aq) ---> 2 NaF(aq) + Cl2(g)

In this activity you will attempt to determine the relative activities of the halogens by observing some singlereplacement reactions in which they are involved.

The materials you will need include some small (100 mm) test tubes, long- stemmed plastic pipets, smallcorks, and dropper bottles containing the following: TCE (trichloroethylene); freshly-prepared chlorinewater; 0.1M NaCl; 0.1M NaBr; and 0.1M NaI; a waste container for used TCE. The "0.1M" simply indicateshow concentrated the solutions are. CAUTION: Wear safety glasses and aprons during this activity!

7-7 ©1997, A.J. Girondi

Procedure Part A.

1. Obtain and clean (if needed) two small (100 mm) test tubes. The tubes do not need to be dry. Placeabout 20 drops of 0.1M solutions of NaBr and NaI in separate tubes. Label the tubes with a marker astubes 1 and 2, respectively. Note that neither of the solutions has any color.

2. Add enough TCE to each tube to roughly double the volume of liquid in each tube. Caution: TCEshould not come into contact with your skin. Do not breathe the vapors. TCE does not mix with watersolutions. You will see two layers in each tube. TCE is more dense, so it goes to the bottom of each tube.

3. Your teacher will prepare some fresh chlorine water for you by adding some concentrated hydrochloricacid (HCl) to a 5% chlorine bleach solution until the evolution of gas ceases. Chlorine water is simply waterwhich has some chlorine gas, Cl2, dissolved in it. Do not allow chlorine water to contact your skin or eyes.

4. Add 3 to 5 drops of the fresh Cl2 water to the contents of both test tubes. Tightly cork each tube andhold away from your face. Shake each tube for 5 to 10 seconds to mix the substances well. Allow thecontents of the tubes to rest for 10 or 20 seconds, and the observe the contents.

5. Observe tube 1 which contains NaBr solution and TCE. The reaction which occurred in this tube was:2 NaBr(aq) + Cl2(aq) ----> 2 NaCl(aq) + Br2(aq). Chlorine replaced bromine, and when this tube wasshaken, the free bromine moved to the TCE layer. The TCE layer should now have an amber color. Theamber color is caused by the presence of bromine in TCE. Note that chlorine must be more active thanbromine since it was able to replace it.

Amber Color in TCE Indicates the Presence of Bromine

6. Next, observe tube 2 which contains NaI solution and TCE. The reaction which occurred in this tubewas: 2 NaI(aq) + Cl2(aq) ----> 2 NaCl(aq) + I2(aq). Chlorine replaced iodine, and when this tube wasshaken, the free iodine moved to the TCE layer. The TCE layer should have a pinkish color. The pinkcolor is caused by the presence of iodine in TCE. Note that chlorine must be more active than iodinesince it was able to replace it.

Pink Color in TCE Indicates the Presence of Iodine

7. Finally, obtain another 100 mm test tube and place about 20 drops of 0.1M NaCl solution and 20 dropsof TCE into it. Add 3 to 5 drops of fresh Cl2 water and shake. When the tube was shaken, the chlorinemoved from the NaCl solution to the TCE layer. There should be little or no color in the TCE (bottom)layer. This shows us that when TCE contains chlorine, Cl2, no color (or only a very pale color) is seen inthe TCE.

Chlorine has no significant color when present in TCE

8. Enter the colors you observed into Table 7.3 before going on to part B.

9. Since TCE should not be poured into your sink, empty the contents of the test tubes into the containerprovided by your instructor. Rinse the tubes well before use in part B below.

Procedure Part B.

10. To a small (100mm) test tube (#1) add about 20 drops of NaBr solution, 5 drops of NaI solution, andthen add enough TCE to roughly double the volume of liquid in the tube. Next, add 1 or 2 drops of freshchlorine water. Tightly cork and shake the tube. Wait about 10 seconds and observe the TCE (bottom)layer. Repeat this procedure of adding a drop or two of Cl2 water and shaking, until a color change isobserved in the TCE layer. What color is eventually observed in the TCE layer?____________________ . This color indicates the presence of which halogen in the TCE layer?

________________

7-8 ©1997, A.J. Girondi

The Cl2 which you added to this tube reacted with one of the halogen compounds already present in the

tube (NaBr and NaI). The halogen which was replaced then entered the TCE layer. Based on the color in

the TCE, which halogen compound did the Cl2 react with? __________________ Which halogen (Br2 or

I2) is now found in the TCE layer? ______________ This change involved a single replacement reaction.

Complete the equation for that reaction in the space below:

{9} __________ + Cl2(aq) ----> ____________ + _____________

11. With a long-stemmed bureal plastic pipet, remove the remaining NaBr and NaI solutions in the toplayer of the test tube and place into another test tube (#2). Now add enough TCE to test tube #2 toroughly double the volume of liquid in the tube. Then, add Cl2 water 1 or 2 drops at a time, shake andobserve until a color is observed in the TCE layer - following the same procedure as you did in step 10. Isthe color eventually observed in the TCE layer the same as that observed in step 10? ________ If so,the same reaction is again occurring as did in step 10. If this is the case, repeat this step with yet anothertest tube (#3). (Remove the top layer in tube #2 and put it in tube #3, add TCE, add Cl2 water, shake, etc.)This step must be repeated until you finally see a different color in the TCE layer.

What color is eventually observed in the TCE? _______________ This indicates that the added Cl2 isnow reacting with what halogen compound (NaBr or NaI)? Complete the equation for that reaction in thespace below: {10} __________ + Cl2(aq) ----> ____________ + _____________

In steps 10 and 11, Cl2 was added to a mixture of NaBr and NaI solutions. Since chlorine was able toreplace both bromine and iodine, we can conclude that chlorine is more "active" than bromine and iodine.

Chlorine first replaced the lesser active of these two halogens, however. Which halogen (bromine or

iodine) was replaced first by the Cl2? ______________ Which halogen (bromine or iodine) is least active?

__________________.

Note: Use the same caution with bromine water that you did with chlorine water!

12. Place about 20 drops of NaCl solution in one small test tube and about 20 drops of NaI solution in

another small tube. Add about 20 drops of TCE to each tube. Next, you will be adding bromine, Br2, to

each tube. If the added bromine is able to replace the chlorine in the NaCl tube, what color will appear in

the TCE layer of that tube? _________________ If the added bromine is able to replace the iodine in

the NaI tube, what color will appear in the TCE layer of that tube? __________________ Now add 3 to 5

drops of bromine water (Br2) to each of these tubes. Stopper, shake, and allow the contents of the

tubes to rest for 10 to 20 seconds. Observe the TCE layer. If the TCE in both tubes is colorless,

repeat the addition of bromine water. Continue this process until the TCE in one of the tubes has a color.

Based on your observations, was bromine able to replace the chlorine in the NaCl, or was it able to replace

the iodine in the NaI tube? _________________________________________________________

Complete and balance the equation below for the reaction which you observed.

{11} __________ + Br2(aq) ----> ____________ + _____________

Again, since TCE should not be poured into your sink, discard the contents of the tubes into thecontainer provided by your instructor.

7-9 ©1997, A.J. Girondi

The most active elements are capable of replacing elements of lesser activity. Based on your

results, list the order of activity of the three halogens tested - most active first, etc.________________

Do your results agree with the activity of the halogens listed in Table 7.1? ____________

Table 7.3 Activities of the Halogens

Color of Cl2 in TCE _____________________

Color of Br2 in TCE _____________________

Color of I2 in TCE _____________________

(Chlorine water itself has a pale yellow color.)

SECTION 7.4 Balancing Equations of Double Replacement Reactions

The fourth type of reaction known as double replacement (also known as double displacement).In this type of reaction, the metals (which have positive oxidation numbers) switch places.

For example: NaCl + AgNO3 ----> AgCl + NaNO3

Note that in the double replacement reactions which you will be studying, there are two reactants and twoproducts. Both reactants and both products are compounds. In the example above, do you see how thesodium (+1) and the silver (+1) switch "partners"? That's how double replacement works. Polyatomic ionsdo not come apart in these reactions, so just treat them as single elements.

Problem 3. Balance the double replacement reactions below by adding the needed coefficients. Allformulas are already written correctly. Compare the formulas of the products to those of the reatants tosee if you understand how double replacement reactions work.

a. _____PbCl2 + _____H2SO4 ----> _____PbSO4 + _____ HCl

b. _____BaCO3 + _____Ca(NO3)2 ----> _____Ba(NO3)2 + _____CaCO3

c. _____Pb(NO3)2 + _____HI ----> _____PbI2 + _____HNO3

d. _____FeBr3 + _____Ba(OH)2 ----> _____Fe(OH)3 + _____BaBr2

e. _____Ag2SO4 + _____Cu(ClO3)2 ----> _____AgClO3 + _____CuSO4

f. _____AgNO3 + _____AlCl3 ----> _____AgCl + _____Al(NO3)3

7-10 ©1997, A.J. Girondi

ACTIVITY 7.5 Observation of Some Double Replacement Reactions

When two dissolved substances react to form a solid, the solid is called a precipitate . Look at the equationbelow.

AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)

The precipitate formed is {12}_______________. The substances which are dissolved in water (aqueous)

are {13}________________________________________. In this activity, you will observe several

double replacement reactions in which precipitates are formed. Follow directions carefully.

Chemical Reaction 1:

Obtain dropper bottles of 0.1M BaCl2 and 0.1M K2SO4 solutions. Place a few drops of BaCl2 in a well on aspot plate. To the BaCl2, add a few drops of K2SO4. The precipitate which is produced is due to theformation of BaSO4. Rinse and dry the spot plate.

Balance the equation that represents this reaction:

{14} _____BaCl2 (aq) + _____K2SO4(aq) ----> _____BaSO4(s) + _____KCl(aq)


Obtain dropper bottles of 1% (NH4)2S and 0.1M CuCl2 . (Caution: keep these solutions off skin andclothes.) Place a few drops of (NH4)2S in a well on a spot plate, and add a few drops of CuCl2. The suddendark appearance is due to the formation of copper (II) sulfide, CuS. Clean the spot plate as you didabove.

Balance the equation that represents this reaction:

{15} _____(NH4)2S(aq) + _____CuCl2(aq) ----> _____NH4Cl(aq) + _____CuS(s)


Next, we will look again at a reaction you saw in Chapter 3. Obtain dropper bottles of 0.25M NaI and 0.25MPb(NO3)2 solutions. Place a few drops of NaI in a well on a spot plate. Add a few drops of Pb(NO3)2solution. Observe the result and rinse the spot plate. The colored product is PbI2. Balance the equationbelow that represents this reaction:

{16} _____Pb(NO3)2(aq) + _____NaI(aq) ----> _____PbI2(s) + _____NaNO3(aq)

SECTION 7.6 More Work With Equations for Double ReplacementReactions

Double replacement reactions are similar to single replacement in the sense that sometimes theyhappen and sometimes they don't. However, you don't use the activity series to help you with double

replacement. Instead, you will need to remember that double replacement reactions will happen if at leastone of the products meets at least one of the following three requirements:

7-11 ©1997, A.J. Girondi

1. a precipitate is formed

2. a gas is formed

3. a slightly ionized product is formed (example: H2O or NH4OH)

In order for a doublereplacement reaction tooccur, at least one of theseconditions must be met.

Water can also be written as HOH

Remember: only one of the products of a double replacment reaction has to be a precipitate, gas, or slightly ionized substance in order for the reaction to occur.

In other words, if at least one of the products is a precipitate, a gas, or a slightly ionized substance, then the reaction will happen. Three examples are listed below:

1. NaCl(aq) + AgNO3(aq) ---> NaNO3(aq) + AgCl(s) Precipitate, AgCl, forms

2. Na2S(aq) + 2 HCl(aq) ----> 2 NaCl(aq) + H2S(g) A gas, H2S, forms

3. NaOH(aq) + HCl(aq) ----> NaCl(aq) + HOH(l) Water, HOH, forms

If none of these three conditions are met, then the result is no reaction:

NaNO3(aq) + Cu(ClO3)2(aq) ----> NR

At this point, you can recognize double replacement reactions as those in which two compounds arereacting with each other. Actually, this rule is too general, but it will suit your purposes for now. You mayhave noticed that single replacement reactions can be recognized as having an element reacting with acompound.

You may be asking yourself, "How am I supposed to know whether or not the products of a doublereplacement reaction are precipitates, slightly ionized substances, or gases?" Glad you asked! Theanswers are as follows.

1. To identify a product as a precipitate, you must check a solubility table or solubility rules.

2. The only slightly ionized substances you need to worry about in your study of ALICE are water (H2O or HOH) and ammonium hydroxide (NH4OH). There are others, but don't be concerned at this point.

3. Since it would be difficult for you to identify a product as a gas, sufficient information will have to be given to you. For example, you should look for the subscript (g) written after one of the products.

The solubility rules are listed in Table 7.4, and a solubility table is found in Table 7.5. Copies of thesetables are also available in the reference section of your ALICE materials which you will be permitted to useduring tests and quizzes. The information in these tables is valid for the dilute solutions (0.1M) which arenormally found in the laboratory. It is recommended that you use the solubility table (Table 7.5) as yourprimary source of information. If what you are looking for is not there, then you can resort to the solubilityrules (Table 7.4). The rules are rather general, while the table is quite specific.

7-12 ©1997, A.J. Girondi

SECTION 7.7 More Practice on Balancing Double ReplacementEquations

Sample Problem: Complete and balance the equation below:

Ba(OH)2 + Al2(SO4)3 ----> ___???___

Solution: The metals (barium and aluminum) will change places, meaning barium will combine withhydroxide and aluminum will combine with sulfate:

Ba(OH)2 + Al2(SO4)3 ----> BaSO4 + AlOH not done yet!

Finally, we have to use oxidation numbers to make sure that all of the formulas are written correctly. (Note that the formula AlOH in the example above is not yet written correctly.) Oxidation numbers are shown above the elements in the equation below. Note that the formulas for reactants and products are now all correct. We are now ready to determine if the reaction happens and, if it does, balance it.

Ba(OH)2 + Al2(SO4)3 ----> BaSO4 + Al(OH)3+2 –1 +3 –2 +2 –2 +3 –1 Formulas are now correct,

but the equation is notyet balanced.

The reaction does happen since precipitates are formed. (This example is unusual in that both products are precipitates.) We add the needed coefficients to balance the equation:

3 Ba(OH)2(aq) + Al2(SO4)3(aq) ----> 3 BaSO4(s) + 2 Al(OH)3(s) DONE!

Problem 4. Complete and balance the equations below. Check first to determine whether or not thereaction happens. If not, answer with NR. If a precipitate forms, identify it with the subscript "(s)." Informulas for products, write the symbol of the element with the positive oxidation number first in eachformula.

a. _____MgCl2 + _____KOH ----> ________________________________________

b. _____Mg(NO3)2 + _____NaCl ----> ________________________________________

c. _____Sr(OH)2 + _____HCl ----> ________________________________________

d. _____Pb(NO3)2 + _____K2SO4 ----> ________________________________________

e. _____H2SO4 + _____Ca(OH)2 ----> ________________________________________

f. _____HCl + _____K2S ----> H2S(g) + _________________________________

g. _____BaCl2 + _____K2CrO4 ----> ________________________________________

h. _____H3PO4 + _____ NaOH ----> ________________________________________

i. _____KCl + _____Ba(C2H3O2)2 ----> ________________________________________

7-13 ©1997, A.J. Girondi

TABLE 7.4SOLUBILITY RULES

Type of Compound Precipitate?

1. Most compounds of nitrates (NO31-), acetates (C2H3O21-),and chlorates (ClO31-) ARE soluble.- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO(Some silver and mercury compounds in this category are not soluble.)

2. All common compounds of Na, K, and NH41+ ARE soluble. - - - - - - - - - - - - - - - - - - - - NO

3. Most chloride compounds ARE soluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO(But those of Ag, Hg, and Pb* are NOT soluble) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - YES

4. Most sulfate compounds ARE soluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO(But those of Pb, Ba, Sr, and Ca** are NOT soluble) - - - - - - - - - - - - - - - - - - - - - - - - - - - YES

5. Most carbonate (CO32-), phosphate (PO43-),and sulfide (S2-) compounds are NOT soluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - YES(But those of Na, K, and NH41+ ARE soluble) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO

6. Most hydroxides are INsoluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - YES(But those of all Group1A metals, and Ca, Ba and Sr ARE soluble) - - - - - - - - - - - - - - NO

*Lead chloride, PbCl2, is fairly soluble in hot water. Consider it a precipitate.**Calcium sulfate, CaSO4, is slightly soluble. Consider it a precipitate.

SECTION 7.8 Classifying And Balancing Equations for Single and Double Replacement Reactions

Problem 5. In Part 1 of this problem you will be working with word equations for single and doublereplacement reactions. In word equations, you must write formulas for reactants and products. Be sure towrite the formulas correctly using subscripts, and then balance the equation using coefficients. If noreaction occurs, answer with "NR." Identify the type as single replacement (SR) or double replacement(DR) in the space at the left side.

Part 1. Word Equations

_____a. fluorine + sodium bromide; __________________ -----> ___________________

_____b. calcium + silver nitrate; __________________ -----> ___________________

_____c. sodium sulfide + lead (II) nitrate; __________________ -----> ___________________

_____d. barium chloride + silver acetate; __________________ -----> ___________________

_____e. potassium chloride + sodium nitrate; __________________ -----> ___________________

_____f. iodine + iron (II) chloride; __________________ -----> ___________________

_____g. copper (II) sulfate + hydrogen sulfide; __________________ -----> ___________________

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_____h. zinc chloride + sodium sulfide; __________________ -----> ___________________

_____i. hydrogen gas + copper (II) oxide; __________________ -----> ___________________

_____j. lead (II) nitrate + barium chloride; __________________ -----> ___________________

_____k. silver + copper (II) nitrate; __________________ -----> ___________________

In Part 2 of this problem, complete and balance each equation. If no reaction occurs, answer with "NR."

Part 2. Formula Equations.

l. _____ FeSO4(aq) + _____ BaCl2(aq) -------> _____________________________________________

m. _____ Al(s) + _____ AuNO3(aq) -------> _____________________________________________

n. _____ K2S(aq) + _____ H2SO4(aq) -------> H2S(g) +_____________________

o. _____ HC2H3O2(aq) + _____ Ca(OH)2(aq) -------> _______________________________

p. _____ HNO3(aq) + _____ Ba(OH)2(aq) -------> _______________________________

q. _____ Pt(s) + _____ MgCl2(aq) -------> _______________________________

r. _____ Cl2(g) + _____ KI(aq) -------> _______________________________

s. _____ Br2(l) + _____ ZnF2(aq) -------> _______________________________

t. _____ NaNO3(aq) + _____ AlCl3(aq) -------> _______________________________

u. _____ Cr(III)(s) + _____ NiSO4(aq) -------> _______________________________

v. _____ HCl(aq) + _____ KOH(aq) -------> _______________________________

w. _____ Na(s) + _____ Ni(ClO3)2(aq) -------> _______________________________

ACTIVITY 7.9 Predicting the Formation of Precipitates

Suppose you have solutions of the following three compounds: KOH, Fe(NO3)3, and CuCl2. Ifyou were to mix these solutions - two at a time - in all possible combinations (of two), how many reactionswill result in the formation of a precipitate? {17}__________ What are the correct chemical formulas ofthese precipitates? {18}_________________________________________________

Obtain a dropping plate, 0.1M Fe(NO3)3, 0.1M CuCl2, and 0.5M KOH. These three solutions are indropper bottles. Using 5 to 10 drops of each solution, mix all possible pair combinations of thesesolutions. Look for the formation of any precipitates. Was your prediction regarding the number ofprecipitates correct? ___________ In the space below, write balanced chemical equation(s) thatrepresent the formation of each precipitate. Identify each precipitate with the subscript "(s)."

{19}____________________________________________________________________________

______________________________________________________________________________

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Table 7.5A Solubility Table

For our purposes, the combinations below designated with the letter P, are considered to be insoluble enough to form precipitates. If a P is not listed for a substance, it is soluble in water (it dissolves). If youcannot find the data you need in this table, check the solubility rules. P = PRECIPITATE

1 - acetate, C2H3O21- 6 - chromate, CrO42- 11 - phosphate, PO43-

2 - bromide, Br1- 7 - hydroxide, OH1- 12 - silicate, SiO42-

3 - carbonate, CO32- 8 - iodide, I1- 13 - sulfate, SO42-

4 - chlorate, ClO31- 9 - nitrate, NO31- 14 - sulfide, S2-

5 - chloride, Cl1- 10 - oxide, O2-

1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4

aluminum - - - - - - P - - P P P - Pammonium - - - - - - - - - - - - - -barium - - P - - P - - - P P P P Pbismuth - P P - - - P P - P P - P Pcadmium - - P - - - P - - P - - - Pcalcium - - P - - - P - - P P P - Pcopper(II) - - P - - - P - - P P P - Phydrogen - - - - - - - - - - - P - -iron(II) - - P - - - P - - P P P - Piron(III) - - - - - P P - - P P P - Plead(II) - P P - P P P P - P P P P Pmagnesium - - P - - - P - - P P P - -manganese(II) - - P - - - P - - P P P - Pmercury(I) P P P - P P - P - P P - P Pmercury(II) - P P - - P P P - P P - P Pnickel - - P - - - P - - P P - - Ppotassium - - - - - - - - - - - - - -silver P P P - P P - P - P P - P Psodium - - - - - - - - - - - - - -strontium - - P - - P - - - - P P P -tin(II) - P - - - P P - - P P - - Ptin(IV) - - - - - - P - - P - - - Pzinc - - P - - P P - - P P P - P

(Note that sodium, potassium, and ammonium never form precipitates.)

ACTIVITY 7.10 (Optional) The Reaction Between NaOH and Ni(NO3)2

(Optional activities such as this are those which can be skipped if you are short of time or behind most of the other groups in your class.)

Chemicals react in the simple whole number ratios shown by the coefficients in balancedequations. Stoichiometry is that branch of chemistry in which we relate quantities of reacting substances.The exact quantities of reactants and products shown by a chemical equation are called stoichiometricquantities. For example, barium nitrate reacts with sulfuric acid in the following way:

7-16 ©1997, A.J. Girondi

Ba(NO3)2(aq) + H2SO4(aq) ---> BaSO4(s) + 2 HNO3(aq)

Note that one unit of Ba(NO3)2 reacts with one unit of H2SO4 to form one unit of BaSO4 and two units ofHNO3. This ratio represents stoichiometric quantities. If you happen to mix together anything differentfrom a stoichiometric ratio of the two reactants, you will find that when the reaction is over, a certain amountof one of the reactants will be left unreacted. We say that it was present "in excess." The “units” referredto here are actually quantities known as “moles.” You will study much more about moles in Chapters 8, 9and 10.

The purpose of this activity is to study the stoichiometry (reacting ratio) in the reaction betweensodium hydroxide and nickel nitrate. ? NaOH(aq) + ? Ni(NO3)2(aq) ---> ? NaNO3(aq) + ? Ni(OH)2(s)

In this reaction, one of the products, Ni(OH)2, is a green precipitate which is easy to see. You are going tomix the two reactants together in different ratios. Having done this, you will notice that different amountsof the Ni(OH)2 precipitate will be formed. By determining the ratio of reactants which produces a maximumamount of the product, you will be able to experimentally determine the stoichiometric ratio which isreflected by the coefficients in the balanced equation. In other words, you are going to balance theequation by experiment, rather than by simple inspection. You will need 9 test tubes (150 mm), a testtube rack, 0.1M Ni(NO3)2 solution, 0.1M NaOH solution, 2 pipets (10 mL), phenolphthalein solution, 1 corkto fit a 150 mm test tube. Handle NaOH solution with care. Be sure to wear safety glasses and an apronduring this activity.

Procedure:

1. Obtain nine 150 mm test tubes (Important: the tubes must all have the same internal diameter! Checkthis carefully!) Label the test tubes with a grease pencil or marker with the numbers 1 through 9. Placethem in numerical order in a test tube rack. If possible, use a rack which allows you to see the contents ofthe bottom third of each tube.

2. Ask the instructor or lab assistant to demonstrate the use of a volumetric pipet. Using a volumetricpipet, put 2 mL of Ni(NO3)2 solution in tube 1; put 4 mL in tube 2; put 6 mL in tube 3, etc. Continueincreasing the volumes by 2 mL until you have put 18 mL of Ni(NO3)2 into tube 9.

3. You will now work with the test tubes in reverse order. Put 2 mL of NaOH solution into tube 9; 4 mL intotube 8, etc. Continue this process until you have added 18 mL of NaOH solution to tube 1.

4. Add 3 to 5 drops of phenolphthalein solution to each of the 9 tubes.

5. Place a cork into tube 1 and mix the contents by inverting the tube several times. Rinse the cork andrepeat this process for tube 2. Continue until the contents of all tubes have been mixed.

6. Allow the tubes to rest undisturbed for one day before making any decisions concerning the amount ofprecipitate in each tube.

(Next day)

You should note that the tubes contain varying amounts of Ni(OH)2 precipitate, and that thesolution in some of the tubes has turned pink. Phenolphthalein reacts with NaOH to form a pink color.Therefore, those tubes in which the solution has a pink color contain an excess of (unused) NaOHreactant. That means that those tubes did not contain enough Ni(NO3)2 reactant to use up all of the NaOHreactant. Which tube numbers fall into this category? ______________ Therefore, do you think thatthose tubes contained stoichiometric quantities of the reactants? _____________

7-17 ©1997, A.J. Girondi

Now look at those tubes which do not reveal the pink color. In these tubes, all of the NaOHreactant was consumed. However, you will note that these tubes contain different amounts of precipitatebecause they contained different amounts of Ni(NO3)2 reactant. The greater the amount of Ni(NO3)2reactant in each successive tube, the greater the amount of precipitate product which formed. This is trueup to a point where the amount of precipitate formed reaches a maximum. Of ALL the tubes, which onecontains the maximum amount of precipitate?__________ This is the tube in which the two reactantsmost closely approximate a stiochiometric ratio. What is the ratio of 0.1M NaOH to 0.1M Ni(NO3)2 in thistube? ______________

For tube # ________:

volume 0.1 M NaOHvolume 0.1 M Ni(NO3)2

= _ _ _ _ _ _ _ _ _ _ _

Round this ratio to the closest whole number. Result: _____________________________________Now balance the equation below by inspection:

_____NaOH(aq) + ______Ni(NO3)2(aq) ---> ______NaNO3(aq) + ______Ni(OH)2(s)

What's the value of the NaOH / Ni(NO3) 2 ratio in the balanced equation:________________________

Does this ratio agree with the experimental NaOH / Ni(NO3) 2 ratio which you obtained? _____________If so, then you should now understand how you can determine - by experiment - the stoichiometric ratio ofcoefficients needed to balance an equation.

(None of the tubes in the activity contained the exact stoichiometric ratio for this reaction – not even thetube with the maximum amount of precipitate.)

Empty and clean all tubes, and return all materials to the proper places.


This is the end of chapter 7. Before you continue on to the next chapter, review the conceptsintroduced in chapter 7. Mark a check in front of each learning outcome below when you are certain thatyou have mastered it. Arrange to take the quizzes or exam on chapter 7, and move on to chapter 8.

_____1. Distinguish between the reactants and products of a given chemical equation.

_____2. Recognize and name two major types of chemical reactions - single replacement and double replacement.

_____3. Predict whether single and double replacement reactions will occur or not, given the reactants.

_____4. Correctly complete and balance two major types of chemical equations - single replacement anddouble replacement.

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Questions:

{1} Br; {2} Because nonmetals replace nonmetals; {3} H; {4} Because metals replace metals; {5} Zn;{6} K, Zn, Pb, Cu, Ag; {7} Between Zn and Pb; {8} astatine; {9} 2 NaI + Cl2 ---> 2 NaCl + I2; {10} 2 NaBr +Cl2 ---> 2 NaCl + Br2;{11} 2 NaI + Br2 ---> 2 NaBr + I2; {12} AgCl; {13} AgNO3, NaCl, NaNO3; {14} 1,1,1,2; {15} 1,1,2,1{16} 1,2,1,2; {17} 2; {18} Fe(OH)3 and Cu(OH)2;{19} Fe(NO3)3 + 3 KOH ---> Fe(OH)3(s) + 3 KNO3 and CuCl2 + 2 KOH ---> Cu(OH)2(s) + 2 KCl

Problems:

1. a. 2,2,2,1; b. 1,2,1,1; c. 2,3,2,3; d. 3,1,3,2; e. 3,2,2,3; f. 3,2,3,2; g. 1,2,2,1; h. 2,1,2,1; i. 1,2,1,1;j. 2,1,1,1

2. a. 2 Al + 3 H2SO4 -----> Al2(SO4)3 + 3 H2

b. 2 KI + Cl2 -----> 2 KCl + I2

c. 2 Fe + 3 Cu(NO3)2 -----> 2 Fe(NO3)3 + 3 Cud. NRe. NRf. CaSO4 + 2 K -----> K2SO4 + Cag. NRh. Fe + CuCl2 -----> FeCl2 + Cui. NRj. 3 Cl2 + 2 AlBr3 -----> 2 AlCl3 + 3 Br2

3. a. 1,1,1,2; b. 1,1,1,1; c. 1,2,1,2; d. 2,3,2,3; e. 1,1,2,1; f. 3,1,3,1

4. a. MgCl2 + 2 KOH -----> Mg(OH)2(s) + 2 KClb. NRc. Sr(OH)2 + 2 HCl -----> SrCl2 + 2 HOHd. Pb(NO3)2 + K2SO4 -----> PbSO4(s) + 2 KNO3

e. H2SO4 + Ca(OH)2 -----> CaSO4 + 2 HOHf. 2 HCl + K2S -----> H2S(g) + 2 KClg. BaCl2 + K2CrO4 -----> BaCrO4(s) + 2 KClh. H3PO4 + 3 NaOH -----> Na3PO4 + 3 HOHi. NR

5. a. SR 2 NaBr + F2 -----> 2 NaF + Br2b. SR 2 AgNO3 + Ca -----> Ca(NO3)2 + 2 Agc. DR Na2S + Pb(NO3)2 -----> PbS(s) + 2 NaNO3

d. DR BaCl2 + 2 AgC2H3O2 -----> 2 AgCl(s) + Ba(C2H3O2)2e. DR NRf. SR NRg. DR CuSO4 + H2S -----> CuS(s) + H2 SO4

h. DR ZnCl2 + Na2S -----> ZnS(s) + 2 NaCli. SR H2 + CuO -----> Cu + H2Oj. DR Pb(NO3)2 + BaCl2 -----> Ba(NO3)2 + PbCl2(s)

k. SR NR

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l. FeSO4 + BaCl2 -----> FeCl2 + BaSO4(s)

m. 3 AuNO3 + Al -----> Al(NO3)3 + 3 Aun. K2S + H2SO4 -----> H2S(g) + K2SO4

o. 2 HC2H3O2 + Ca(OH)2 -----> Ca(C2H3O2)2 + 2 HOH(l)

p. 2 HNO3 + Ba(OH)2 -----> Ba(NO3)2 + 2 HOH(l)

q. NRr. Cl2 + 2 KI -----> 2 KCl + I2

s. NRt. NRu. 2 Cr + 3 NiSO4 -----> Cr2(SO4)3 + 3 Niv. HCl + KOH -----> KCl + HOH(l)

w. 2 Na + Ni(ClO3)2 -----> 2 NaClO3 + Ni

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NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 8

THE MOLECONCEPT

(Part 1)

Avogadro's NumberAtomic and Molecular Mass

Percentage Composition8-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


8-2 ©1997, A.J. Girondi

ACTIVITY 8.1 Determining The Mass of Extremely Small Objects

In chapter 5 you became familiar with some of the most common elements and their chemicalsymbols. You will recall that an element is made up of extremely small particles called atoms. The questionarises as to precisely how small (or how large) atoms are. What is more important, you should know themass of the atoms so that you can compare, for example, the mass of boron and carbon atoms.

From an experimental standpoint, it would be impossible to isolate one atom in order to measureits mass; but, there is a way to accurately estimate the mass of a very small object. This activity will help toillustrate this.

1. Get the vials marked 8.1 from the materials shelf. Use a balance to get the information needed tocomplete Table 8.1. Look at Table 8.1 now, so that you will know what data is needed. Use proper unitswhen recording your data. Most balances are not sensitive enough to measure the mass of something aslight as one grain of rice. To calculate the mass of 1 grain of rice, you will need to know exactly how manythere are in each vial. Fortunately, these have already been counted for you. The number of objects ineach vial is marked on the vial.

2. Weigh each vial carefully and record the collected data in Table 8.1. Knowing the total mass of vial andcontents, calculate the total mass of the contents only. Record this information, too.

Knowing the total mass of the pieces in each vial, calculate the mass of one grain of rice, onestaple, and one lead shot. Express this number in scientific notation in Table 8.1.

Table 8.1Determining the Mass of Small Objects

Mass of an empty vial _________g

Mass of vial + rice _________g Mass of vial + lead shots_________g

Mass of rice grains _________g Mass of lead shots _________g

Mass of 1 rice grain _________g Mass of 1 lead shot _________g

Number of rice grains __________ Number of shots __________

Mass of vial + staples _________g

Mass of staples _________g

Mass of 1 staple _________g

Number of staples __________

It is impossible to determine the mass of one grain of rice, or one staple, or one lead shot usingthe laboratory balance. However, by measuring the mass of a large number of these objects, it is possibleto accurately estimate their individual masses. The mass of a large number of atoms can be determinedjust as easily as the mass of a vial of rice can be determined. No problems arise, however, until you try tocount the number of atoms in your sample! It takes much more than a magnifying glass and forceps to

8-3 ©1997, A.J. Girondi

count atoms. To give you an idea of the size of atoms, a hydrogen atom has a diameter of only 1.06 X 10-8

cm and a mass of only 1.67 X 10-24 grams.

The problem of counting atoms was solved in 1811 by an Italian physicist named AmedeoAvogadro. He proposed that:

equal volumes of any two gases contain equal numbers of moleculesunder the same conditions of temperature and pressure.

Shortly, you will see how Avogadro came to this conclusion and why it is important.

Figure 8.1 Equal Volumes of 4 Gases

H2

1 Volume

N2

1 Volume

O2

1 Volume

Cl2

1 Volume

Use Avogadro's proposal and the information in Table 8.2 below to answer the next several questions.

Which gas contains the same number of molecules as gas A? {1}__________

Which gas contains the same number of molecules as gas B? {2}__________

If gases A and C have equal numbers of molecules, how can you account for their different masses?

{3}____________________________________________________________________________

Table 8.2Temperature, Pressure, Volume and Mass of Four Gases

Temp. Pressure Volume Mass

Gas A 0oC 760 mm 44.8 L 28 gGas B 0oC 760 mm 22.4 L 1 gGas C 0oC 760 mm 44.8 L 32 gGas D 0oC 760 mm 22.4 L 14 g

To understand why "Avogadro's hypothesis" (as it has come to be known) is reasonable, youhave to keep in mind that most of the volume of matter is empty space. This is especially true of gases.The size of the molecules has nothing to do with the volume that gases occupy. It is the amount of spacebetween the molecules that determines the volume that the gas occupies. Thus, if gas A consists of "big"molecules and if gas B consists of smaller ones, this does not affect their volumes in any way. On theother hand, temperature and pressure do influence the distance between the molecules. Therefore,temperature and pressure have important effects on the volumes of gases.

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SECTION 8.2 The Law of Combining Volumes of Gases

In 1808, Gay-Lussac published his law of combining volumes of gases. He found that whengases react with each other, their volumes are always in a ratio of small whole numbers. In other words, 2liters of gas A combine with 1 liter of gas B to form 2 liters of gas C. He never saw evidence that 2 liters ofgas A combine with 1.6 liters of gas B to form 3.6 liters of gas C. The combining ratios always involvedsmall whole numbers. He did not know why this was true, nor how significant it might be. Gay-Lussac wasable to show that 2 liters of hydrogen combined with 1 liter of oxygen to form 2 liters of water vapor. Canyou believe it? In this case, two plus one equals two!

+ =

2 L hydrogen + 1 L oxygen = 2 L water vapor

Figure 8.2 Combining Volumes of Hydrogen and Oxygen

This discovery (of a 2:1:2 ratio) troubled John Dalton who could not make sense of it. (The diatomic natureof some gases had not yet been discovered.) For that reason, Dalton tried to disprove Gay-Lussac's work,but failed to do so. He thought that the formula for water was HO, and so, he would have liked to haveseen evidence that the ratio was 2:2:2, because then his equation for the formation of water would havebeen balanced:

2 Liters H + 2 Liters O ----> 2 Liters HO

However, Gay-Lussac's 2:1:2 ratio was not to be denied. But, how could a balanced equation be writtento explain this? 2 H + 1 O ----> 2 HO <--- This doesn't balance!

SECTION 8.3 Avogadro's Discovery

The answer to this puzzle was proposed by Avogadro in 1811. He pointed out that Dalton wasconfusing the concepts of atoms and molecules. Accepting the formula of water as HO, Avogadrosuggested that the oxygen involved in the reaction was composed of diatomic molecules, because thenGay-Lussac's 2:1:2 ratio could be explained as follows:

+ +

hydrogen shown as monatomic oxygen shown as diatomic

WATER VAPOR

2 L H(g) + 1 L O2(g) ----> 2 L HO(g)

+

Figure 8.3 Avogadro Proposes that Oxygen is Composed of Diatomic Molecules

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Thus, Avogadro is also given credit for first having suggested the idea of diatomic molecules.Since hydrogen was still assumed to be monatomic (H instead of H2), the "HO" formula for water wasaccepted in order to retain the necessary 2:1:2 ratio of liters found in the equation (Figure 8.3).

Since this idea seemed to work, what eventually led to the conclusion that the formula for waterwas H2O and not HO? The answer came from the study of the reaction between hydrogen gas andchlorine gas to form hydrogen chloride gas. The formula of hydrogen chloride was assumed to be HCl. Inthe reaction between these two gases, it was found that 1 volume of hydrogen combined with 1 volume ofchlorine to form 2 volumes of hydrogen chloride:

+

1 volume hydrogen + 1 volume chlorine -------------> 2 volumes hydrogen chloride

1 L hydrogen gas + 1 L chlorine gas -------------> 2 L hydrogen chlorine gas

Figure 8.4 Combining Volumes of Hydrogen and Chlorine

If the two gases were monatomic, then a 1:1:1 ratio with only one volume of HCl product would have beenexpected:

1 H + 1 Cl ----> 1 HCl

The 1:1:2 volume ratio which was found by experiment only makes sense if you assume that bothhydrogen and chlorine gases are composed of diatomic molecules (as shown in figure 8.4 and in theequation below:

1 L H2 + 1 L Cl2 -----> 2 L HCl

Both H2 and Cl2 molecules split up to form the two volumes of HCl molecules. At this point, at least threegases were assumed to be diatomic: H2, O2, and Cl2.

Since hydrogen was now considered to be diatomic (H2), the equation representing the formationof water had to be reexamined. Instead of this:

2 L H(g) + 1 L O2(g) ----> 2 L HO(g)

the reaction must be this:

2 L H2(g) + 1 L O2(g) ----> 2 L H2O(g) Bingo!

Note that if hydrogen were written as H2 instead of H, then water had to be written as H2O instead of HO.

8-6 ©1997, A.J. Girondi

+

diatomic hydrogen + diatomic oxygen ----------> water (H2O)

Figure 8.5 Formation of Water from H2 and O2

Now if H2 and O2 combine in a 2 to 1 molecule ratio, and knowing from Gay-Lussac's work thatthey also combine in a 2 to 1 volume ratio, it follows that the two volumes of H2 must contain twice as manymolecules as does the one volume of O2. So, one volume of H2 must contain the same number ofmolecules as does one volume of O2. This is what lead Avogadro to conclude that apparently:

EQUAL VOLUMES OF GASES CONTAIN EQUAL NUMBERS OF MOLECULES(IF THEY ARE AT THE SAME TEMPERATURE AND PRESSURE)

This was his major contribution, and it remains one of the most important discoveries in all of science.Avogadro's hypothesis (or Avogadro's law) went relatively unnoticed, however, for almost 50 years!

Oxygen molecules are much larger than hydrogen molecules. How is it possible then, that one liter of

oxygen can contain the same number of molecules as one liter of hydrogen? {4}____________________________

____________________________________________________________________________________________________________________

SECTION 8.4 The Relative Scale Of Atomic Masses

Indeed, it was the study of gases that gave the emerging science of chemistry a real boost. Ifequal volumes of gases (under the same conditions) contained equal numbers of molecules, then itbecame obvious that H2 had the least mass of any known molecule. It was found that a volume ofhydrogen gas, H2, had only 1/16 the mass of an equal volume of oxygen gas, O2. Helium gas, He, wasfound to have a mass equal to 1/8 that of an equal volume of O2. These discoveries allowed scientists tobegin the development of a scale of relative masses of atoms and molecules. We say relative massesbecause in those early days it was impossible to determine the actual mass of a single atom or molecule.However, by determining the masses of equal volumes of gases (at the same T and P), it was possible todevelop a relative scale which allowed the comparison of the masses of atoms of different elements.(Thanks to Avogadro's hypothesis.)

Oxygen gas was used as the first standard to which other gases were to be compared. The unitsthat were used for this relative scale were called atomic mass units (amu). Oxygen molecules, O2, wereassigned a relative mass of 32 amu. Why 32? The reason is simple. Since it was now known thathydrogen molecules, H2, had 1/16 the mass of oxygen molecules, O2, then assigning a value of 32 amu toO2 would give what value on the relative scale to H2 molecules? {5}__________ This, in turn, would givethe hydrogen atom, H, (the one with least mass) a relative mass of {6}______amu. It all seemed very niceto assign a value of 1 to hydrogen. Since H2 had 1/16 the mass of O2, it was safe to assume that hydrogenatoms, H, had 1/16 the mass of oxygen atoms, O. The helium atom, He, had 4 times the mass of hydrogenatoms, H, and 1/4 the mass of oxygen atoms and, so, was assigned a relative mass of {7}_______ amu.

8-7 ©1997, A.J. Girondi

Thus was born the scale of relative atomic and molecular masses. Although the original standardof comparison was oxygen, O, with a relative mass of 16, (O2 was 32), it has since been changed tocarbon with a relative mass of 12. Roughly speaking, how does the mass of an average carbon atom

compare to that of an average hydrogen atom? {8}_________________________________________

If you check the periodic table, you will note that the atomic masses of the elements hydrogen,helium, carbon, and oxygen are not exactly 1, 4, 12, or 16. This is because of the existence of isotopeswhich are atoms of the same element which have different masses. Isotopes are discussed in more detailin a later chapter. The atomic masses on the periodic table are average masses of the atoms of theelements as they occur in nature – including their isotopes. The most common isotope (or form) of carbonhas a relative mass of 12 amu and it is referred to as carbon-12. Less common forms of carbon haverelative masses of 13 and 14. As mentioned above, carbon-12 is the form of carbon now used as themodern standard for atomic masses.

The atomic mass of an atom is defined as the relativemass of an atom (in amu) compared to that of carbon-12.

The atomic mass of an element refers to the average relative mass of the atoms of an element, as theyoccur in nature, compared to carbon-12. Atomic masses of each element appear on the periodic table. Itis an average mass and does not represent the atomic mass of any particular atom of the element. Forexample, check a periodic table for the atomic mass of sulfur and enter that value in the space below.

Atomic mass of the element sulfur: ___________ amu

A number of isotopes of sulfur are found in nature. The most common (in order of abundance) are: sulfur-32, sulfur-34, sulfur-33 and sulfur-36. The atomic mass you wrote in the space above is the averagerelative mass of sulfur atoms in the proportion in which you would find them in nature.

We now have much more accurate methods for determining relative atomic masses, as well as theactual atomic masses of atoms, even though the real masses are very tiny. For example, we know that theactual mass of a hydrogen-1 atom is 1.67 X 10-24 g. That's small! You will recall that a hydrogen-1 atomalso has a relative mass of 1 amu. Therefore:

1 amu = 1.67 X 10-24 g.

Thus, the actual mass – in grams – of an atom of carbon-12 can be calculated:

12 amu X

1.67 X 10-24 g

1 amu = 2.0 X 10-23 g

Using this relationship between amu and grams, you can now calculate the actual masses of helium- 4 andoxygen-16 atoms. Do that in the space provided for problems 1 and 2 below, and show your work:

1 helium-4 atom = 4 amu; 1 oxygen-16 atom = 16 amu; 1 amu = 1.67 X 10-24 g

(If necessary, review how to enter numbers into your calculator using scientific notation (Chapter 2).

Problem 1. Calculate the actual mass in grams of an atom of helium-4.

8-8 ©1997, A.J. Girondi

Problem 2. Calculate the actual mass in grams of an atom of oxygen-16.

SECTION 8.5 Gram-Atomic and Gram-Molecular Masses

Working with individual atoms was not practical because of their size and tiny real masses. So,scientists devised a way of talking about larger quantities of an element. They invented a unit which theycalled the gram-atomic mass of an element.

One gram-atomic mass (1 gam) of an element is defined as theatomic mass of an element expressed in grams instead of in amu.

To determine how much of an element is 1 gam, simply look up its atomic mass on the periodic table andexpress it in grams. (When using atomic masses from the periodic table, we will generally round them totwo decimal places.)

What is the mass in grams of 1 gam of element number 24, chromium? {9}_____________ g

Grams are huge units compared to amu's and can be used as a practical unit of measure in the lab.Thus, 16.00 g of O is 1 gram-atomic mass, abbreviated as 1 gam, of oxygen atoms. 32.00 g is 1 gram-molecular mass, abbreviated as gmm, of oxygen molecules, O2. (Molecular mass is the sum of the massesof the atoms in the formula of a molecule; so if O = 16.00 then O2 = 32.00.) 1.01 g of H atoms = 1 gam of Hatoms, while 2.02 g of H2 molecules = 1 gmm of H2 molecules. The atomic mass units (amu) used on therelative scale of atomic masses is too small to be useful in the lab. A gram-atomic masses (gam) is ameasurable quantity of an element with which experiments can be performed.

One gram-atomic mass of an element (1 gam) is equal to the atomic mass ofthe element (from the periodic table) expressed in grams instead of in amu.

The atomic mass of an element as found on the periodic table is the relative mass of an atom expressed in atomic mass units (amu).

Examples: atomic mass hydrogen (H) = 1.01 amu

atomic mass nitrogen (N) = 14.01 amu

One gram-atomic mass (gam) of an element is the atomic mass expressed in grams rather than in amu.

Examples: 1 gam hydrogen (H) = 1.01 grams H 1 gam nitrogen (N) = 14.01 grams N

The formula mass (or molecular mass) of a compound is the sum of the atomic masses (in amu) of theatoms in the formula of the compound.

Examples: molecular mass (or formula mass) of H2 = 2.02 amu molecular mass (or formula mass) of N2 = 28.02 amu

8-9 ©1997, A.J. Girondi

One gram-molecular mass (gmm) of a compound is the molecularmass of the compound expressed in grams, rather than in amu.

Examples: 1 gram-molecular mass (1 gmm) of H2 = 2.02 grams H2

1 gram-molecular mass (1 gmm) of N2 = 28.02 g N2

The early experimenters found that 1 gmm of all gaseous elements had a volume of 22.4 liters (atSTP). Hey! That was an incredibly important discovery! 32.00 g of O2, 2.02 g of H2, 4.00 g of He, all hadvolumes of 22.4 L at STP. You will soon try to verify this volume, experimentally. According toAvogadro's hypothesis, these equal volumes contain equal numbers of atoms or molecules,

Thus, 1 gmm H2 gas = 2.02 g H2 = 22.4 L H2

Thus, 1 gam Ar gas = 39.95 g Ar = 22.4 L Ar

Thus, 1 gmm N2 gas = 28.01 g N2 = 22.4 L N2

For elements that are gases: 1 gam of gas (at STP) = 22.4 L

For compounds that are gases: 1 gmm of gas (at STP) = 22.4 L

22.4 LitersOXYGEN

O2

32.00 g

22.4 LitersHYDROGEN

H2

2.02 g

22.4 LitersHELIUM

He

4.003 g

22.4 LitersARGON

Ar

39.95 g

22.4 LitersNITROGEN

N2

28.02 g

All of these containers represent 22.4 L of gas at STP

Figure 8.6 Mass Versus Volume of 5 Gases at STP

Experimentally, we have found that 1 gam of any element contains 6.02 X 1023 atoms of that element. Wewill not be discussing how this was discovered, but it has been done using a number of differentexperimental techniques.

1 gam of an element = 6.02 X 1023 atoms of the element

1 gmm of a compound = 6.02 X 1023 molecules of the compound

Thus, the number of H2 molecules in 2.02 g of H2 is {10}_________________, a very large number to besure! Furthermore, since 2 grams of H2 have a volume of 22.4 L, then 6.02 X 1023 is the number ofmolecules of H2 found in {11}___________ L at STP. You could do similar calculations, and you would findthat 4.00 g of He and 32.00 g of O2 (both of which also have volumes of 22.4 L at STP) also contain 6.02 X1023 molecules. Remember, Avogadro told us that equal volumes of gases under similar conditionscontain equal numbers of molecules.

For all elements: 1 g-am = 6.02 X 1023 atomsFor all compounds: 1 gmm = 6.02 X 1023 molecules

For gas elements: 1 gam = 6.02 X 1023 atoms = 22.4 L at STPFor gas compounds: 1 gmm = 6.02 X 1023 molecules = 22.4 L at STP

8-10 ©1997, A.J. Girondi

SECTION 8.6 The Avogadro Number and the Mole

Since atoms are so small, chemists have to work with extremely large numbers of them. For thisreason, a very large unit was needed to express numbers of atoms. 1 dozen = 12 (too small); 1 gross =144 (too small); 1 ream = 500 (too small). It was decided that the number of particles in 22.4 L of a gas atSTP would be a good number to use: 6.02 X 1023.

The new unit was called a mole, which was derived from the Latin meaning "heap" or "pile." Thenumber, 6.02 X 1023, of anything is 1 mole of that thing, regardless of whether it's pencils, or oranges, oratoms. Avogadro was never aware of this number. He had already gone to that great chem lab in the skywhen the number was determined. This number has come to be known as Avogadro's number - named inhis honor.

Since at STP, 22.4 L of a gas contains the Avogadro number, or 1 mole, of gas particles, thequantity 22.4 L of a gas has come to be known as the molar volume of a gas. The Avogadro number andthe molar volume are so important in chemistry, that you should immediately commit them to MEMORY.

The Avogadro Number = 6.02 X 1023 = 1 mole of atoms or molecules

1 gam of an element = 6.02 X 1023 atoms = 1 mole of the element

1 gmm of a compound = 6.02 X 1023 molecules = 1 mole of the compound

22.4 L of any gas at STP = 1 mole of gas = 6.02 X 1023 atoms or molecules

Let's use helium gas as an example:

4.00 g He = 6.02 X 1023 helium atoms = 1 mole of helium = 22.4 L He gas @ STP

This value comes from the periodic table

Since helium is a gas, this is also true

Here was a convenient way for chemists to count atoms or molecules! By simply measuring a volume of agas under known conditions of temperature and pressure, they could indirectly count out any number ofparticles they wanted! WOW! For example, how many atoms are found in 6.8 liters of helium at STP?

6.8 L He X

1 mole He

22.4 L He X

6.02 X 1023 atoms

1 mole He = 1.8 X 1023 He atoms

22.4 L = 1 mole ONLY if the substance is a GAS at STP

And, since 1 gam = 1 mole, we can also convert a mass of any element to the corresponding number ofatoms of that element. For example, how many atoms are there in 8.30 grams of helium?

8.30 grams He X

1 mole He

4.00 g He X

6.02 X 10 23 atoms

1 mole He = 1.25 X 1024 atoms He

4.00 g is 1 gam of helium

What a breakthrough! It was now possible to count atoms! Chemistry was finally well on its way tobecoming a more quantitative field of science.

8-11 ©1997, A.J. Girondi

SECTION 8.7 The Molar Volume of a Gas

Problem 3. Keeping in mind what you now know about molar volume, determine the number of molespresent in each of the volumes of gases given below. Assume that all are at STP. Use dimensionalanalysis as shown in the example below. Show your work in the spaces provided.

94.1 Liters Cl2 gas X

1 mole Cl222.4 L Cl 2

= 4.20 moles Cl2

a. 22.4 L gas X

= ____________ moles gas

b. 11.2 L gas X

= ____________ moles gas

c. 44.8 L gas X

= ____________ moles gas

d. 89.6 L gas X

= ____________ moles gas

Problem 4. Use what you know about the mole to make the conversions below. Use dimensionalanalysis as shown in the examples below. Assume all substances used are gases. Show your work in thespaces provided.

4.20 moles Cl 2 X

22.4 L Cl 2

1 mole Cl2 = 94.1 L Cl2 gas

0.871 moles F2 X

6.02 X 10 23 molecules F2

1 mole F2 = 5.24 X 1023 molecules F2 gas

Note: We will use the abbreviation "mcls" to represent "molecules."

a. 4.32 moles O 2 X

= ___________ Liters O2 @ STP

b. 4.32 moles O 2 X

= ________________ molecules O 2

c. 2.00 moles CO 2 X

= ___________ Liters CO2 @ STP

d. 2.00 moles CO 2 X

= ________________ molecules CO 2

e. 8.30 Liters NO2 (@ STP) X

= ___________ moles NO 2

8-12 ©1997, A.J. Girondi

f. 0.433 mole NO 2 X

= ___________ molecules NO 2

g. 6.82 X 1021 mcls SO 2 X

mole SO 2 mcls SO 2

X L SO 2

moles SO 2 = ________ L SO 2 @ STP

h. 35.8 Liters H2 (@STP) X

mole H2 X

mole H2 = _________ molecules H 2

SECTION 8.8 Gram-Atomic Mass

You will recall that our original problem, when we began this chapter, was to determine the mass ofone atom. Earlier, you were told that one atom of hydrogen-1 has a mass of 1.67 X 10-24 g.

Chemical reactions do not occur one atom at a time. Large groups of atoms react with one anothersimultaneously. Because of this fact, it makes much more sense to talk about 1 mole of hydrogen atoms,rather than about one individual hydrogen atom. One mole of hydrogen atoms has a mass of 1.01 grams.One gram is certainly a much more manageable number than 1.67 X 10-24 grams. To simplify matters,instead of measuring and recording masses of individual atoms, masses of moles of atoms are used.

The mass of one mole of substance is called its molar mass. Remember that 1 g-at. mass = 1 mole.The term molecular mass is used when molecules are considered. These used to be known as atomicweight and molecular weight, and there are still many books which still use these older terms. Atomic massand molecular mass are more appropriate, but old habits are hard to break! Since these measuresrepresent the mass of 1 mole of atoms or molecules, the units of molar masses are always expressed asgrams per mole (grams/mole).

Problem 5. Refer to the periodic table to find the atomic masses (in amu) of the elements listed below.Round the masses off to the hundredths column. Change these atomic masses to gram-atomic massesby changing the units from amu to grams. Place this unit (grams) after each of the masses.

Element Gram-Atomic Mass (gam)

a. oxygen ______________ = 1 mole oxygen atoms

b. zinc ______________ = 1 mole zinc atoms

c. magnesium ______________ = 1 mole magnesium atoms

d. calcium ______________ = 1 mole calcium atoms

e. hydrogen ______________ = 1 mole hydrogen atoms

f. lead ______________ = 1 mole lead atoms

Since the atomic masses above are expressed in grams, they each represent 1 gam or 1 mole of eachelement.

8-13 ©1997, A.J. Girondi

SECTION 8.9 The Gram–Molecular Mass of Compounds

In Chapter 5 you learned that atoms can combine to form compounds. If you add the atomicmasses of all the atoms in the formula of a compound, you obtain the formula mass or the molecular mass.Some compounds consist of simple units called molecules and some do not. Each molecule consists oftwo or more atoms. If a substance does consist of molecules, then the term molecular mass can besubstituted for the term formula mass. The term formula mass can be used with any compound. However,in practice, the term molecular mass is used most frequently. The procedure below shows how todetermine the molecular mass of water (H2O). One mole of H2O has a molecular mass of 18.02 amu.When expressed in grams instead of amu's, the molecular mass is called the gram-molecular mass orgmm) and represents the mass of one mole of a compound. The gmm of H2O is 18.02 grams.

Elements atomic mass of elements

H 1.01 amuH 1.01 amu O 16.00 amu

H2O 18.02 amu <--- Molecular mass of H2O18.02 g <--- gram–molecular mass (gmm) of H2O

When calculating molecular masses, it is often easier if you use the [SUM] key found on mostcalculators that have a memory. On some calculators, the [SUM] key may be given the symbol [M+] oranother designation. To illustrate the use of the sum key, let's use a calculator to determine the formulamass of the compound Ba(NO2)2. (Caution: calculators which have more than one memory may not

operate as described in the following discussion.) The "formula unit" contains 1 atom of Ba, 2 atoms of N,and 4 atoms of O. Let's begin by making sure that the calculator's display and memory are clear. Do this bypressing the "clear all" or [CA] key. Begin the problem by entering the atomic mass of Ba into thecalculator: 137.34. Press the [M+] or [SUM] key. This value is now stored in the calculator's memory. Tocheck this, clear the calculator and then press the recall memory key which may have the designation [RM]or [RCL]. The atomic mass of Ba should now appear in the display. Be careful not to press the "clear all"[CA] key, because that will not only clear the display, but it will also clear the memory! Next, enter theatomic mass of N into the calculator: 14.01. Since there are 2 atoms of N in the formula, multiply by 2, andpress the [=] key. the display should now read 28.02. We need to add this product to the quantity in thememory. Do this by pressing the [M+] key. The subtotal in the memory should now be 165.36. To checkthis, press the [RM] or [RCL] key. Next, enter the atomic mass of oxygen: 16.00. (Actually, you only needto enter 16. Zeros to the right of the decimal point do not need to be entered into the calculator.) Nowmultiply by 4 since there are 4 atoms of O represented in the formula. Press the [=] key. The displayshould reveal the product: 64. Again press the [M+] key to add this quantity to the sum in the memory. Tosee the formula mass of Ba(NO2)2, press the [RCL] or [RM] key. The answer should be 229.34. That's allthere is to it! Before starting another problem, be sure to press the [CA] key to clear both the display andthe memory. Use this same procedure with a calculator to determine the formula masses of thecompounds listed on the next page. Use a periodic table and round off all atomic masses to thehundredths column.

Note: a molecule such as (NH4)3PO4 contains a total of {12}____________ atoms.

Problem 6. For the following compounds, express gram-molecular masses (gmm) rounded to twodecimal places.

a. NaOH gmm = _____________ g = 1 mole NaOH

b. P2O5 gmm = _____________ g = 1 mole P2O5

8-14 ©1997, A.J. Girondi

c. (NH4)3PO4 gmm = _____________ g = 1 mole (NH4)3PO4

d. Fe(C2H3O2)3 gmm = _____________ g = 1 mole Fe(C2H3O2)3

e. Ca(NO3)2 gmm = _____________ g = 1 mole Ca(NO3)2 f. CuSO4 •5 H2O gmm = _____________ g = 1 mole CuSO4 •5 H2O

g. (NH4)2O gmm = _____________ g = 1 mole (NH4)2O

h. Na2SO4 •10 H2O gmm = _____________ g = 1 mole Na2SO4 •10 H2O

i. 1 gmm CCl4 = ____________ g CCl4 = 1 mole CCl4

j. 1.00 mole Barium (Ba) = 1 gam Ba = __________ g Ba

k. 1.00 mole SO2 = 1 gmm SO2 = ____________ g SO2

SECTION 8.10 Percentage Composition of Compounds

The formula of a compound reveals the relative number of atoms of each element that composethe substance. It can also be used to determine the mass percents of each element in the compound.When we describe the composition of a compound according to the percent by mass of each element init, that is known as the percentage composition of the compound.

The percentage composition of a compound can be determined by using its formula and theatomic masses from the periodic table. It can also be determined in the laboratory using experimental data.An example of each type of determination is shown below.

Example 1. Calculate the percentage composition by mass of each element in the compound, K2CrO4.Atomic masses: K = 39.10; Cr = 52.00; O = 16.00. The molecular mass of K2CrO4 is calculated as follows:

K = 39.10 X 2 = 78.20 Cr = 52.00 X 1 = 52.00 0 = 16.00 X 4 = 64.00 Molecular mass = 194.20

%K in K2CrO4 = 78.20 amu

194.20 amu X 100% = 40.30% K

%Cr in K2CrO4 = 52.00 amu

194.20 amu X 100% = 26.80% Cr

%O in K2CrO4 = 64.00 amu

194.20 amu X 100% = 33.00% O

Total of Percents = 100.10%(Due to rounding, the total percent may fall between 99 and 101.)

8-15 ©1997, A.J. Girondi

Example 2. A 5.00 gram sample of NaOH is decomposed and it is found that the sample contained 2.88grams of Na. Calculate the percentage by mass of Na in NaOH.

2.88 g Na

5.00 g sample X 100% = 57.6% Na

(Notice that you divide the mass of the element you are trying to find the percentage of by the mass of theentire sample, then you multiply by 100%.)

Problem 7. Determine the percent by mass of each element in the following compounds. Show yourwork. Round atomic masses to the hundredths column. Show work.

a. KClO3

K = ________%

Cl = ________%

O = ________%

Total = ________%

b. Mg(ClO3)2 Mg = _________%

Cl= _________%

O = _________%

Total = _________%

c. Decomposition of a 5.00 g sample of ethyl alcohol yields 2.61 g of carbon, 0.658 gram of hydrogen,and 1.73 grams of oxygen. Calculate the mass percent of each element in ethyl alcohol.

C = ________%

H = ________%

O = ________%

Total = ________%

ACTIVITY 8.11 Percentage of Oxygen In Potassium Chlorate

In this activity you will experimentally determine the percent by mass of oxygen in the compoundcalled potassium chlorate, KClO3. Use the equipment set-up shown in Figure 8.7 . When KClO3 isheated it decomposes, giving off the oxygen as oxygen gas, O2.

2 KClO3(s) ----------> 2 KCl(s) + 3 O2(g)

∆

8-16 ©1997, A.J. Girondi

Figure 8.7Heating KClO3 in a Crucible

Procedure:

1. Clean a crucible and cover. Place the crucible in the clay triangleas shown in Figure 8.7. Heat the empty crucible and cover for a fewminutes to burn off any impurities. Be sure to tilt the cover asillustrated. Balance it carefully to avoid breakage. Put out the flameand allow the crucible and cover to cool.

2. Measure out exactly 2.00 g of the dry KClO3. Add thiscompound to the crucible. Measure out exactly 0.50 g of MnO2 andadd this to the KClO3 in the crucible. Mix the contents of thecrucible. (The MnO2 is a catalyst that helps the reaction along. Itdoes not participate or decompose.)

3. With the cover on the crucible, measure the mass of the crucible,cover, and contents. Record this mass as (a).

4. To remove oxygen gas from the compound, place the cover onthe crucible so that it covers the opening completely. It is importantto begin by heating gently for several minutes (your teacher willshow you how to heat gently). Slowly increase the intensity of theheating over a period of 10 minutes so that the last couple ofminutes involve high heat. Remove the cover and heat strongly foran additional 2 minutes. (If the cover seems stuck, ask your teacherfor help.)

5. Allow the crucible to cool. Measure and record the total mass of the crucible, cover, and contents afterheating (b). The crucible and cover can be cleaned by placing them in a boiling water bath (which yourteacher may have available for you to use – ask).

Table 8.3Percentage of Oxygen in KClO3

a. mass of crucible + cover + KClO3 + MnO2 __________ g (this is the total mass before heating)

b. mass of crucible + cover + KCl + MnO2 __________ g (this is the total mass after heating)

c. mass of KClO3 used 2.00 g

d. mass of oxygen given off __________ g (a minus b)

8-17 ©1997, A.J. Girondi

6. Using only the experimental data from Table 8.3, calculate the percent by mass of oxygen in KClO3. Ifyou aren't sure how to do this, refer back to the first sample problem in Section 8.10. Show your work.

Oxygen = ____________%

7. In part "a" of problem 7, you used formula and atomic masses to calculate the accepted or theoreticalvalue for the percent of oxygen in KClO3. Use that accepted value and your observed experimental valuecalculated in Activity 8.11 above to determine your percentage error.

Error = __________%

Typical amount of error on this activity range from 1 to 10 percent. Comment on your amount of error:

______________________________________________________________________________

If your error was greater than 5%, what part of the procedure do you think was responsible for most of it?

______________________________________________________________________________

SECTION 8.12 Chapter Review

Before leaving Chapter 8, some review will be helpful. Remember that 22.4 liters of any gas (atSTP) contains 6.02 X 1023 atoms, and that many particles of any substance is one mole. We alsodiscovered in this chapter that 1 gram-atomic mass (gam) of any element contains 6.02 X 1023 atoms.

Therefore:

1 gam = 1 mole = 6.02 X 1023 atoms = 22.4 L @ STP

Keep in mind that 1 mole of any gas = 22.4 L only if the gas is at STP.

For substances which are not gases, the 22.4 liter volume does not apply.

So, for solid or liquid elements: 1 gam = 1 mole = 6.02 X 1023 atoms

For example, 12.01 g carbon = 1 mole carbon = 6.02 X 1023 C atoms

For solid or liquid compounds: 1 gram-molecular mass (gmm) = 1 mole = 6.02 X 1023 molecules

For example, 18.02 grams H2O = 1 mole (or H2O = 18.02 g/mole).

8-18 ©1997, A.J. Girondi

So, we can make use of ratios like :

18 .02 g H 2O1 mole H 2O

OR 18 .02 g H 2O

6.02 X 10 23 molecules


This is the end of Chapter 8 which covers Part I of your study of the mole. Chapter 9 will offermore information on the use of the mole concept. Carefully review what you have learned in this chapter,and go over the learning outcomes listed below. Check off each one after you are certain that you havemastered the concepts and skills. Arrange to take any quizzes or exams on Chapter 8, and then go on toChapter 9.

_____1. List the sequence of experiments which led to the developmentof Avogadro's hypothesis.

_____2. Define the mole as applied to atoms and molecules.

_____3. Find atomic masses on the periodic table and calculate formulamasses.

_____4. Determine the number of atoms in 1 gram-atomic mass of an elementor the number of atoms in a given number of moles of an element.

_____5. Perform mole calculations using the relationship 1 mole = 22.4 L @ STP.

_____6. Calculate the percentage composition of compounds given the formula or experimental data.

8-19 ©1997, A.J. Girondi


Questions:

{1} C; {2} D; {3} Molecules of gas A have a different mass than molecules of gas C; {4} Most of a givenvolume of a gas is empty space, so the size of the molecules can vary; {5} 2 amu; {6} 1; {7} 4;{8} 12 times greater; {9} 52.00; {10} 6.02 X 1023; {11} 22.4; {12} 20

Problems:

1. 6.68 X 10-24 g2. 2.67 X 10-23 g3. a. 1 mole; b. 0.5 mole; c. 2 moles; d. 4 moles4. a. 96.8; b. 2.60 X 1024; c. 44.8; d. 1.20 X 1024; e. 0.371; f. 2.61 X 1023; g. 0.254; h. 9.62 X 1023

5. a.16.00; b. 65.38; c. 24.31; d. 40.08; e. 1.01; f. 207.206. a. 40.00; b. 141.94; c. 149.12; d. 233.00; e. 164.10; f. 249.71; g. 52.10; h. 322.24; i. 153.81;

j. 137.33; k. 64.067. a. 31.91% K, 28.93% Cl, 39.17% O, Total 100.01%

b. 12.71% Mg, 37.08% Cl, 50.21% O, Total 100.00%c. 52.2% C, 13.16% H, 34.6% O, Total 99.96%

8-20 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 9

THE MOLECONCEPT

(Part 2)

Unit ConversionsStoichiometry

Mass-Mass & Mass-Volume9-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


9-2 ©1997, A.J. Girondi

SECTION 9.1 The Law of Conservation of Matter

In an earlier chapter you experimented with the law of conservation of matter (or mass). The Lawstates that "matter is neither created nor destroyed in a chemical reaction." Therefore, the total mass ofthe reactants must equal the total mass of the products in any chemical reaction. In addition to provingthis law experimentally using real masses, it can also be demonstrated using atomic masses from theperiodic table. Problem 1 will illustrate this.

Problem 1. Follow the directions below.

a. Finish balancing the equation shown below.

3 NO2 + _____H2O -----> _____HNO3 + _____NO

b. Now, add the atomic masses of all the atoms on the reactant (left) side of the equation:

Total mass of reactants:__________

c. Next, add the atomic masses of all the atoms on the product (right) side of the equation:

Total mass of products:__________

Do your answers support the law of conservation of mass?__________ If not, check to see if you

balanced the equation properly. For the reaction 2 H2 + O2 ----> 2 H2O, if 2.80 g of H2 combine with

22.2g of O2, how many grams of water should be produced if the law of conservation of mass is true?

{1}__________

SECTION 9.2 Converting Grams to Moles and Moles to Grams

Using your knowledge of unit analysis, determine which equation below is correct for changingmoles to grams. Circle the correct equation:

moles X molesgrams

= grams OR moles X gramsmoles

= grams

Sample Problem: Change 7.65 moles of sulfur to grams of sulfur.

Solution:

7.65 moles S X 32.07 g S1 mole S

= 245.34 g S (rounds to 245 g)

9-3 ©1997, A.J. Girondi

Sample Problem: Change 1.43 moles of NO2 to grams of NO2.

Solution:

1.43 moles NO2 X 46.01 g NO2 1 mole NO2

= 65.8 g NO2

Problem 2. Use the equation you just circled to calculate the mass in grams of each of the following.Use the periodic table, and round atomic masses to the hundredths column. Show your work.

a. 3.00 moles Na X =

b. 3.50 moles CaCO3 X =

c. 1.50 moles Ba(OH)2 X =

d. 0.800 mole Fe3O4 X =

e. 0.00200 mole Pb(C2H3O2 )2 X =

Once again use your knowledge of dimensional analysis to determine which equation below iscorrect for calculating the number of moles. Circle the correct equation.

grams X

gramsmole

= moles OR grams X mole

grams = moles

Sample Problem: Convert 345 grams of Ca to moles of Ca.

Solution:

345 g Ca X 1 mole Ca 40.08 g Ca

= 8.61 moles Ca

Sample Problem: Convert 288 grams of NaNO3 to moles of NaNO3 .

Solution:

288 g NaNO3 X 1 mole NaNO3 85.01 g NaNO3

= 3.39 moles NaNO3

9-4 ©1997, A.J. Girondi

Problem 3. Use the equation you just circled to calculate the number of moles for each of the following.Use the periodic table, and round atomic masses to the hundredths column. Show your work.

a. 200. g F2 X =

b. 180. g Ca X =

c. 10.0 g Na2S X =

d. 68.6 g H2SO4 X =

e. 216 g P2O5 X =

Problem 4. This problem will give you additional practice working with mole problems. Use atomicmasses rounded to the hundredths column, use dimensional analysis, show your work, and assume allsituations involving gases are at STP. Show complete set-up for each problem.

a. What is the mass in gramsof 4.2 moles of Na2CO3

b. Calculate the number ofmoles in 14.5 g of C4H10.

c. Calculate the mass in grams of 0.00400moles of KMnO4.

d. How many moles are presentin 24.5 g of K2Cr2O7?

9-5 ©1997, A.J. Girondi

e. Calculate the number of molesin 15.5 g of Ca3(PO4)2.

ACTIVITY 9.3 Determination of the Gram–Atomic Mass (GAM) of Silver

There are several methods for determining the gram atomic mass (gam) of an element. You willrecall that the gram-atomic mass of an element is the mass in grams of one mole of that element. In thisactivity, the gram-atomic mass of silver will be determined using a compound (silver oxide) of knowncomposition, Ag2O. Gather the following equipment: crucible and cover, ring stand, iron ring, claytriangle, crucible tongs, spatula, burner, balance, goggles or safety glasses, apron.

1. Clean a crucible and cover. Place the crucible in the clay triangle as shown in Figure 9.1 below. Heatthe crucible and cover for about 2 minutes. Be sure to tilt the cover as illustrated. Balance it carefully toavoid breakage. Put out the flame and use your crucible tongs to put the crucible and lid on the base ofthe ring stand to cool.

2. Measure the mass of the crucible + cover.Record this mass as (a) in Table 9.1.

3. Obtain about 1.75 g of silver oxide powder.Add this compound to the crucible. With thecover on the crucible, measure the mass of thecrucible and its contents. Record this mass as(b).

4. To remove oxygen from the oxide compound,tilt the cover as before and strongly heat thecrucible, cover and contents in the hottest part ofthe flame for 10 minutes. Allow the crucible tocool. The crucible should now contain only silvermetal. It may look white, but it is silver metal.Measure and record the mass of the crucible,cover, and silver metal (d).

Figure 9.1 Heating a Crucible

crucible

pipestemtriangle

lid

5. If time permits, reheat strongly for an additional 5 minutes. After cooling, again measure the mass of thecrucible, cover, and metal to check for constancy of mass.

9-6 ©1997, A.J. Girondi

Table 9.1The Atomic Mass of Silver Metal

a. mass of crucible + cover __________ g

b. mass of crucible + cover + Ag2O __________ g (this is the total mass before heating)

c. mass of oxide compound (Ag2O) used (b–a) __________ g

d. mass of crucible + cover + Ag metal __________ g (this is the total mass after heating)

e. mass of silver metal. (d–a) __________ g

6. Do not discard the silver in the crucible. Your instructor will have a container available in which you canput it. Try scratching the silver with a spatula or other metal object. This should reveal the shiny, silveryluster of the metal. If there is a magnifying glass or binocular scope in your lab, use it to observe the metalmore closely. If you have trouble removing the silver from the crucible, give it to your instructor. The silvercan be recycled for use in other experiments.

1. Find the mass of oxygen in your sample of Ag2O. (b–d)

__________ g O

2. Convert the number of grams of oxygen in your sample (from calculation 1 above) to moles of oxygen.(Oxygen was not in its diatomic form (O2) here, since it was combined with silver.) 16.00 g O = 1 mole O.

__________ mole O

3. Using the result of calculation 2 above, find the number of moles of silver metal in your sample ofcompound (Ag2O). Since the formula is known to be Ag2O, this means that the compound contains 2moles of Ag for every one mole of O. To find moles of Ag in your sample, just multiply the result ofcalculation 2 above, by two.

__________ mole Ag

4. Find the molar mass (g/mol) of the silver metal. (Divide the mass in grams of silver metal in Table 9.1 (e)above by the number of moles of Ag found in calculation 3 above.)

__________ g/mole

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5. Obtain the accepted value for the atomic mass of silver from the periodic table. Compare it to yourexperimentally determined value and determine your percentage of error. Check your reference materialsif you forget the formula for percent error. Show calculations below.

Accepted Value from Periodic Table = ____________ g/mole Ag = 1 gram-atomic mass (gam) of Ag

Observed (experimental) Value = ____________ g/mole Ag = 1 gram-atomic mass (gam) of Ag

Calculation of % Error. Show Work.

% Error = _____________

This experiment usually yields results with a higher percentage error than most of the activities you will

perform in this course. Nevertheless, this procedure should produce less than 15% error. Does your %

error fall within this range? __________ If not, how do you account for the excess error? ____________

______________________________________________________________________________

ACTIVITY 9.4 Balancing An Equation By Experiment

In this activity you will observe the reaction of iron nails with a solution of copper (II) chloride anddetermine the number of moles involved in the reaction. By converting grams of iron metal consumed andgrams of copper metal produced to moles, you will see how experimental data can be used to balance anequation.

Procedure:

1. Find the mass of a clean, empty, dry 250 mL beaker which you have marked with your initials. Recordthe mass to the nearest 0.01 g in Table 9.2 below.

2. Add about 8.00 g of copper (II) chloride crystals to the beaker (note that when a word like about is used,it means that you do not have to use exactly that amount. However, it is important to know precisely howmuch you are using.) Record the mass of CuCl2 in Table 9.1. Add about 50 mL of distilled water to thebeaker, and swirl or stir the solution to dissolve all of the solid.

3. Obtain two nails. Clean the nails using a piece of fine sandpaper. Find and record the mass of the twonails (together). If you have at least 25 minutes left in your class, place the nails into the solution and allowthe reaction continue for at least 20 minutes. Occasionally swirl the solution in the beaker. (If your classtime is short, you can allow the reaction to run longer, but do not allow it to continue for more than a fewhours.) During that time, you will see the formation of copper in the beaker. At the same time the iron nailswill be partially consumed. The solution will slowly change color from blue to green as some of the blueCuCl2(aq) is consumed and replaced by pale yellow FeCl3(aq). This is a single replacement reaction, andthe balanced equation is:

Fe(s) + CuCl2(aq) ----> Cu(s) + FeCl2(aq)

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4. After about 20 minutes, remove any remaining pieces of nail using your forceps or crucible tongs. Ifany copper is sticking to the nails, use your wash bottle to rinse as much of the copper as you can backinto the beaker. Dry the nails on a paper towel. When the nails are dry, find their mass and record in Table9.1. The nails can then be discarded.

5. Decant means to pour off only the liquid from a container that contains both solid and liquid. Carefullydecant the liquid from the solid. (No filtering is necessary.) Pour the liquid into another beaker so that incase you lose any solid, you can recover it. (You will always lose a little - don't worry about that.) Afterdecanting, rinse the solid copper in your beaker with some distilled water from your wash bottle. Decantthe rinse water. Repeat this rinsing three more times.

Table 9.2The Iron - Copper (II) Chloride Reaction

1. Mass of empty, dry, labeled beaker __________ g (Before the Reaction)

2. Mass of beaker + CuCl2 __________ g

3. Mass of two iron nails(before reaction) __________ g

4. Mass of two iron nails(after reaction) __________ g

5. Mass of beaker + dry copper __________ g

6. Mass of copper produced (5 – 1) __________ g

6. Next, wash the solid copper with about 25 mL of 1.0 M hydrochloric acid (HCl). Decant again, and rinseonce more with distilled water. Place the beaker containing the copper in an oven to dry until the next day.

7. When the copper is completely dry, find and record the mass of the beaker plus the copper. Scrapethe copper out of the beaker and discard it in the waste can. Clean and return all equipment. Perform thefollowing calculations, rounding atomic masses to the hundredths column. Show your work.

Calculations:

1. Referring to items 3 and 4 in Table 9.2, calculate the mass in grams of iron used in the reaction.

2. Using your answer to calculation 1 above, calculate the number of moles of iron used in the reaction.

3. Using items 1 and 5 in Table 9.2, calculate the number of grams of copper metal produced and enterthe result in the data table.

9-9 ©1997, A.J. Girondi

4. Using your answer to calculation 3 above, determine the number of moles of copper metal produced.

5. Use your results from calculations 2 and 4 above to determine a simple whole–number ratio of moles ofiron reacted to moles of copper produced. (Hint: divide each quantity of moles by the smallest of the twoand round to the closest whole number.) If this confuses you , consider the following. Suppose youhave a ratio such as 0.64 to 0.16. To change this to an equivalent whole-number ratio we will divide eachnumber the the smallest of the two which is 0.16. Since 0.64 ÷ 0.16 = 4, and since 0.16 ÷ 0.16 = 1, theequivalent whole-number ratio is 4 to 1.

Experimental whole–number ratio of Fe to Cu is: _______ to _______

6. The balanced equation for this reaction was shown previously in the activity. What Fe to Cu mole ratio ispredicted by the balanced equation for this reaction? ______ to ______

7. The equation for this reaction is Fe(s) + CuCl2(aq) ----> Cu(s) + FeCl2(aq) How does your experimentalratio compare to the Fe – Cu mole ratio predicted by the equation?

______________________________________________________________________________

SECTION 9.5 Introduction To Mass-Mass Problems

You have seen that the coefficient numbers in a balanced chemical equation may represent therelative number of moles or the relative number of molecules involved in a reaction. These reacting ratiosdo not change. The reaction below is a balanced chemical equation. Two moles of Li2O decompose toform four moles of Li and one mole of O2. Note the 2:4:1 mole ratio in the following equation:

2 Li2O ----> 4 Li + O2

The ratio of moles of reactant (Li2O) to moles of Li produced is 2:4 in this balanced equation. Knowingthis relationship can help to solve a wide variety of problems such as the next sample problem.

Sample Problem: Referring to the equation above, determine the number of moles of Li producedwhen 3.00 moles of Li2O decompose.

Begin this problem by writing the information given and the information you are hoping to find as follows:

3.00 moles Li2O X ? ? ?

? ? ? = ???? moles Li

Once you have the basic framework of your equation, you simply build the rest of the equation using dataoffered in the problem or from the balanced equation. Enter the data in the correct positions, so that yourunits cancel properly and leave you with only the units you want in the answer:

9-10 ©1997, A.J. Girondi

4:2 ratio from balanced equation

3.00 moles Li2O X 4 moles Li

2 moles Li2O = 6.00 moles Li

Because these problems begin with a mass (grams or moles) and end with a mass (grams or moles), theyare sometimes called "mass-mass" problems. Notice that complete units are very, very important. Wedidn't just use moles, but we used moles Li and moles Li2O. In chemistry, when you do calculations whichrelate quantities of two substances used or produced in a chemical reaction, you are doing what is knownas stoichiometry. Stoichiometric problems give you information about one substance, and ask forinformation about a different substance in a chemical reaction. For this reason, you need to refer to thechemical equation for the reaction. Most of the problems in the rest of this chapter involve stoichiometry.Your calculations should always include complete units. Your teacher will be looking for them! Try thisnext problem, yourself. Show your work.

Problem 5. Using the equation from the last sample problem, determine the number of moles of O2

produced when 3.00 moles of Li2O decompose.

3.00 moles Li2O X

=

Other varieties of mass-mass problems are shown below. Examine the examples and then try theproblems which follow.

Sample Problem: In the reaction 2 H2 + O2 ----> 2 H2O, how many grams of H2O will be formed if we use4.5 grams of O2?

Note that you are asked to change information about O2 into information about H2O. (Change 4.5grams O2 into grams of H2O) Hmmmm. This is stoichiometry. In order to change information about one

substance into information about another substance, you must first know a relationship between thosetwo substances. Where do we find such a relationship? In the equation for the reaction! Whatrelationship do we already know between O2 and H2O? Well, thanks to the balanced equation, we knowthat 1 mole of O2 is needed to produce 2 moles of H2O. The relationship that we already know is a moleratio. Therefore, in order to change information about O2 into information about H2O, we must do it in theterms of the relationship that we already know. In other words, we must do it in terms of moles in this case.

Since we cannot change grams of O2 directly into grams of H2O, we will first change the 4.5 g of O2

into moles of O2. Then, we can change moles of O2 into moles of H2O. Finally, we will change moles ofH2O into grams of H2O. This will all be done using unit analysis. Note that the part of the solution wherewe change from one substance to the other (O2 to H2O) occurs, is the mole-to- mole ratio:

Relating Two Different Substances

Relationship From Balanced Equation unit conversionunit conversion

4.5 g O2 X 1 mole O232.00 g O2

X 2 moles H2O1 mole O2

X 18.02 g H2O1 mole H2O

= 5.1 g H2O

9-11 ©1997, A.J. Girondi

Remember, in order to establish a relationship between two substances, you must first know a relationshipbetween them. Balanced chemical equations provide such relationships in the form of mole-to-moleratios. Thus, problems which change information about one substance to information about anothersubstance must make use of the mole-to-mole ratio in the balanced equation. If the information given inthe problem is not in the form of moles, it has to be changed to moles inorder to then use the mole to moleratio. Thus, the mole to mole ratio is the "heart of" a stoichiometry problem. When the problem is set up,there may be any number of "unit conversions" before or after the mole-to-mole ratio, depending on theinformation given and requested in a particular problem. Note the two unit conversions used in theillustration above. The diagram in Figure 9.2 outlines the process. There are many other unit conversionsbesides the ones listed in Figure 9.2.

The problem below is simpler than the last sample problem since the information is given in moles. Fill inthe missing parts of the set-up so that all units divide out except grams of Al2O3 and solve the problem.

Problem 6. In the reaction: 4 Al + 3 O2 ---> 2 Al2O3 how many grams of Al2O3 will be formed if we startwith 5.0 moles of Al?

5.0 moles Al X moles Al

X 1 mole Al2O3

= __________ g Al2O3

Use as many unit conversions as needed to change the starting information to moles.

Use mole-to-mole ratio in the balanced equation to convert to moles of the substance in the answer.

Is the information you are going to start with given in moles?

NO

YES

Do you want your answer to be expressed in moles?

YES

You are Done!

NO Use as many unit conversions as needed to change moles to the units requested in the answer.

SOME UNIT CONVERSIONS1 g-atomic mass = 1 mole1 g-formula mass = 1 mole22.4 L = 1 mole gas @ STP6.02 X 1023 atoms = 1 mole6.02 X 1023 molecules = 1 mole

START

Stoichiometry?

Figure 9.2 Flow Diagram for Solving Stoichiometry Problems

9-12 ©1997, A.J. Girondi

Solve the following problems using dimensional anaylsis. All measurements must include units!

Problem 7. In the reaction: C + 2 Cl2 ----> CCl4, how many grams of CCl4 can be prepared starting with8.6 grams of Cl2?

Problem 8. In the reaction: FeCl3 + 3 NaOH ---> Fe(OH)3(s) + 3 NaCl, how many grams of Fe(OH)3 canbe produced if we start with 0.50 moles of NaOH?

Problem 9. In the reaction 2 Al + 3 Pb(NO3)2 ---> 2 Al(NO3)3 + 3 Pb, how many grams of Al are neededto produce 68 grams of Pb?

Problem 10. In the reaction in problem 9 above, how many moles of Pb will be formed if 23.2 grams ofAl(NO3)3 are formed?

Problem 11. In the reaction: BaCO3 ----> BaO + CO2, how many grams of CO2 can be produced fromthe decomposition of 45 grams of BaCO3?

9-13 ©1997, A.J. Girondi

ACTIVITY 9.6 The "Silver Tree" Reaction

The purpose of this activity is to use mass-mass calculations to predict the outcome of anexperiment. In order to accomplish this, you will be measuring the masses of both the reactants and theproducts. A single replacement reaction is involved in which copper replaces the silver in silver nitrate:

Cu(s) + 2 AgNO3(aq) ----> Cu(NO3)2(aq) + 2 Ag(s)

Notice that one of the products will be pure silver metal. This activity requires that you perform a number oflaboratory procedures over a period of 3 days.

Procedure for Day 1: (About 15 minutes required)

1. Obtain a clean, dry 100 or 150 mL beaker and use a magic marker to label it as "Beaker A" and includeyour name(s).

Figure 9.3 Copper Coil in Beaker

2. Weigh the beaker and record the mass in the Table9.3. (Note: all masses in this activity should bemeasured to the nearest 0.01 or 0.001 g dependingon the balances available. Use the most precisebalance available, and use balances of the sameprecision throughout the activity.)

3. Measure about 25 to 30 centimeters of copper wireand coil it around a test tube. Measure and record itsmass.

4. Leave some wire uncoiled to make a hook which willfit over the edge of the beaker. (See Figure 9.3)

5. Measure between 2 and 3 grams of silver nitratecrystals. Record the exact mass of the crystals.Caution: silver nitrate crystals can create dark stains orspots on your skin and clothing. The stained area willnot darken for several hours. Use care in handlingAgNO3. Clean up any spilled crystals with a damptowel.

6. Fill beaker A about two-thirds full of distilled water. Add the AgNO3 crystals. Stir with a glass stirring roduntil most of the crystals have dissolved. The coils of your copper wire should not be too close together.When the coil is immersed in the solution, most of the coils should be submerged. (See Figure 9.3)

7. Place the copper coil into the solution and observe the beginning of the reaction. Store the beaker inyour drawer until the next day.


8. Lift the copper wire from beaker A and shake off any silver residue which has formed. Rinse the wireusing a wash bottle with distilled water, allowing the rinse water to fall into beaker A. Set the wire aside todry until the next day. DO NOT discard the wire until you have weighed it and recorded its mass!

9. Obtain a second clean, dry 100 or 150 mL beaker. Decant most of the solution in beaker A into thesecond beaker. Try not to lose too many of the solid silver crystals as you decant the solution. You do not

9-14 ©1997, A.J. Girondi

need to decant all of it. Discard the solution in the second beaker rinse it with water.10. Squirt a little distilled water over the silver metal in beaker A. Decant most of the rinse water into thesink, retaining the silver metal.

11. Place beaker A containing the wet silver metal into a warm drying oven until the next day. Any waterleft in this beaker will evaporate.


12. Weigh the dry copper wire, which can then be discarded in the waste can. Record the mass in the datatable.

13. Weigh beaker A (with contents) and record. Empty the silver metal in beaker A into the containerprovided by your instructor. It can be purified and used in other experiments. Remove any markings fromthe beaker with alcohol and a towel.

Table 9.3The “Silver Tree” Reaction

Item(s) Mass in Grams

1. Beaker A __________

2. Copper Wire (day 1) __________

3. Copper Wire (day 3) __________

4. Copper Metal Reacted (2-3) __________

5. Silver Nitrate (day 1) __________

6. Beaker A + Silver Metal (day 3) __________

7. Silver Metal Formed (6-1) __________

Calculations:

1. Here is where your expertise in mass-mass problems will come in handy! Based on the number ofgrams of copper that were consumed (item 4 in Table 9.3), predict how many grams of silver should havebeen produced. Show work.

__________ g Ag should be produced

2. The actual amount of silver produced is item 7 in Table 9.3. Is it close to your predicted value incalculation 1 above? __________ Calculate your percentage error using the formula for %E. You canfind it in your reference notebook. The accepted value (A) is your prediction from calculation 1. Theobserved value is item 7 in Table 9.3. Show work.

9-15 ©1997, A.J. Girondi

3. Calculate the number of moles of copper which reacted. Be sure to use as many significant figures asyou are entitled to. Show work.

4. Calculate the number of moles of silver produced. Be sure to use as many significant figures as you areentitled to.

5. Using the answers to calculations 3 and 4 above, determine the whole number ratio of moles of Cuused to moles of Ag produced. Do this by dividing moles of copper (calculation 3 above) and moles ofsilver (calculation 4 above) by the smallest of the two values. Show work.

According to the balanced equation for this reaction, what should the mole to mole ratio be for Cu to Ag?

______ Cu(s) + 2 AgNO3(aq) ----> Cu(NO3)2(aq) + ______ Ag(s)

How does your experimentally–determined Cu:Ag ratio in calculation 5 compare to that in the balanced

equation?______________________________________________________________________

SECTION 9.7 Introduction to Mass–Volume Problems

The important unit conversion which you know that relates mass to volume (moles to liters) is:1 mole of a gas = 22.4 L @ STP. In mass-volume problems, the same basic procedure is followed nomatter how complex a problem may become. The key to success is to always include all units when youset up the problem. In that way, you can be sure to set it up so that all units cancel properly.

Problem 12. Determine the volume in liters (at STP) of nitrogen gas, N2, needed to produce 92.0

grams of product, NO2, in the reaction: N2(g) + 2 O2(g) ----> 2 NO2(g)

The two relationships which you need to use which are not given in the problem are:

22.4 L = 1 mole gas and NO2 = 46.0 g/mole.

Furthermore, the ratio of moles of N2 to moles of NO2 = 1:2. The basic framework has been drawn for youbelow. (We can call this framework a "fencepost" because it begins to look a bit like one. In that event,doing dimensional analysis is called "fenceposting.") The problem has been partially completed for you

9-16 ©1997, A.J. Girondi

below. You should now fill in the blanks and calculate the answer. Make sure to use complete units and tobe certain that all units divide out except those you want in the answer. (Problems which attempt toestablish a relationship between a mass and a volume are called mass-volume problems.)

92.0 g NO2 X g NO2

X moles N2 X

moles N2 = _______ L N2

The next four problems refer to the following equation. Use the same approach as you did to solve theprevious problems. Show all of your work, including complete units. Assume STP for gases. 2 KClO3(s) ----> 2 KCl(s) + 3 O2(g)

Problem 13. How many moles of KClO3 are required to form 2.5 liters of O2?

Problem 14. How many grams of KClO3 are needed to produce 11.2 liters of O2?

Problem 15. How many liters of O2 are formed when 367.8 g of KClO3 decompose?

Problem 16. How many moles of KCl are produced when 2.3 L of O2 are produced?

SECTION 9.8 A Variety of Stoichiometry Problems To Solve

Use the equation below to solve problems 17 through 20. Assume STP conditions for any gases. Showyour work, and include units with all measurements.

3 H2(g) + N2(g) ----> 2 NH3(g)

9-17 ©1997, A.J. Girondi

Problem 17. How many moles of ammonia (NH3) would be produced by reacting 6.00 moles ofhydrogen (H2) with an adequate amount of nitrogen (N2)?

Problem 18. If 56.0 grams of N2 are reacted with a sufficient amount of H2, how many moles of product(NH3) are formed?

Problem 19. How many grams of NH3 are formed when 5.60 liters of N2 react with a sufficient amount ofH2?

Problem 20. How many liters of NH3 will be formed when 3.01 X1023 molecules of H2 react with asufficient amount of N2?

The following problems involve all of the information you have gathered thus far about moles. Use

unit analysis to set up these problems and to be sure your units cancel properly. Show all your work, andassume STP conditions for any gases.

Problem 21. What is the volume in liters occupied by 10.0 grams of CO2 gas at STP? (Hint: you mustfirst find the molecular mass of CO2.)

9-18 ©1997, A.J. Girondi

Problem 22. If 1.00 mole of a gas has a mass of 18.0 grams, and you have 15.0 grams of the gas, whatwould its volume be at STP?

Problem 23. If the molecular mass of a gas is 34.0 g/ 1 mole, what would be the mass of 30.0 L at STP?

Problem 24. What is the volume in liters of 6.40 g of O2 gas at STP?

Problem 25. A gas sample with a mass of 50.0 grams has a volume of 40.0 L. What is the gram-molecular mass (GMM) of the gas? (GMM is epxressed in units of grams/mole.) Hint: This problem is a bitdifferent form the others. Notice that you are given grams and liters but asked for grams/mole. Why notstart your set-up with 50.0 grams/40.0 liters and then just convert liters to moles?

Problem 26. Calculate the gram-molecular mass, GMM, (in g/mole) of a gas if 11.5 g of the gas has avolume of 5.60 L at STP? (You want to end the problem with units of grams/moles.)

Problem 27. Is the gas mentioned in problem 26 above H2O, SO2, or NO2? Show how you know.

9-19 ©1997, A.J. Girondi


By now you must realize that the mole is a central concept in chemistry and is very useful whenworking with chemical equations. When you are certain you understand each one, place a check mark inthe space preceding each learning outcome below. Then, move on to chapter 10.

_____1. Use dimensional (unit) analysis to make conversions between moles, grams, liters, and molecules.

_____2. Calculate mole–mole relations from balanced chemical equations.

_____3. Calculate mass–mass relations from balanced chemical equations.

_____4. Calculate mass–volume relations from balanced chemical equations.

_____5. Use the molar volume concept in problems involving gases.

9-20 ©1997, A.J. Girondi


Questions: {1} 25.0 g

Problems:

1. a. 3, 1, 2, 1; b. 156.05 amu; c. 156.05 amu2. a. 68.97 g (69.0 rounded); b. 350.32 g (350. rounded); c. 257.03 g (257 rounded)

d. 185.24 g (185 rounded); e. 0.651 g3. a. 5.26 moles; b. 4.5 moles; c. 0.128 mole; d. 0.699 mole; e. 1.52 moles4. a. 445.16 g rounds to 450 g; b. 0.249 mole; c. 0.632 g; d. 0.0833 mole; e. 0.0500 mole5. 1.50 moles O2

6. 254.9 g Al2O3 (250 g rounded)7. 9.3 g CCl48. 18 g Fe(OH)39. 5.9 g Al10. 0.163 mole Pb11. 10. g CO2

12. 22.4 L N2

13. 0.074 mole KClO3

14. 40.9 g KClO3

15. 100.8 L O2

16. 0.068 mole KCl17. 4.00 moles NH3

18. 4.00 moles NH3

19. 8.52 g NH3

20. 7.47 L NH3

21. 5.09 L CO2

22. 18.7L23. 45.5 g24. 4.48 L O2

25. 28.0 g/mole26. 46.0 g/mole27. NO2

9-21 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


“ALICE”

CHAPTER 10

THE MOLECONCEPT

(Part 3)

Empirical FormulasMolecular FormulasThe Ideal Gas Law

10-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


10-2 ©1997, A.J. Girondi

SECTION 10.1 Laboratory Determination of Empirical Formulas

An empirical formula is one which identifies which elements are present in a compound, and itgives the simplest ratio of atoms of those elements in the compound. Examples of empirical formulasinclude NaCl, CaCl2, and Al2O3 in which the simplest ratios are 1:1, 1:2, and 2:3, respectively. You canwrite empirical formulas using oxidation numbers, but they can also be determined using experimentaldata from experiments. The following example will illustrate how this is done.

Sample Problem: A laboratory analysis found that a compound contained 36.5 g Na, 25.4 g S, and38.1 g of O. What is the empirical formula of the compound?

Note: In these kinds of problems, you should attempt to calculate the mole quantities to at least 2 decimal places;otherwise, some problems may be solved incorrectly.

Solution:

Step 1: Change the mass data to moles:

36.5 g Na X

1 mole Na22.99 g Na

= 1.59 mole Na

25.4 g S X 1 mole S32.06 g O

= 0.79 mole S

38.1 g O X 1 mole O16.00 g O

= 2.38 mole O

Step 2: Change the numbers of moles to a simple whole number ratio. You do this by dividing each of themole quantities by the smallest of them. In our example, the smallest is 0.79. Thus, the calculations are:

1.59 mol Na0.791

= 2.01 mol Na; 0.791 mol S

0.791= 1.00 mol S;

2.38 mol O0.791

= 3.01 mol O

These results are very close to 2:1:3, so the answer is Na2SO3.

In some problems of this kind, a third step is necessary. This is illustrated in the next sample problem.

Sample Problem: What is the empirical formula of a compound that contains 53.73% iron, Fe, and 46.27% sulfur, S?

(When the data are given in percentages, they can be changed to grams simply by assuming that you have 100 grams of the sample.)

Step 1: Change the mass data to moles:

53.75 g Fe X 1 mole Fe55.83 g Fe

= 0.9620 mol Fe

46.27 g S X 1 mole S32.07 g S

= 1.443 mol S

10-3 ©1997, A.J. Girondi

Step 2: Change the numbers of moles to a simple whole number ratio. You do this by dividing each of the mole quantities by the smallest of them.

0.9620 mol Fe0.9620

= 1.000 mol Fe and 1.443 mol S

0.9260= 1.50 mol S

After completing step 2, the mole ratio is 1:1.5. The 1.5 is not close enough to a whole number to be rounded. Therefore, we go to step three.

Step 3: Multiply the mole ratio by the smallest integer which will give a whole number ratio. In this case, that integer is 2:2 X 1.000 = 2.000; and 2 X 1.50 = 3.00

Thus, the answer is: Fe2S3

Note: In order to be rounded to a whole number in step 2, a mole quantity should be within one-tenth of that number.

For example, 2.1 can be rounded to 2, or 3.9 can be rounded to 4 and step 3 is not needed. However, 2.2 and 3.8

should not be rounded to 2 and 4. In such a case, step three is necessary.

Problem 1. It is found that a sample of magnesium oxide consists of 4.04 g of magnesium (Mg) and2.66 g of oxygen (O). Calculate the empirical formula of the magnesium oxide using this information.

Problem 2. Experimental evidence reveals that a compound between iron (Fe) and oxygen (O)consists of 2.24 g Fe and 0.96 g O. Find the empirical formula.

Problem 3. An investigation reveals that the percentage composition of a compound is 75.0% carbon(C) and 25.0% hydrogen (H) by mass. Find its empirical formula.

10-4 ©1997, A.J. Girondi

Problem 4. Calculate the empirical formula of a compound which contains 32.4% Na, 22.6% S, and45.0% O.

Problem 5. A laboratory analysis of an unknown compound revealed that 100 g of it consisted of26.60 g K, 35.40 g Cr, and 38.08 g O. What is the empirical formula of the compound?

ACTIVITY 10.2 Determination of the Formula of an Oxide of Tin

In an earlier chapter you learned how to write chemical formulas for compounds by using what areknown as oxidation numbers. Now that you know something about the mole concept, you are ready to tryto determine the chemical formula of a compound experimentally in the lab. In this activity you are going tomake a compound of tin and oxygen, and you are going to gather data which will allow you to, hopefully,determine its formula.

1. Clean and dry an evaporating dish and a watch glass cover (see figure 10.1). Determine the mass ofthe dish with watch glass cover to the nearest 0.01 g.

Figure 10.1 Heating an Evaporating Dish

iron ringwire screen

watch glassring stand

evaporating dish

2. Place about 2 g of 30-mesh granulated tin inthe dish, cover with the watch glass, andmeasure the mass. If the balance in your labdisplays masses to three decimal places, thenread the mass of the tin to the nearest 0.001 g orto the nearest 0.01 g if your balance is accurateto only two decimal places. Enter all this data inTable 10.1.

3. While your instructor is watching, under afume hood, carefully add 5 mL of 8 M nitric acid(HNO3), and replace the watch glass. (The 8 Mindicates the concentration of the acid.) Handlethe acid with great care, and be sure to wearsafety glasses! Do not allow the acid to drip onanything.

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4. A reaction should begin which will result in the production of a reddish toxic gas, NO2. When thisreaction has subsided, you can take the dish back to your lab station for the rest of this activity. Briefly,here is what is happening in the dish. Some of the oxygen from the nitric acid, HNO3, combines with thetin to form a compound of tin and oxygen. At the same time, the acid decomposes to form H2O and NO2.The unfinished equation is:

? HNO3 + ? Sn ----> ? Sn?O? + ? H2O + ? NO2

So as the reaction goes on, you should see the formation of water, NO2 gas, and the white compound oftin and oxygen. In order to complete and balance the equation, you need to determine the correctformula for the tin–oxygen compound.

5. Position the dish, watch glass and the contents on a wire gauze supported by a ring stand as shown inFigure 10.1. Begin heating the dish with a low flame. Hold the burner in your hand and rotate it slowlyunder the gauze. Excessive heating at this point will result in spitting and popping in the dish. Continue toheat slowly and carefully until the contents are dry.

6. When popping and spattering no longer occur, remove the watch glass with tongs and place it upsidedown on your lab table. Break up the solid with a stirring rod, and heat with a hot flame until the solidbecomes a pale yellow. Be careful not to unnecessarily lose any material when you remove the watchglass or use the stirring rod.

7. After the dish has cooled, replace the watch glass cover and determine the mass of the dish, cover,and contents. Reheat the dish – without the cover – for a few minutes, allow to cool again, and determinethe mass again as before. If the two masses are not very close, reheat again until there is no further loss ofmass.

8. Discard the product in the waste container. Clean and return all equipment.

Table 10.1

1. Mass of dish and watch glass __________ g

2. Mass of dish, watch glass, and tin __________ g

3. Mass of tin (subtract 1 from 2 above) __________ g

4. Mass of dish, watch glass, and product __________ g (after heating to a constant mass)

5. Mass of oxygen in product __________ g (subtract 2 from 4 above)

Calculations:

1. Since all of the tin you used ends up in the product, you should be able to calculate the number ofmoles of tin in the product. Express your answer to three decimal places.

10-6 ©1997, A.J. Girondi

2. From the mass of oxygen gained, calculate the number of moles of oxygen, O, in the product. (Thisoxygen is not in the gaseous diatomic form.) Express your answer to three decimal places.

3. Enter your results from calculations 1 and 2 above in the blanks below.

Sn___________ O___________

This is the formula of the compound you made. However, we want to change these results to a wholenumber ratio. Do this by dividing each of the subscripts by the smallest of the two. For example, if you gota ratio such as 0.33 to 0.97, you would divide each value by 0.33. The result would be a ratio of 1 to 2.9.This is close enough (within ±0.1) to be rounded to 1 to 3. What if this process does not result in valueswhich are close (within ±0.1) to whole numbers? For example, what if the division process describedabove gives you a ratio such as 1 to 2.5? The 2.5 cannot be rounded to a whole number. You shouldthen multiply the values by the smallest integer which will give you a whole number ratio. In the case of 1to 2.5, that integer would be 2. The resulting ratio would be 2 to 5. Enter the whole number ratio resultingfrom your lab data into the blanks below.

The correct formula for the compound which you made is SnO2. The unbalanced equation representingthe reaction which you used to make the SnO2 is shown below. Balance it by adding the correctcoefficients.

_____ HNO3 + _____ Sn ----> _____ SnO2 + _____ H2O + _____ NO2

In this compound the oxidation number of tin is +4 and of oxygen is -2. What are the two possible names

for the tin–oxygen product?

{1}_____________________________ and {2}_________________________________.

SECTION 10.3 Molecular Formulas

The empirical formula of a compound gives the simplest ratio of the atoms in the molecule. Amolecular formula reveals the exact number of atoms of each element in the molecule. For example, theempirical formula of a certain gas is NO2, while its molecular formula is N2O4. Its empirical formula mass is46, while its molecular formula mass is 92. Note that the molecular formula's mass is a multiple of theempirical formula's mass. That's because the molecular formula is always a multiple of the empiricalformula. If you know the empirical formula mass and the molecular formula's mass, you can calculate themolecular formula as shown below:

(empirical formula mass) X (M) = (molecular formula's mass)

The letter M represents the multiple that relates the two formulas. For our example above:

(46) X M = (92), so M = 2 Therefore, 2 X (NO2) = N2O4

With this information, you should now be able to do problem 6.

10-7 ©1997, A.J. Girondi

Problem 6. Your answer to problem 3 (earlier in this chapter) represents the empirical formula of thecompound. If the molecular mass of that compound is 48, find its molecular formula. (Begin by calculatingthe formula mass of your answer to problem 3.)

Problem 7. Butane is a compound of 82.7% carbon (C) and 17.3% hydrogen (H). Its molecular mass is58 amu. What are the empirical formula and molecular formulas of butane?

Problem 8. An organic (carbon) compound if found to contain 92.25 g of carbon and 7.75 g ofhydrogen. If the molecular mass is 78, what is the molecular formula?

SECTION 10.4 The Ideal Gas Law: PV = nRT

You may remember that when you were studying the gas laws, it was mentioned that there wasone other law that you would learn about later. Now's the time!

It's called the ideal gas law, because it describes the behavior of an ideal gas, which is defined as agas that always obeys the gas laws under all conditions. Of course, there is no such gas. Real gases, likeH2 or O2 do not strictly obey the gas laws under extreme conditions like very high pressures or very lowtemperatures. Nonetheless, we don't have to be concerned about that, because our work is not usuallydone under such conditions.

The ideal gas law is different from the others you have learned in that it deals with only one set ofconditions. The problems you solved using the other gas laws included an "old" and "new" temperature,and "old" and "new" pressure, etc. Maybe you were given an "original" volume and asked to find a "new"volume. However, you will use the ideal gas law to solve problems involving only one pressure,temperature, volume, etc. The formula for the ideal gas law is:

PV = nRT

10-8 ©1997, A.J. Girondi

where P = pressure, V = volume, n = moles, T = temperature, and R is called the ideal gas constant. R hasa value of 0.0821 and its units are really crazy. They are shown below.

R = 0.0821 L. atm.mole K

These units are "liter-atmospheres per mole-Kelvin." Scientific constants usually have strange units, butdon't worry. They cancel when you use R in problems. R, like all constants, was experimentallydetermined, and is used to make the equation "work." We often use such constants. For example,consider the equation: P = C x A. P = number of people, and A = number of arms. What's the value ofthe constant C? Well, if you got O.5, that's right. In other words, if you have 16 arms and multiply by 0.5,you get 8 people. C is used to make the equation work. Its units would be people/arms, right?

Anyway, PV=nRT is the only gas law that you have studied that uses such a constant. When aconstant is used in an equation, the units of the other variables in the equation must be the same as thoseused in the constant. For that reason, when using the 0.0821 value in the ideal law you MUST expressvolume in liters, pressure in atmospheres, temperature in K (as always), and mass (n) in moles. Learn therelationships below which will allow you to make any needed conversions.

1 atmosphere = 760 mm Hg; 1 L = 1000 mL; oC + 273 = K

To solve a variety of problems, you will have to rearrange the ideal gas equation. How’s your algebra? For example, of we solve the equation for n, V, and T, respectively, we get:

n = PVRT

V = nRT

PT =

PVnR

Sample Problem: What volume would 34 g of CO2 gas have at 23oC and 800. mm pressure?(Notice that only one set of conditions is given? This is a hint that the use of the ideal gas law is appropriate.)

Solution: We must change 34 g of CO2 to moles, oC to K, and mm to atm. These are the results of thoseconversions: 34 g CO2 = 0.77 mole; 23oC = 296 K; 800 mm = 1.05 atm. When we do the substitutionswe get:

(1.05 atm)(V) = (0.77 mole)(0.0821 L.atm./mol.K.)(296 K)

Solving for V we get:

V = (0.77 mole)(0.0821 L.atm./Mole K)(296 K)

1.05 atm.

And the answer is: 17.8, or 18 Liters

Solve the problems below. Show your work, and include units with all measurements.

Problem 9. What is the volume of 2.77 moles of H2 gas at a pressure of 700. mm and a temperature of10.0oC?

10-9 ©1997, A.J. Girondi

Problem 10. What is the temperature in oC of 1.06 moles of O2 gas at a volume of 670. mL and apressure of 900. mm?

Problem 11. What is the pressure (in mm of Hg) exerted by 0.80 moles of methane (CH4) gas at 67oCand if it has a volume of 2.3 liters?

In the ideal gas law, the letter "n" represents moles of gas. Moles of a substance can be calculated bydividing the number of grams (g) of the substance that you have by its molecular mass (MM). For example,4.04 g of H2 (molecular mass = 2.02) is 2.00 moles:

4.04 g H2 X 1 mole H22.02 g H2

= 2.00 moles H2

In other words, moles = grams

molecular mass or n =

gMM

Now if n = g

MM, then the ideal gas law can also be written as PV =

gMM

RT

An alternate form of the ideal gas law is: PV = g

MMRT

This form of the ideal gas equation can be rearranged to solve for any of the variable it contains.

For example: MM = gRTPV

or g = (MM)PV

RT

Remember this form of the ideal gas law because it is particularly useful in problems which involve themass in grams of a gas or the molecular mass of a gas.

Problems which have you solved before using dimensional analysis can also be solved using theideal gas law. Problems 11 through 14 below are identical to problems 21, 22, and 23 in chapter 9. Solvethem again, but use the ideal gas law this time, and check Chapter 9 to see if you get the same answersyou got before.

Problem 12. A gas sample with a mass of 50.0 grams has a volume of 40.0 L at STP. What is themolecular mass of the gas? (You want to end with units of g/moles.)

10-10 ©1997, A.J. Girondi

Problem 13. What is the volume in liters occupied by 10.0 grams of CO2 gas at STP? (Hint: you mustfirst find the molecular mass of CO2.)

Problem 14. If 1.00 mole of a gas has a mass of 18 grams, and you have 15.0 grams of the gas, whatwould its volume be at STP?

At this point, go back and compare the three problems above to problems 21, 22, and 23 in Chapter 9.Compare the methods of solving the problems as well as the answers.

Now let's try a problem in which the gas is not at STP:

Problem 15. At 600. mm Hg pressure and 35.0oC, 2.70 L of a gas has a mass of 3.80 grams. What isits molecular mass? (Since you are using 0.0821 for the constant R, remember that you must expresspressure in atm and temperature in K.)

ACTIVITY 10.5 Determination of the Molar Volume of a Gas

In this activity you will learn one method for determining the molar volume of a gas, and you willlearn how to collect a gas in the lab by a procedure known as water displacement. You will be makingmeasurements at room temperature and pressure, but conversion to standard conditions (STP) will benecessary to complete the calculations. Fill a large beaker (600 mL or larger) about 3/4 full of tap water.Next, ask your instructor or others in your class whether you should follow procedure A or B below (do notfollow both). You will use procedure B if you have a balance in your lab which can measure mass to threedecimal places.

10-11 ©1997, A.J. Girondi

Procedure A: Obtain a piece of magnesium ribbon about 3.5 cm long. On ananalytical balance which is accurate to three decimal places , determine themass of your piece of Mg. It must have a mass between 0.044 and 0.046.Trim with scissors as needed. Record the mass in Table 10.3.

Procedure B: (Follow this only if you cannot use procedure A.) Cut a piece ofmagnesium ribbon about 3.5 cm long. Carefully measure the length of thepiece of ribbon to the nearest millimeter, mm. Record the length in Table10.3.

1. Obtain a piece of fine copper wire about 15 cm long and tie it to the Mgribbon which has been rolled into a coil that will fit inside the gas-measuringtube (see Figure 10.2). Obtain a ring stand with a utility clamp to support thegas-measuring tube.

2. Slowly pour about 10 to 15 mL of 6 M HCl (hydrochloric acid) into the gastube. Use caution. Incline the tube slightly so air may escape and slowly fill itto the brim with tap water from a beaker. Try not to mix the acid and water anymore than necessary.

3. With the tube completely full of water, insert the magnesium ribbon about3 or 4 cm into the tube. With the wire against the side of the tube, insert a 1-hole or 2-hole stopper. The stopper should force a little water out of the tubeand should hold the wire in place.

4. With your finger over the hole in the stopper, invert the tube and place thestoppered end into the beaker 3/4 full of water. Clamp the gas-measuringtube in place so that the bottom of the rubber stopper in slightly above thebottom of the beaker. The reaction will not start immediately, since the acidhas to make its way down to the metal. The acid will react with the magnesiumto produce hydrogen gas:

Mg(s) + 2 HCl(aq) ----> MgCl2(aq) + H2(g)

5. When the Mg has reacted completely, tap the tube with your finger toremove any bubbles you may see on the side of the tube. Adjust the heightof the gas- collecting tube so that the level of the water inside the tube isequal to that in the beaker. This will make the pressure on the gas inside thetube equal to the pressure inthe room. If you have troubledoing this, ask your teacherfor advice. Now read thelevel of the water inside thetube as carefully as possible.This will be the volume ofgas in the tube. Record thevolume in Table 10.3.

6. Record the temperatureof the water in the beaker(oC) which we will assume tobe the temperature of the gas in the tube. In addition, measure thebarometric pressure (mm Hg). Your teacher will help you to read the barometer in the lab. Thetemperature and pressure must be measured immediately after the gas has been collected. Record thesevalues in the table.

10-12 ©1997, A.J. Girondi

1 or 2 holed stopper Cu wire

acid + water

Figure 10.2

Mg

Table 10.2Vapor Pressures of Water

T (oC) P (mm) T (oC) P (mm)

16 13.6 22 19.817 14.5 23 21.2 18 15.5 24 22.419 16.5 25 23.820 17.5 26 25.2 21 18.6 27 26.7

7. Empty the contents of the tube and beaker and rinse both. Return all equipment to the proper place.(Ifyou followed procedure B earlier in this experiment, ask your teacher for the precise mass of 1.000 meterof magnesium ribbon. Record this value in Table10.3. If you followed procedure A, you can disregardthis value.)

Table 10.3Molar Volume of a Gas

*(Skip entries 1, 2, and 3 below if you followed procedure A. Skip entry 4 if youfollowed procedure B.)

*1. Length of Mg ribbon used ____________ mm

*2. Mass of 1.000 meter of Mg ribbon ____________ g/m

*3. Calculated mass of Mg ribbon used ____________ g (should be between 0.040 and 0.045 g)

*4. Mass of Mg ribbon used ____________ g (from analytical balance)

5. Volume of gas in tube ____________ mL

6. Temperature ____________ oC

7. Atmospheric pressure ____________ mm Hg

(Do calculation 3 before completing entries 8 and 9 below.)

8. Water vapor pressure ____________ mm Hg

9. Pressure of dry H2 ____________ mm Hg

Calculations:

1. Calculate the mass of Mg ribbon used to the nearest 0.001 g by using the mass of 1.000 meter of theribbon, which you got from the teacher. Use unit analysis. (Skip this calculation if you followed procedureA.)

2. Based on the mass of Mg used, calculate the number of moles of Mg used. Express your answer tothree decimal places. That amount of precision is needed!

3. The pressure of the gases in the tube is equal to the atmospheric pressure in the room. The H2 gas inthe tube was mixed with water vapor. This is a problem which results from collecting a gas in this way.Therefore, the pressure in the tube is the sum of the pressures exerted by H2 gas and H2O vapor.Dalton's law of partial pressures reveals that the total pressure exerted by a mixture of gases is equal to thesum of the partial pressures exerted by each gas.

10-13 ©1997, A.J. Girondi

We need to know the pressure exerted by the H2 only. We call this the pressure of the "dry" H2. To getthis, we will subtract the partial pressure of water vapor from the total pressure. The formula is:

Ptotal = Pwater + Phydrogen.

Ptotal is the atmospheric pressure. The vapor pressure of water, Pwater, is dependent on temperature andcan be found in Table 10.2. (Your ALICE reference notebook also contains a table of water vaporpressures which covers a broader temperature range.) Use the atmospheric pressure and the partialpressure of water vapor to calculate the pressure of the "dry" hydrogen: Phydrogen = Ptotal - Pwater

Pressure of "dry" hydrogen = _____________ mm Hg

4. The pressure you just calculated above is the pressure to be used in your calculations. You now havethe pressure, the temperature, and the volume of the H2 under these conditions. Now, you mustcalculate what the volume of the H2 which you collected in the tube would be (in liters) at STP. Use thecombined gas law (from Chapter 4 as given below).

P1V1T1

= P2V2

T2

__________ L H2 at STP

5. In calculation 2, you determined the number of moles of Mg used in this reaction. Enter this value inthe blank below. According to the equation for the reaction, the number of moles of Mg used is the sameas the number of moles of H2 produced. Fill in the blanks in the mole-to-mole ratio below, and solve formoles of H2 produced.

Mg + 2 HCl ----> MgCl2 + H2

______ mole Mg X mole H2mole Mg

= ________ mole H2

How many moles of H2 were produced in your experiment? _____________ moles H2 produced

6. With your results from steps 4 and 5 above, you have established a relationship between moles of H2

and liters of H2 at STP. Now you must use this relationship to determine what volume the hydrogen gaswould have occupied if you had collected 1.00 mole of it (instead of the very small number of moles of H2

which you actually did collect). You can do this by unit analysis. The "fencepost" is set up for you below.Copy the data from calculations 4 and 5 above into the ratio below and solve the problem.

1.00 mole H2 X L H2

moles H2 = __________ L H2

Because of the very small volume of gas collected in this activity, there is usually a fair amount of error.How does your answer for the volume of one mole of a gas at STP, compare to the generally accepted

value of 22.4 L? _________________________________________________________________

10-14 ©1997, A.J. Girondi

Problem 16. Assume that 0.0200 mole of a gas is collected in the lab by water displacement. If thetemperature of the gas is 26.0oC and the total pressure in the gas tube is 745 mm Hg (including watervapor), what is the volume in liters of the dry gas? (Use the ideal gas law.)

Caution: you can’t use the total pressure in your calculations. You must first find the pressure of the “dry”gas by subtracting the water vapor pressure at 26.0oC from the total pressure. It is the pressure of the“dry” gas that is used in the calculations. Water vapor pressures can be found in Table 12 in yourreference notebook.

SECTION 10.6 Optional Review Problems

Problem 17. Using the ideal gas law, determine the number of moles of gas which will have a volume of2.5 liters at 1.3 atm and 303 K.Problem 18. Using the ideal gas law, determine the temperature (in oC) of 1.8 moles of a gas if it has avolume of 50.9 L at 705 mm Hg.Problem 19. Using a form of the ideal gas law, calculate the number of grams of helium gas, He, whichare present in an 8.00 liter container at 345oC and 1.33 atm.Problem 20. A 14 liter container holds 50.0 grams of a gas at 2.3 atm and 345 K. The gas is known tobe one of the following: O2, Cl2, Br2, or CO2. What is the molecular mass and the identity of the gas?Problem 21. How many moles of KClO3 are required to form 37.3 g of KCl?

2 KClO3(s) ----> 2 KCl(s) + 3 O2(g)

Problem 22. Analysis of a sample of propane gas (also known as LP gas) from a backyard grill is shownto contain 36.7 grams of C and 8.244 grams of H. Calculate the empirical formula of propane gas.Problem 23. Calculate the empirical formula of vitamin C if it is composed of 40.9% Carbon, 4.58%hydrogen, and 54.5% oxygen.Problem 24. A sample of a hypothetical compound contains 2.44 g of element X and 1.02 g ofelement Z. The atomic mass of hypothetical element X is 12.2 grams/mole, while that of hypotheticalelement Z is 2.04 grams/mole. The molecular mass of this hypothetical compound is 69.2. Calculate boththe empirical formula and the molecular formula of this compound.Problem 25. 245 mL of a “wet” gas is collected by water displacement at 27.0oC and a total pressure of805 mm Hg. How many moles of the “dry” gas does this represent?. (Answer using 3 sig figs.)

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This is the end of Chapter 10. Check the learning outcomes below to be sure that you havemastered them. Take the Chapter 10 exam, and move on to Chapter 11.

_____1. Calculate the empirical formula of a compound from experimental data.

_____2. Calculate the molecular formula of a compound from experimental data.

_____3. Solve problems using both forms of the ideal gas law, PV = nRT and PV = gRT/MM

_____4. Solve problems related to the collection of a gas by water displacement.


Questions:

{1} tin (IV) oxide; {2} stannic oxide

Problems:

1. MgO2. Fe2O3

3. CH4

4. Na2SO4

5. K2Cr2O7

6. C3H12

7. C2H5 and C4H10

8. C6H6

9. 69.9 L10. -264oC11. 7379 mm Hg (rounds to 7400 mm Hg)12. 28.0 g/mole13. 5.09 L14. 19 L15. 45.1 g/mole16. 0.518 L17. 0.13 mole18. 47oC19. 0.84 g20. 44 g/mole, CO2

21. 0.50 mole22. C3H8

23. C3H4O3

24. X2Z5, X4Z10

25. 0.0102 mole

10-16 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 11

PERIODICCLASSIFICATION

OF THEELEMENTS

The Periodic TableFamilies of Elements

11-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


11-2 ©1997, A.J. Girondi

SECTION 11.1 Introduction to the Periodic Table

So far in your study of chemistry, you have spent most of your time learning about the physicalprinciples which determine the behavior of atoms and molecules. These principles allow us to predict howtemperature, volume, and pressure changes affect gases. They also allow us to solve mass-mass andmass-volume problems using balanced chemical equations. We have also been able to use physicalprinciples to determine atomic masses and chemical formulas using experimental data. However, they donot tell us much about the actual physical and chemical properties of specific chemical substances.

A working knowledge of chemistry should include familiarity with the actual properties of chemicalsubstances, as well as an understanding of the laws that govern chemistry. Hopefully, you will begin todevelop this working knowledge as you study this chapter. Each element has its own characteristic set ofproperties that help to distinguish it from other elements. As early as the 1800's, chemists began to lookfor similarities among elements that would allow them to be classified into groups. Early chemists knewthat it would be much easier to study groups of elements rather than each of the elements individually.

The task at hand was to develop a way of arranging the elements so they could be groupedaccording to common traits. The Russian chemist Dmitri Mendeleev was the first person to successfullyarrange the elements in an orderly fashion. He arranged the elements in order of increasing atomicmasses in 1869. He then developed a table in which elements with similar chemical properties wereplaced in the same vertical columns. This table was later called the "Periodic Table of the Elements."

Mendeleev received a considerable amount of ridicule for his arrangement. His colleaguesjokingly suggested that he try arranging the elements alphabetically! Mendeleev ignored the ridicule andremained committed to his method of arrangement. In the preparation of his table, if the element with thenext highest atomic mass did not fit a particular group, Mendeleev left a blank space on his table andmoved the element up to the next higher group. This resulted in several blank spaces in his table ofelements. He believed that these blank spaces would eventually contain elements that had not beendiscovered yet. He predicted the properties of three of these unknown elements. All three werediscovered in Mendeleev's lifetime and were found to have properties very close to what he hadpredicted! This proved that Mendeleev's arrangement of elements was, indeed, useful.

From his studies, Mendeleev concluded that both the chemical and physical properties of theelements vary in a periodic (repeating) fashion with increasing atomic mass. The horizontal rows in thetable are called periods or rows, while the vertical columns are called families or groups. Look at a periodictable now, and distinguish the periods (rows) from the families (groups). Table 11.1 is a copy of theperiodic table which Mendeleev developed in 1871. The periodic table that was developed by Mendeleevenabled chemists to classify knowledge and concentrate their studies on physical and chemical propertiesof groups of elements rather than on each element individually. This arrangement of elements enableschemists to make predictions. These predictions are based on the repeating nature of chemicalproperties.

In problem 1 you will be asked to organize a periodic table using some fictional elements. Thefictitious elements and properties are listed in Table 11.2. You are going to enter these elements into theblocks of Table 11.3 Some of the elements have already been filled in for you. As you record eachsymbol on the periodic table, also record its mass in the lower right corner and its formula with chlorine inthe upper left corner of each block. This will make it easier to see the periodic (repeating) properties of theelements. You should use the two rules below when arranging the fictitious elements listed in Table 11.2.

RULE 1: Elements are arranged in order of increasing atomic mass from left to right in each row row.

RULE 2: Elements in each vertical column all form compounds with similar chemical formulas.

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Table 11.1Mendeleev's Periodic Table (1871)

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Problem 1. Fictitious elements, atomic masses, and chemical formulas (that show how each elementcombines with chlorine) are given in Table 11.2. On the basis of rule 1 and rule 2 mentioned earlier, youshould now arrange the elements in Table 11.2 in their proper order in the rows and columns of the"Periodic Table of Fictitious Elements" (Table 11.3). Note that some of the fictitious elements havealready been put into Table 11.3 to get you started.

Table 11.2List of Hypothetical Elements

Element Formula of Compound Atomic Mass with Chlorine (g/mole)

A A2Cl3 22B (none) 36D DCl3 44F F2Cl3 5G GCl4 31H HCl4 11J (none) 15M MCl3 9N NCl4 49P P2Cl3 39R RCl2 7S SCl2 41T (none) 52U ????? ??Z ZCl3 28

Table 11.3 A Table of Hypothetical Elements

F2Cl3

F 5

- - - -

J 15

ZCl3

Z 28

SCl2

S 41

??

? ??

Atomic Mass

Symbol

Formula with Cl

11-5 ©1997, A.J. Girondi

Look again at the completed blocks in Table 11.3. One space contains the symbol U. Study theother elements in the same column and those in the same row as U. In the spaces below give your bestestimate for the atomic mass of this unknown element and give your prediction of the formula of thecompound it will form with chlorine:

Atomic Mass of U is in the range:{1}__________________; Formula of U with Cl is: {2}______________

SECTION 11.2 Development of the Periodic Table

In science we develop models or theories to explain our observations. These models are notperfect, and most have exceptions. Mendeleev's arrangement of elements is no different. WhenMendeleev arranged elements according to similar properties, their atomic masses increased in numberfrom left to right. However, there were exceptions to this rule. Study the periodic table in your classroomor in your notebook.

Problem 2. Find two pairs of elements that appear to have their atomic masses reversed from theaccepted increasing order:

________________ and ________________; ________________ and _________________

These exceptions cast some doubt on the reliability of using atomic mass as a means of organizing theperiodic table.

Henry Moseley found the reason for this apparent exception to Mendeleev's rule. Moseleyaltered the periodic table to base it on atomic numbers rather than on atomic masses, and he restated theperiodic law so that it stated that "the properties of elements are a periodic function of their atomicnumbers." This law has no exceptions. You will learn the reason for this when you study atomic structurein an upcoming chapter.

Once again look at a periodic table in your classroom or in your notebook. There are severalthings concerning atomic numbers that are important for you to notice.

Problem 3. Answer the questions below .

a. How do atomic numbers change as you move through the periodic table? ____________________

b. Are atomic numbers whole numbers or decimal numbers? _________________

c. What element has the smallest atomic number? _________________________

d. As of June, 1995 there are 111 elements (officially). How many are listed on your table? __________

All matter, from concrete to human skin, iscomposed of either pure elements or elements incombination with each other. The proportions andkinds of elements that are combined determine whatthe particular substance is. Actually, only the first 92elements are considered natural. Those elements withatomic numbers larger than 92 are artificial elements,produced by scientists working in laboratories. Theseelements are referred to as the transuranium elements.When a new element is "discovered," other scientistsmust conduct experiments to confirm the discovery.

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Element Name Symbol

104 Unnilquadium Unq105 Unnilpentium Unp106 Unnilhexium Unh107 Unnilseptium Uns108 Unniloctium Uno109 Unnilennium Une

Table 11.4Naming New Elements

The discoverer then receives the honor of naming the new element. Precisely who discovered elements104 through 107 is questionable at this time. Russian scientists reported the production of element 104in 1964, element 105 in 1970, element 106 in 1974, and element 107 in 1976. However, Americanscientists have claimed that they produced element 104 in 1968, element 105 in 1970, and element 106in 1974. The Soviet Union has proposed the name Kurchatovium (Ku) for element 107 and Bohrium (Bh)for element 105. The United States has proposed the following names and symbols for elements 104through 109: 104 = Rutherfordium (Rf); 105 = Hahnium (Ha); 106 = Seaborgium (Sg); 107 =nielsbohrium (Ns); 108 = hassium (Hs); 109 = meitnerium (Mt).

Until the official names for elements have been agreed upon, it has been suggested that thoseelements be named according to a system which makes names out of combined Greek prefixes. (nil = 0;un = 1; quad = 4; pent = 5; hex = 6; sept = 7; oct = 8; enn = 9) The names are given a suffix of "ium."

The periodic table allows us to study groups of elements as well as individual elements. Onebroad classification of elements is by metals and nonmetals. If we study this further, we find that all metalshave several properties in common that differ significantly from the properties of nonmetals.

Problem 4. During 1994 and 1995, scientists in Germany produced elements 110 and 111. Namethem using the system described above, and give their symbols.

a. 110 name: _________________________ symbol: ___________

b. 111 name: _________________________ symbol: ___________

ACTIVITY 11.3 Metals, Nonmetals, & Metalloids - Comparing Properties

Obtain the vials from the materials shelf labeled 11.3. Three of the elements are aluminum,copper, and tin. The remaining 2 elements listed, sulfur and silicon, are classified as a nonmetals. (Siliconis also a metalloid.) Examine each sample for the properties called for in Table 11.5. The melting points ofthe substances may be found in the Handbook of Chemistry and Physics or in a similar reference book.The battery-powered device found with the vials may be used to test the electrical conductivity of eachsample. Simply touch the two wires to one piece of each element. If the device indicates that current isbeing conducted through the element, you should classify the element as having high conductivity. If itdoes not light, classify the substance as having low conductivity. The wires of the device should nottouch each other. "Phase" in Table 11.5 refers to solid, liquid, or gas. "Luster" refers to dull or metallic.

Study the information you have collected from Table 11.5 and react to the four statements below.

1. Make a general statement about the properties of a metallic substance.

______________________________________________________________________________

______________________________________________________________________________

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Table 11.5Properties of Selected Metals and Nonmetals

Element Phase Luster Conductivity Melting Point (oC)

copper(metal)

tin(metal)

aluminum(metal)

sulfur(nonmetal)

silicon(nonmetal-metalloid)

Propose a hypothesis about why frying pans are usually made of metals while their handles are usually

made of nonmetals like wood or plastic. ________________________________________________

_____________________________________________________________________________

The melting point of a substance is related directly to its hardness. Elements with high melting points aregenerally hard substances. With this is mind, list the substances in Table 11.5 from hardest to softest.

hardest---> ____________________________________________________________<------softest

Do the hardest materials appear to be metals or nonmetals? {3}________________________________

Locate the "staircase" on the right side of the periodic table. The elements that are located to theleft of this staircase are classified as metals, while those to the right of it are nonmetals.

Are most elements metals or nonmetals?{4}______________________________________________

Name 3 metals:__________________________________________________________________

Name 3 nonmetals:_______________________________________________________________

The elements that border the staircase on the periodic table are known as metalloids. Theseelements have some properties that make them resemble metals, but other properties that make themresemble nonmetals. They include the elements just above and just below each step of the staircasestarting with aluminum (#15). A good example is silicon. Silicon is known as a "semiconductor" because itconducts electricity better than nonmetals, but not as good as metals. It is the major component ofcomputer chips.

Obtain the sample of silicon from the box of materials for Activity 11.3. While silicon is a metalloid,it falls on the nonmetal side of the staircase. Describe one property of silicon that makes it remind you of ametal:

______________________________________________________________________________

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SECTION 11.4 The Modern Periodic Table

Dividing the elements into metals and nonmetals is only one way in which they are categorized.As mentioned earlier, another way to classify them is by families (or groups) and periods (or rows).Elements such as boron, carbon, and neon are in the same horizontal row and are said to be in the sameperiod. Elements in the same vertical column, such as lithium and sodium, belong to the same family orgroup. To distinguish the families of elements, the families (or groups) have names as well as numbers.Note in Table 11.6 that the families are numbered 1A through 8A, while the periods (rows) are numbered1,2 3, etc.

1A

2A 5A 6A 7A

8A

3A 4A

Table 11.6The “A” Families on the Periodic Table

107 108 109 110 111Unh Uns Uno Une Uun Uuu

*Some authors do not consider aluminum (#13) or Astatine (#85) to be metalloids.

The families used to be broken up into two subgroups which were labeled the "A" families and the"B" families. The transition elements in the center of the table were included in this system. You probablywill see periodic tables in your classroom or elsewhere that still use this system. However, the mostmodern labeling system now includes 18 families of elements including the transition elements. It isshown in the following partial periodic table (Table 11.7).

Some families are often named according to the first element in the family. For example, family 16above is the oxygen family, family 15 is the nitrogen family, family 14 is the carbon family, and family 13 isthe boron family. These families also have other names which are not in common use. Other familieshave commonly-used special names. Family 18 is known as the noble gases, family 17 elements areknown as the halogens, family 1 elements are called the alkali metals, while family 2 elements are known asthe alkaline-earth metals. See Table 11.8. The metals in the center block of the table are collectivelyknown as the transition metals. The two rows at the bottom of the table (the lanthanide and actinideseries) are collectively known as the rare earth elements. In Table 11.8, write in the "names" of families 13through 16.

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Rows 6 and 7 of the periodic table are not shown here (to save space).

You have seen that metals have properties that help to distinguish them from nonmetals. In thesame way, each chemical family has a set of characteristic properties. We will be taking a closer look atsome of the properties associated with each chemical family.

Problem 5. Refer to Table 11.8 and a periodic table to help you with the following.

a. Write the symbols for the alkali metals: _______________________________________________

b. Write the symbols for the alkaline-earth metals: _________________________________________

c. Write the symbols for the halogens: _________________________________________________

d. Write the symbols for the noble gases: _______________________________________________

e. Name the elements in group IIIA (or group 13): _________________________________________

______________________________________________________________________________

f. Name the elements in group VA (or group 15):__________________________________________

______________________________________________________________________________

You have been told that the members of a family of elements have similar properties. The alkalimetals, for example, all react with water in the same proportions to form compounds that have similarchemical formulas.

Problem 6. The chemical equations below illustrate reactions between a few alkali metals and water.Complete and balance each equation. Note the similarity of the equations for the various family 1Aelements:

a. 2 Li + 2 HOH ----> 2 LiOH + 1 H2

b. _____Rb + _____HOH ------> _____RbOH + _____H2

c. _____Cs + _____HOH ------> _____CsOH + _____H2

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Transition Metals

1A 2A 3A 4A 5A 6A 7A 8A

1 2 13 14 15 16 17 18

<----- 3 THROUGH 12 ------>

Lanthanide Series

Actinide Series

Rare-Earth Elements

Table 11.8Names of Classes and Families of Elements

transition metals

Sc Zn

ACTIVITY 11.5 Properties of Alkaline-Earth Element Compounds

One chemical family that is more easily and safely studied is the alkaline-earth metals. Find thematerials labeled 11.5 on the materials shelf. Because elements in the same families have strong familyresemblances, you can predict the chemical behavior of other family members if you know how one or twofamily members act in a chemical reaction. The experiment presented here will help to illustrate this point.Be sure to wear glasses! Use standard 150 mm test tubes.Procedure:

1. Place 5.0 mL of 0.1 M magnesium nitrate, Mg(NO3)2, solution into a test tube. Into a second test tube,place 5.0 mL of 0.1 M calcium nitrate, Ca(NO3)2, solution.

2. Add 5.0 mL of 0.1 M sodium carbonate, Na2CO3, solution to each test tube. Stopper and shake bothtubes well. Describe what happens in each tube.

3. Strontium, barium, magnesium, and calcium are all members of the same family, the alkaline-earthmetals. Suppose you had solutions of strontium nitrate, Sr(NO3)2, and barium nitrate, Ba(NO3)2. Predictwhat would happen if you mixed each of these solutions with sodium carbonate, Na2CO3, solution.

______________________________________________________________________________

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4. What knowledge allows you to make the above prediction? ________________________________

______________________________________________________________________________

5. Test your prediction by repeating the procedure in steps 1 and 2, but this time use 0.1 M strontiumnitrate, Sr(NO3)2, solution and 0.1 M barium nitrate, Ba(NO3)2, solution. Describe the changes in detail.

______________________________________________________________________________

______________________________________________________________________________

6. Suppose you were given a test tube containing a clear liquid. How could you test it to determine if an

alkaline-earth metal were in the liquid? _________________________________________________

______________________________________________________________________________

7. Wash all tubes and put the materials and solutions back in their proper places.

ACTIVITY 11.6 The Properties of Halogen Compounds

Another group of elements that have family resemblances are the halogens. You will observesome of those similarities in this activity. Find the materials labeled 11.6 on the materials shelf. Be sure towear glasses!

Since the halogens are a family, they all react in a similar fashion with silver nitrate, AgNO3, to forma solid (precipitate). Each precipitate which forms, however, is slightly different from the others. We will betaking advantage of these differences to try to identify which halogen is present in an "unknown"solution. By comparing the reaction of the unknown halogen to the reactions of the known halogens, it ispossible to identify the unknown as being a compound of bromine, chlorine, or iodine.

1. Carry out the reactions outlined in Table 11.9. In each situation, add 1 mL of each halogen solution to 1mL of AgNO3 solution, stopper, and shake each test tube.

2. Observe any reaction that occurs, and then add 2 to 3 mL of 6.0 M NH4OH solution, stopper, and shakefor a minute. Observe any change.

3. Complete and balance each equation. (The first one has been done for you.) Identify any precipitate.Write your observations such as color of a precipitate, etc. Be certain that your test tubes and otherequipment are clean before you begin any of the combinations listed in Table 11.9.

4. Wash and rinse several times with a little distilled water if you have any doubt. Caution: handle AgNO3

with care. It can cause harmless, but dark, temporary stains on skin which do not appear until hours aftercontact.

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Table 11.9Halogen Reactions

Reactants Products

NaCl(aq) + AgNO3(aq) ----> ____AgCl(s)____ + ___NaNO3(aq)__

Observations: ____________________________________________________________

NaBr(aq) + AgNO3(aq) ----> ______________ + ______________

Observations: ____________________________________________________________

NaI(aq) + AgNO3(aq) ----> ______________ + ______________

Observations: ____________________________________________________________

Unknown + AgNO3; Observations: ____________________________________________

Identity of unknown: _____________________________(Note: the "unknown" is one of the following: NaCl, NaBr, or NaI)

The addition of NH4OH solution caused which of the precipitates to dissolve?_____________________

1. On what evidence did you base your conclusion about the identity of the unknown? _____________

_____________________________________________________________________________

2. Based on what you learned in this activity, how might you determine if your tap water at home has

chlorine in it?____________________________________________________________________

Give it a try! Using your answer above, test some tap water and test some distilled water. Did your

hypothesis work? ________________

The reactions in Table 11.9 fall into which of the 4 major reaction types?{5}________________________ (choose either direct combination, decomposition, single replacement or double replacement)

SECTION 11.7 Other Classes Of Elements On The Periodic Table

So far, we have looked closely at two chemical families, halogens and alkaline-earth metals. Theremaining chemical families also have distinguishing properties. The large middle portion of the periodictable contains the transition metals. In general, these transition metals have higher melting points and arephysically stronger than the alkali or alkaline-earth metals. Because of this property, many of the transitionmetals such as iron, titanium, and vanadium are used in construction of buildings, aircraft, and specialtyproducts.

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The noble gas family is composed of elements that are all gases at room temperature and are alsononmetals. The noble gases are so named because of their very nonreactive character. It was longbelieved that the noble gases would not react with any other elements. For that reason, they used to becalled the "inert" gases. But in the 1960's scientists managed to force xenon and fluorine to formcompounds under extreme conditions. Nevertheless, the noble gases are the least reactive elements onthe periodic table. In the next chapter you will learn why they behave this way.

The elements in the lanthanide series and the actinide series are also metals. Together they areknown as the rare-earth elements. The lanthanides all have similar chemical and physical properties.These elements are relatively common, but because of the difficulty in separating them from each other,they are not commonly available. Among the actinides, all of which are radioactive, only uranium andthorium exist in any significant amounts in nature. The other actinides were discovered and observed asthe products of controlled nuclear reactions and, in many cases, have been produced in only very smallamounts. Uranium and thorium are used as fuel in nuclear reactors and in the warheads of nuclearweapons.

Both the lanthanide and the actinide series actually belong up in rows 6 and 7 of the periodictable. If you follow the atomic numbers of the elements, you will see where these two series actually fit.However, if we put them there, they would make the periodic table rather long. So, to make the wholething more compact, we take these two series out and put them at the bottom.

The lanthanides and actinides actually fit into the periodic table between what two families or

categories of elements? {6}_________________________________________________________

Within the major chemical families, chemists have had some difficulty in finding a proper place forone element in particular. This element is hydrogen. In most periodic tables, hydrogen is placed in Family1A (or group 1) above lithium. But, you have seen that the elements in that family are metals. Hydrogen iscertainly not a metal, so this placement seems a bit odd. Hydrogen shares some of its physicalcharacteristics with the halogen family. But, hydrogen has little in common with the halogens in terms of itschemical properties. To help resolve this problem, hydrogen is often placed in both family 1A and 8A(groups 1 and 17). Some periodic tables show hydrogen above family 1A, but it is detached as if to pointout its uniqueness.

ACTIVITY 11.8 Making, Collecting, and Studying Hydrogen Gas

This experiment will familiarize you with some of the properties of hydrogen. It is hazardous,however, and you must be closely supervised and observed by the teacher as you do this. Be sure tonotify the teacher before you begin, and have your set-up checked before you actually generate any gas.

1. Get the materials labeled 11.9 from the materials shelf. Set up the apparatus as shown in Figure 11.1below. Be sure to wear safety glasses and an apron.

2. Have the teacher check your set-up before going any further. All connections must be snug to preventhydrogen gas from escaping.

3. Fill three or four 150 mm (regular size) test tubes with water, put your thumb on top and carefully invertthem into the trough so that little or no water is lost. Keep all flames and sparks away from your set-up!!Add 10 mL of 3 M hydrochloric acid (HCl) to the flask containing 4 to 5 grams of mossy zinc, and then insertthe rubber stopper assembly.

11-14 ©1997, A.J. Girondi

water level

150 mm test tube filled with water

HCl + Zn

bubbles of H2 gas

metal shelf

pneumatic trough

Figure 11.1 Collection of Hydrogen Gas by Water Displacement

Overall reaction: 2 HCl + Zn ---> ZnCl2 + H2

4. When the hydrogen gas starts to form in the flask, allow the gas to bubble for a minute or so to force airout of the flask and rubber tubing before attempting to collect any gas in a test tube. Then, collect gas inthe test tubes by placing the end of the tubing under the mouth of each tube until all of the water hasbeen forced out. If the gas stops before you have collected enough, add more HCl to the flask.

5. Store the tubes upside down inverted in the water or on your desk after they are full of gas. When youhave collected enough gas to fill the test tubes, you can stop the generator by adding water to the flask.

6. Pour the diluted acid into a sink, and throw the zinc into the refuse can. To test the tubes for thepresence of hydrogen, get away from your hydrogen generator and light a wooden splint (don't light anyburners close to your equipment!). Quickly move the flaming splint next to the mouth of one of yourinverted test tubes. (Keep it away from the others.) Repeat the procedure with the remaining tubes ofgas.

1. Describe what happens._________________________________________________________

2. Why did you store the tubes of H2 gas upside down? {7}__________________________________

3. Hydrogen explodes by reacting with oxygen gas in the air to form water vapor. Write the balancedequation for the reaction which occurred when the contents of the tube met the burning splint.

{8}__________________------>__________________

Return all of your equipment to the materials shelf. Rinse the tubes and flask.

SECTION 11.9 Using the Periodic Table to Predict Properties and Formulas of Compounds

One of the factors that gave validity to the arrangement of elements on the periodic table was thatit enabled chemists to predict the properties of several undiscovered elements. In the table below youcan see how well Mendeleev was able to predict the properties of an undiscovered element which hecalled "ekasilicon," which turned out to be what we now call germanium. The property which Mendeleev

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described as combining power in the table above refers to the number of bonds an atom forms with otheratoms. This predictive power can be very useful in determining the chemical formulas of substances.

Look at the formulas of the Group 1 (Family 1A) elements combined with chlorine and brominebelow.

Chlorides: LiCl NaCl KCl RbCl CsCl FrClBromides: LiBr NaBr KBr RbBr CsBr FrBr

Notice how all group 1 (family 1A) elementscombine with group 17 (family 7A) elements in a1:1 ratio. It is possible to use the periodic table tomake predictions in the same way thatMendeleev did. Once you know the formulas ofcompounds of an element, you can predict theformulas of similar compounds. Look at theformulas of chlorides formed by elements infamilies 1A - 7A (groups 1,2 & 13-17) as shown inTable 11.11.

Table 11.10Mendeleev's Predictions About

"Ekasilicon"

Property Prediction Actual Value

atomic mass 72 72.59density (g/cm3) 5.5 5.32 combining power 4 4specific heat (J/g.oC) 0.305 0.322

How many Cl's are attached to each of the column 1A elements shown above? {9}__________________

How many Cl's are attached to each of the column 4A elements? {10}____________________________

We can state this general principle:

Given the formula of a compound, the replacement of an element in the compound byanother element from the same family will often give the correct formula for anothercompound.

For example, beryllium chloride has the formula BeCl2. What is the formula for magnesium chloride? Sincemagnesium is in the same family as beryllium, the formula will be MgCl2.

Note also that when chlorine is combined with chlorine, it is shown as Cl2. You should recall thatthere are seven diatomic elements. All of the halogens (F2, Cl2, Br2, I2) and elemental gases (H2, N2, O2)do this. Two atoms are bonded together when these elements exist in a free form. You will soon learnwhy this is.

Table 11.11Chorine Compounds of the “A” Family Elements

(1) (2) (13) (14) (15) (16) (17) (18)1A 2A 3A 4A 5A 6A 7A 8A

LiCl BeCl2 BCl3 CCl4 NCl3 OCl2 FCl - - - -NaCl MgCl2 AlCl3 SiCl4 PCl3 SCl2 Cl-Cl - - - -KCl CaCl2 (Cl-Cl can also be written as Cl2)

Note: Table 11.11 will make more sense to you when you learn more about atomic structure in the next few chapters.

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Problem 7. Bromine belongs to the same family as chlorine. Based on the principle given above,complete Table 11.12 below by writing formulas for bromide compounds of the "A" family elements which"belong" in the blank spaces. A few have already been written for you.

1A

2A 3A 4A 5A 6A 7A

8A

LiBr

CaBr2

CBr4

Table 11.12Bromine Compounds of the "A" Family Elements

Zinc, cadmium, and mercury all belong to the same family in the periodic table. The formulas and names ofcertain zinc compounds are: Formula Name

ZnO zinc oxide Zn(NO3)2 zinc nitrate ZnS zinc sulfide ZnCl2 zinc chloride

Problem 8. Since cadmium and mercury are in the same family as zinc, you should be able to predict theformulas of similar compounds formed with mercury and cadmium. Do this below.

a. cadmium oxide __________ mercury (II) oxide __________

b. cadmium nitrate __________ mercury (II) nitrate __________

c. cadmium sulfide __________ mercury (II) sulfide __________

d. cadmium chloride __________ mercury (II) chloride __________

The periodic table gives us a useful method for predicting chemical formulas. You will have abetter understanding of why this all works after you study atomic structure in the upcoming chapter.(Since atomic structure was unknown in his time, Mendeleev never did understand why this predictingpower of the periodic table works!)

The periodic table, however, is not flawless as a predictor. For example, if carbon monoxide, CO,were used as a reference, you would predict that the compound carbon sulfide, CS, would also exist - itdoes not. (Oxygen and sulfur are in the same family.) Even though your predictions may not be 100%correct, this general principle is very useful. It allows you to make predictions based on a small amount ofinformation.

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Problem 9. In this problem, you are given several chemical names and formulas. Predict the formula ofthe compound asked for. In order for predictions to be possible, however, elements must be members ofthe same family on the periodic table. In the first example, Na and Cl are in the families 1A and 7A. K andBr are also in families 1A and 7A. This means they will form compounds with similar chemical formulas.

Given: Predict the formula for:

a. sodium chloride: NaCl potassium bromide: ____KBr____

b. magnesium chloride: MgCl2 calcium iodide: ___________

c. calcium sulfate: CaSO4 magnesium sulfate: ___________

d. barium hydroxide: Ba(OH)2 calcium hydroxide: ___________

e. chlorine: Cl2 iodine: ___________

f. silver nitrate: AgNO3 copper (I) nitrate: ___________

g. carbon dioxide: CO2 silicon dioxide: ___________

h. beryllium sulfide: BeS strontium oxide: ___________

Problem 10. For more practice, underline the substance in the right column that would be more likelyto exist after comparing the choices to a known substance in the left column.

Known Substance Possible Substances

a. SF6 ClF6 or TeF6 b. PCl5 AsCl5 or GeCl4c. WCl6 WS6 or WBr6d. Na2SiF6 K2SiCl6 or NaPF6

e. K2SO4 Mg2SO4 or Na2SO4

f. Na2CO3 K2CO3 or Ca2CO3

Problem 11. Given the chemical formulas below, predict the likely chemical formula for the followingsubstances.

Given Predict the formula for:

a. CaCO3 a carbonate of magnesium _____________(calcium carbonate)

b. Na3PO4 a phosphate of potassium _____________(sodium phosphate)

c. CO2 a silicon-oxygen compound _____________(carbon dioxide)

d. Na2CrO4 a chromate of potassium _____________(sodium chromate)

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e. InPO4 a phosphate of aluminum _____________(indium phosphate)

f. PbCl2 a bromide of silicon _____________(lead (II) chloride)

g. MgCrO4 a calcium chromate compound _____________(magnesium chromate)

h. K2SO4 a sulfate of sodium _____________(potassium sulfate)

i. SiF4 a chloride of carbon _____________(silicon tetrafluoride)

j. Na3PO4 a lithium phosphate compound _____________(sodium phosphate)

ACTIVITY 11.10 Reactions of Sodium and Potassium With Water

-> -> -> -> TEACHER DEMONSTRATION <- <- <- <-

Hydrogen is very explosive and will ignite when heated sufficiently. The reactions between thealkali metals and water (see problem 6) occur very rapidly and release a great deal of heat. In fact, the heatgiven off by some of these reactions is enough to ignite the gas. These metals are often stored in oil orkerosene to prevent them from reacting with air or water. Even the moisture in air may be sufficient tocause an explosion under certain circumstances.

The point is that these metals all behave in a similar way. There is, however, a difference in the"activity" of these metals in water. Some react more vigorously than others. To demonstrate this, yourteacher will demonstrate for you the reaction of sodium and potassium with water. He will put somephenolphthalein in the water and then add the metal. The phenolphthalein will turn pink if a hydroxidecompound forms (such as NaOH or KOH). A fizzing action will illustrate the production of the hydrogen.Stand back! The equations for the reactions are:

2 Na(s) + 2 HOH(l) ----> 2 NaOH(aq) + H2(g)

2 K(s) + 2 HOH(l) ----> 2 KOH(aq) + H2(g)

1. How were the reactions of sodium and potassium with water similar?

______________________________________________________________________________

______________________________________________________________________________

2. How were the two reactions different? _______________________________________________

______________________________________________________________________________

11-19 ©1997, A.J. Girondi


Knowing how to use the periodic table is a must for students of chemistry. To check yourfamiliarity with the use of information contained in the periodic table, review the learning outcomes below.When you are sure you have mastered an outcome, check the box in front of it. Arrange to take the examon Chapter 11, and go on to Chapter 12.

_____1. Describe the general organization of the modern periodic table based on Moseley's periodic law.

_____2. Explain what is meant by periods (also called rows or series) and groups (also called families).

_____3. Describe the general properties of metals and nonmetals, and show their location on the periodic table.

_____4. Indicate the position on the periodic table of the alkali metals, the alkaline-earth metals, the halogens, the noble gases, the transition metals, the lanthanide series, and the actinide series.

_____5. Predict the physical properties of an element given those of its neighbors and "family members"on the periodic table.

_____6. Predict the chemical formulas of compounds given those of similar compounds.

11-20 ©1997, A.J. Girondi


Questions:

{1} between 22 and 28; {2} UCl2; {3} metals; {4} metals; {5} double replacement; {6} between thealkaline-earth metals and the transition metals; {7} hydrogen is less dense than air;{8} 2 H2(g) + O2(g) ----> 2 H2O(g); {9} one; {10} four

Problems:

Table 11.3 A Table of Hypothetical Elements

F2Cl3

F 5

- - - -

J 15

ZCl3

Z 28

SCl2

S 41

A2Cl3

22

AP2Cl3

P 39

RCl2

R 7

MMCl3

9

DCl3

D 44

- - - -

B 36

- - - -

T 52

HCl4

H 11

31

GCl4

GNCl4

N 49

UCl2

Ubetween 22 and 28

1.

2. Cobalt (#27) and Nickel (#28); Tellurium (#52) and Iodine (#53); and there are more.3. a. increase; b. whole; c. hydrogen; d. (varies)4. a. ununnilium, Uun; b. unununium, Uuu5. a. Na, Li, K, Rb, Cs, Fr; b. Be, Mg, Ca, Sr, Ba, Ra; c. F, Cl, Br, I, At; d. He, Ne, Ar, Kr, Xe, Rn;

e. boron, aluminum, gallium, indium, thallium; f. nitrogen, phosphorus, arsenic, antimony, bismuth6. a. (given)

b. 2 Rb + 2 HOH ----> 2 RbOH + H2

c. 2 Cs + 2 HOH ----> 2 CsOH + H2

7. LiBr, BeBr2, BBr3, CBr4, NBr3, OBr2, FBr, - - -;NaBr, MgBr2, AlBr3, SiBr4, PBr3, SBr2, ClBr, - - -;KBr, CaBr2;

8. a. CdO, HgO; b. Cd(NO3)2, Hg(NO3)2; c. CdS, HgS; d. CdCl2, HgCl29. a. KBr; b. CaI2; c. MgSO4; d. Ca(OH)2; e. I2; f. CuNO3; g. SiO2; h. SrO10. a. TeF6; b. AsCl5; c. WBr6; d. K2SiCl6; e. Na2SO4; f. K2CO3

11. a. MgCO3; b. K3PO4; c. SiO2; d. K2CrO4; e. AlPO4; f. SiBr2; g. CaCrO4; h. Na2SO4; i. CCl4;j. Li3PO4

11-21 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 12

MODERN VIEWOF ATOMICSTRUCTURE

(PART 1)

Atomic Structure & SpectraNuclear NotationOrbital Notation

Configuration Notation12-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


12-2 ©1997, A.J. Girondi

SECTION 12.1 Early Discoveries About the Atom

You have already learned about the concept of atoms and how they can combine to formcompounds. You have also seen that each element has its own characteristic set of properties which helpto distinguish it from all other elements. In this chapter we will study the structure of atoms and the lawsgoverning the behavior of the particles that make up atoms. This knowledge will lead to an explanation ofthe properties of the elements and of their tendencies to form compounds.

John Dalton regarded the atom as a particle with no internal parts. He believed an atom to be thesmallest possible particle. However, certain experiments were being performed which gave definiteindications that Dalton's view was not correct and that there was some sort of internal structure to the atom.It became apparent that atoms consisted of particles that had electrical charges and that these particlesinteracted according to the laws of electromagnetism. Charged particles carry either a positive (+) ornegative (-) charge. We call two negative charges or two positive charges "like" charges. The laws ofelectromagnetism state that two like charges {1}_______________ each other, while unlike charges,{2}_____________ each other.

Experiments performed during the late 1800's and early 1900's by chemists and physicists madeit clear that atoms could, indeed, be broken into smaller parts, contrary to the ideas of {3}_____________.In 1897, J.J. Thomson discovered that atoms could be "taken apart" when he studied the effects ofelectrical discharge on atoms of various gases. In his experiments he concluded that atoms were comingapart by yielding a stream of negatively charged particles with very small masses (compared to the massesof the atoms). These small negative particles became known as electrons. Thomson is credited with thediscovery of electrons which were present in the atoms of all of the different gases that he examined.

Another scientist, Ernest Rutherford, and his students performed experiments in England duringthe first decade of the 20th century in an attempt to determine the size of atoms. In 1906, Rutherford hadhis students direct a beam of positively-charged subatomic "alpha" particles at a very thin sheet of goldmetal. It was known as the alpha scattering experiment." Since they believed that matter was mostlyempty space, they expected the particles to pass through the thin sheet unhindered. To their surprise,they found that a small fraction of the particles bounced right back! This led Rutherford to believe that inthe center of the atom was a small but very dense "nucleus" with which some of the alpha particles musthave collided. He concluded that most of the mass of the atom was contained in the {4}__________________.He also concluded that the nucleus was {5}______________ charged since it repelled the positively-charged alpha particles. He later said that "It was quite the most incredible event that has ever happenedto me in my life. It was almost as if you fired a 15- inch shell into a piece of tissue paper and it came backand hit you."

Rutherford realized that electrons were located at a considerable distance from the nucleus. If thiswere an accurate description of an atom and we could inflate the nucleus of a hydrogen atom to the size ofa basketball, the electron would orbit this "basketball" nucleus at a distance of more than 15 miles away!Visualizing the atom like this enables you to realize that most of the atom is, indeed, nothing more thanempty space! So much for the athlete who thinks he is "solid muscle!"

With regard to the nucleus itself, it became obvious to scientists that the nucleus was composedof small particles. One of the particles in the nucleus is the proton. In 1914, Rutherford was given creditfor discovering protons. A proton carries a charge equal to the charge of an electron, but opposite incharacter. The electron carries a charge of -1 while the proton carries a charge of {6}_______. In thenucleus of a neutral atom (that is, an atom with no overall electric charge), there must be an equal numberof protons to balance the charges carried by the electrons. Unlike the electron, the proton is a particle witha relatively large mass, in atomic terms. The proton has a mass equal to 1,836 times that of the electron!Using the common unit of the gram to measure masses, the electron has a mass equal to 9.1 X 10-28

grams, and the proton has a mass equal to 1.673 X 10-24 grams.

12-3 ©1997, A.J. Girondi

A second component of the nucleus was discovered in1932 by another Englishman, James Chadwick. This particlebecame known as the neutron. A neutron is an electrically neutralparticle, meaning that it carries no electric charge. The neutron wasvery difficult to discover. Because it has no charge of its own, it isneither attracted to nor repelled by an electrical charge. A neutronhas a mass slightly larger than that of the proton, equal to 1.675 X10 -24 grams. The presence of this particle accounted for theobserved masses of atoms, which were found to be greater thanthat predicted if only protons were present in the nucleus. Figure12.1 and the accompanying table present an overall summary of thelocations and properties of the components of an atom.

Each element differs from all others in that atoms of eachelement contain a specific number of electrons, protons, and{7}_______________. Indeed, the number of protons in thenucleus determines the actual identity of an element. Determiningthe number of electrons, protons, and neutrons in any givenelement is a relatively simple process. The atomic number of anelement is equal to the number of protons found in the nucleus.The element with atomic number 19, potassium, has 19 protons.For atoms to be neutral, they must have equal numbers of positive(protons) and negative (electrons) charges. This means thatpotassium must have 19 protons and {8}________ electrons.

Dalton’s Atomic Theory (1803)

Thomson Discovers the Electron (1897)

Rutherford Discovers the Nucleus (1906)

Bohr Develops the PlanetaryModel of the Atom

(1913)

Rutherford Discovers the Proton (1914)

Chadwick Discovers the Neutron (1932)

Table 12.1Location and Properties of Subatomic Particles

Particle Charge Comparative Mass Location

electron -1 1/1836 outside nucleus

proton +1 1 inside nucleus

neutron 0 1 inside nucleus

If an element has 10 protons in its nucleus, how many positive charges does it have?{9}________.

If an atom with 10 protons is neutral, how many electrons must it have?{10}__________

The atomic number of oxygen is 8. How many protons does it have?{11}__________

How many electrons does oxygen have?{12}___________

SECTION 12.2 Mass Number, Atomic Mass, and Isotopes

As you learned in an earlier chapter, the atomic masses of the elements that appear on theperiodic table are really average masses. We will use the element hydrogen as an example. Mosthydrogen atoms (99.85% of them) have no neutrons. A few hydrogen atoms (0.15% of them) have oneneutron. These two slightly different kinds of hydrogen are often referred to as light and heavy hydrogen.

12-4 ©1997, A.J. Girondi

This situation is fairly typical of most elements. Elements exist in nature as a mixture of isotopes.Isotopes are atoms that have the same number of electrons and protons, but different numbers of{13}______________. As a result, {14}________________ of an element are atoms that have differentmasses. The atomic masses that appear on the periodic table are merely "weighted averages" of themasses of all the naturally-occurring isotopes. Weighted averages are determined by including therelative amounts of each isotope found in nature when calculating the average mass.

On the periodic table the atomic mass of carbon is 12.011. Actually, no carbon atoms have thismass. Some have more mass than this, and some have less. The 12.011 is the {15}_____________ massof carbon atoms in nature. The number of protons in the nuclei of atoms of a specific element is constant,but the number of neutrons can vary. Therefore, the mass numbers (the sum of the protons andneutrons) of atoms of an element can vary. Atoms of a particular isotope of an element are identified bywriting the mass number of the isotope after the name of the atom. For example, the two naturally-occurring isotopes of hydrogen are hydrogen-1 and hydrogen-2. The most common form of carbon iscarbon-12, while less common isotopes include carbon-13 and carbon-14. The difference in these massnumbers is a result of differences in the number of {16}_______________ in these atoms. The sum of the{17}________________ and {18}_________________ in the nucleus of a particular isotope of an atom isknown as its {19}_______________________.

While the atomic mass of an atom of a particular isotope of an element is not found on the periodictable, it is very close to the sum of the masses of the neutrons and protons in the atom.{20}______________ have such small masses that they are not even considered when atomic masses arecalculated. This is the case even if an atom contains 100 or more electrons! The mass of a proton isexpressed as 1 amu (atomic mass unit). Since the proton and the neutron have almost the same mass,the approximate mass of a neutron must also be 1 amu. (Actually, it is a tiny bit more than 1.) So, what isthe mass of an isotope such as oxygen-16?

Oxygen-16 is an atom of a particular kind of oxygen containing a total of 16 protons and neutrons.Remember, the sum of the protons and neutrons in the nucleus of an atom is called its mass number.Thus, the mass number of this atom is 16. Since each proton and each neutron has a mass of about 1amu, you would expect that an atom of oxygen-16 would have a mass of about {21}_________amu.However, the actual mass of oxygen-16 atoms is 15.99491 amu. Why this tiny difference? This differenceis a result of the fact that when protons and neutrons combine to form a nucleus, some matter is convertedinto energy. This is a kind of nuclear reaction. Therefore, whenever a nucleus forms, a bit of matter will belost. Thus, chlorine-37 has an atomic mass of 36.96590 amu.

You can determine the mass number of a specific isotope of an element if you round off its atomicmass to the closest integer (whole number). So, to get the mass number of chlorine-37 just round36.96590 to {22}_______. This gives you the sum of the protons and neutrons in the nucleus of thisspecific form (isotope) of chlorine.

Since the atomic mass of an element given on the periodic table is an average, rounding it off willusually give you the mass number of the most common isotope of the element. For example, on theperiodic table the average mass given for carbon is {23}____________. Rounding will yield 12, which isthe mass number of the most common isotope of carbon, which is carbon-12. Using this method, what ifthe mass number of the most common isotope of element #12, magnesium? {24}___________. What isthe name of the most common isotope of element #15, phosphorus?{25}_______________________.

Since the mass number of an atom is equal to the total number of protons and neutrons itcontains, you can determine the number of neutrons in an atom by subtracting the atomic number(number of protons) from the mass number (number of protons + neutrons).

atomic number = number of protons mass number = atomic mass rounded to closest integer neutrons = (mass number — atomic number)

12-5 ©1997, A.J. Girondi

The "nuclear notation" of an atom includes both the atomic number and the mass number of theatom. The atom's symbol is written with its mass number expressed as a superscript at the upper-leftlocation and the atomic number written as a subscript at the lower-left location. For example:

Nuclear notation of sodium -23: 1123 Na means mass number = 23 and atomic number = 11

Here are a couple of others:

hydrogen-1: 11H means 1 proton and 0 neutrons in the nucleus

(atomic number = 1, and mass number = 1)

hydrogen-1: 12H means 1 proton and 1 neutron in the nucleus

(atomic number =1, mass number = 2)

Problem 1. For practice, determine the number of electrons, protons, and neutrons in the atoms ofaluminum and cesium listed below.

a. aluminum–27;

b. cesium–133;

1327Al protons: electrons: neutrons :

55133Cs protons: electrons: neutrons :

Problem 2. The element uranium has three different isotopes. In the blanks below, write the number ofelectrons, protons, and neutrons for each isotope of uranium.

a. 92234U protons: electrons: neutrons :

b. 92235U protons: electrons: neutrons :

c. 92238U protons: electrons: neutrons :

Problem 3. Supply the missing information in the table below.

NuclearIsotope Notation At.No. Mass No. No. e's No. p's No. n's

a. aluminum-27 _____ _______ _____ _____ _____

b. bismuth-209 _____ _______ _____ _____ _____

c. calcium-40 _____ _______ _____ _____ _____

d. copper-64 _____ _______ _____ _____ _____

e. ____________ __2__ ___4___ _____ _____ _____

f. ____________ _____ __207__ __82_ _____ _____

g. ____________ __8__ _______ _____ _____ __8__

h. ____________ _____ _______ _____ __50_ __69_

i. ____________ _____ _______ __30_ _____ __36_

12-6 ©1997, A.J. Girondi

Here is an example which shows how atomic masses on the periodic table are calculated:

Sample Problem: Oxygen exists in nature in three forms. Oxygen-16 (99.759%), oxygen-17(0.037%), and oxygen-18 (0.204%). Oxygen-16 atoms have a mass of 15.99491 amu; oxygen-17 atomshave a mass of 16.99914 amu, and the mass of oxygen-18 atoms is 17.99916 amu. What is the averagemass of oxygen atoms in nature?

To solve this problem, assume that you have a 100 atom sample of oxygen. In that case, sincepercent means parts per hundred, 99.759 atoms in the sample would be oxygen-16 (even though atomscannot exist in fractional parts); 0.037 atoms would be oxygen-17, and 0.204 atoms would be oxygen-18.

The total mass of the oxygen-16 atoms in the sample is:

99.759 atoms X

15.99491 amu

1 atom = 1595.64 amu


0.037 atom X

16.99914 amu

1 atom = 0.63 amu


0.204 atom X

17.99916 amu

1 atom = 3.67 amu

Atomic masses such as these are determined by experiment.

The total mass of all 100 atoms in the sample would be:

1595.64 amu + 0.63 amu + 3.67 amu = 1599.94 amu.

To get the average mass, we divide by the number of atoms in the sample:

1599.94 amu

100 = 15.9994 amu

Does this answer agree with the atomic mass of oxygen on the periodic table?____________

In the sample problem above, the mass of an atom of oxygen-16 is listed as 15.99491 amu. Assuming

that protons and neutrons weigh 1 amu each, why doesn't oxygen-16 weigh 16 amu? {26}____________

______________________________________________________________________________

Problem 4. In nature, copper is found to exist in two forms: copper-63 and copper-65. Copper-63atoms have a mass of 62.93 amu, while copper-65 atoms have a mass of 64.93 amu. Naturally-occurringcopper contains 69.40% copper-63. Calculate the atomic mass of naturally-occurring copper atoms.Show all of your work.

12-7 ©1997, A.J. Girondi

Does your answer to Problem 4 compare favorably to the atomic mass of copper appearing on the periodictable? _________

SECTION 12.3 The Spectrum of Visible Light

Now that you know something about the makeup of atoms in terms of electrons, protons, andneutrons, we will take time to examine some of the experimental evidence that has contributed to ourpresent understanding of atomic structure. In particular, we will be studying the light emitted by atomswhen we add energy to them. Such light is called the "emission spectrum" of an atom. To understand thesignificance of emission spectra from atoms, we need to learn more about light spectra in general.

In Figure 12.1, is a box that represents a continuous light spectrum. The numbers above thenames of the colors represent the wavelengths of light in units called angstroms. Angstroms are denotedby the symbol Å which is a capital A with a small circle above it. An angstrom is a unit of length equal to 1 X10 -10 m (which is very small). The wavelengths in Figure 12.1 range from 3900 Å to 7700 Å and comprisethe continuous spectrum of visible light. These are the wavelengths of light that our eyes can see. Whenall of the colors (wavelengths) are present, they appear to us as white light. The numbers below thespectrum in Table 12.1 represent nanometers (nm). A nanometer is 1 X 10-9 meter ; therefore, ananometer is ten times larger than an Angstrom. Both units are commonly used to measure wavelengthsof light.

3900 Å 4400 Å 5100 Å 5600 Å 6300 Å 7700 Å

violet blues greens yellow reds

390 nm 440 nm 510 nm 560 nm 630 nm 770 nm

Figure 12.1 Components of a “Continuous” Spectrum of Visible White Light

<----- shorter wavelengths (higher energy) (lower energy) longer wavelengths ----->

A spectroscope is a laboratory instrument that separates the continuous spectrum of white lightinto its component wavelengths (colors). The spectrum is said to be continuous because all of thewavelengths from violet to red are present in white light.

Light is a form of energy. The violet end of the spectrum represents light of higher energy thanlight in the red region, which represents visible light of the lowest energy. With this in mind, looking againat Figure 12.1, you can conclude that as the wavelength of the light gets {27}_______________, theenergy content of the light gets smaller.

What does all this have to do with atomic structure? We shall soon see. When atoms absorbenergy we describe them as being "excited." Excited atoms are capable of emitting light. However, atomsdo not just give off light all by themselves. Something has to be done to them, first. Fluorescent lights donot emit light until an electrical current is passed through them. The sun emits light through the process ofnuclear fusion which causes hydrogen atoms to combine to form helium. Electricity and nuclear fusionboth involve energy changes in a system. When energy is added to a system (that is, through fusion orpassing an electrical current through it), the atoms absorb that added energy. This "overload" of energycauses the electrons in the atoms to become very unstable. By unstable we mean that at least some ofthe electrons in the atoms have too much energy and they "want" to change their position or their motionin some way in order to get rid of the excess energy and, thereby, become more stable. Because excitedelectrons are so unstable, they will give off this added energy at the first opportunity. This energy that isemitted by the excited electrons may at least partly be in the form of visible light.

12-8 ©1997, A.J. Girondi

In fluorescent bulbs, a stream of electrons flows between the metal ends of a glass tube whichcontains a mixture of argon gas and mercury vapor. The electron stream excites the {28}______________of the mercury atoms which then emit energy as invisible ultraviolet rays. These rays excite the atoms inthe coating of the glass tube which then emit visible light. The light emitted by specific atoms (or moreaccurately, by excited electrons in specific atoms) does not compose a continuous spectrum (containingall wavelengths). The light emitted by excited electrons of a particular atom is composed of certain specificwavelengths and is called a bright-line spectrum, since only certain wavelengths of light appear as coloredlines. Each element emits a different characteristic set of colored bright lines (wavelengths). No twoelements will emit the same set of wavelengths of light. Atoms of different elements have{29}____________________ bright-line spectra. Figures12.2 through 12.6, show the visible portion ofthe emission spectra of 5 different elements. Each black line represents a wavelength of light which isemitted by atoms of that element. If this document were printed in color, the lines would have a differentcolors.

ACTIVITY 12.4 The Emission Spectra of Elements

(This activity may be done by all lab groups in your class at the same time . Ask your instructor.)

Procedure - Part A. Now you are going to actually look at the light emitted by excited electrons in atomsof several metallic elements. A dark or dim room should be available for this activity.

1. Obtain a burner and a platinum or nichrome wire with a small loop at the end. Pour some of each of thesix solutions provided into separate wells of a dropping plate. Each solution contains a different metal ionincluding Sr2+, Li1+, Ba2+, Ca2+, Cu,2+ and Na1+. Pour some dilute hydrochloric acid (HCl) into a seventhwell on the plate. Make note of which solution is in each well.

2. Dip the loop of the wire into the lithium solution and then touch it to the edge of the flame of the burner.This should impart some color to the flame. The burst of color may last only a fraction of a second. Forsome solutions it may last longer. What color of light does the lithium produce?{30}________________Clean the wire well by running tap water over it and then dipping it into the hydrochloric acid. Follow this byrinsing the wire with tap water again. Touch the clean wire to the flame. If it is clean, the color will havedisappeared. If not, repeat the cleaning procedure.

3. Repeat the procedure above with the other solutions, noting the color emitted by each solution (testthe sodium solution last). The metal solutions which may include copper, strontium, calcium, barium, orothers. Note your observations below. Since sodium is not easy to remove from the wire, clean the loopwell several times before returning it to the materials shelf. The color which you saw emitted by eachmetallic element in the flame was actually a blend of various wavelengths of light. These wavelengths oflight compose what is known as the bright-line spectrum or emission spectrum of each element.

Metal in Solution Color Emitted

_________________ _______________________

_________________ _______________________

_________________ _______________________

_________________ _______________________

_________________ _______________________

sodium _______________________

4. Discard the solutions and rinse the dropping plate. Return the loop and all materials to the properplace.

Scientists have made quantitative observations of bright-line spectra by measuring thewavelengths of the light waves emitted by elements. This is how we got the information revealed inFigures 12.2 through 12.6. Other references are also available that can provide you with the spectra of

12-9 ©1997, A.J. Girondi

other elements. There may be a "spectrum chart" available in your classroom. Use the chart in yourclassroom or Figures 12.2 through 12.6 to complete problem 5 below.

Problem 5. Study the emission spectrum of sodium vapor in Figure 12.6, and provide a rough estimateof the seven wavelengths (in nm) emitted in the spaces provided below. Each line in Figure 12.6represents a different wavelength of light being emitted.

a. __________ b. ___________ c. __________ d. __________

e. __________ f. ___________ g. __________

If you were to actually observe the emission spectrum of sodium vapor through a spectroscope, youwould probably see only one or two bright lines instead of seven. The reason for this is that the yellowlines located near 570 nm are so bright, that they "wash out" the other lines. They are so bright, in fact,that sodium vapor lamps are commonly used as street lights in areas where strong illumination is neededsuch as high-crime neighborhoods.

Procedure - Part B.

Every element emits characteristic wavelengths, and no twoelements emit the same set of wavelengths. In this way, spectra arelike fingerprints. By studying atomic spectra, we can identifyelements. This is how scientists know which elements are presentin the sun. Wavelengths of the sun's light have been measured,separated, and matched with spectra of known elements to identifythe contents of the sun. Your teacher may have equipment which isdesigned to "excite" atoms of gases, so that their bright-line spectracan be observed using small hand-held spectroscopes. Beforegoing on, ask your teacher if this equipment is available in your lab. Ifso, your teacher will assist you in its use. Caution: high voltageequipment will be used. (See Figure 12.7)

Most light is a blend of many wavelengths. Thespectroscope you will use is a device that contains a plastic "grating"which separates the wavelengths emitted by a light source so thatyou can observe them separately. List below the gases (or vapors)that you observed using this high-voltage equipment. Figure 12.7

Spectra Observed:_______________________________________________________________

_____________________________________________________________________________

Ask your teacher if any kinds of "vapor" lights are used to illuminate any of the rooms in yourschool. If so, use a spectroscope to examine the light emitted by them, and see if you can identify theelement in the lights by comparing the spectra they emit to those in Figures 12.2 through 12.6 or thosefound on a spectrum chart in your classroom. If you do have such lights in your school, go to the properlocation and observe the light using your hand-held spectroscope. Supply the information requestedbelow.

Location of lights in school:__________________________.

Symbol of element revealed by emission spectrum:________.

Check with your instructor to see if you identified the unknown element(s) correctly.

12-10 ©1997, A.J. Girondi

Figure 12.2 Emission Spectrum of Hydrogen

4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å

400 nm 450 nm 500 nm 550 nm 600 nm 650 nm

Figure 12.3 Emission Spectrum of Mercury

4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å

400 nm 450 nm 500 nm 550 nm 600 nm 650 nm

Figure 12.4 Emission Spectrum of Calcium

4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å

400 nm 450 nm 500 nm 550 nm 600 nm 650 nm

Figure 12.5 Emission Spectrum of Neon

4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å

400 nm 450 nm 500 nm 550 nm 600 nm 650 nm

Figure 12.6 Emission Spectrum of Sodium

4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å

400 nm 450 nm 500 nm 550 nm 600 nm 650 nm

SECTION 12.5 The Bohr Model of the Atom

12-11 ©1997, A.J. Girondi

Niels Bohr developed a theory to account for the locationof electrons around the nucleus of an atom. He believed his theorywould explain the bright-line spectra emitted by excited atoms. Heproposed that electrons followed specific paths or orbits aroundthe nucleus. These paths or energy levels, as they are also called,are numbered starting with the lowest one (closest to the nucleus)as 1, the next farther from the nucleus as 2, the next as 3, and soforth. Figure 12.8 illustrates 4 of the energy levels of a n atom.

Bohr's model resembled a planetary system like our solarsystem in which he suggested that the electrons revolve aroundthe nucleus. The first energy level, nearest the nucleus, isrepresented as number 1. Each level thereafter is increased byone. A total of 7 energy levels are needed to explain the structureof all of the elements.

1 2 3 4

Figure 12.8 Energy Levels

The orbits around the nucleus are called energy levels because there are different and veryspecific energies associated with each level. Bohr knew that energy was being added to an atom when itwas being heated or when an electrical current was passed through an element. This extra energy has togo somewhere. The added energy is absorbed by the electrons that are in the outermost orbit (farthestfrom the nucleus). Since there are specific amounts of energy associated with each energy level, theelectron that absorbs all of this extra energy can no longer stay in the orbit (energy level) in which itnormally belongs which is its "ground state." Instead, it will move to another energy level. It is now in an"excited state" and is what we earlier referred to as an "excited" electron.

This electron is not doomed to spend the rest of its time in this higher energy level. Indeed, theexcited electron is now very unstable. Because this is an unstable state, the electron will soon return to itsground state. This is where the atomic spectra enters into the picture. The energy that the electron emitswhen it returns to its ground state is in the form of light and heat. Figure 12.9 gives a general picture ofthis process. In "A" , the electron is excited and jumps to a new energy level that is further from thenucleus. As the electron falls back to its ground state in "B", energy is given off in the form of light.

Figure 12.9 Emission of Light from an Excited Electron

LIGHTEMITTED

A B

As electrons move to lower energy levels, the energy they emit is given off in a "burst" or quantityof energy with a well-defined wavelength. The quantities of emitted energy are called quanta. Bohr'stheory of the atom gave birth to the quantum theory, a name that reflects the notion that atoms mustabsorb and emit energy in specific amounts. Therefore, we say that the energy is "quantized." Aquantum of energy can be defined as the amount of energy needed to move an electron from one energylevel to the next higher one. Similarly, it can be defined as the amount of energy emitted when an

12-12 ©1997, A.J. Girondi

electron moves from its present energy level to a lower one.

In his theory, Bohr proposed that electrons were only "allowed" to exist at certain distances fromthe nucleus. These distances became his energy levels. He believed that electrons were not "allowed"to exist between these levels. Although the reasons for this behavior were unclear, the idea did explainthe existence of bright-line spectra. The spectrum of hydrogen contains a specific set of bright lines. Ofthese, only 3 or 4 are clearly visible to the naked eye. (See Figure 12.2) So Bohr asked, why only this setof lines? Why does the excited hydrogen electron emit only this specific set of wavelengths?

According to Bohr, the fact that there was always the same set of bright lines in the hydrogenspectrum was evidence that only certain energy changes were possible for the hydrogen electron. Theelectron could only make certain "jumps" and, therefore, could only emit certain wavelengths of light. Thenumber of lines was limited because there were only a few "excited" energy levels to which the electronwas "allowed" to move. He explained that the electron was - for whatever reason - not "permitted" to existbetween these levels.

Since every element has a specific number of electrons, different wavelengths of light will beemitted by excited electrons of atoms of different elements. Even without a spectroscope you can seethat the glow of a neon sign is quite different from the fluorescent lights in your classroom. You probablydid not realize that what you were seeing was the result of excited electrons at work.

To better understand how an electron behaves as it returns to its stable ground state, considerhow you would jump down a staircase. Just as an electron can only stop at certain energy levels, you canonly stop at certain levels (steps). The electron cannot exist between energy levels, and you cannot stopbetween the steps.

To get down to the bottom of the staircase,you could jump all the way down at once; or onestep at a time; or one step followed by a two-stepjump; or perhaps a two-step jump followed by onestep, and then by a three-step jump. A variety ofdifferent jumps is possible, each yielding a differentamount of energy.

Likewise, suppose an electron jumps out tolevel 4 in Figure 12.10. There are ten different sizesof jumps possible as it returns to its ground state.So how many different wavelengths of light could beemitted by this electron?{31}__________ A samplecontaining many excited atoms of this elementwould emit all possible wavelengths.

4

3

2

1

Ground State

Figure 12.10 Energy Level Diagram

SECTION 12.6 The Theory of Quantum Mechanics

Bohr's model worked well in explaining the emission spectra of hydrogen. But, it really did notwork well for atoms with more electrons in them. It even failed to explain some of the complexities of thespectrum of hydrogen atoms. In addition, it could not explain why electrons are not pulled into thenucleus of the atom (which would cause the atom to collapse).

We know, today, that electrons do not travel around the nucleus as planets do around the sun.The failure of Bohr's model led to the development of the theory of quantum mechanics. Quantummechanics is a very complex theory. It involves describing the approximate location of any electron in anyatom. The theory provides an explanation for the behavior of electrons in atoms. The details of quantummechanics, while beyond the scope of this course, will be discussed briefly later in this chapter. This will

12-13 ©1997, A.J. Girondi

provide you with an appreciation and a general knowledge of the principles of the theory.

In these next sections, you will see how each of the energy levels is divided into sublevels. Wewill concern ourselves with how the electrons distribute themselves among the energy levels andsublevels of atoms.

Each energy level can hold a maximum number of electrons. The number of electrons in anygiven energy level is limited to 2n2. This expression means 2(n2), not (2n)2! For example, the energy level3 can be occupied by 2(3)2, or 18, electrons at the most.

Problem 6. Calculate the maximum number of electrons that could occupy the first four energy levels in

Energy Level Maximum No. Electrons (2n2)

1 __________

2 __________

3 __________

4 __________

Energy levels are sometimes called "shells" in which case each one is designated by a letter, Kthrough Q (see Figure 12.13). You may see this notation in your future studies in chemistry, so it isimportant that you know something about it.

Each main energy level consists of one or more sublevels. The number of sublevels is equal tothe number of the energy level. Therefore, the third energy level has three sublevels, and the fifthenergy level has five sublevels. Which energy level has four sublevels?{32}_________. Which energylevel has two sublevels? {33}__________. The energy level diagram can now be redrawn. The mainenergy levels and the sublevels within each energy level are labeled in Figure 12.11. Each sublevel isdesignated by a small letter (s,p,d,f).

By overlapping we mean that a higher energy level begins before a lower one ends. Notice onfigure 12.11 how the 4th energy level begins before the third is full. Notice the overlapping of the mainenergy levels that occurs as you go beyond the third energy level. This means that a higher energy levelbegins before a lower one ends. Clearly, atoms are more complicated than a staircase. Certainly, the thirdstep on a staircase cannot begin before the second step ends! However, you must keep in mind thatenergy levels are not "physical things," whereas staircases are real physical things. Energy levels areconceptual things – regions of space. The complexity of overlapping energy levels becomes morenoticeable as you progress from the third energy level to the seventh.

The first sublevel of each energy is designated as an "s" sublevel. This is followed by a "p"sublevel, the a "d", then an "f", depending upon how many electrons there are.

Note Figure 12.13, in which the energy sublevels s, p, d, f are each shown with one or morecircles. These circles represent orbitals. The number of circles represents the number of orbitals in aparticular sublevel. At this point, you can think of an orbital as a region of space where up to two electronscan coexist. We think of energy levels as a staircase, and electrons can exist only in the steps, notbetween them, right? We think of orbitals as "houses" where electrons live. Each of these orbital"houses" can contain a maximum of two electrons, and – as you can see – there are different numbers oforbital "houses" on different steps.

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6d

6p

6s

7s5f

5d

5p

5s

4f

4d

4p

4s3d

3p

3s

2p

2s

1s

Nucleus

1st Energy Level

2nd Energy Level

3rd Energy Level

4th Energy Level

5th Energy Level

6th Energy Level

7th Energy Level

Figure 12.11 Overlapping of Energy Levels

We will learn more about orbitals later in this chapter, and we willchange the definition somewhat at that time. Looking at Figure 12.13,notice that every s sublevel has only one orbital. Look at the p sublevelsin figure 12.13. How many orbitals does the p sublevelhave?{34}__________. How many orbitals does the d sublevelhave?{35}__________. How many orbitals does the f sublevelhave?{36}__________ Since each orbital can hold a maximum of twoelectrons, what is the capacity of an "s" sublevel?{37}__________ Of a"p" sublevel?{38}__________ Of a "d" sublevel?{39}__________ Of an"f" sublevel?{40}__________. To help you remember the order in whichthe energy levels and sublevels fill, you should become familiar with thediagonal rule. It is shown in Figure 12.12. By simply following the arrowsalong the diagonals, you have the correct order of filling. Use thisdiagonal rule to see if the levels and sublevels are properly arranged inFigure 12.13. The order of filling is:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6dIn theory there are other orbitals such as 5g, 6f, etc., but they are neverused since there are no atoms that have enough electrons to fill them.

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7sFigure 12.12

The Diagonal Rule

12-15 ©1997, A.J. Girondi

5f

6d

7s

5d

5p

5s

3d

3p

3s

1s

2s

2p

4s

4p

4d

6s

4f

6p

Energy Level 7 (Q Shell)

Energy Level 6 (P Shell)

Energy Level 5 (O Shell)

Energy Level 4 (N Shell)

Energy Level 3 (M Shell)

Energy Level 2 (L Shell)

Energy Level 1 (K Shell)

ORBITALS

Figure 12.13 Atomic Energy Levels and Sublevels

SECTION 12.7 Orbital Notation of Electrons

Now that we have the sublevels and orbitals identified, we can use this information to describewhere the electrons are located in the various atoms represented on the periodic table. To do this, youmust keep several basic rules in mind:

1. Electrons fill up the energy sublevels.

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2. The lowest energy sublevel must be completely filled before the next higher sublevel can begin to befilled. For example, according to the diagonal rule, the 4s sublevel must be filled before the 3d sublevelcan be used. Caution: note that this rule refers to sublevels, not energy levels.

3. Each orbital (represented by a circle in figure 12.13) can hold a maximum number of two electrons. Wewill represent electrons as slashed lines. The symbol (/) represents a single electron in an orbital. A fullorbital would contain two electrons and would be represented like this: (X). This method of representingthe electrons in an atom is called orbital notation.

4. Electrons repel each other and will not pair up in a single orbital in any given sublevel until all orbitals inthat sublevel are half full (have one electron in them). The incorrect and correct way to represent a 2porbital containing three electrons is shown below:

Incorrect 2p Orbital Notation Correct 2p Orbital Notation 2p 2p (X)(/)( ) (/)(/)(/)

Problem 7. You should now fill in the correct orbital notation to the right of the incorrect notation shownbelow for a 3d sublevel with 5 orbitals that contain 6 electrons:

Incorrect 3d Orbital Notation Correct 3d Orbital Notation 3d 3d

(X)(X)(X)( )( ) ( )( )( )( )( )

Orbital notation can be easily written for each element on the periodic table. For example, to writethe orbital notation for an atom of iron, we must first look up the atomic number of iron on the periodictable. The atomic number of iron is {41}_________. How many electrons does iron have?{42}__________.The orbital diagram shown below illustrates the order in which the sublevels and orbitals are filled for ironaccording to the diagonal rule.

Iron: 1s 2s 2p 3s 3p 4s 3d(X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (X)(/)(/)(/)(/)

Each "X" represents two slashed lines. Is the number of slashed lines shown equal to the number ofelectrons in iron?{43}__________ In Table 12.2, the orbital notations for the first ten elements areillustrated using this set of rules. Notice that even empty orbitals are shown if a sublevel contains at leastone electron (see boron and carbon).

Table 12.2Orbital Notation of the First Ten ElementsElement 1s 2s 2p

Hydrogen (/) Helium (X) Lithium (X) (/)Beryllium (X) (X)Boron (X) (X) (/)( )( ) Carbon (X) (X) (/)(/)( ) Nitrogen (X) (X) (/)(/)(/) Oxygen (X) (X) (X)(/)(/)Fluorine (X) (X) (X)(X)(/)Neon (X) (X) (X)(X)(X)

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Table12.3 summarizes much of the important information about the location of electrons in thefirst four energy levels.

Table 12.3Data Summary for Energy Levels 1 Through 4

Energy Level 1st 2nd 3rd 4th

Shell K L M N Shell Capacities 2 8 18 32 No. of Sublevels 1 2 3 4 Sublevel Types s s,p s,p,d s,p,d,f Sublevel Capacities 2 2,6 2,6,10 2,6,10,14

No. of orbitals per sublevel: s = 1; p = 3; d = 5; f = 7

Problem 8. Now it is time for you to use what you have learned. Fill in the electrons for the elementsbelow. Follow the rules listed in section 12.7 as well as the diagonal rule. The finished product will be theorbital notation for an atom of each element.

Element 1s 2s 2p 3s 3p 4s 3d

a. beryllium ( ) ( ) b. magnesium ( ) ( ) ( )( )( ) ( )c. nitrogen ( ) ( ) ( )( )( )d. silicon ( ) ( ) ( )( )( ) ( ) ( )( )( )e. calcium ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )f. potassium ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )g. argon ( ) ( ) ( )( )( ) ( ) ( )( )( )h. sodium ( ) ( ) ( )( )( ) ( )i. chlorine ( ) ( ) ( )( )( ) ( ) ( )( )( )j. copper ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( )k. phosphorus ( ) ( ) ( )( )( ) ( ) ( )( )( )l. nickel ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( )

Note: The 3d sublevel is a "higher energy" level than the 4s because electrons in the 3d have moreenergy than those in the 4s. Electrons fill the "lower energy" levels first. However, the lower energy 4ssublevel is farther from the nucleus than the higher energy 3d. A "higher" energy sublevel is notnecessarily farther from the nucleus than a "lower" energy sublevel. The distance of the energy sublevelsfrom the nucleus is:

Increasing Distance from Nucleus----->1s 2s,2p 3s,3p,3d 4s,4p,4d,4f 5s,5p,5d,5f 6s,6p,6d,etc. (closest) (farthest)

SECTION 12.8 Electron Configuration Notation

There is another form of notation which reveals which energy level and sublevel the electrons ofan atom are in, but it does not show the individual orbitals. This method is known as electron configurationnotation. A superscript is used to reveal the number of electrons in each sublevel. For example, theconfiguration notation for lithium, element number 3, is: 1s22s1. This means that there are 2 electrons in

12-18 ©1997, A.J. Girondi

sublevel "s" of energy level 1, and there is one electron in sublevel "s" of energy level 2. Configurationnotation does not reveal whether the electrons are alone or paired in their orbitals like orbital notationdoes. Note that the sum of the superscripts should equal the number of electrons in the atoms. (A lithiumatom contains three electrons.)

If an atom has a configuration notation of 1s22s22p3, how many electrons does the atom contain?{44}_________ How many of these electrons are found in the second energy level?{45}___________How many are found in the first energy level?{46} __________ More examples of configuration notationare shown below.

Element Configuration Notation

hydrogen 1s1

helium 1s2

lithium 1s22s1

beryllium 1s22s2

boron 1s22s22 p 1

manganese 1s22s22 p 63s23 p 64s23 d 5

selenium 1s22s22 p 63s23 p 64s23 d 104 p 4

antimony 1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 3

Problem 9. Give the electron configuration notation for the elements listed below.

Element Atomic Number Configuration Notation

a. C 6 ________________________________________________

b. N 7 ________________________________________________

c. O 8 ________________________________________________

d. F 9 1s22s22 p 5

e. Ne 10 ________________________________________________

f. P 15 ________________________________________________

g. Ar 18 ________________________________________________

h. Ca 20 1s22s22 p 63s23 p 64s2

i. Mn 25 ________________________________________________

j. Zn 30 ________________________________________________

k. Kr 36 ________________________________________________

12-19 ©1997, A.J. Girondi


Problem 10. Naturally-occurring iridium (Ir) consists of two isotopes. Iridium–191 (atomic mass =190.9992 composes 37.58% of it, while iridium–193 (atomic mass = 192.9913) composes the remaining62.7%. Calculate the average atomic mass of iridium atoms.

Problem 11. Write the nuclear notation for the following isotopes:

a. iridium–191 b. chlorine–37 c. chromium–50

Problem 12. Indicate the number of protons, electrons and neutrons in each of the following:

protons electrons neutrons

iridium-191 ______ _______ _______

chlorine-37 ______ _______ _______

chromium-50 ______ _______ _______

Problem 13. Write the orbital notation for:

a. sulfur: 1s 2s 2p 3s 3p 4s 3d( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( )

b. vanadium: 1s 2s 2p 3s 3p 4s 3d( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( )

Problem 14. Write the configuration notation for:

a. sulfur: __________________________________________________________

b. vanadium: _______________________________________________________

ACTIVITY 12.10 An Optional Enrichment Activity -The Colors of Transition Metal Chemistry

The elements with atomic numbers 21 through 30 exhibit many properties different from those ofother elements in the same period (row). The elements directly below in the next two periods, however,have similar properties. Together these elements are known as the transition metals. The purpose of thisactivity is to compare the properties of some substances which contain transition metals with those which

12-20 ©1997, A.J. Girondi

do not contain non-transition metals. We will then try to relate any unique properties of the transitionmetals to the electrons which they contain.

Sc Ti V Cr Mn Fe Co Ni Cu Zn

Figure 12.14 The Transition Metals of Row Four

Materials: 96-well microplate; plastic micropipets; toothpicks; 0.1M solutions of compounds containingperiod 4 metals: KNO3, Ca(NO3)2, NH4VO3, Cr(NO3)2, Mn(NO3)2, Co(NO3)2, Fe(NO3)3, Ni(NO3)2, Cu(NO3)2,Zn(NO3)2; 6.0M ammonia (NH3); 1M potassium thiocyanate, KSCN; 6M hydrochloric acid (HCl); one sheetof white paper.

Procedure:

1. Place a 96-well microplate on a sheet of white paper. The numbered columns should be at the top withthe lettered rows to your left.

2. Place 5 drops of KNO3 in each of wells A1, B1, C1, and D1.

NH3(aq) -----> A

KSCN ------> B

HCl ------> C

Control ------> D

1 2 3 4 5 6 7 8 9 10

1 = KNO3 4 = Cr(NO3)3 7 = Fe(NO3)3 10 = Zn(NO3)22 = Ca(NO3)2 5 = Mn(NO3)2 8 = Ni(NO3)23 = NH4VO3 6 = Co(NO3)2 9 = Cu(NO3)2

Figure 12.15 Microplate

3. Place 5 drops of Ca(NO3)2 in each of wells A2, B2, C2, and D2.

4. Continue to repeat this process of placing 5 drops of solutions of metal compounds in subsequent

12-21 ©1997, A.J. Girondi

columns of the microplate. Use chemicals in the order in which they are given in the materials list above.The last drops will be in column 10.

5. Add 5 drops of 6M NH3 to each well in row A. Mix well with a toothpick. Rinse the toothpick with waterbefore moving from well to well. CAUTION: ammonia solution is caustic and has strong vapors.

6. Add 5 drops of KSCN solution to each well in row B. Mix well with a toothpick, rinsing the toothpickbetween uses as before.

7. Finally, add 5 drops of HCl to each well in row C. CAUTION: Do not mix HCl and KSCN. Clean up allspills immediately. Mix well. Use row D as a control for comparing with the other rows.Data and Observations:

1. Observe row D of your microplate. Record your observations - particularly colors, if any - in Table 12.4.

2. Compare the solutions in the wells in rows A, B, and C in each column to the solution in row D in thatcolumn. Where there was a change, record what you observe.

A1_______________ A2_______________ A3_______________ A4_______________

A5_______________ A6_______________ A7_______________ A8_______________

A9_______________ A10_______________

B1_______________ B2_______________ B3_______________ B4_______________

B5_______________ B6_______________ B7_______________ B8_______________

B9_______________ B10_______________

C1_______________ C2_______________ C3_______________ C4_______________

C5_______________ C6_______________ C7_______________ C8_______________

C9_______________ C10_______________

D1_______________ D2_______________ D3_______________ D4_______________

D5_______________ D6_______________ D7_______________ D8_______________

D9_______________ D10_______________

TABLE 12.4 Transition Metal Chemistry

Analysis and Conclusions:

Comment on the order of the metal ions in the wells from left to right in the rows of your microplate whencompared to the positions of the metals in the periodic table:{47}_______________________________

Did you observe any colors in columns 1 and 2 of your microplate? {48}_________________________

The metals used in columns 1 and 2 of your microplate are not transition metals. With respect to sublevel

12-22 ©1997, A.J. Girondi

(s,p,d,f), what kind of electrons do the transition metals have that these metals do not have?{49}________

Review what you observed in columns 3 through 9 of your microplate. What do the results in thesecolumns have in common?{50}________________________________________________________

Did the metal in column 1 and 2 produce any colors?{51}__________ Did the metal in column 10 produceany colors?{52}___________ Is the "d" sublevel of electrons completely filled in the metal atoms found incolumns 3 through 9 of your microplate?{53}___________

What is true about the distribution of electrons in the transition metals in columns 3 through 9 that is

responsible for the fact that they produce colored products? {54}_______________________________

______________________________________________________________________________

With respect to color, the results in column 10 should have been different from the results in columns 3

through 10. Why? {55}______________________________________________________________

What property would help to identify a compound as one which contains a transition metal ion? { 56}____________________________________________________________________________________


This is the end of chapter 12. Check off each of the learning outcomes on the next page whichyou have completed. When you have mastered them all, arrange to take the chapter 12 exam, and go tochapter 13 which is part 2 of your study of atomic structure.

_____1. List the subatomic particles and describe their location and properties.

_____2. Determine the number of electrons, protons, and neutrons in an atom of any element given information such as its atomic number, mass number, or atomic mass.

_____3. Define isotope, give an example of one, and calculate the atomic mass of an element given the masses of its isotopes and their abundance in nature.

_____4. Describe the Bohr model of the atom.

_____5. Explain how atomic spectra help to explain the behavior of electrons.

_____6. Explain what is meant by energy levels, sublevels, orbitals, and electron configurations.

_____7. Give the capacities of energy levels and sublevels.

_____8. Write the electron configuration notation of atoms.

_____9. Write the orbital notation for atoms.

12-23 ©1997, A.J. Girondi


Questions:

{1} repel; {2} attract; {3} Dalton; {4} nucleus; {5} positively; {6} +1; {7} neutrons; {8} 19; {9} 10{10} 10; {11} 8; {12} 8; {13} neutrons; {14} isotopes; {15} average; {16} neutrons; {17} protons;{18} neutrons; {19} mass number; {20} electrons; {21} 16; {22} 37; {23} 12.01; {24} 24;{25} phosphorus-31; {26} Some matter is converted into energy when an atom of oxygen forms;{27} longer; {28} electrons; {29} different (or unique); {30} reddish; {31} 10; {32} fourth; {33} second;{34} 3; {35} 5; {36} 7; {37} 2; {38} 6; {39} 10; {40} 14; {41} 26; {42} 26; {43} yes; {44} 7; {45} 5;{46} 2; {47} the same; {48} no; {49} d; {50} they have color; {51} no; {52} no; {53} no;{54} They have an incomplete (unfilled) "d" sublevel; {55} Elements in column 10 on the microplate havea filled "d" sublevel; {56} color

Problems:

1. a. 13,13,14 b. 55, 55, 78

2. a. 92, 92, 142 b. 92, 92, 143 c. 92, 92, 146

NuclearIsotope Notation At.No. Mass No. No. e's No. p's No. n's

a. aluminum-27 Al _13__ __27___ _13__ _13__ _14__

b. bismuth-209 Bi _83__ __209__ _83__ _83__ _126_

c. calcium-40 Ca _20__ __40___ _20__ _20__ _20__

d. copper-64 Cu _29__ __64___ _29____ _29__ _35__

e. _helium-4____ He __2__ ___4___ __2__ __2__ __2__

f. _lead-207____ Pb _82__ __207__ __82_ _82__ _125_

g. _oxygen-16__ O __8__ __16__ __8__ __8__ __8__

h. _tin-119_____ Sn _50__ __119__ __50_ __50_ __69_

i. _zinc-66_____ Zn _30__ __66___ __30_ __30_ __36_

27

13209

8340

20

29

64

4 2

82

207

16

8

50

119

3066

3.

4. 63.54

5. These are estimates: 490 nm, 495 nm, 568 nm, 572 nm, 592 nm, 595 nm, 615 nm

6. 2, 8, 18, 32

7. 3d (X)(/)(/)(/)(/)

12-24 ©1997, A.J. Girondi

Element 1s 2s 2p 3s 3p 4s 3d

a. beryllium (X) (X) b. magnesium (X) (X) (X)(X)(X) (X)c. nitrogen (X) (X) (/)(/)(/)d. silicon (X) (X) (X)(X)(X) (X) (/)(/)( )e. calcium (X) (X) (X)(X)(X) (X) (X)(X)(X) (X)f. potassium (X) (X) (X)(X)(X) (X) (X)(X)(X) (/)g. argon (X) (X) (X)(X)(X) (X) (X)(X)(X)h. sodium (X) (X) (X)(X)(X) (/)i. chlorine (X) (X) (X)(X)(X) (X) (X)(X)(/)j. copper (X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (X)(X)(X)(X)(/)k. phosphorus (X) (X) (X)(X)(X) (X) (/)(/)(/)l. nickel (X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (X)(X)(X)(/)(/)

8.

9. a. 1s22s22p2

b. 1s22s22p3

c. 1s22s22p4

d. (given)e. 1s22s22p6

f. 1s22s22p63s23p3

g. 1s22s22p63s23p6

h. (given)i. 1s22s22p63s23p64s23d5

j. 1s22s22p63s23p64s23d10

k. 1s22s22p63s23p64s23d104p6

10. Check periodic table for atomic weight of Iridium (#77). Answer may be a little off due to rounding.

11. a. 77191Ir b. 17

37Cl c. 2450Cr

12. iridium-191: 77 protons, 77 electrons, 114 neutronschlorine-37: 17 protons, 17 electrons, 20 neutronschromium-50: 24 protons, 24 electrons, 26 neutrons

a. sulfur: 1s 2s 2p 3s 3p 4s 3d(X) (X) (X(X)(X) (X) (X)(/)(/) ( ) ( )( )( )( )( )

b. vanadium: 1s 2s 2p 3s 3p 4s 3d(X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (/)(/)(/)( )( )

13.

14. sulfur: 1s22s22p63s23p4

vanadium: 1s22s22p63s23p64s23d3

12-25 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 13

MODERNVIEW

OFATOMIC

STRUCTURE(PART 2)

NOTICE OF RIGHTS




[email protected]


13-2 ©1997, A.J. Girondi

SECTION 13.1 The Valence Shell and Valence Electrons

Now that you know how to write electron configurations for atoms, we can examine chemicalfamilies in terms of electron configurations. You will recall that chemical families are the vertical columns onthe periodic table. One such family consists of the elements Li, Na, K, Rb, Cs, and Fr. These elementscomprise column 1A. They are called the alkali metals. Their electron configurations are listed in Table13.1 below.

Table 13.1Electron Configurations for Alkali Metals

At. No. Configuration Notation

3 Li 1s22s1

11 Na 1s22s22 p 63s1

19 K 1s22s22 p 63s23 p 64s1

37 Rb 1s22s22 p 63s23 p 64s23 d 104 p 65s1

55 Cs 1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 66s1

87 Fr 1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 66s24 f 145 d 106 p 67s1

There is an important similarity in all of these electron configurations. Can you see what it is? Circle theoutermost energy level (1,2,3,4,etc.) of each of the elements in Table 13.1. How many electrons arethere in the outermost energy level of each of these elements? {1}__________. The shell or energy levelthat contains the outermost electrons for an element is called the valence shell. The electrons in thatenergy level are called valence electrons. Because these electrons are among the ones that are farthestfrom the nucleus, they are relatively easy to remove. The chemical activity of an element is determined inpart by the number of valence electrons that it has.

The alkali metals all have one valence electron. This similarity in number of valence electronscauses the chemical behavior of these elements to be very similar. Sometimes hydrogen, which has a 1s1configuration, is also listed with this family since it also has one valence electron. However, it behavesquite differently from the other elements in the family because of its small size and the fact that there areno other electrons in levels beneath the valence shell, as there are in the rest of the family. On someperiodic tables, hydrogen is placed in a class by itself.

When you attempt to identify the valence electrons in an atom, be sure to count all the electronsin the entire outermost energy level, not just the outermost sublevel. For example, for germanium (#32)the configuration notation is 1s 22s22 p 63s23 p 64s23 d 104 p 2. How many valence electrons are there?The answer is 4. That's right, 4. The outermost energy level is 4 (the N shell), and there are 4 electronsthere. They are the 4s2 and the 4p2 electrons. Many students would make the mistake of thinking thatonly the 4p2 electrons are in the valence shell. All of the electrons in the outermost energy level (energylevel 4 in this case) must be included. Reread this paragraph until you are sure you understand.

Let's see if you've mastered this idea. How many valence electrons are there in the atom whoseconfiguration notation is listed below?

1s22s22 p 63s23 p 64s23 d 104 p 6 Number of valence electrons = {2}_____________

What is the configuration notation of only the valence electrons? {3}____________________________

13-3 ©1997, A.J. Girondi

Write the electron configuration notation for only the valence electrons in aluminum (#13): {4}__________

O.K., now write the valence shell electron configuration notation for Te (#52): {5}___________________

Problem 1. In Table 13.2, write the electron configuration notation for only the valence shell of theindicated families. Be careful, only the valence shell configuration is being asked for, but be sure toinclude all of the electrons in the valence shell!

Table 13.2Valence Shell Configuration for Families 2A & 3A

Family 2AElement At. No. Valence Shell Configuration

Be 4 _____________________

Mg 12 _____________________

Ca 20 _____________________

Sr 38 _____________________

Ba 56 _____________________

Ra 88 _____________________

Family 3AElement At. No. Valence Shell Configuration

B 5 _____________________

Al 13 _____________________

Ga 31 _____________________

In 49 _____________________

Tl 81 _____________________

The chemical activity of the elements depends on their electronic structure.Elements in the same group or family have the same valence shell structure. For example, in Group 3Athe element B has the valence shell: 2s22p1, while in the same family Al has the valence shell: 3s23p1.Mendeleev was not correct in arranging the elements according to increasing atomic masses, whileMoseley was correct in arranging them according to increasing atomic number. Why? Well, the atomicnumber indicates the number of protons in the nucleus as well as the number of electrons in a neutralatom. When arranged according to increasing atomic number, elements with the same valence shellstructure can be lined up in a vertical group (family). Thus, it is the valence shell configurations that allowus to arrange elements periodically in the table. Mendeleev did his work prior to the discovery of theinternal structure of atoms. It is a tribute to his insight that he developed his arrangement based onchemical similarities. Arranging the atoms according to increasing atomic masses was a very logical movegiven the state of knowledge at his time. The fact that he was incorrect does not detract from thecontribution he made to chemistry.

13-4 ©1997, A.J. Girondi

Problem 2. For further practice in working with valence shell electron configuration notation, completeTable 13.3 (show valence electrons only).

Table 13.3Electron Configuration and Valence Electrons of Selected Elements

Element At. No. Valence Shell Electron Configuration Notation

Si 14 __________________________________

Br 35 __________________________________

V 23 __________________________________

As 33 __________________________________

S 16 __________________________________

Y 39 __________________________________

At 85 __________________________________

Au 79 __________________________________

Cl 17 __________________________________

SECTION 13.2 The Use of "Core" Notation

The task of writing notations can get cumbersome for atoms that have a lot of electrons. Ashortcut method, core notation, has been developed that is quicker and more convenient. Suppose youwant to write the configuration and orbital notation for chlorine. To write core notation, find the noble gason the periodic table that has the atomic number closest to but less than that of chlorine. The atomicnumber of chlorine is 17. The closest noble gas is Neon (#10). Then write the notations of chlorine asfollows:

3s 3pChlorine: 17Cl = [Ne] 3s23p5 or: [Ne] (X) (X)(X)(/)

This indicates that chlorine atoms have all the electrons that neon has (the neon core) plus those shown inthe orbitals that follow. Here's lead:

Lead = 82Pb = [Xe] 6s24 f 145 d 106 p 2

6s 4f 5d 6p or: [Xe] (X) (X)(X)(X)(X)(X)(X)(X) (X)(X)(X)(X)(X) (/)(/)( )

Lead has all the electrons that xenon has (the xenon core) plus those shown in the notation. The coreform of configuration notation for 100 elements is shown in Table 13.4. This table follows the diagonalrule. There are a number of exceptions to the diagonal rule, but they are not shown in table 13.4. Thetable of electron configurations in Appendix B of your ALICE materials shows the actual electronconfigurations of the elements including exceptions to the diagonal rule.

Refer to Tables 13.4 and 13.5 as you study examples of the "core form" of orbital notation. Table13.5 shows the configuration of the last electron to enter each atom as orbitals fill up according to thediagonal rule. It really helps when writing core notation. Examine Table 13.5 for a moment now.

13-5 ©1997, A.J. Girondi

Element Orbital and Configuration Notation Using Core Notation

2s 2pLithium (#3) [He] (/) or [He] 2s1

boron (#5) [He] (X) (/)( )( ) or [He] 2s22 p 1

oxygen (#8) [He] (X) (X)(/)(/) or [He] 2s22 p 4

3s 3pmagnesium (#12) [Ne] (X) or [Ne] 3s2

silicon (#14) [Ne] (X) (/)(/)( ) or [Ne] 3s23 p 2

4s 3d 4pgallium (#31) [Ar] (X) (X)(X)(X)(X)(X) (/)( )( ) or [Ar] 4s23 d 104 p 1

bromine (#35) [Ar] (X) (X)(X)(X)(X)(X) (X)(X)(/) or [Ar] 4s23 d 104 p 5

Problem 3. Try writing the core form of configuration notation for the elements below. Check youranswers using Table 13.4.

Element Orbital Notation Using Core Notation

Sulfur (#16) ___________________________________________________________

Scandium (#21) ___________________________________________________________

Niobium (#41) ___________________________________________________________

Tellurium (#52) ___________________________________________________________

SECTION 13.3 Ionization Energy

You will recall that when energy is absorbed by an electron, it is said to be "excited." It is thevalence electrons that are easiest to "excite." Look at the two electron configurations below. There is anobvious difference between them. Both electron configurations represent eleven electrons. Both aresodium atoms (assuming the elements represented have 11 protons).

1s22s22 p 63s1 1s22s22 p 65s1 (sodium atom) (unstable sodium atom)

The 3s1 electron has been "excited"and jumped up to a higher energy level

Since sodium has an atomic number of 11, these configurations represent a sodium atom. What hashappened is that the electron in the 5s orbital is excited and is extremely unstable. It can emit that extraenergy as light to get back down to its ground state (the 3s level). But, when you excite the electron in thefirst place, what if the 3s1 electron absorbs more energy than it can handle? Indeed, this is a commonoccurrence. In this case, the 3s1 electron would simply leave the atom altogether. What is left is called asodium ion. An ion is an atom with an unequal number of electrons and protons. This means that all ionshave either a positive or negative charge. To see whether sodium would become a positive or negativeion, we must look at its electron configuration with and without its 3s1 electron.

Neutral Na Atom = 1s22s22 p 63s1 Na1+ Ion = 1s22s22 p 6

The sodium ion has a plus 1 (+1) charge. It is important for you to remember that ions can only be made byeither adding or removing electrons. The number of protons is never altered in a chemical change.

13-6 ©1997, A.J. Girondi

Table 13.4 "Core Notation" of 100 Elements According to the Diagonal Rule

1 H 1s1 51 Sb [Kr]5s24d105p3

2 He 1s2 52 Te [Kr]5s24d105p4

3 Li [He]2s1 53 I [Kr]5s24d105p5

4 Be [He]2s2 54 Xe [Kr]5s24d105p6

5 B [He]2s22p1 55 Cs [Xe]6s1 6 C [He]2s22p2 56 Ba [Xe]6s2 7 N [He]2s22p3 57 La [Xe]6s24f1 8 O [He]2s22p4 58 Ce [Xe]6s24f2 9 F [He]2s22p5 59 Pr [Xe]6s24f3 10 Ne [He]2s22p6 60 Nd [Xe]6s24f4 11 Na [Ne]3s1 61 Pm [Xe]6s24f5 12 Mg [Ne]3s2 62 Sm [Xe]6s24f6 13 Al [Ne]3s23p1 63 Eu [Xe]6s24f7 14 Si [Ne]3s23p2 64 Gd [Xe]6s24f8 15 P [Ne]3s23p3 65 Tb [Xe]6s24f9 16 S [Ne]3s23p4 66 Dy [Xe]6s24f10 17 Cl [Ne]3s23p5 67 Ho [Xe]6s24f11

18 Ar [Ne]3s23p6 68 Er [Xe]6s24f12 19 K [Ar]4s1 69 Tm [Xe]6s24f13 20 Ca [Ar]4s2 70 Yb [Xe]6s24f14 21 Sc [Ar]4s23d1 71 Lu [Xe]6s24f145d1

22 Ti [Ar]4s23d2 72 Hf [Xe]6s24f145d2

23 V [Ar]4s23d3 73 Ta [Xe]6s24f145d3

24 Cr [Ar]4s23d4 74 W [Xe]6s24f145d4

25 Mn [Ar]4s23d5 75 Re [Xe]6s24f145d5

26 Fe [Ar]4s23d6 76 Os [Xe]6s24f145d6

27 Co [Ar]4s23d7 77 Ir [Xe]6s24f145d7

28 Ni [Ar]4s23d8 78 Pt [Xe]6s24f145d8

29 Cu [Ar]4s23d9 79 Au [Xe]6s24f145d9

30 Zn [Ar]4s23d10 80 Hg [Xe]6s24f145d10

31 Ga [Ar]4s23d104p1 81 Tl [Xe]6s24f145d106p1 32 Ge [Ar]4s23d104p2 82 Pb [Xe]6s24f145d106p2

33 As [Ar]4s23d104p3 83 Bi [Xe]6s24f145d106p3

34 Se [Ar]4s23d104p4 84 Po [Xe]6s24f145d106p4

35 Br [Ar]4s23d104p5 85 At [Xe]6s24f145d106p5

36 Kr [Ar]4s23d104p6 86 Rn [Xe]6s24f145d106p6

37 Rb [Kr]5s1 87 Fr [Rn]7s1 38 Sr [Kr]5s2 88 Ra [Rn]7s2 39 Y [Kr]5s24d1 89 Ac [Rn]7s25f1 40 Zr [Kr]5s24d2 90 Th [Rn]7s25f2 41 Nb [Kr]5s24d3 91 Pa [Rn]7s25f3 42 Mo [Kr]5s24d4 92 U [Rn]7s25f4 43 Tc [Kr]5s24d5 93 Np [Rn]7s25f5 44 Ru [Kr]5s24d6 94 Pu [Rn]7s25f6 45 Rh [Kr]5s24d7 95 Am [Rn]7s25f7 46 Pd [Kr]5s24d8 96 Cm [Rn]7s25f8 47 Ag [Kr]5s24d9 97 Bk [Rn]7s25f9 48 Cd [Kr]5s24d10 98 Cf [Rn]7s25f10 49 In [Kr]5s24d105p1 99 Es [Rn]7s25f11 50 Sn [Kr]5s24d105p2 100 Fm [Rn]7s25f12

13-7 ©1997, A.J. Girondi

1s1 1s2

2s1

3s1

4s1

5s1

6s1

7s1

2s2 2p1 2p2 2p3 2p4 2p5 2p6

3p1

4p1

5p1

6p1

3p2

4p2

5p2

6p2

3p3

4p3

5p3

6p3

3p4

4p4

5p4

6p4

3p5

4p5

5p5

6p5

3p6

4p6

5p6

6p6

3s2

4s2

5s2

6s2

7s2

3d1 3d2 3d3 3d4 3d5 3d6 3d7 3d8 3d9 3d10

4d1

5d1

6d1

4d2

5d2

6d2

4d3

5d3

6d3

4d4

5d4

6d4

4d5

5d5

6d5

4d6

5d6

6d6

4d7

5d7

6d7

4d8

5d8

6d8

4d9

5d9

6d9

4d10

5d10

4f1 4f2 4f3 4f4 4f5 4f6 4f7 4f8 4f9 4f10 4f11 4f12 4f13 4f14

5f1 5f2 5f3 5f4 5f5 5f6 5f7 5f8 5f9 5f10 5f11 5f12 5f13 5f14

Table 13.5Last–Entering Electrons According to the Diagonal Rule*

*Since there are many exceptions to the diagonal rule, see Appendix B of your reference notebook for the actual electron distributions in atoms.

The energy required to remove electrons is called the ionization energy. The amounts ofionization energy required to remove the outermost electrons of sodium, magnesium, and aluminum arelisted in Table 13.6.

After the first electron is removed from an atom, the size of the atom decreases. The remainingelectrons are "pulled" closer to the nucleus and become even harder to remove. This is evident in Table13.6. In the table, valence electrons and the energies required to remove valence electrons are enclosedin boxes. Notice that after the valence electron(s) are removed from the atoms, the amount of energyrequired to remove any more electrons goes up dramatically. The reason for this is that after the valenceelectrons are removed, the atoms are left with the electron configuration of a noble gas and additionalelectrons would have to come from another energy level which is much closer to the nucleus. Suchelectrons are much more strongly attracted to the nucleus. Notice that to get a second electron out ofsodium (Na) we would have to move to another energy level which is closer to the nucleus.

Because the noble gas configurations are so stable, it is very difficult to remove more electronsfrom an atom after it has achieved an "octet" of eight electrons in the outermost energy level. In Table13.6, the biggest increase in ionization energy comes after which electron is removed from Al? (1st, 2nd,3rd, etc.): {6}_________ What accounts for this big increase in ionization energy in aluminum?{7}______________________________________________________________________________________

Notice in Table 13.6 that the big increase in ionization energy for Mg comes between the second and third

electrons. This means that it is much harder to remove the third electron from an atom of Mg than it was to

remove the first and second ones. This is evidence that the third electron in much closer to the nucleus.

We conclude that the third electron belongs to a different energy level.

13-8 ©1997, A.J. Girondi

Table 13.6Energy Needed to Remove Electrons

Element Configuration Ionization Energy (Kilojoules/mole)Notation 1st Elec. 2nd Elec. 3rd Elec. 4th Elec.

Na 1s22s22p63s1 498 4560

Mg 1s22s22p63s2 736 1445 7730

A l 1s22s22p63s23p1 577 1815 2740 11600

◊◊◊

◊◊◊

◊◊◊

◊◊◊ indicates a change in energy levels

So, when the big increase in ionization energy occurs after the first electron, the atom probably belongs to what family on the periodic table? {8}__________ When the big increase occurs after two electrons have been removed, the atom probably belongs to family {9}__________.

The valence shell arrangement of electrons in noble gases makes them very stable. They havewhat we call an octet of electrons in their outer (valence) shells. For neon the octet is in the 2nd energylevel: 2s22p6. For argon it is in the 3rd energy level: 3s23p6. Helium is the only noble gas that does nothave an octet. Instead, it has a full outer energy level (1s2) which is also very stable. (Remember, the firstenergy level holds a maximum of 2 electrons.)

The large "gaps" in ionization energies serve as great evidence for the existence ofenergy levels, sublevels, and the stability of certain electron arrangements.

Atoms gain or lose electrons in order to obtain a stable arrangement. The two most stable arrangementsare:

1. an octet (s2p6) in the outermost energy level (as seen in the noble gases except for helium)2. the full outermost energy level of helium (1s2)

Other arrangements that are "slightly" stable include:

3. a full sublevel (such as s2, or p6, or d10, etc.) 4. a half-full sublevel (such as s1, p3, d5, etc.)

In order to become more stable, atoms often lose, gain, or share electrons when they react witheach other. They lose, gain, or share as many electrons as needed to establish a stable arrangement(especially an octet). To accomplish this, atoms generally tend to lose or gain enough electrons toachieve the electron configuration of the closest noble gas. (By closest we mean according to atomicnumber.)

Sodium loses one electron and ends up with the arrangement of neon:

Na atom: 1s22s22 p 63s1 ---> Na1+ ion: 1s22s22 p 6

Magnesium, [Ne]3s2, loses two electrons and becomes like neon:

Mg atom: 1s22s22 p 63s2 ---> Mg2+ ion: 1s22s22 p 6

Aluminum, [Ne]3s23p1, loses three electrons to become like neon:

Al atom: 1s22s22 p 63s23 p 1 ---> Al3+ ion: 1s22s22 p 6

Let's go to row four and consider the elements potassium (19K) and calcium (20Ca). Have a periodic table infront of you as you read this. Element 19, potassium, becomes an ion by losing one electron to achievethe stable argon octet (3s23p6):

13-9 ©1997, A.J. Girondi

K atom: 1s22s22 p 63s23 p 64s1 ---> K1+ ion: 1s22s22 p 63s23 p 6

Calcium loses two electrons to achieve the argon octet:

Ca atom: 1s22s22p63s23p64s2 ---> Ca2+ ion: 1s22s22p63s23p6

SECTION 13.4 The Order in Which Atoms Lose Their Electrons

Before going on, a very important point has to be made. When electrons enter atoms, they fill theenergy levels and sublevels according to the diagonal rule (with exceptions). However, after the electronsare in an atom their distances from the nucleus do not follow the diagonal rule. For example, the order offilling for titanium (#22)is:

1s22s22 p 63s23 p 64s23 d 2

however, the actual distances of the sublevels from the nucleus is:

1s22s22 p 63s23 p 63 d 24s2

Another example is rhenium (#75) for which the order of filling is:

1s22s22 p 63s23 p 64s23 d 104 p 65s24 d 105 p 66s24 f 145 d 5

while the distance of the sublevels from the nucleus is:

1s22s22 p 63s23 p 63 d 104s24 p 64 d 104 f 145s25 p 65 d 56s2

Remember, the order of the energy levels and sublevels according to distance from the nucleus is:

1s,2s,2p,3s,3p,3d,4s,4p,4d,4f,5s,5p,5d,5f,6s,6p,6d,6f*,7s

*The 6f sublevel is never used, because there are only 111 elements as of the year 1995.

When atoms lose their electrons, they do not lose them in the reverseorder of the diagonal rule. Rather, they lose the ones that are farthestfrom the nucleus, first.

This is because the nuclear attraction is weaker at greater distances. The electrons in the highest energylevel of an atom may not be farthest from the nucleus. For example, electrons enter the 4s sublevelbefore they enter the 3d. The 3d is described as a "higher" sublevel because electrons in 3d orbitalscontain more energy than those in the 4s. The electrons in the 4s orbital, however, are farther from thenucleus.

Example: The configuration notation for 30Zn is 1s22s22 p 63s23 p 64s23 d 10

The orbital notation for 30Zn is:

1s 2s 2p 3s 3p 4s 3d (X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (X)(X)(X)(X)(X)

The electrons in 30Zn are lost in the following order: 4s, 3d, 3p, 3s, 2p, 2s, and finally, 1s

In ordinary chemical reactions an atom normally loses electrons only from its valence shell. Forzinc those electrons would be in the 4s sublevel. Sometimes atoms can also lose some "d" electrons inthe sublevel immediately below the valence shell. For zinc those would be the electrons in the 3dsublevel. Zinc does not lose any of its 3d electrons, however, because the full 3d sublevel is too stable.

13-10 ©1997, A.J. Girondi

As a result, the only oxidation number that zinc has is +2, indicating that it can lose two electrons (the 4s2).

Many of the transition metals (21Sc through 30Zn) in row 4 can lose electrons and form ions in atleast three ways:

1. They may lose their 4s and 3d electrons and be left with the stable argon octet. For example, titanium(#21) can lose 4 electrons to achieve the stable argon octet:

Ti atom: 1s22s22 p 63s23 p 64s23 d 2 ---> Ti4+ ion: 1s22s22 p 63s23 p 6

2. They may lose only their 4s electrons and be left a highest energy level which is full: (3s23p63d10).Remember the third energy level has a capacity of 18. Zinc (30Zn) loses two electrons to achieve a highest(3rd) energy level which is full with 18 electrons (3s23p63d10):

Zn atom: 1s22s22 p 63s23 p 64s23 d 10 ---> Zn2+ ion: 1s22s22 p 63s23 p 63 d 10

3. Or, they may lose their 4s electrons and be left with arrangements such as half-full sublevels:

Manganese (25Mn) loses 2 electrons and is left with a half-full outer sublevel:

Mn atom: 1s22s22 p 63s23 p 64s23 d 5 ---> Mn2+ ion: 1s22s22 p 63s23 p 63 d 5

Manganese (25Mn) can also lose its 3d electrons in addition to its 4s electrons thus forming an Mn7+ ion:

Mn atom: 1s22s22 p 63s23 p 64s23 d 5 ---> Mn7+ ion: 1s22s22 p 63s23 p 6

It can also lose electrons in other ways resulting in the Mn6+ and the Mn4+ ions.

Remember! It is always the electrons in the valence shell that are lost first.

Let's consider element #31, gallium. It is a family 3A element. You should recall that elements inthis family have an oxidation number of +3. The elements above gallium in this family, B and Al, form ionsby losing three electrons to achieve stable octets. However, gallium is in row four and contains 3delectrons. It also loses three electrons, but that does not leave gallium with a stable octet, but it doesleave it with a full outermost energy level:

Ga atom: 1s22s22 p 63s23 p 64s23 d 104 p 1 ---> Ga3+: 1s22s22 p 63s23 p 63 d 10

You see, the 4s2 and 4p1 electrons are the ones that are lost first. Remember, again, that the outermostelectrons are the ones that go. In this case, the electrons in the highest sublevel (3d) are not members ofthe outermost (4th) energy level. Let's briefly get back to that third kind of stable arrangement thatinvolves half-full sublevels. For example, the electron configuration of copper is listed below according tothe diagonal rule. Also listed, is the actual electron arrangement in copper as determined by experiment.

Cu (from diagonal rule): 1s22s22 p 63s23 p 64s23 d 9

Cu (actual configuration) : 1s22s22 p 63s23 p 64s13 d 10

Why does the copper atom deviate from the trend represented by the diagonal rule? If one of copper's 4selectrons moves to the 3d level, the next-to-the-highest sublevel will be half-full (4s1), while the highestone will be full (3d10). The stability of these half-full and full sublevels makes this the preferred state forcopper. Similar reasoning explains the deviation of chromium from the diagonal rule as seen below.

Cr (from diagonal rule): 1s22s22 p 63s23 p 64s23 d 4

Cr (actual configuration): 1s22s22 p 63s23 p 64s13 d 5

13-11 ©1997, A.J. Girondi

Thus, you see that the information you get from the diagonal rule is not perfect. In fact, there are quite afew other deviations from the diagonal rule involving other elements. Some of them are not wellunderstood. If you want the best information regarding the electron configuration of electrons in an atom,you must check a reference source such as the tables found in Appendix B of your ALICE materials. Lookthere now, and find at least two other elements that deviate from the diagonal rule. Name them:

{10}_____________________________________________________________________________

SECTION 13.5 Trends in Ionization Energies and Atomic Radii

The amounts of ionization energy required to drive out the first electron from atoms is summarizedin Table 13.7. You should notice that certain generalizations can be made regarding trends in theionization energies as you go down a family or across a row.

Generally speaking, how do the values change as you go down a family?

{11}____________________________________________________________________________

Generally speaking, how do the values change as you go across a row or series from left to right?

{12}____________________________________________________________________________

As you move down a family, the atoms are adding energy levels and are getting larger. Theoutermost electrons are farther and farther from the nucleus, which makes them easier and easier toremove. As you follow the trends in Table 13.7, some of the values seem too large and out of place.These numbers reflect the extra stability that results from full and half-full outer sublevels such as the 1013kJ/mole for phosphorous (15P) resulting from its half-full p3 outer sublevel. Therefore, the ionizationenergies get smaller as you move down a family.

As you move from left to right across a row, the atoms are not adding energy levels, but they areadding sublevels and (you might think) gaining a little in size. However, protons are added to the nucleusas you move across a row, and the increasingly positive nuclei pull the electrons in tighter. So generallyspeaking, as you go across a row from left to right the atoms actually get a bit smaller. This puts electronscloser to the nucleus, and ionization energies tend to increase.

Table 13.8 contains data about atomic radii which will give you an idea of how the sizes of theatoms are changing. Look at that table now and answer the following:

Generally speaking, summarize how atomic sizes vary as you go down a family:

{13}____________________________________________________________________________

Why? {14}_______________________________________________________________________

Generally speaking, summarize how atomic sizes vary as you go across a row:

{15}____________________________________________________________________________

Why? {16}_______________________________________________________________________

13-12 ©1997, A.J. Girondi

H

1314

Li

519

Na

498

K

418

Rb

402

Cs

377

Fr

- - - -

Be

900

Ca

590

Sr

548

Ba

502

Ra

510

Sc

632

Y

615

La

540

Ac

665

Ti

657

Zr

661

Hf

674

Unq

V

649

Cr

653

Mn

715

Fe

757

Co

757

Ni

736

Cu

745

Zn

908

Ga

577

Ge

761

As

946

Se

941

Br

1138

Kr

1351

Nb

665

Mo

686

Tc

703

Ru

711

Rh

720

Pd

803

Ag

732

Cd

866

In

556

Sn

707

Sb

833

Te

870

I

1008

Xe

1172

Ta

761

W

770

Re

761

Os

841

Pd

803

Pt Au Hg

1008

Tl Pb

715

Bi

703

Po

820

At

- - - -

Rn

1038

Unp Unh Uns Uno Une Uun Uuu

Al

577

Mg

736

Si

787

P

1013

S

1000

Cl

1251

Ar

1519

B

799

C

1088

N

1402

O

1314

F

1682

Ne

2079

He

2372

Ce

527

Pr

523

Nd

531

Pm

536

Sm

544

Eu

548

Gd

594

Tb

565

Dy

573

Ho

582

Er

590

Tm

598

Yb

602

Lu

523

Th

665

Pa U Np Pu

561

Am

577

Cm Bk Cf Es Fm Md No Lr

Table 13.7First Ionization Energies of Atoms (kJ/Mole)

1A

2A 3A 4A 5A 6A 7A

8A

First ionization energies generally increase from left to right, and generally decrease from top to bottom.

---------------------------------------------------------------------------------------------------------------------

H

0.32

Li

1.23

Na

1.54

K

2.03

Rb

2.16

Cs

2.35

Fr

- - - -

Be

0.89

Ca

1.74

Sr

1.91

Ba

1.98

Ra

2.20

Sc

1.44

Y

1.62

La

1.69

Ac

2.0

Ti

1.32

Zr

1.44

Hf

1.44

Unq

V

1.22

Cr

1.18

Mn

1.17

Fe

1.17

Co

1.16

Ni

1.15

Cu

1.17

Zn

1.25

Ga

1.26

Ge

1.22

As

1.20

Se

1.17

Br

1.14

Kr

1.12

Nb

1.34

Mo

1.30

Tc

1.27

Ru

1.25

Rh

1.25

Pd

1.28

Ag

1.34

Cd

1.48

In

1.44

Sn

1.40

Sb

1.40

Te

1.36

I

1.33

Xe

1.31

Ta

1.34

W

1.30

Re

1.28

Os

1.26

Pd

1.27

Pt

1.30

Au

1.34

Hg

1.48

Tl

1.48

Pb

1.47

Bi

1.46

Po

1.46

At

1.45

Rn

- - - -


Al

1.18

Mg

1.36

Si

1.11

P

1.06

S

1.02

Cl

0.99

Ar

0.98

B

0.82

C

0.77

N

0.74

O

0.74

F

0.68

Ne

0.67

He

0.31

Ce

1.65

Pr

1.64

Nd

1.64

Pm

1.63

Sm

1.62

Eu

1.85

Gd

1.62

Tb

1.61

Dy

1.60

Ho

1.58

Er

1.58

Tm

1.58

Yb

1.70

Lu

1.56

Th

1.65

Pa U

1.42

Np Pu Am Cm Bk Cf Es Fm Md No Lr

Table 13.8Atomic Radii (Size) of Atoms (Å)

1A

2A 3A 4A 5A 6A 7A

8A

1 Å = 1 X 10-8 cm

The size of atoms generally decreases across a row from left to right, and increases down a family from top to bottom.

13-13 ©1997, A.J. Girondi

SECTION 13.6 Configuration and Orbital Notation of Ions

Problem 4. In Table 13.9 determine the charge of the most common ions formed by each element.The electron configuration of the ions should be the same as the nearest noble gas (according to atomicnumber) to each element. Some atoms gain electrons to become ions, while others will be losingelectrons. Some atoms can do both. Complete Table 13.9.

Table 13.9Electron Configurations of Selected Atoms and Ions

Element Electron Configuration Electron Configuration Charge of the Atom of the Ion on Ion

N ______________________ ______________________ ______

Na ______________________ ______________________ ______

Mg ______________________ ______________________ ______

Li ______________________ ______________________ ______

Be ______________________ ______________________ ______

O ______________________ ______________________ ______

F ______________________ ______________________ ______

C ______________________ ______________________ +4

Problem 5. Complete Table 13.10 below. Do not use core notation here.

Table 13.10Electron Configurations of Some Common Ions

Ion Electron Configuration No. of Protons No. of Electrons of Ion in Ion in Ion

Si4+ __________________________ ______ ______

C4- __________________________ ______ ______

N3- __________________________ ______ ______

Al3+ __________________________ ______ ______

Cl1- __________________________ ______ ______

K1+ __________________________ ______ ______

P3- __________________________ ______ ______

S2- __________________________ ______ ______

13-14 ©1997, A.J. Girondi

Problem 6. Table 13.11 below, asks you to write the orbital notation for the IONS for which you wroteconfiguration notation in Table 13.10. Do not use core notation here.

Table 13.11Orbital Notation of Some Common Atoms and Ions

Atom Ion Orbital Notation

1s 2s 2p 3s 3p 4s

Si ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )Si4+ ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

C ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )C4- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

N ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )N3- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

Al ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )Al3+ ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

Cl ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )Cl1- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

K ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )K1+ ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

P ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )P3- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

S ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )S2- ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

Problem 7. Knowing what you now do about how and why atoms become ions, can you see whyelements in the same family often form ions that have a common charge or charges? Look back over thework you have done and determine the most common negative and positive charge on ions for each ofthe families listed in Table 13.12.

Table 13.12Ion Charges in Families of Elements

Number of Most Common Most CommonFamily Valence Electrons Positive Charge Negative Charge

1A _______ ________ __XXXX__

2A _______ ________ ________

3A _______ ________ ________

4A _______ ________ ________

5A _______ ________ ________

6A _______ ________ ________

7A _______ ________ ________

8A _______ __XXXX__ __XXXX__

XXXX = these ions are either rare or nonexistant, so we will ignore them.

13-15 ©1997, A.J. Girondi

SECTION 13.7 Quantum Mechanics and Atomic Orbitals

Quantum mechanics is a set of laws which describe the behavior of very small particles such aselectrons. These laws provide a logical basis for the periodic arrangement of the elements on the periodictable. The similarities in numbers of valence electrons accounts for the existence of chemical families.The energy level diagrams and shorthand electron configurations we have examined are good"bookkeeping" devices for keeping track of electrons in elements.

Quantum mechanics also allows us to "draw pictures" of the probable locations of electronsaround the nucleus. The theory is based partially on the fact that small particles, electrons in this case,which are confined to a very small space exhibit very unusual behavior. The electrons in an atom do notfollow any predictable pathway. In fact, sometimes they behave like particles and sometimes they behavelike waves. When an electron has wave characteristics, it acts as though it is spread out over a large space.Because of their dual nature, electrons cannot be accurately represented as being little particles that orbitthe nucleus like planets orbit the sun. Such a picture ignores the wave characteristics of the electron.

What quantum mechanics theory does allow us to do is to construct three-dimensional shapesarranged about the nucleus of an atom to represent the space in which a particular electron spendspractically all of its time. The analogy below might be helpful in understanding what these three-dimensional shapes mean.

Suppose that you were to take a map of your city and use it to mark points on it to represent yourlocation at particular times during the day and night. Further, suppose that each hour you mark a point onyour map for your given location and that you do this for one month. At the end of the month, your mapwould show a large number of points clustered around the location of your school and your home. That is,there would be a large "density" of points at these particular locations. There would be other locations atwhich there might be a considerable density (number) of points such as a favorite shopping mall or eatingplace. However, the density of these locations would not be as great as that around your school andhome. It is likely that there would be locations on your map at which no points would be placed (city dump,police station, etc.).

Now suppose someone wanted to locate you for some reason. If they had a copy of your map,they could see where the "probable" places were in which they might locate you. They would probably goto the place with the highest density of points, and, if you were not there, they would travel to the nextmost probable location. This map would not tell them exactly where you were, but it certainly would savethem time in trying to find you. Such a map could be called a "probability density plot" for your locationover a long time period.

Quantum mechanics can give us similar information about the location of electrons. It allows us toconstruct probability density plots for the various electrons in an atom. The equations which have beendeveloped to do this are too complex to be presented here. Unlike your map, which is a two-dimensionalrepresentation, the probability density plots for electrons must be constructed in three dimensions.(Although we can only draw two- dimensional plots on these flat pages, we must hope that your mind willallow you to think in three dimensions.) The plots that we draw for electrons represent the space in whichparticular electrons spend almost all their time. These are orbitals. They do not represent the space inwhich electrons spend 100% of their time, because this would require plots with infinite extent. You mighttake a trip out of the city during the month and would not be able to place any points on your map torepresent your location, because your map only encloses the boundaries of your city. Your map hasbecome a probability density plot of your location for almost all of the time (but not 100%). Just as youmight spend some of your time out of town, electrons can and do spend some time outside of the three-dimensional spaces that we plot for them.

What do these three-dimensional probability density plots look like? Drawings appear in Figures13.1 through 13.3. The size and shape of these plots depend on the orbital that an electron occupiesabout the nucleus. There is some regularity among them. For example, all "s" orbitals (like 1s, 2s, 3s, etc.)

13-16 ©1997, A.J. Girondi

have the same basic shape - a hollow sphere. (Actually their description is not this simple, but it's goodenough for our purposes.) The radius for the sphere representing the 1s orbital is smaller than that for the2s orbital, which is smaller than that for the 3s orbital, and so on.

The three orbitals in the "p" sublevel for atoms all have the same shape. The "p" orbitals areshaped like a long balloon squeezed together in the middle. Since there are three of these in every major"p" sublevel, they are found to point along the directions of a set of x, y, z axes that intersect at thenucleus. Just as is the case for "s" orbitals, the 3p orbitals extend farther out from the nucleus than the2p, and so on. The shapes of the "s" and "p" orbitals are shown in Figures 13.1 & 13.2. The shapes thatappear in Figures 13.1 through 13.3 are actually depictions of three-dimensional waves. In a sense, thesurfaces of these figures correspond to the boundaries of your city on the map in the analogy that weused. The electrons in a particular orbital are very likely to be within the boundary surfaces most of thetime, just as you were very likely to be located within the boundaries of your city most of the time. In yourprobability density map there would be a concentration of points around those locations where you spendmost of your time. Within the boundaries of the probability density plots for electrons shown in figure11.1, there are regions in which the "s" electrons spend most of their time. The same thing is shown the"p" electrons in Figure 13.2.

The shape of "d" and "f" probability density plots become more complicated. The shape of justone of the "d" orbitals is shown in Figure 13.3. There are five "d" orbitals, since there are five "d"sublevels. Electrons are important in understanding chemical reactions and chemical bonding.Knowledge about quantum mechanics is helpful in understanding the kinds of ions formed by atoms andthe shapes of molecules that atoms form.

1s 2s 3s

Figure 13.1Depictions of 1s, 2s, and 3s Orbitals

px py pz

Figure 13.2Depictions of the Three 2p Orbitals

13-17 ©1997, A.J. Girondi

Figure 13.3One of the Five "d" Orbitals

The subject of atomic structures and electron behavior asrevealed by the use of quantum mechanics is a very difficultsubject. You have been given much information, but don't feelthat anything is wrong if this leaves you somewhat confused.You have only scratched the surface of the subject, and a realunderstanding would require much more study (perhaps incollege–level chemistry or physics). You have discovered thatelectrons do not behave in a manner that makes it easy to draw"pictures" of atoms. In this chapter, you have been "let in" on avery clever way that scientists have developed to deal withelectrons. While this way is at times confusing and difficult, it isalso very fascinating and quite useful, as you will see in the nextchapter.

ACTIVITY 13.8 Creating a Probability Density Plot

The position of an electron in an atom at a given moment cannot be predicted. The region ofspace in which the electron can most probably be found is, however, predictable. This region is oftencalled an "electron cloud" or "orbital" and is represented by a fuzzy shape with the nucleus at the center.The shape is determined by mathematical calculations using what is known as the wave mechanical modelof the atom. The electron cloud is not absolute. Sometimes, the electron may be found outside thecloud.

The purpose of this activity is to create a probability distribution of locations (a probability densityplot) around a central point. You will need a crayon and a paper "bull's eye" target. A copy of the targetcan be found within the next few pages. The procedure follows:

1. Place the target in front of you on your lab table or on the floor.

2. Obtain the crayon provided and hold it about 2 feet above the target. Drop the crayon onto the target,trying to hit the center. Repeat this at least 50 times. (See Figure 13.4 below.)

3. Count the number of marks in each numbered region of the target and record the numbers in data table13.13 below.

Figure 13.4Creating a Probability Density Plot

Table 13.13Distribution of Hits

Area Number of Hits

1

2

3

4

5

Out

13-18 ©1997, A.J. Girondi

Plot your results on the grid on the next page. Plot the number of hits on the Y axis and thenumbered areas on the X axis.

Based on your results, which numbered area of the target had the highest probability of a hit?_______

According to your data, the probability that a given drop of the crayon will hit region 1 is: 1 in _______. Of

the various shapes of plots shown in figures 13.1, 13.2, and 13.3 would you say that your plot most

resembles an s, p, or d orbital?_________

Have your lab partner make a copy of your plot by placing another copy of the target under yours,and holding it up to a window. Using a pen or crayon, duplicate the hit marks. Be sure to include yourtarget when submitting this chapter to your instructor for approval.


The learning outcomes for Chapter 13 are listed below. Check each one when you are certainthat you have mastered it. Arrange to take any quizzes or exams, and then move on to chapter 14.

_____1. Write the electron configuration of atoms and of ions.

_____2. Write the orbital notation for atoms and of ions.

_____3. Write the core notation (configuration and orbital) for atoms and ions.

_____4. Define valence electrons and determine the number of them in an atom.

_____5. Describe trends in ionization energy across a row and down a column on the periodic table.

_____6. Describe trends in the size of atoms across a row and down a column on the periodic table.

_____7. Explain what is meant by ionization energy.

_____8. List the common charges of ions of elements in columns 1A - 8A of the periodic table.

_____9. Be able to identify probability density plots of s, p, and d orbitals.

_____10. Define a probability density plot and explain how it is useful in describing the location of an

electron in an atom.

13-19 ©1997, A.J. Girondi


Questions:

{1} one; {2} eight; {3} 4s24p6; {4} 3s23p1; {5} 5s25p4; {6} third; {7} Fourth electron must come from thesecond energy level which is closer to the nucleus; {8} 1A; {9} 2A; {10} copper, niobium, molybdenumand many more; {11} decrease; {12} increase; {13} get larger; {14} each successive one has one moreenergy level; {15} get smaller; {16} increased attraction for electrons by nuclei with more protons;

Problems:

1. 2s2, 3s2, 4s2, 5s2, 6s2, 7s2; 2s22p1, 3s23p1, 4s24p1, 5s25p1, 6s26p1

2. 3s23p2, 4s24p5, 4s2, 4s24p3, 3s23p4, 5s2, 6s26p5, 6s2, 3s23p5

3. See Table 13.4 for answers.4. N 1s22s22p3 1s22s22p6 -3

Na 1s22s22p63s1 1s22s22p6 +1Mg 1s22s22p63s2 1s22s22p6 +2Li 1s22s1 1s2 +1Be 1s22s2 1s2 +2O 1s22s22p4 1s22s22p6 -2F 1s22s22p5 1s22s22p6 -1C 1s22s22p2 1s2 +4

5. Si4+ 1s22s22p6 14 10C4- 1s22s22p6 6 10N3- 1s22s22p6 7 10Al3+ 1s22s22p6 13 10Cl1- 1s22s22p63s23p6 17 18K1+ 1s22s22p63s23p6 19 18P3- 1s22s22p63s23p6 15 18S2- 1s22s22p63s23p6 16 18

1s 2s 2p 3s 3p 4s

Si (X) (X) (X)(X)(X) (X) (/)(/)( ) ( )Si4+ (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )

C (X) (X) (/)(/)( ) ( ) ( )( )( ) ( )C4- (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )

N (X) (X) (/)(/)(/) ( ) ( )( )( ) ( )N3- (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )

Al (X) (X) (X)(X)(X) (X) (/)( )( ) ( )Al3+ (X) (X) (X)(X)(X) ( ) ( )( )( ) ( )

Cl (X) (X) (X)(X)(X) (X) (X)(X)(/) ( )Cl1- (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )

K (X) (X) (X)(X)(X) (X) (X)(X)(X) (/)K1+ (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )

P (X) (X) (X)(X)(X) (X) (/)(/)(/) ( )P3- (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )

S (X) (X) (X)(X)(X) (X) (X)(/)(/) ( )S2- (X) (X) (X)(X)(X) (X) (X)(X)(X) ( )

6.

7. 1A 1 +1 XXXX2A 2 +2 -63A 3 +3 -54A 4 +4 -45A 5 +5 -36A 6 +6 -27A 7 +7 -18A 8 XXX XXX

13-20 ©1997, A.J. Girondi

1

2

3

4

5

Figure 13.5Target for Probability Density Plot

13-21 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 14

CHEMICALBONDING

(PART 1)

14-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


14-2 ©1997, A.J. Girondi

SECTION 14.1 Introduction to Chemical Bonding

The major topic in this chapter is chemical bonding. The study of bonding will help you to realizehow important the information is that you learned concerning the internal structure of atoms. The study ofchemical bonding will help to bring all that information together in a comprehensive way. All chemicalcompounds contain atoms that are fastened together by what are called chemical bonds. When we writechemical equations, they usually describe the overall effect of the "making" and "breaking" of chemicalbonds that take reactants and turn them into products. These equations contain chemical formulas, butthey do not give descriptions of the bonds that are broken and formed when reactions occur.

We can use what we have learned about the structure of atoms and the behavior of electrons todetermine why atoms of elements combine to form compounds as they do. We will also look at thestructure of these compounds in terms of the chemical bonds that are formed. We will be studying twomajor categories of bonds: ionic bonds and covalent bonds. There are also molecules which containbonds which have properties somewhere in between those of covalent and ionic bonds, and those arecalled polar covalent bonds. You cannot tell what kind of bonding a compound has simply by looking at itsformula; however, you will learn how to determine the nature of the bonding in compounds.

One of the key concepts in chemical bonding theory involves the behavior of electrons as theyorganize themselves into some sort of "stable" arrangement in the orbitals of atoms. You studied thispreviously, and you should recall that certain arrangements of electrons are more stable than others.

The most stable arrangement is an octet of electrons (8 of them) arranged in an s2p6 fashion in theoutermost energy level of an atom or ion. This arrangement is characteristic of the noble gases.Remember? A full outer energy level is another very stable arrangement. This is characteristic of heliumwhich has only one energy level and it is full with two electrons (1s2). Other arrangements exist which areless stable than these, but still represent above average stability. They are full and half-full sublevels.Examples include the s2p3 arrangement characteristic of the elements in family 5A (note that s2 is full andp3 is half-full), and the s2 (a full outer sublevel) arrangement characteristic of the elements in family 2A.

Atoms lose or gain electrons in an attempt to achieve one of these stable arrangements. By far,the most important of these is the "octet" arrangement of the noble gases. These atoms are so stable,that it is very difficult, indeed, to get them to bond to other atoms. They "like" being the way they are.

Atoms attempt to obtain a noble gas configuration by either gaining electrons from other atoms orby losing electrons to other atoms. (A noble gas configuration is either the 1s2 arrangement of helium oran octet, s2p6, such as the other noble gases have.) An atom that accepts electrons may require only afew electrons to obtain a noble gas configuration. An atom that loses electrons may only need to lose afew to achieve a noble gas configuration. In this situation, both atoms would attain stability by losing orgaining a few electrons.

You studied ion formation previously, but let's review it here in Chapter 14 before we study theformation of chemical bonds. The following orbital diagrams describe the electron configuration in thesodium atom and ion.

1s 2s 2p 3s

Na atom: (X) (X) (X)(X)(X) (/) 1s22s22p63s1

Na1+ ion: (X) (X) (X)(X)(X) ( ) 1s22s22p6

Notice that the sodium atom loses one electron to achieve the stable 2s22p6 arrangement. The result isthe formation of a Na1+ ion which has the same electron configuration as a noble gas. The sodium ion,Na1+, has the electron configuration of what noble gas?{1}________________________

14-3 ©1997, A.J. Girondi

The electron configurations of the fluorine atom and the fluoride ion are shown below. 1s 2s 2p

F atom: (X) (X) (X)(X)(/) 1s22s22p5

F1- ion: (X) (X) (X)(X)(X) 1s22s22p6

Notice that the fluorine atom gains one electron to achieve a stable "octet" arrangement. The fluoride ion,F1-, has the electron configuration of what noble gas?{2}_______________________ Because theyhave stable noble gas configurations, F1- and Na1+ are very stable ions. The formation of stable ions isvery important in the formation of certain types of chemical bonds. Before we begin to study the types ofbonding, let's review more information about ion formation.

Problem 1. Complete the exercise below. In the orbital diagrams, use slashes to place the correctnumber of electrons in the orbitals for the neutral atoms. Then, determine the most probable charge onthe ions and place the correct number of electrons in the orbitals of the ions. You may want to refer to aperiodic table. Remember, the atoms "want" the arrangement of the nearest noble gas (going eitherforward or backward on the periodic table). Remember that metal atoms tend to lose electrons in order togain stability, while nonmetal atoms tend to gain electrons. For example, magnesium metal tends to loseelectrons, while nitrogen (a nonmetal) tends to gain them.

1s 2s 2p 3s 3pa. nitrogen atom: ( ) ( ) ( )( )( ) ( ) ( )( )( )

1s 2s 2p 3s 3p b. negative nitrogen ion: ( ) ( ) ( )( )( ) ( ) ( )( )( )

1s 2s 2p 3s 3pc. magnesium atom: ( ) ( ) ( )( )( ) ( ) ( )( )( )

1s 2s 2p 3s 3p d. positive magnesium ion: ( ) ( ) ( )( )( ) ( ) ( )( )( )

As a very general rule (recognizing that there are exceptions), you can assume that if the outervalence shell contains less than 4 electrons, an atom will more often lose electrons and become apositively charged ion. If the atom has a valence shell that contains more than 4 electrons, it will moreoften accept electrons and become a negatively charged ion. The atoms in family 4A have 4 electrons inthe valence shell (half of an octet). You might expect these atoms to be somewhat "confused" as towhether they "should" gain or lose electrons. In theory they could either lose four electrons or gain fourelectrons to achieve an octet and a noble gas configuration. In addition, they could do something else,too. They could share electrons with other atoms. In fact, this actually is the way that elements such ascarbon and silicon (in family 4A) most frequently do achieve stability. This sharing results in the formationof covalent or polar covalent bonds.

Problem 2. Sometimes, atoms which normally lose electrons can gain them (and vice versa) to achievestability under the right circumstances. For example, sulfur normally forms the S2- ion. It can also form S4+

and S6+ ions. Do you see how that results in stability? Explain in the spaces below, why each ion listedmay have more stability than normal.

S2-:______________________________________________________________________

S4+:______________________________________________________________________

S6+:______________________________________________________________________

14-4 ©1997, A.J. Girondi

You should remember that atoms in a family have the same arrangement of electrons in theirvalence shells. Therefore, they often lose or gain the same number of electrons to achieve a noble gasconfiguration. As a result, they often form ions that have the same charge. The most common ion chargefor each family is listed in Table 14.1 below. Keep in mind, though, that depending on what other elementthey are reacting with, elements can sometimes assume charges other than those listed in the table.

Table 14.1Common Ion Charges of Family 1A – 8A Elements

Family: 1A 2A 3A 4A 5A 6A 7A 8A

Charge: +1 +2 +3 ±4 –3 –2 –1 0

SECTION 14.2 Using Oxidation Numbers

The ion charges listed above are commonly referred to as oxidation numbers. The oxidationnumber of an atom is usually the charge it would have if the atom assumed the electron configuration ofthe nearest noble gas.

Another way to define oxdiation number is as:

the number of electrons lost, gained, or shared by an atom when it forms bondswith other atom(s). Therefore, atoms that are not bonded to atoms of otherelements are assigned an oxidation number of zero.

Problem 3. Using only a periodic table, try to predict a common oxidation number of the elementsbelow:

a. 20Ca:________ 31Ga:________ 36Kr:________

b. Name two elements that have the same oxidation number as nitrogen:

__________________ and __________________

You were actually using oxidation numbers in an earlier chapter when you were learning to writechemical formulas. There is a table of oxidation numbers in your ALICE reference notebook. However,keep in mind that your reference tables only include the common oxidation numbers of the elements.They have others which are less common, but nevertheless are possible.

Remember that ions of opposite charge bond in proportions that create neutral compounds. Forexample, if barium, which has an oxidation number of +2 and can be represented as Ba2+, were tocombine with O2-, the compound BaO would be formed, and it would be neutral. However, if the Ba2+

combines with Cl1-, it would form BaCl2 which is neutral in this 1 to 2 ratio. Aluminum ions with a charge of+3 bond to oxygen ions with a -2 charge to form Al2O3 which is neutral. Hopefully, this is all familiar to youfrom your previous work with chemical formulas.

14-5 ©1997, A.J. Girondi

Problem 4. Complete the table below as a review. One example is done for you.

Table 14.2Formulas for Selected Compounds

Element 1 Ion Symbol Element 2 Ion Symbol Compound Formula

sodium Na1+ chlorine Cl1- NaCl

barium ______ oxygen ______ __________

aluminum ______ bromine ______ __________

boron ______ chlorine ______ __________

hydrogen ______ oxygen ______ __________

silicon ______ oxygen ______ __________

hydrogen ______ nitrogen ______ __________

The element with the positive oxidation number is written first in the formula of a compound. Anexception to this rule (for whatever reason) is the compound between hydrogen and nitrogen. In thatcase, the negative element is written first. Why? Good question! If necessary, go back and change youranswer for the hydrogen – nitrogen compound in Table 14.2.

If you know the oxidation numbers of some atoms in the formula of a compound, they can oftenbe used to determine the oxidation numbers of other atoms in the same formula. Let's use the compoundK2CrO4 to illustrate this point. Potassium has an oxidation number of +1 and oxygen has an oxidationnumber of -2. So, what's the oxidation number of the chromium atom (Cr) in the formula K2CrO4?

2 K1+ ions = +24 O2- ions = – 8

Total = – 6

Note that in the calculations shown in the box at right, the total of the charges is –6. So, to make the compound neutral we need a charge of +6. Thus, the one chromium atom must have a charge of +6.

Problem 5. Try the same procedure to determine the oxidation number of the single manganese (Mn)atom in potassium permanganate, KMnO4. Show how you got your answer.

In Table 14.3, the oxidation numbers of some elements are given, and you are asked to calculate theoxidation numbers of the other elements. Try to use only the information given. Do not refer to anyreference source. An example is given in the table.

14-6 ©1997, A.J. Girondi

Problem 6. Complete Table 14.3 below.

Table 14.3Balancing Electron Charges

Compound Oxidation Numbers Unknown Element and Ox. No.

FeCl3 Cl = -1 Fe = +3

Fe2O3 O = -2 ________________

FeS S = -2 ________________

KClO3 K = +1; O = -2 ________________

Cu2O O = -2 ________________

NaNO3 Na = +1; O = -2 ________________

An ion, either positive or negative, that is composed of more than one atom is known as apolyatomic ion or radical. You were first introduced to these ions when you learned to write formulas.These ions are interesting from a chemical standpoint in that they tend to behave as a single atom. A list ofsome of these polyatomic ions is found in your ALICE reference notebook.

The total charge of a polyatomic ion is equal to the sumof the oxidation numbers of all of the atoms in the ion.

Note that the sum of the charges here is not zero! For example, what is the charge of a chromium atom inthe dichromate ion: Cr2O72-? The charge on the ion is -2. The charge on an oxygen atom is also -2. Sincethere are seven oxygen atoms in the formula of the ion, their total charge is: (-2) X 7 = -14. Since the ionhas a charge of -2, then the chromium atoms must have a total charge of +12: (+12) + (-14) = -2. Now,since there are two chromium atoms in the formula of the ion, each one must have a charge of +6:(+12) ÷ 2 = +6. Thus, the oxidation number or charge on each chromium atom in the dichromate ion is+6.

Problem 7. Complete Table 14.4. Find the oxidation number of each atom whose oxidation number isnot given. Note that one has been done for you as an example.

Table 14.4Oxidation Numbers of Atoms in Polyatomic Ions

Polyatomic Ion Oxidation Numbers Unknown Element and Ox. No.

MnO41- O = –2 Mn = +7

NO21- O = –2 __________________________

S2O32- O = –2 __________________________

NH41+ H = +1 __________________________

HSO41- H = +1; O = –2 __________________________

H2PO41- H = +1; O = –2 __________________________

AsO43- O = –2 __________________________

14-7 ©1997, A.J. Girondi

SECTION 14.3 Ionic and Covalent Bonding

Now that you know more about ions, you are capable of understanding ionic bonding. An ionicbond exists between two ions which are held together by the attraction of a positive and a negativecharge. The series of drawings below illustrate the formation of an ionic bond between sodium andfluorine.

According to our criteria for bonding, atoms with a noble gas configuration in their outer shells arevery stable. When Na atoms and F atoms encounter each other, both may attain the outer shellconfiguration of neon (s2p6). The Na atom donates its 3s1 electron to the F atom. In the process, the Naatom becomes a Na1+ ion and the F atom becomes an F1- ion. Since unlike charges attract, we expectthe two ions to attract each other. This type of bond is an example of an ionic bond (that is, a bond formedbetween two ions).

Ionic bonds are always formed by two or more ions being held together by attraction of oppositecharges. One or more electrons are always moved from one atom to another in ionic bonding. In covalentbonding, electrons are not moved but are, instead, shared between atoms.

Atoms that are held together by sharing electrons are said to have covalent bonds. This situationmay seem a bit more complex, but it is still based on the prediction that a noble gas configuration is stable.In some instances, two atoms may have identical electron configurations that are short of the number ofelectrons needed to achieve a noble gas arrangement. Since neither atom has a greater tendency toaccept or to give away electrons to the other atom, the only way that they can both attain the noble gasconfiguration is by "equal sharing" of some electrons. In situations where this occurs, the two atomsremain close to each other in order to share the available electrons.

loses electron to become a

gains electron to become a

Na

F F1-

sodium atom sodium ion

fluorine atom fluoride ion

to produce: Na1+ F1-

sodium fluoride

Figure 14.1 Formation of an Ionic Bond

Na1+

In order to have a 100% "pure" covalent bond, the two atoms involved in the bond must share thebonding electrons equally. Such equal sharing occurs only if the bonded atoms have an equal attractionfor the shared electrons. This situation exists in molecules such as the 7 diatomic gases (H2, etc.).

Ionic and covalent bonds represent extremes in types of chemical bonding. Most chemicalbonds fall into a category that can be described as being somewhere in between the two extremes.

14-8 ©1997, A.J. Girondi

These bonds are calledpolar covalent bonds and represent a situation in which two atoms have notcompletely exchanged electrons as they would in ionic bonding. However, in polar covalent bonding, theatoms are not sharing the valence electrons in an equal way as they would in 100% covalent bonding.The result is a bond with characteristics somewhere in between ionic and covalent. Since the electronsthat are being shared are more strongly attracted to one of the atoms in the bond, this atom then assumesa partial negative charge character (∂ –). The atom from which the atoms have shifted assumes a partialpositive charge character (∂ +). These slightly positive and negative charge "characters" do not amount toa "full" positive or negative charge such as would characterize a positive or negative ion in an ionic bond.Still, one end of the bond is more positive and the other is more negative. The bond has charged

H2 + Cl2 H Cl∂ + ∂ –

polar covalent bond

poles (like a magnet has north and southpoles), and we describe it as a polar bond.The phrase polar covalent describesbonding in which some of the valence shellelectrons between two atoms are beingshared unequally to form a polar bond.

2

SECTION 14.4 Introduction to Lewis Electron Dot Notation

An excellent way to help visualize covalent and polar covalent bonding is by the use of electrondot representations of bonds in molecules. This method of notation was devised by a prominentAmerican chemist by the name of Gilbert N. Lewis, and, hence, it is called the Lewis electron dot method.This will be the third form of electron notation that you have learned. You learned orbital notation andconfiguration notation in previous chapters, and now you will learn about dot notation.

Since there are eight total electrons in the outer valence shell of the noble gases (s2p6), these canbe arranged in pairs around the nucleus of an atom. The nucleus of the atom and its inner electron shellsare represented in dot notation by the chemical symbol for the atom. The valence electrons are thenarranged around the symbol.

The reason that the electrons are paired is due to the fact that electrons occupy orbitals around anucleus in pairs. (Recall that each orbital was represented by parentheses or a circle in which we placedone or two slashes to represent the electrons existing in them.) Before trying to illustrate bonds using dotnotation, let's first learn how to represent the valence electrons of single atoms. The steps outlinedbelow will guide you.

Z Z 21

4 356

7 8

Step 1. Write the symbol of the atom. There are eight positions for dots aroundthe symbol. These eight positions are often numbered starting with position 1 atthe "four o'clock" position and moving counterclockwise around the symbol. Thedots represent electrons that are found in the valence shell of the atom. The lettersrepresent the atom's nucleus and inner core of electrons.

Step 2. When you put dots around the symbol, you should generally start at position 1 and continueadding dots counterclockwise until all valence electrons have been represented. In addition, you mustmake note of whether or not the electrons are paired or unpaired. They are paired if they are in the sameorbital. They are unpaired if they are alone in an orbital. If they are paired, then they should be shown as apair around the symbol in the dot notation. If they are not paired, they should be shown alone at one sideof the symbol. The following examples will help. The orbital notation of the atoms is shown first and isfollowed by the dot notation.

14-9 ©1997, A.J. Girondi

Element Orbital Notation Dot Notation

1s 2s 2p 3s 3p 4s1H (/) ( ) ( )( )( ) ( ) ( )( )( ) ( ) H

2He (X) ( ) ( )( )( ) ( ) ( )( )( ) ( ) He

3Li (X) (/) ( )( )( ) ( ) ( )( )( ) ( ) Li

4Be (X) (X) ( )( )( ) ( ) ( )( )( ) ( ) Be

5B (X) (X) (/)( )( ) ( ) ( )( )( ) ( ) B

6C (X) (X) (/)(/)( ) ( ) ( )( )( ) ( ) C

7N (X) (X) (/)(/)(/) ( ) ( )( )( ) ( ) N

8O (X) (X) (X)(/)(/) ( ) ( )( )( ) ( ) O

9F (X) (X) (X)(X)(/) ( ) ( )( )( ) ( ) F

10Ne (X) (X) (X)(X)(X) ( ) ( )( )( ) ( ) Ne

You should note that the dot notation shows the correct number of valence electrons, and it shows whichones are paired and which are unpaired. Remember that dot notation only shows the valence shellelectrons.

Problem 8. Since all atoms in a particular family (1A to 8A) have the same valence shell arrangement,they should then all have the same dot notation. Fill in the blanks for the elements listed on the next pageto take your first "shot" at dot notation.


1s 2s 2p 3s 3p 4s11Na ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

12Mg ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

13Al ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

14Si ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

15P ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

16S ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

17Cl ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

18Ar ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

19K ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

20Ca ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) _____

Problem 9. Follow the directions from problem 8 for the two elements below, but be careful. Theseelements contain 3d electrons, but they are not part of the outermost (valence) energy level. We will useorbital core notation this time to keep things more manageable.


4s 3d 4p 5s

21Sc [Ar] ( ) ( )( )( )( )( ) ( )( )( ) ( ) _____

22Ti [Ar] ( ) ( )( )( )( )( ) ( )( )( ) ( ) _____

14-10 ©1997, A.J. Girondi

If you did these last two (Sc and Ti) correctly, the dot notation should show two dots on the right side ofeach symbol. Many students ask, "What happens when you have more than eight electrons in thevalence shell?" You should see by now that it is never going to happen. Well, almost never. There is oneelement for which the dot notation cannot be written because it has too many electrons in its valenceshell. If you are curious, you can check the electron distributions in Appendix B of ALICE to see if youcan find that mysterious element. Hint: check #46.

Problem 10. To give you a feel for the similarities in families, complete the table below with the dotnotation of the elements through #54. You should follow the diagonal rule (even though there are manyexceptions to it).

Sc

1A

2A 3A 4A 5A 6A 7A

8A

Li

Rb

Mg

Sr Zr

Mn

Rh

Zn

Al

Sn

N

Se

Cl

Xe

He

Table 14.5Dot Notation of 54 Elements - According to the Diagonal Rule*

(There are numerous exceptions to the Diagonal Rule)

In practice, it doesn't really matter where you put the dots around the symbol. You don't alwayshave to start at "position 1" and move counterclockwise. But, it is very important to always use the correctnumber of dots and to have the correct number of paired and unpaired dots.

For example, nitrogen can be illustrated as N N Nor as or as

NNitrogen has onen o n b o n d i n gelectron pair

Nitrogen has threeunpaired (bonding)electrons

The electrons represented by the dotstructures are categorized into two types:nonbonding electron pairs and single (orbonding) electrons. Where singleelectrons appear, an ordinary covalentbond can be formed. Study the dotstructure of nitrogen below. The bondingand nonbonding electrons are identified.

Next, let's consider the Lewis structure of the F2 (fluorine) molecule to illustrate how atoms formcovalent bonds. This molecule consists of two fluorine atoms bonded together. Each fluorine atom hasseven electrons in its valence shell (2s22p5). This is one less than the Ne configuration: (2s22p6).

Since both fluorine atoms are identical, we would not expect them to form an ionic bond in whichone atom donates an electron to the other. If this were to happen, one atom would have eight electrons,while the other would have only {3}_______________________, which is not stable. If each atom were allowed toshare one pair of electrons, both atoms would essentially have eight electrons. This is illustrated in thefollowing Lewis dot structures.

14-11 ©1997, A.J. Girondi

By sharing a pair of electrons, each fluorine atom assumes a noble gas configuration. The onepair that is shared represents a covalent bond. The other electron pairs surrounding each nucleus (thereare three of them around each nucleus) are not involved in the bonding of the molecule and are callednonbonding electrons.

F F F F+

1 bonding pair

nonbonding valence pairs

7 valence electrons each

fluorine atoms fluorine molecule

We also know that oxygen exists in the form of O2 molecules. Each oxygen atom has only six valenceelectrons (2s22p4). This means that there are a total of twelve valence electrons in the O2 molecule.Below is the dot structure showing how two oxygen atoms can share electrons so that each atom has astable octet .

You will notice that in order for each of the oxygen atoms to have eight electrons (the neonconfiguration), they must share two pairs of electrons. These two pairs represent two covalent bonds (adouble bond). Interestingly, we find that a certain amount of energy is required to split the F2 moleculeapart. To do the same with the oxygen molecule requires much more energy, which is consistent with theidea that there are two bonds between the oxygen atoms while there is only one between the fluorineatoms.

O O O O+

2 bonding pairs

nonbonding valence pairs


oxygen atoms oxygen molecule

each oxygen now has 8 valence electrons

The element nitrogen exists in the form of N2 molecules. Using the same approach as above, wecan include the ten valence electrons in this molecule (five each from the nitrogen atoms that have 2s22p3

configurations). (See below) This molecule must have three pairs of bonding electrons representingthree covalent bonds. This is called a triple bond. It requires much more energy to separate the N2

molecule into atoms than it does to separate O2 or F2 because of the number of covalent bonds.

N N N N+

3 bonding pairs

nonbonding valence pair


nitrogen atoms nitrogen molecule

each nitrogen now has 8 valence electrons

The H2 molecule is also the stable form of the element hydrogen. Since this element is in the firstperiod (with He being the noble gas in that row), it only requires two electrons to fill its outer (1s) level. Theelectron dot formula for this molecule only has one pair of electrons in it, which are the bonding electronsbetween the hydrogen nuclei. There are no nonbonding valence pairs as there have been in theprevious examples.

14-12 ©1997, A.J. Girondi

Problem 11. Draw the dot notations for the H2 , Cl2, and Br2 molecules.

SECTION 14.5 Electron Dot Notation of Ions

Electron dot notation can also be used to represent ions. Positive ions are known as cations(pronounced "cat ions"), while negative ions are referred to as anions (pronounced "an ions"). Earlier inthis chapter you saw the configuration notation and orbital notation of the fluorine atom and ion. For yourconvenience, those notation are repeated below along with the appropriate dot notations. (Note that thenames of anions like the fluoride ion end with the suffix "ide.")

1s 2s 2p

F atom: (X) (X) (X)(X)(/) 1s22s22p5 F

F1- ion: (X) (X) (X)(X)(X) 1s22s22p6 F1-

You can see that the fluorine atom gains one electron to gain the stable neon octet. Thus, the dotnotation of the fluoride ion contains one more dot than that of the fluorine atom. Oxygen atoms gain twoelectrons to become oxide ions which have the stable neon octet. Examine the three forms of corenotation for the oxygen atom and the oxide anion below:

1s 2s 2p

O atom: (X) (X) (X)(/)(/) 1s22s22p4 O

O2- ion: (X) (X) (X)(X)(X) 1s22s22p6 O2-

Note that the charge is always shown with the ion's dot notation. The phosphorus atom and thephosphide anion have the following core notations:

3s 3p

P atom: [Ne] (X) (/)(/)(/) [Ne] 3s23p3 P

P3- ion: [Ne] (X) (X)(X)(X) [Ne] 3s23p6 or simply [Ar] P 3-

Problem 12. Complete the notations below for the sulfur atom and the sulfide anion:

3s 3p

S atom [Ne] ( ) ( )( )( ) [Ne] __________ S

S2- ion [Ne] ( ) ( )( )( ) [Ne] __________ or ________ S

14-13 ©1997, A.J. Girondi

The electron dot symbols for cations (positive ions) do not show any dots! The forms of core anddot notation for the sodium atom and cation are found below. Note that sodium loses one electron to gainthe stable neon configuration. You should also note again that the charge is always shown on the ion.

3s

Na atom: [Ne] (/) [Ne]3s1 Na

Na1+ ion: [Ne] ( ) [Ne] Na1+

Why is the empty 3s orbital shown for the sodium ion? Well, the complete electron configuration of thesodium ion is 1s22s22p6. Note that although it has no electrons in it, energy level 3 is still considered to bethe valence shell of the sodium ion, and so it is the energy level used to write the notations.

Let's study another example using a cation. The core and dot notations for the aluminum atom and cationare illustrated below. Again, note that no dots are used to write the dot notation of the cation, and theempty orbitals in the valence shell are shown. It works this way for all cations.

3s 3p

Al atom: [Ne] (X) (/)( )( ) [Ne]3s23p1 Al

Al3+ ion: [Ne] ( ) ( )( )( ) [Ne] Al3+

Problem 13. Now it's your turn! Complete 3 the forms of notation below for the calcium atom andcalcium cation.

4s

Ca atom: [Ar] ( ) [Ar]_________________ Ca

Ca2+ ion: [Ar] ( ) _____________________ Ca

Problem 14. Complete the three forms of notation for the silicon atom and the silicon cation below.

3s 3p

Si atom: [Ne] ( ) ( )( )( ) [Ne]_________________ Si

Si4+ ion: [Ne] ( ) ( )( )( ) _____________________ Si

Problem 15. Complete the three forms of notation for the chlorine atom and the chlorine anion below.

3s 3p

Cl atom: [Ne] ( ) ( )( )( ) [Ne]_________________ Cl

Cl1- ion: [Ne] ( ) ( )( )( ) _____________________ Cl

14-14 ©1997, A.J. Girondi

You can write electron dot notations for compounds as well as for individual atoms and ions. To do this,you simply combine the electron dot notations of the atoms or ions involved. Let's look at examples of thedot notations of a few compounds which are ionic (composed of ions). In the first example below, notethat to write the dot notation of NaCl, all we need to do is write the dot notation for the sodium cation rightnext to the dot notation for the chloride anion:

Na1+

Cl1-

dot notation of sodium cation:

dot notation of chlorine anion:

dot notation of NaCl: Na Cl1+ 1-

To write the dot notation for an ionic compound such as CaCl2, we combine the dot notation for onecalcium cation with two dot notations for two chloride ions. The negatively-charged chloride ions areattracted to the positively-charged calcium ion. However, since they have the same charge, the chlorideions repel each other. Therefore, they stay as far apart as possible, and we write them on opposite sidesof the calcium ion:

Ca

Cl1-

dot notation of calcium cation:


dot notation of CaCl2:

2+

CaCl1-

Cl1-2+

Examine the dot notations of the aluminum cation and the fluoride anion below. Note that in the dotnotation for the AlF3 compound, the negative fluoride ions are attracted to the positive aluminum ion. Thethree fluoride ions are spaced as far apart as possible. Why?{4}________________________________________________

_____________________________________________________________________________________________________________________

F1-

dot notation of aluminum cation:


dot notation of AlCl3:

Al3+

Al3+

F1-

F1-

F1-

Problem 16. Write the dot notation for the ionic compound KF:

Problem 17. Write the dot notation for the ionic compound BaF2:

Problem 18. Write the dot notation for the ionic compound GaCl3:

14-15 ©1997, A.J. Girondi

SECTION 14.6 Polar Covalent Bonds

All of the examples we have used so far to illustrate covalent bonding have involved bondsbetween two identical atoms. These are cases in which the bonds are 100% covalent because neitheratom has a stronger tendency to gain or lose electrons. Next, we will look at the large category of chemicalbonds that we have previously referred to as polar covalent bonds. These are bonds can be described intwo ways (which really mean the same thing). These descriptions are:

1. The bonds involve incomplete electron transfer between atoms; that is, they are only partially ionicbonds.

2. The bonds involve unequal sharing of electron pairs; that is, they are only partially covalent bonds.

Both descriptions are useful and, as mentioned, amount to the same thing. These bonds are somewherebetween the two extremes of 100% ionic bonds and 100% covalent bonds. It is helpful to use Lewis dotstructures to describe molecules that contain these bonds. Check the dot notation for the NH3 moleculein the second equation below. Sometimes, equations make use of electron dot structures. In thoseequations which involve diatomic gases, the diatomic gas molecules must break apart into individual atomsbefore they can react with other elements. In this chapter, diatomic gases will often be shown as individualatoms with valence electrons shown as dots.

H2 2 H H H 2 H or

Because nitrogen has three bonding electrons, it can form three bonds. Nitrogen combines with threehydrogen atoms by sharing three electrons.

N3 H + N HH

H The NH3 molecules has 3 polar covalent bonds.

By sharing electrons with hydrogen, nitrogen now has what is called a stable octet configuration or a stablenoble gas configuration. This means it has the electron configuration of a noble gas, and that makes itvery stable. The nitrogen atom is surrounded by eight electrons (an octet).

H O+H

If hydrogen atoms combine with oxygenatoms, they combine in a ratio of 2:1.

2 H O

Note that oxygen ends up with a stable octet, and each hydrogen ends up with a stable heliumconfiguration. This is why water is a stable molecule. When not using electron dot structures, thisequation can be written as: 2 H2 + O2 -----> 2 H2O. Oxygen combines with two hydrogens to formwater; thus, we can be fairly certain that oxygen has two bonding electrons.

14-16 ©1997, A.J. Girondi


This is the end of Chapter 14 and the first half of our discussion of chemical bonding. Theremainder of the discussion on chemical bonding follows in chapter 15. Check the learning outcomes.When you feel that you have mastered them, take the exam or quizzes and move on to Chapter 15.

_____1. Determine the ion charge (common oxidation number) of an element based on its electron configuration.

_____2. Give a common ion charge (oxidation number) for each family of elements (1A through 8A).

_____3. Predict the common oxidation numbers of elements and use them to write the correct formulas for compounds containing those elements.

_____4. Write Lewis electron dot symbols for atoms and ions, and electron dot structures for simple molecules.

_____5. Distinguish between nonbonding and bonding electrons in dot structures.

14-17 ©1997, A.J. Girondi


Questions:

{1} neon; {2} neon; {3} six; {4} The negative fluoride ions repel each other.

Problems:1.

1s 2s 2p 3s 3pa. Nitrogen atom: (X) (X) (/)(/)(/) ( ) ( )( )( )b. Nitrogen ion (N3-): (X) (X) (X)(X)(X) ( ) ( )( )( )c. Magnesium atom: (X) (X) (X)(X)(X) (X) ( )( )( )d. Magnesium ion (Mg2+): (X) (X) (X)(X)(X) ( ) ( )( )( )

2. S2-: stable because it has the argon configuration)S4+: stable because it has a full outer sublevel (3s2)S6+: stable because it has the neon configuration

3. a. 20Ca:___+2_____ 31Ga:___+3_____ 36Kr:__0______

b. Name two elements that have the same oxidation number as nitrogen: ____phosphorus__ and ___arsenic___. (Elements like Sb and Bi in family 5A are considered metals rather than nonmetals, and since metals tend to lose rather than gain electrons, they assume different oxidtion numbers.)

4.sodium Na1+ chlorine Cl1- NaClbarium Ba2+ oxygen O2- BaOaluminum Al3+ bromine Br1- AlBr3boron B3+ chlorine Cl1- BCl3hydrogen H1+ oxygen O2- H2Osilicon Si4+ oxygen O2- SiO2

hydrogen H1+ nitrogen N3- NH3

5. Mn = +7

6.FeCl3 Cl = -1 Fe = +3 Fe2O3 O = -2 Fe = +3FeS S = -2 Fe = +2KClO3 K = +1; O = -2 Cl = +5Cu2O O = -2 Cu = +1NaNO3 Na = +1; O = -2 N = +5

7.MnO41- O = -2 Mn = +7 NO21- O = -2 N = +3S2O32- O = -2 S = +2NH41+ H = +1 N = -3HSO41- H = +1; O = -2 S = +6H2PO41- H = +1; O = -2 P = +5 AsO43- O = -2 As = +5

14-18 ©1997, A.J. Girondi

8.

11Na (X) (X) (X)(X)(X) (/) ( )( )( ) ( ) _____

12Mg (X) (X) (X)(X)(X) (X) ( )( )( ) ( ) _____

13Al (X) (X) (X)(X)(X) (X) (/)( )( ) ( ) _____

14Si (X) (X) (X)(X)(X) (X) (/)(/)( ) ( ) _____

15P (X) (X) (X)(X)(X) (X) (/)(/)(/) ( ) _____

16S (X) (X) (X)(X)(X) (X) (X)(/)(/) ( ) _____

17Cl (X) (X) (X)(X)(X) (X) (X)(X)(/) ( ) _____

18Ar (X) (X) (X)(X)(X) (X) (X)(X)(X) ( ) _____

19K (X) (X) (X)(X)(X) (X) (X)(X)(X) (/) _____

20Ca (X) (X) (X)(X)(X) (X) (X)(X)(X) (X) _____

Na

Al

Mg

SiP

S

Cl

Ar

K

Ca

9.

21Sc [Ar] (X) (/)( )( )( )( ) ( )( )( ) ( ) _____

22Ti [Ar] (X) (/)(/)( )( )( ) ( )( )( ) ( ) _____

Sc

Ti

10. All elements within a family has the same dot notation. Exception is helium in family 8A. All transitionelemtns have the same notation. (Remember this does not account for exceptions to the diagonal rule.)

H H Cl Cl Br Br11.

12.

S atom [Ne](X) (X)(/)(/) [Ne]3s23p4

S2- ion [Ne](X) (X)(X)(X) [Ne]3s23p6 or just [Ar]

S

S2-

13.

Ca atom: [Ar](X) [Ar]4s2 Ca

Ca2+ ion: [Ar]( ) [Ne]3s23p6 or just [Ar] Ca2+

14.

Si atom: [Ne](X) (/)(/)( ) [Ne]3s23p2 Si

Si4+ ion: [Ne]( ) ( )( )( ) [He]2s22p6 or just [Ne] Si4+

Cl1-

15. 3s 3p

Cl atom: [Ne] (X) (X)(X)(/) [Ne]3s23p5

Cl1- ion: [Ne] (X) (X)(X)(X) [Ne]3s23p6 or just [Ar]

Cl

Ba FF17.2+

16. K F1+ 1- 1-1-

Ga3+

Cl1-

Cl1-

Cl1-

18.

14-19 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 15

CHEMICALBONDING

(PART 2)

15-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


15-2 ©1997, A.J. Girondi

SECTION 15.1 sp3 Hybridization of Atoms

In this chapter we will continue the discussion from Chapter 14 concerning Lewis dot structures.The electron dot structures for NH3 and H2O were discussed at the end of the last chapter. The formulasof these compounds are in agreement with the numbers of bonding electrons in nitrogen and oxygen.Nitrogen has three bonding electrons, while oxygen has two. The dot structures are presented again(below) for your review.

H N H O

N HH

H H OH

and and The atoms

The molecules

The next example presents us with a problem. It involves the compound CH4, which is methaneor natural gas. In this molecule there are four identical C–H covalent bonds. Each involves a shared pair ofelectrons between a H atom and the C atom. Since all four bonds in this molecule are exactly alike, thebonding electrons must be located in four identical orbitals. However, the dot notation of the carbon atomindicates that it has only two bonding electrons. How can carbon bond to four hydrogen atoms?

C + 4 H -----> CH4 ?????

bonding nonbonding

Solving this problem was a challenge for scientists. They knew that carbon almost always forms fourbonds, yet the orbital notation of the carbon atom showed that it had only two bonding electrons. In aneffort to explain this discrepancy, scientists devised the concept of hybrid orbitals. Figure 15.1 belowillustrates an ordinary carbon with the valence electron configuration: 2s22p2.

2 electrons in the 2s orbital

2 of the 2p orbitals contain 1electron, while the third 2porbital is empty

The unhybridized carbon atom has 2electrons in the 2s orbital and 1 unpairedelectron in 2 of its 2p orbitals

+ =

+ =

px

py

pz

py

pz

px

Figure 15.1 The Unhybridized Carbon Atom

The word hybrid is usually used in connection with things such as hybrid plants and animals like amule. In biology, the term hybrid refers to the offspring of two animals or plants of different species. Thehybrid animal or plant exhibits some characteristics of both "parents," yet it is different from both.

15-3 ©1997, A.J. Girondi

In this same sense, we can mix the 2s and 2p orbitals in the carbon atom in order to end up withhybrid orbitals. The hybrids are unique and different from the s and p orbitals from which they came. Theyhave a different shape, a different distance from the nucleus, etc. When two or more hybrid orbitals form,however, they are identical to each other in every way.

The orbital notation below illustrates what happens when hybridization occurs in carbon.

Unhybridized Carbon Electron Moves Hybridized Carbon

2s 2p 2s 2p sp3 hybrids (X) (/)(/)( ) (/) (/)(/)(/) (/)(/)(/)(/)

Note that unhybridized carbon has two "paired" electrons in its 2sorbital and two of its three 2p orbitals contain a single "unpaired"bonding electron. Its other 2p orbital is empty. Whenhybridization occurs, the one 2s and three 2p orbitals changetheir shapes and become four identical "hybrid" orbitals. Thesehybrid orbitals are located at a distance from the nucleussomewhere between where the 2s and 2p orbitals used to be.

You already know that when there is more than oneorbital in a sublevel, the electrons won't pair up in the orbitals untileach orbital has one electron. Well, since these new hybrids areall in the same new sublevel, one of the two 2s electrons leavesits orbital to enter one of the new empty hybrid orbitals.

C

Figure 15.2The Hybridized Carbon Atom

CH4

Since these four hybrid orbitals were "born" out of one s and three p orbitals, they areknown as sp3 hybrid orbitals. The hybridized carbon atom now has four bondingelectrons, which explains the existence of molecules like CH4.

C C

C

H

H

H

H

First and 2 H2 4 H

then finally, C + 4 H

15-4 ©1997, A.J. Girondi

C

Figure 15.3The Methane Molecule, CH4

HH

H

H

The electron density plot of the CH4 molecule is shown at right.Notice that the hybrid orbitals are directed toward the corners(vertices) of a geometric figure called a tetrahedron whichmeans four-sided. The angles between the bonds are 109.5o.Name an element in row 3 of the periodic table that you wouldexpect to form a tetrahedral shape when bonded to four otheratoms: {1}_______________________.

C Cunhybridized carbon atom hybridized

carbon atom

SECTION 15.2 sp and sp2 Hybridization of Atoms

Below is a diagram of the orbitals of the beryllium atom. Beryllium is a family 2A element on theperiodic table, indicating that it has two valence electrons. In the unhybridized Be atom, these electronsare found paired in the 2s orbital. Based on this information, how many bonds can the unhybridized Beatom form using unpaired electron(s)?{2}________.


All three of the 2porbitals are empty

The unhybridized beryllium atom has 2electrons in the 2s orbital and noelectrons in any of its three 2p orbitals

+ =

+ =

px

py

pz

py

pz

px

Figure 15.4 The Unhybridized Beryllium Atom

This baffled scientists because compounds such as BeCl2 and BeF2 were known to exist, andthese compounds should be forming two identical bonds. The hybridization theory can be expanded inan effort to explain this.

In order to form two identical bonds, the Be atom needs two unpaired (lone) electrons eachpresent in a separate but identical orbital. So, while carbon needed four hybrid orbitals, Be needs onlytwo. Thus, the theory of hybridization dictates that the 2s orbital and only one of the three 2p orbitals in

15-5 ©1997, A.J. Girondi

the Be atom, hybridize and become identical, assuming a new shape and distance from the nucleus. Thetwo 2s electrons become unpaired with each one entering one of the new hybrid orbitals. The tworemaining 2p orbitals remain unchanged. Since these two new hybrid orbitals were "born" out of one sand one p orbital, they are called sp hybrid orbitals.

Unhybridized Be Atom Electron Moves Hybridized Be Atom

2s 2p 2s 2p sp hybrids 2p (X) ( )( )( ) (/) (/)( )( ) (/)(/) ( )( )

How many unpaired bonding electrons does this give the hybridized Be atom? {3}_________ Berylliumand other elements in family 2A of the periodic table can form two bonds that are 180o apart. Themolecules formed with family 2A elements at their center are thus linear in shape. Name an element fromrow 3 of the periodic table that you would expect to form a linear molecule when bonded to two otheratoms: {4}____________________.

Figure 15.5 The Hybridized Beryllium Atom

py

pza hybrid orbital

a hybrid orbital

Cl ClBeThe Linear BeCl2 molecule:

Family 3A elements can form hybrid orbitals, too. It is known that molecules such as BH3 exist inwhich the B atom (from Family 3A) forms three identical bonds with hydrogen. Yet, if you consider thenotation of the unhybridized B atom (1s22s22 p 1), it reveals one pair of non-bonding electrons (2s2) in itsvalence shell and only one unpaired bonding electron (2p1). (See Figure 15.6.) So, how could the boronatom form three bonds with hydrogen? After all, each hydrogen atom with one valence electron is"looking" for a single unpaired (lone) electron from another atom that it can share and, thereby attain thestable helium configuration: 1s2.

Again, the theory of hybridization can be expanded to provide an explanation. Since the boronatom already has one bonding electron, it needs only two more to explain the three bonds. The theoryholds that the 2s orbital and two of the three 2p orbitals undergo hybridization and become identical. Thethird 2p orbital is left unchanged.

Unhybridized B Atom Electron Moves Hybridized B Atom

2s 2p 2s 2p sp2 hybrids 2p (X) (/)( )( ) (/) (/)(/)( ) (/)(/)(/) ( )

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1 of the 2p orbitals contains 1electron, while the other two2p orbitals are empty

The unhybridized boron atom has 2electrons in the 2s orbital and 1 unpairedelectron in one of its 2p orbitals

+ =

+ =

px

py

pz

py

pz

px

Figure 15.6 The Unhybridized Boron Atom

How many unpaired electrons are available for forming bonds in the unhybridized B atom?{5}_________.

How many unpaired electrons are available to form bonds in the hybridized B atom?{6}__________. The

new hybrid orbitals were "born" out of one s and two p orbitals, so they are called sp2 hybrid orbitals.

Boron and the other elements in family 3A of the periodic table all form three bonds which are120oC apart. The molecules formed with family 3A elements as their central atom are referred to as beingplanar or trigonal planar in shape. Trigonal means three-sided, and planar means flat.

Figure 15.7 The Hybridized Boron Atom and the BH3 Molecule

py

hybrid orbital

hybrid orbitals B

H H

H

Name an element from row 3 of the periodic table that you would expect to form a trigonal planar molecule

when bonded to three other atoms.{7}_______________________

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SECTION 15.3 Summary Of sp, sp2 , and sp3 Hybridization

Study Table 15.1. It summarizes much of the information about hybridization theory that youhave learned up to this point.

Table 15.1Bonding Electrons in Elements of Families 1A – 8A

Family 1A 2A 3A 4A 5A 6A 7A 8A

Dot Notation Li Be B C N O F Ne (Unhybridized)

Dot Notation Li Be B C N O F -- (Hybridized)

No. ofValence 1 2 3 4 5 6 7 8 Electrons

No. ofBonds 1 2 3 4 3 2 1 0

Our discussion of hybridization has not included elements in families 5A to 8A. Actually, they canbe considered to undergo hybridization, but that is a subject that will be left for a more advanced course.The theory of hybridization does not consider family 1A elements. You will note in Table 15.1 that infamilies 5A - 8A, the number of bonds which atoms will usually form is equal to their number of unpairedvalence electrons.

SECTION 15.4 The Concept of Electronegativity

As you can see, electron dot formulas are quite useful for illustrating the polar covalent bonds inmolecules. There is another way of describing just how polar a bond is. To do this, we can use a conceptcalled electronegativity. The idea of electronegativity was first developed by Nobel prize winner LinusPauling, perhaps the foremost American chemist of the 20th century. The electronegativity of an atomreflects its tendency to "attract" electrons to itself.

A numerical value for electronegativity is assigned to atoms of every element. The value assignedto atoms of each element was determined by scientists who observed the behavior of the atoms in variouschemical reactions. Don't be concerned about the details of how the numbers were determined. You willuse these numbers to estimate the degree of ionic and covalent character in certain bonds.

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Problem 1. Use the information in Table15.1 to complete and balance the equations below, written inelectron dot form. The products should also be written in electron dot form. To save you some work, twosteps have already been done. They are: (a) any diatomic or polyatomic elements involved have alreadybeen broken into individual atoms which are ready to bond to other atoms; and, (b) the dot notation showsatoms in the hybridized form, where necessary, so that they are ready to bond to other atoms. The first equation is done for you as an example.

a. __2__ H + __1__ O ---------->

b. _____ H + _____ C ---------->

c. _____ Br + _____ Be ---------->

d. _____ F + _____ Al ---------->

e. _____ H + _____ Cl ---------->

f. _____ H + _____ S ---------->

g. _____ F + _____ P ---------->

H O

H

Since electronegativity reflects an atom's "attracting power" for electrons, the higher theelectronegativity value, the stronger the attraction. Consider a reaction between two elements. Theirrelative attraction for electrons determines how they react. We can use the electronegativity scale todetermine this attraction. See the scale in Table 15.2 (or Table R-8 in your ALICE reference notebook).Refer to those electronegativity values when answering the questions that follow.

1. In general, which group (metals or nonmetals) has the lowest electronegativity values? {8}____________

2. Arrange these elements in order of increasing electronegativity: bismuth (#83), chlorine (#17),tellurium (#52), gallium (#31), thallium (#81): {9}____________________________________________

3. Which element on the periodic table has the highest electronegativity? {10}_____________________

4. Which two elements has (have) the lowest electronegativity?{ 11}_____________________________

5. Which family of elements has the highest electronegativities? {12}____________________________

6. Which family of elements has the lowest electronegativity values? {13}_________________________

When the difference in electronegativity between atoms is high, electrons are considered to betransferred from one atom to another. In questions 3 and 4 above, you located the elements with thehighest and lowest electronegativities. Now let's see what happens when two of these atoms, cesium andfluorine, combine:

Cs + F -----> Cs1+F1-

electronegativity values: 0.7 4.0

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Because fluorine's electronegativity is so great, it "steals" an electron from cesium. The result is thatcesium becomes a positive ion and fluorine becomes a negative ion. The attraction of opposite chargesholds the two ions together. This force forms what we have called a 100% ionic bond. Ionic bonds involvethe actual transfer of electrons between atoms.

When two atoms with identical electronegativities combine to form a bond, we describe such abond as being 100% covalent. Our previous examples of H2, F2, N2, and so forth are cases where thisoccurs. These molecules have atoms in them with equal electron–attracting power, and we expect theatoms to share valence electron pairs equally in the bonds that are formed.

We can use the scale of electronegativity values to estimate the degree (percent) of ionic andcovalent character in polar covalent bonds which are formed between atoms not having identicalelectronegativities. Below Table 15.2 is a scale showing electronegativity differences. Below eachelectronegativity difference is a listing of percent ionic character.

The higher the percent ionic character, the more ionic the bond. Thelower the percent ionic character, the more covalent the bond.

Study the procedure below for determining the percent ionic and percent covalent character for NaCl.

Electronegativity of Na = 0.9Electronegativity of Cl = 3.0Difference = 3.0 – 0.9 = 2.1

Referring to Table 15.2 (or Table R-8 ), a difference of 2.1 corresponds to a 67% ionic bond. The percentionic and covalent characters must add up to 100%, so the percent covalent character is 100% – 67% =33% covalent. The bond between Na and Cl is 67% ionic and 33% covalent.

Problem 2. Using this same procedure, calculate the percent ionic and percent covalent characters ofthe compounds listed below. In the compounds with multiple atoms, AlCl3, for example, it is onlynececssary to find the electronegativity difference between one Al atom and one Cl atom. Remember,you are determining the character of each bond separately.

Percentages of Ionic and Covalent Characters in Selected Compounds

1 = Value of most electronegative element 4 = Percent Ionic character2 = Value of least electronegative element 5 = Percent covalent character 3 = Electronegativity Difference

Compound 1 2 3 4 5

MgBr2 Br = 2.8 Mg = 1.2 1.6 47 53

NaF

CO

NH3

FrCl

PBr3

15-10 ©1997, A.J. Girondi

Sodium chloride (NaCl) is an ionically bonded substance. You have seen sodium chloride manytimes - it is table salt. Each grain of salt that you pour out of the shaker is certainly more than one ion ofsodium and one ion of chlorine. From our studies of atomic size, you found that it takes a great number ofatoms or ions for us to be able to see them. The salt crystals are actually huge numbers of Na1+ and Cl1-

ions bonded together to form an ionic bond. The diagram below helps to show that ionic solids formcrystals that are held together by attraction of opposite charges. There are no NaCl molecules as such.The ionic crystal is just an array of ions. The arrangement of ions is called a crystal lattice. The nature of alattice varies from one ionic solid to another, depending on the ratio and size of positive and negativeions in the compound.

Figure 15.8NaCl Crystal Lattice

Ionic and polar covalent compounds have properties that arecharacteristic of their bond types. In Activity 15.5 you will be examining theproperties of an ionic compound and of a compound with polar covalentbonds. You may assume that the properties of the ionic and covalentsubstances which you will examine are typical of ionic and covalentcompounds.

ACTIVITY 15.5 Comparing Properties Of Ionic & Covalent Compounds

We will use rock salt, NaCl, as a typical ionic compound for all of the tests. For the first test,naphthalene will be used as a typical covalently–bonded substance. For the second test, paraffin willserve as the covalent substance. Table sugar (sucrose), C12H22O11, will be used as the covalentsubstance in tests three and four. Get small samples of each of these from the materials shelf. Subjecteach substance to the test outlined below and record your observations in Table 15.3. Be sure to wearsafety glasses!

Use NaCl (ionic) and naphthalene (covalent) for test 1:

Test 1. Smell each compound. If you detect an odor, you may assume the substance is volatile (whichmeans it evaporates easily at room temperature).

Use NaCl (ionic) and paraffin (covalent) for test 2:

Test 2. Test the hardness of each compound by grinding a small piece of each with a metal file. Keepeach substance dry as you do this. Try to distinguish between the two. There is a difference.

Use table sugar (sucrose) for the covalently–bonded substance in tests 3 and 4:

Test 3. Place a very small sample of NaCl and sugar (sucrose) in separate piles on a piece of metal such asa can lid (from the materials shelf). DO NOT use naphthalene! It is highly flammable! Place the metal on aniron ring on a ring stand and heat the materials by applying heating with your burner under the metal. Notewhich material melts first. Stop heating as soon as one material begins to melt; otherwise, you will causeburning to occur.

Test 4. Dissolve a little NaCl (about enough to cover a quarter) in a 100 mL beaker about 1/2 full of tapwater. Stir well and test the solution for electrical conductivity using the apparatus designed for thatpurpose. (Your teacher will guide you.) Dip the electrodes of the apparatus into a beaker of water to rinsethem and repeat this procedure using a little sugar instead of NaCl. Indicate whether the substance, whendissolved, conducts a current.Results and Conclusions:

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From your results, state the properties that are characteristic of ionic substances: _________________

______________________________________________________________________________

State the general properties of a covalent substance: ______________________________________

______________________________________________________________________________

Using your observations from this activity and the table of electronegativity values, determine which ofthese compounds, BaCl2 or CO2, would conduct more electric current when dissolved in water. Explain.

{14}____________________________________________________________________________

______________________________________________________________________________

Drawing on your observations in this activity and Table 15.2, which substance would you expect to havethe lowest melting point: AlBr3, CS2, or CaCl2? Explain your reasoning.

{15}____________________________________________________________________________

______________________________________________________________________________

Table 15.3Properties of Ionic and Covalent Compounds

Property Ionic Substance Covalent Substance

Volatility (high or low)

Hardness

Melting Point (high or low)

ElectricalConductivity

SECTION 15.6 More About Molecular Geometries

You already know something about the shapes of certain molecules. Knowing the shapes canhelp you to determine many of the chemical properties of molecules. Lewis electron dot structures arevery useful in helping us to predict geometries of molecules. Ordinary chemical formulas don't help muchin this respect. Let's now expand your knowledge of molecular shapes.

In a previous chapter, we described electron probability density plots as a means of indicating the

15-13 ©1997, A.J. Girondi

three-dimensional spaces in which electrons spend most of their time. The shapes and arrangement ofthese orbitals determine the overall shapes of molecules. We can use the compound called ammonia,NH3, as an example to illustrate how the probability plots can be used to explain the geometry (shape) of amolecule. Observe the electron dot structures in the equation below. (Assume N2 and H2 have beenbroken apart into N and H.)

3 HN N HH

H+

According to the equation above, how many valence electrons does H have?{16}_________

How many valence electrons does N have?{17}____________

How many bonding electrons does N have?{18}____________

How many nonbonding electron pairs does N have?{19}__________

HN

N HH

H

Valence Orbitals: 2s 2p 1s(X) (/)(/)(/) (/)

Lewis Dot Symbols:

Lewis Dot Structure:

Molecular Geometry: (pyramidal)

N atom H atom

The shape of the NH3 molecule can be illustrated by using lines to represent bonds. The shape of amolecule depends on the bond angle between the bonds which join the H atoms to the central N atom.The shapes of molecules can often be determined by looking at the valence electrons of the central atom.Remember, because electrons are all negatively charged, they repel each other. To achieve a conditionof maximum stability, electrons locate themselves as far away from each other as possible. This serves tominimize the electrostatic repulsion between the pairs of electrons. The position of the electronsdepends of the number of electrons present. The bond angle for the ammonia, NH3, molecule has beendetermined to be 107o, forming what is referred to as a pyramidal–shaped molecule. The elements infamily 5A of the periodic table (such as nitrogen) all tend to form pyramidal-shaped molecules when theyare bonded to three other atoms. The unshared (nonbonding) electron pair on nitrogen tends to repelthe three shared pairs, causing the pyramidal arrangement.

Bond angles are determined by studying molecules with x-rays. Name another element (from row3) on the periodic table that you would expect to be the center of pyramidal–shapedmolecules:{20}_______________________.

Now we will follow a similar procedure as we study the shape of molecules which have elements

15-14 ©1997, A.J. Girondi

from family 6A at their centers. Oxygen will be used as an example of a typical family 6A element. Water'sformula is H2O. The formula equation and the dot structure equation for the formation of water are shown

below.

2 H2 + O2 -----> 2 H2O OR 2 H + O -----> OHH

O

H H

104.5o bond angle

The two unshared (nonbonding) pairs of electrons and the two shared pairs in the molecule all tend torepel each other. The result is a bent–shaped molecule with a bond angle of 104.5o. Bent molecules aretypically formed when family 6A elements become bonded to two other atoms. Name an element from row3 on the periodic table that you would expect to form the center of a bent molecule:{21}______________.

If you take a more advanced course in chemistry, you will learn that it is possible for unsharedelectron pairs of electrons to become involved in bonding, and thereby, become shared pairs. Considerthe formation of the ammonium, NH41+, ion. A hydrogen ion (H1+) forms when a hydrogen atom loses itselectron. The hydrogen atom's electron configuration is 1s1. The hydrogen ion (H1+) configuration is 1s0.It has no electrons, but it "wants" the helium configuration (1s2) to gain stability. So, the hydrogen ion islooking for two electrons to share. When this H1+ ion meets a molecule of NH3, it bonds to the unsharedpair on the N atom as follows:

H1+ + N

H

H

H

N

H

H

H

H

1+

Since the hydrogen ion carries a positive charge, the entire particle assumes a positive charge when theammonium ion is formed. (Don't confuse ammonia, NH3, with ammonium, NH41+.) We can show this byenclosing the whole structure inside brackets with the plus sign outside. This also happens when the H1+

ion reacts with a water molecule. The product (below) is an ion (the hydronium ion) that is very important inthe study of acids.

H1+ +

1+

There are many other examples. Whenever an atom or ion that needs a pair of electrons in order tobecome stable meets another particle which has one or more unshared pairs of electrons, this kind ofreaction is possible. Table 15.4 summarizes the shapes of the molecules you have studied in this chapter.Notice in the molecular drawings that a dash is often used to represent a pair of shared electrons (a bond).Also notice that while all unshared pairs are shown in the dot structure, in the drawings of the moleculesonly the unshared pairs around the central atoms are shown. Any other unshared pairs are "assumed" tobe there - as needed.

Problem 3 requires that you complete the blanks in Table 15.5. The drawings of the moleculesuse dashes for bonds, and unshared pairs are only shown if they exist around the central atom.

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Table 15.4Geometries of Molecules Containing Family 1A – 8A Elements

Family 1A 2A 3A 4A 5A 6A 7A Shape linear linear planar tetrahedral pyramidal bent linear

H Cl I – Be – I

B

ClCl

Cl

C

ClCl

Cl

Cl

N

HH H

O

HH

Cl ClDrawing ofMolecule

ClH

BClCl

Cl

C

Cl

ClCl

Cl

N

H

HH ClCl DotStructure

II Be

Formula HCl BeI2 BCl3 CCl4 NH3 H2O Cl2

Problem 3. Complete Table 15.5.

Table 15.5Dot Structures and "Stick" Drawings of Molecules

Compound Dot Structure Stick Drawing Geometry

CHCl3 tetrahedral

LiI

PCl3

GaF3

CH2Cl2

H2S

C

Cl

Cl

Cl

H C

HCl

Cl

Cl

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ACTIVITY 15.7 Determining the Formula of a Hydrate

In this activity you will be determining the formula of a hydrated compound. Remember that youhave seen the formulas of hydrates before. An example is CuSO4•5H2O. This formula states that forevery 1 mole of CuSO4 in the crystals of the compound, 5 moles of water are bonded to it. (The 1 in frontof the CuSO4 is assumed to be there.) The raised period means "plus". The water is bonded to the metalin hydrates by a special kind of covalent bond which is called a coordinate covalent bond. We will save thediscussion of these bonds for a more advanced course.

When you heat a hydrate, you break the covalent bonds which bond the water to the compound,and the water leaves in the form of a vapor. The product that is left is the anhydrous (without water) form ofthe compound. The formula of the anhydrous form is obtained simply by leaving the water out of theformula of the hydrate.

You are going to heat a sample of the hydrated form of barium chloride. We will represent theformula temporarily as BaCl2 •?H2O. Your task is to determine the coefficient that should go in front of theH2O in the formula. Remember, there is already an assumed "1" in front of the BaCl2. You will be lookingfor the mole ratio between the BaCl2 and the H2O in the compound.

Procedure:

1. Obtain a clean porcelain crucible and cover. Measure the mass of the empty crucible and cover to thenearest 0.01 g. Add approximately 3 g of BaCl2 •?H2O crystals to the crucible, replace the cover, andmeasure the mass to the nearest 0.01g. Calculate the mass of BaCl2 •?H2O crystals in the crucible, andrecord this data in Table 15.7.

2. Begin heating slowly. Increase the heat until you have heated the crucible strongly for about 10minutes. Remove the crucible from the triangle, let it cool, and measure the mass of the crucible, cover,and contents. Record data in Table 15.7.

3. Reheat with a hot flame for a few minutes, allow to cool, and remeasure the mass. If the mass is morethan 0.01 g different from what it was after the first heating, reheat and measure again.

4. Dispose of the material in the crucible in the container provided by your teacher. Clean and dry thecrucible and cover.

5. After you have completed Table 15.7, show your results to your instructor who will provide you with the

actual answer. Comment on how your result compares to the actual formula for the hydrate: __________

______________________________________________________________________________

15-17 ©1997, A.J. Girondi

Table 15.7Determination of the Formula of a Hydrate

a. Mass of crucible and cover: __________ g

b. Mass of crucible, cover, & contents before heating: __________ g

c. Mass of BaCl2 •? H2O: __________ g

d. Mass of crucible, cover, & contents after heating: __________ g

e. Mass of H2O driven off (b–d): __________ g

f. Mass of BaCl2 left in crucible (d–a): __________ g

g. Moles of H2O driven off: __________ moles

h. Moles of BaCl2 left in crucible: __________ moles

Whole number mole ratio of moles of BaCl2 to moles of H2O: 1 to ____(To get this whole number ratio, divide the moles of water and the moles ofBaCl2 by the smallest of the two. Round the results to whole numbers)

Experimentally determined formula for the hydrated compound: BaCl2 • ____ H2O


This is the end of Chapter 15. Review the learning outcomes below. Check each one when youthink you have mastered it. Arrange to take any quizzes or exams on Chapter 15 and move on to Chapter16.

_____1. Explain what is meant by the electronegativity of an element.

_____2. Describe the general properties of ionic and covalent compounds.

_____3. Determine the percent ionic and covalent characters of a bond.

_____4. Use hybridization theory to explain why atoms in families 2A, 3A, and 4A form the number of bonds that they do.

_____5. Given the formula of a compound which has an atom from family 2A, 3A, or 4A in the center, identify the type of hybridization that it exhibits.

_____6. Predict the geometry (shape) of simple molecules given the molecular formula.

_____7. Be able to calculate the formula of a hydrate, given the needed experimental data.

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Questions:

{1} silicon; {2} none; {3} two; {4} magnesium; {5} one; {6} three; {7} aluminum; {8} metals;{9} gallium, thallium, bismuth, tellurium, chlorine; {10} fluorine; {11} cesium and francium;{12} family VII - the halogens; {13} family IA - the alkali metals; {14} BaCl2 - it is more ionic, and ioniccompounds form solutions which conduct electricity; {15} CS2 - because it is covalently bonded.Covalent compounds tend to have lower melting points than ionic ones; {16} one; {17} five; {18} three;{19} one; {20} phosphorus; {21} sulfur

Problems:1.

a. __2__ H + __1__ O ---------->

b. __4__ H + __1__ C ---------->

c. __2__ Br + __1__ Be ---------->

d. __3__ F + __1__ Al ---------->

e. __1__ H + __1__ Cl ---------->

f. __2__ H + __1__ S ---------->

g. __3__ F + __1__ P ---------->

H O

H

C HH

HH

Be BrBr

AlF

F

F

ClH

SH

H

PFF

F

2. MgBr2; Mg = 1.2; 1.6; 47; 53NaF; F = 4.0; Na =0.9; 3.1; 91; 9CO; O = 3.5; C = 2.5; 1.0; 22; 78NH3; N = 3.0; H = 2.1; 0.9; 19; 81FrCl; Cl = 3.0; Fr = 0.7; 2.3; 74; 26PBr3; Br = 2.8; P = 2.1; 0.7; 12; 88

3. LiI (same as HCl as shown in Table 15.4, page 15-16)PCl3 (same as NH3 as shown in Table 15.4, page 15-16)GaF3 (same as BCl3 as shown in Table 15.4, page 15-16)CH2Cl2 (same as CCl4 as shown in Table 15.4, page 15-16)H2S (same as H2O as shown in Table 15.4, page 15-16)

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NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 16

SOLUTIONS(PART 1)

16-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




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16-2 ©1997, A.J. Girondi

SECTION 16.1 Introduction To Solutions

Up to this point, the subjects covered have dealt mainly with the structures and properties of puresubstances (elements and compounds) such as sodium chloride (NaCl) and water (H2O). Each has aspecific melting and boiling point, as well as other characteristic properties. When salt and water aremixed, however, their physical properties change. The solid disappears and becomes part of the liquid.The resulting mixture of salt and water is called a solution which is the topic we will study in this chapter andin Chapter 17. You first saw the "classification of matter" in Chapter 5. It is given again below in Figure16.1 so that you can see how solutions fit into the scheme of things.

Matter

Pure SubstancesMixtures

Homogeneous Heterogeneous Elements Compounds

Solutions

Figure 16.1 The Classification of Matter

A solution is defined as a {1}__________________ of two or more substances that arehomogeneously (uniformly) mixed. The formation of a solution does not involve a chemical change;therefore, the chemical properties and identities of the components of the solution do not change. If youadd heat, you can easily separate the mixture and recover the water and salt.

Solutions may exist in any combination of the three phases of matter - solid, liquid, or gas. Whenmost people think of solutions, they usually picture them as liquids. However, not all solutions are liquids.Solid solutions are much more common than you may realize. The nickel coin is a solution containing 25%nickel by mass dissolved in copper. Brass is a solution consisting of a mixture of copper and zinc. Thistype of solid solution consisting of two or more metals is called an alloy. Dentists frequently fill decayedteeth with an alloy of silver and mercury. Alloys containing mercury are called amalgams. The advantage ofalloys is that they can be harder and stronger than the pure metals from which they were prepared. Forexample, tensile strength is defined as a measure of the force required to pull apart a given piece ofmaterial. The tensile strength of pure iron can be increased by 10 times by the addition of 1% of carbonand smaller amounts of nickel and magnesium. This fortified iron is called steel , and there are numerouskinds of steel which contain differing amounts and kinds of additives. Steel is, therefore, a solution!

In addition to liquid solutions and solid solutions, there are also gaseous solutions. Air is anexample of a gaseous solution consisting primarily of nitrogen and oxygen. There are many combinationsthat exist between pure substances of different phases that combine to form solutions. Table 16.1provides examples of solutions made by combining various phases of substances. In order to understandsolution chemistry, you need to learn some new terms. In solutions, the substance that is dissolved orbeing dissolved is called the solute. The substance which does the dissolving is called the solvent. Inseawater, the solid salt is the solute and the water is the solvent. If there is doubt as to which substance isdissolving (such as when a liquid is dissolved in another liquid), it is customary to consider the solvent tobe the substance that is present in the greatest quantity. When 2 mL of alcohol and 1 mL of water aremixed, which one is the solvent?{2}________________________________.

16-3 ©1997, A.J. Girondi

Table 16.1Solutions Containing Various Phases of Matter

Phase o f Major Phase of Added Example Component Component

solid solid metal alloys like steel solid liquid mercury in silver (amalgam)solid gas hydrogen in palladium liquid solid seawater (salt water) liquid liquid gasoline liquid gas carbonated (CO2) soda watergas solid iodine vapor in air gas liquid water vapor in air gas gas air (mostly N2 + O2)

SECTION 16.2 Solutions and the Polarity of Molecules

You are probably aware that if you try to make a solution by pouring two liquids together, you mayhave a problem in that they may not mix. If two liquids don't mix we say that they are immiscible. If they domix, we say that they are miscible. Some liquids mix with each other in any amounts, while others are onlypartially miscible in each other. Ethyl alcohol (C2H5OH) and water, for example, are miscible in anyamounts. It doesn't matter if you use a lot of alcohol with a little water or vice versa. However, the reasonthat you have to shake your salad dressing is because oil are immiscible. They do not form a solution. Youwill experiment with this concept in Activity 16.4.

The reason for the miscibility or immiscibility of liquids involves the phenomenon of polarity ofmolecules. You already know that some chemical bonds are polar and some are not; and, you know thatsome polar bonds are more polar than others. Notice that we mentioned above that the reason involvesthe polarity of molecules, not just the polarity of bonds. You see, some molecules are polar and some arenot. So what makes a molecule polar? It depends on the nature of the bonds in the molecule, the shapeof the molecule, and the formula of the molecule. The bottom line is this: the molecule must have twodifferent ends. In addition, at least part of the molecule must have polar bonds. If a molecule has polarbonds and two different ends, then it will have one end which is more "positive" in charge and a differentend which is more "negative." This is because the electrons in the molecule will spend more time aroundthe more electronegative atoms, making that end of the molecule more negative than the other end.

As we discuss the examples in Table 16.3, you willwant to refer to the table of electronegativity values in yourALICE reference notebook. Remember, the greater theelectronegativity difference between two atoms, the morepolar the bond. As a very general rule, bonds areconsidered to be nonpolar (for practical purposes) if the twoatoms in the bond have an electronegativity difference ofless than 0.5. If the electronegativity difference betweentwo bonded atoms is greater than 1.7, then the bond isconsidered to be ionic, rather than polar (again for practicalpurposes). Table 16.2 summarizes this information. None ofthe examples in Table 16.3 are ionic. They are either polaror nonpolar.

Table 16.2Relationship Between

Electronegativity and Bond Type

Electronegativity Difference Bond Type

0 to 0.5 nonpolar 0.6 to 1. 7 polar covalent 1.7 or more ionic

16-4 ©1997, A.J. Girondi

In problem 2 on page 7, you will be asked to complete Table 16.3. The discussion below will helpyou to complete it. As you read the next couple of pages, you will learn what you need to know tocomplete Table 16.3.

Let's consider the examples found in Table 16.3 one at a time. Does the Cl2 molecule have apolar bond? {3}__________. Does the molecule have two different ends? {4}_________. Therefore, thismolecule does not meet the conditions needed for polarity.

Does the H–Cl molecule contain a polar bond? (Use the guidelines in Table 16.2){5}_____________. Does this molecule have two different ends? {6}__________. Therefore, this is apolar molecule since it meets both criteria. A polar molecule has one end which is more positively chargedand another end which is more negatively charged. This results from an unequal sharing of the electronsin the bond(s). In HCl the chlorine is more electronegative (attracts electrons more) than H, so the Cl endof the HCl molecule is more negative and the H end is more positive. Polar molecules are frequently calleddipoles, which means "two poles."

What is the electronegativity difference of the bonds in the BBr3 molecule (all three areidentical)?{7}__________. So, the bonds are mildly polar. Now, does the molecule have two differentends? {8}__________. Therefore, even though it has mildly polar bonds, it is best to classify BBr3 asnonpolar.Calculate the electronegativity difference for the bonds in the next example, AlF3. Thedifference is {9}_________. So technically, this is an ionic compound which means that it is best toconsider it to be composed of independent ions in a lattice (like NaCl), rather than as being composed ofmolecules. (In polar molecules, the two ends of the molecule have opposite partial charges. In ioniccompounds, the ions carry full charges since electrons are actually transferred.) Furthermore, even if AlF3

were a molecule, would it have two different ends? {10}___________. So this compound will haveproperties more characteristic of ionic compounds than those of polar covalent molecules.What's theelectronegativity difference between Be and I in BeI2? {11}_________. That's good enough to make thebonds polar. Now, does this linear molecule have two different ends?{12}_________ You see, both endsare identical. Since I is more electronegative than Be, both ends of the BeI2 molecule are partiallynegative. The partially positive Be is trapped in the middle. Thus, even though the bonds are polar, themolecule is not.

The next example is H2O. What is the electronegativity difference of an O–H bond?

{13}_________. O.K., so are the bonds polar? {14}________ Note that H2O is a bent molecule. The bentshape results from the mutual repulsion between the shared pairs of electrons and the unshared pairs. Isthe molecule polar? Check Figure 16.2 below. Note that if you look at the figure on the left you mightassume that water has two identical ends. But, if you look to the right, note that the bent shape of themolecule allows us the "see" two other ends which are different. As a result, water is a polar molecule!Since oxygen is more electronegative than hydrogen, the O end is partially negative, while the H end ispartially positive.

H

O

H H

O

H

Figure 16.2 The Polarity of Water

∂ –

∂ +

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CF4 bonds have an electronegativity difference of 1.5 which would make them polar. However,this molecule is tetrahedral in shape, and therefore has no clearly defined different ends. It is nonpolar.However, If the molecule were changed to CF3Cl, that would give it two different ends, and this moleculewould be polar. The carbon–fluorine bond has a greater electronegativity difference than the carbon-chlorine bond does. Fluorine is more electronegative than chlorine, so the fluorine end of the molecule ismore negative than the chlorine end. See Figure 16.3 below. So, you must consider the formula as wellas the shape and the bonds.

Note: to help you to appreciate the three-dimensional structure of molecules such as those shown below, the solidblack line represents a bond within the plane of the paper; the light line represents a bond located behind the plane ofthe paper; and, the lines which get thicker at one end represent bonds which are in front of the plane of the paper(coming out towards the reader).

C

FF F

Cl

C

FF F

F

nonpolar CF4 polar CClF3

∂ +

∂ –

Figure 16.3 Polarity of Two Carbon Compounds

The NH3 molecule has three N–H bonds with an electronegativity difference of 0.9 which allowsus to classify them as polar. The molecule itself has a pyramidal shape resulting from the repulsionbetween the three shared pairs and the single unshared pair. If this molecule were planar, it would nothave two different ends. This was the case with BBr3, if you recall. However, the fact that this is apyramidal molecule allows it to have two different ends if you look at it in the right way. Figure 16.4 below,gives you a top view and a side view for comparison. The side view shows the ends which make themolecule a dipole.

Figure 16.4 Top and Side Views of NH3

N

HH H∂ +

∂ –

N

HH

H

16-6 ©1997, A.J. Girondi

Problem 1. Now look at the next two examples in Table 16.3, BHBr2 and CCl2Br2. See if you candetermine why they are polar. Check Table 16.3, and then redraw these two molecules in the boxesbelow and label the partially positive (∂ +) and partially negative (∂ –) ends of each one. This generalizationmay help you: The more centrally located an atom is on the periodic table, the more centrally located anatom of that element will be in a molecule. This applies to the elements in the "A" families (1A through 8A),and there are, of course, exceptions.

BHBr2 CCl2Br2

Problem 2. Complete Table 16.3 below.

Table 16.3Polarity of Molecules

Molecular Geometric Molecule Has 2 Molecule has MoleculeFormula Structure Different ends? Polar Bonds? is Polar?

(yes / no) (yes / no) (yes / no)

Cl2 _________________ ___________ _______ _______

HCl _________________ ___________ _______ _______

BBr3 _________________ ___________ _______ _______

AlF3 _________________ ___________ _______ _______

BeI2 _________________ ___________ _______ _______

H2O _________________ ___________ _______ _______

CF4 _________________ ___________ _______ _______

NH3 _________________ ___________ _______ _______

BHBr2 _________________ ___________ _______ _______

CF2Br2 _________________ ___________ _______ _______

16-7 ©1997, A.J. Girondi

C C O

H

H

H H

H H

nonpolar portion polar portion

∂ –

∂ +

Figure 16.5 Ethyl Alcohol

Let's consider one other molecule, ethyl alcohol(C2H5OH). The C–H bonds are not considered polar,nor are the C–C bonds. The O–H bond, however, ispolar. Note that the molecule has two different ends,too. Since this polar end molecule has a polar O–Hbond, and since it has two different ends, it is polar. Thenonpolar bonds are concentrated at one end, while thepolar bond is at the other end. The highlyelectronegative oxygen atom makes that end of thepolar bond partially negative. This fact is important, andwe will experiment with this molecule shortly.

ACTIVITY 16.3 A Study Of Molecular Models

Since molecules are three–dimensional objects, it is helpful to see them represented in threedimensions. We will do this by looking at models of the molecules. From the materials shelf, obtain amodel of each of the ten molecules listed in Table 16.3. Look at the models and see if you can determinewhy some molecules have two different ends, while others do not. Put a check mark in after eachobservation below after you have verified it but examining the models.

1. CF4 is not flat.___2. CF4 does not have two different ends, but CF2Br2 does.___3. Note the difference between planar (otherwise known as trigonal planar) and pyramidal molecules.___4. Note the difference between the linear BeI2 and the bent H2O.___

ACTIVITY 16.4 Testing the Miscibility of Liquids

An interesting characteristic of liquids with covalent bonds has to do with how well they mix witheach other. We have already established that because of its bonds and its shape, water is a polarmolecule. On the other hand, gasoline has bonds that have very little electronegativity difference andvery little polarity. Therefore, regardless of its shape, it cannot be a polar molecule since it has no polarbonds. Polar liquids do not mix with nonpolar liquids. In chemistry, there is an old saying: "like dissolveslike." It means that polar liquids dissolve in other polar liquids. Nonpolar liquids dissolve in other nonpolarliquids. Gasoline will mix with oil, because both are nonpolar. They will form a solution. Water will mix withglycerol, which is another polar molecule found in hand lotions and some lubricants. This rule alsoexplains the dissolving of solids in liquids. For example, water (polar) cannot remove grease (nonpolar)from your clothes, but gasoline (nonpolar) can. But, never use it for that purpose! So-called "dry"cleaners use nonpolar solvents in place of water to clean greasy nonpolar dirt from clothes. (Some saythat chewing gum can be removed from hair or clothing by using peanut butter! The oils in peanut butterare nonpolar. What can you assume about chewing gum? {15}_________________________________

Water and glycerol are miscible, but water and salad oil are immiscible. Ethyl alcohol, C2H5OH,which you saw in Figure 16.5, is miscible in both water and in salad oil! Why? If you recall, we saw that ethylalcohol has both a polar and a nonpolar end. Well, the nonpolar end tends to make the molecule dissolvein nonpolar liquids, while the polar end makes the molecule soluble in polar liquids. It goes both ways! Ifyou add alcohol to a mixture of oil and water, it will cause all three liquids to form a solution. That is, tobecome miscible. Soap molecules are similar. They have a polar and a nonpolar end, and therefore soapcan help nonpolar dirt and polar water to mix, and your shower is much more successful!

miscible: describes two substances which are mutually soluble in each other.immiscible: describes two substances which are NOT mutually soluble in each other.

16-8 ©1997, A.J. Girondi

Procedure:

1. Obtain about 5 mL of 2-butanol (nonpolar) and 5 mL of colored water (polar) and mix them in a large(200 mm x 25 mm) test tube. Do not sniff the vapors from the 2-butanol. (The coloring in the water is thereto make it easier to distinguish the 2-butanol from the water.) Stopper the tube and shake. Allow theliquids to settle for five or ten minutes. Are they miscible, or do you see two layers forming (meaning theyare immiscible)? ______________.

2. Next, add about 5 to 10 mL of ethyl alcohol to the mixture which is already in the test tube. Stopper and

shake again. Wait a few minutes. Do you see two layers or are the contents miscible? _______________

Explain: _______________________________________________________________________

______________________________________________________________________________

3. Pour the contents of the tube into the waste container provided by your teacher. Rinse the test tubeseveral times with water.

SECTION 16.5 Predicting Geometries and Polarities of Molecules

Problem 3. Complete Table 16.4, indicating which molecules are polar and which are not. For the polarmolecules, label the partially positive and partially negative ends. (Check the electronegativities of atomsusing Table R–8 in your reference notebook. To help you draw the molecules, keep in mind that:

among the "A" families of elements, the closer an element is to the center of the eight "A" families, the more centrally located it will usually be in the molecule.

For example, elements in family 4A are usually located in the center of simple molecules and usually formfour bonds. Family 5A elements are more likely to be in the center of a molecule than are family 7Aelements, etc.

Table 16.4Polarity of Selected Molecules

Compound Drawing of Molecule Shape Polar/Nonpolar

CFH3 ____________________________ _________________

LiI ____________________________ _________________

PCl3 ____________________________ _________________

GaH3 ____________________________ _________________

CH2Cl2 ____________________________ _________________

H2S ____________________________ _________________

16-9 ©1997, A.J. Girondi

Did you label the partially positive ( ∂ + ) and negative ( ∂ –) ends of those molecules in Table 16.4 whichare polar? If not, do so now.

ACTIVITY 16.6 Testing the Polarities of Solutes and Solvents

In this activity you will be examining how well selected solvents can dissolve certain solutes, andyou will be relating the results to the polarity of the solutes and solvents. Be careful with the materials youwill be using. Several give off toxic fumes, and iodine can stain your skin and clothes. Do not handlechemicals with your fingers. Wear safety glasses and an apron. Be especially careful with iodine. Weargloves if available.

PART A.

1. Obtain four regular–sized (150 mm) test tubes, and label them 1 through 4.

2. Place 3 to 5 mL of water (solvent) into each test tube.

3. Add a "pinch" of sugar to test tube 1, a small piece of styrofoam to test tube 2, 1 small crystal of iodine totest tube 3, and a "pinch" of sodium chloride to test tube 4.

4. Put a stopper in each tube and shake well for about one minute or until the result is obvious.

5. Observe the contents. If it appears that most or all of the solid dissolved in a tube, enter "soluble" in theappropriate blank in Table 16.5. If it does not, enter "insoluble." Dispose of any iodine in the containerprovided by your instructor. The styrofoam can be put into the waste can, while other chemicals can bedisposed of in your sink. Clean the tubes and dry them using towels.

PART B.

1. Add about 2 mL of toluene to each of the four tubes. Keep toluene away from flames! Do not sniff thetoluene vapors. You should have good ventilation in your lab area.

2. To test tubes 1 through 4 add the same materials as you did in part A. Stopper the tubes as in part A,and shake well for one minute each or until the result is obvious. Observe the results and enter "soluble"or "insoluble" in Table 16.5.

3. Dispose of the materials as you did in part A, except be sure to put the toluene in the special containerprovided by your instructor.

Questions:

What conclusion can you make about the solubility of polar solutes in polar solvents?

______________________________________________________________________________

Give specific examples to justify your response. __________________________________________

______________________________________________________________________________

What conclusion can you make about the solubility of nonpolar solutes in polar solvents? ____________

______________________________________________________________________________


______________________________________________________________________________

16-10 ©1997, A.J. Girondi

Table 16.4Polarity of Selected Molecules

Compound Drawing of Molecule Shape Polar/Nonpolar

CFH3 ____________________________ _________________

LiI ____________________________ _________________

PCl3 ____________________________ _________________

GaH3 ____________________________ _________________

CH2Cl2 ____________________________ _________________

H2S ____________________________ _________________

Table 16.5Solubility and Polarity

PART A. POLAR (WATER) SOLVENT

Solute Solute Solvent Solvent Soluble or Insoluble Polarity Polarity

sugar polar water polar ___________________

styrofoam nonpolar water polar ___________________

iodine nonpolar water polar ___________________

NaCl ionic water polar ___________________

PART B. NONPOLAR (TOLUENE) SOLVENT

Solute Solute Solvent Solvent Soluble or Insoluble Polarity Polarity

sugar polar toluene nonpolar ___________________

styrofoam nonpolar toluene nonpolar ___________________

iodine nonpolar toluene nonpolar ___________________

NaCl ionic toluene nonpolar ___________________

What conclusion can you make about the solubility of an ionic solute like NaCl in polar and nonpolar

solvents? ______________________________________________________________________


______________________________________________________________________________

(Some other ionic solutes which have very strong ionic bonds are only slightly soluble or insoluble in water.)

SECTION 16.7 Solubility and Saturation

If you were to add only a small amount of salt (NaCl) to some tap water, you would find that all of itwill dissolve as you stir the solution. If you continue to add the salt, a point will be reached at which the saltwill no longer dissolve no matter how much you stir it. The solvent (H2O) at that point contains all of theNaCl it can "hold," and the solution is said to be saturated. The maximum amount of solute that willdissolve in a given amount of solvent is called the solubility of the solute.

You may wonder why a solvent becomes saturated. When a substance like salt dissolves in water,the polar water molecules are attracted to the positive and negative ions of the salt. The water moleculesthen surround those ions, thereby keeping them apart and "in solution." Eventually you will run out ofwater molecules, and any additional salt will remain in the solid state.

16-11 ©1997, A.J. Girondi

solvent molecule

solute particle

dissolved solute particle

Figure 16.6 Depiction of an Unsaturated Solution

Notice that there are still some "free" solvent molecules available in an unsaturated solution (Figure 16.6).

solvent molecule

solute particle


Figure 16.7 Depiction of a Saturated Solution

undissolved solute particles

Notice that there are no "free" solvent molecules in the saturated solution (Figure 16.7).

There is a limited amount of any solute that will dissolve in a given amount of solvent. Somesolutes are very soluble in the most common solvent (water), while others are almost completely insolublein water. About 200 grams of ordinary table sugar will dissolve in 100 grams of water at 25oC, becausesugar is very soluble in water. By contrast, only 2 X10-4 grams of silver chloride (AgCl) will dissolve underthese same conditions. Silver chloride can be considered to be insoluble in water since so little willdissolve.

Solubility is generally expressed as grams of solute per 100 grams of solvent. You will recall that 1mL of water has a mass of 1 gram, because the density of water is 1 gram / mL.

Sample Problem: Suppose that you do an experiment at 20oC and find that 17.26 g of AlCl3 willdissolve in 25 mL of water to form a saturated solution. Since solubility is expressed in grams of solute per100 g of solvent, the solubility can be calculated in the following manner:

17.26 g AlCl325 mL H2O

X 1 mL H 2O

1 g H2O =

0.69 g AlCl31 g H2O

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So, we have found that 0.69 g of AlCl3 will dissolve in 1 g of H2O at 20oC. However, the definition ofsolubility requires that we calculate the amount of solute that will dissolve in 100 g of solvent (H2O), not 1gram. So, we convert our answer to g AlCl3 per 100 g H2O by multiplying by 100 g H2O:

0.69 g AlCl31 g H2O

X 100. g H 2O = 69 g AlCl3

Thus, the solubility of AlCl3 in H2O at 20oC is expressed as 69 g AlCl3 / 100 g H2O. (Whenever solubilityvalues are reported, the temperature is always listed since solubility depends on temperature.)

Calculate the solubilities (in g solute /100 g solvent) of the compounds described in the problemsbelow.

Problem 4. It is found that 18.4 g of Ba(NO3)2 are needed to saturate 200. mL of water at 20oC.

Solubility = __________g Ba(NO3)2 / 100 g H2O

Problem 5. It is found that 11.31 g of NaBr are needed to saturate 16.5 mL of water at 20oC.

Solubility = __________g NaBr / 100 g H2O

Problem 6. Assume that 2.6 X 10-16 g of Ag2S are needed to saturate 200. mL of water at 20oC.

Solubility = __________g Ag2S / 100 g H2O

Which of the substances in the previous three problems is most soluble? {16}______________________

Which is least soluble?{17}______________________ The next problem is a bit different.

16-13 ©1997, A.J. Girondi

Problem 7. The solubility of MnCl2 at 20oC is 73.9 g MnCl2 per 100. g H2O. How many liters of water arerequired to dissolve 110. grams of MnCl2 at 20oC? (Remember that 1 mL H2O = 1 g H2O)

__________ L H2O

ACTIVITY 16.8 Determining the Solubility of Sodium Chloride

In the following activity, you will be experimentally determining the solubility of common table salt,NaCl. The procedure involves 4 steps:

(1) obtain a sample of saturated salt solution (2) evaporate the water from the solution and find the mass ofthe salt (3) compute the mass of water in the original sample of solution (4) compute the solubility of thesalt (g NaCl / 100 g H2O)

Procedure:

1. Pour approximately 20 mL of saturated salt solution into a 25 mL graduated cylinder. Measure thetemperature of the solution and record it in Table 16.6. Obtain a clean, dry evaporating dish and a watchglass large enough to cover the dish. Find the mass of the dish and watch glass to the nearest 0.01 g.Record. Pour the salt solution into the dish. Measure the mass of the dish, watch glass, and solution tothe nearest 0.01 g. Record.

Table 16.6Solubility of NaCl

1. Temperature of Solution _________oC

2. Mass of dish and watch glass _________ g

3. Mass of dish + watch glass + _________ g solution (before heating)

4. Mass of dish + watch glass + residue _________ g (after heating)

5. Mass of NaCl residue [4 – 2] _________ g

6. Mass of H2O in solution [3 – 4] _________ g

(Use the data from 5 and 6 above for calculations below.)

7. Solubility of NaCl = ____________ g NaCl / 100 g H2O

2. Mount an iron ring on a ring stand andplace a wire gauze with ceramic center onthe ring. Adjust the height of the ring sothat your burner flame will nearly touch thebottom of the wire gauze. Place the dishon the wire gauze with the watch glassserving as a lid. Wear safety glasses!

3. Heat the solution being careful not toallow it to boil too vigorously. Heating forabout 10 to 15 minutes should evaporateall the water including any water on theunderside of the watch glass. When thecontents of the dish are completely dry,allow it to cool and then determine themass of the dish, contents, and watchglass. Record the data in Table 16.6, andcalculate the solubility of the salt (in g NaCl /100 g H2O) from your collected data.

16-14 ©1997, A.J. Girondi

Complete the partial "fencepost" provided below to calculate the solubility of NaCl:

g NaCl / 100 g H2O

g NaCl

g H2O X 100. g H 2O =

The accepted value for the solubility of NaCl at 20oC is 35.9 g NaCl / 100 g H2O. Calculate yourpercentage error. The formula for percentage error can be found in your reference notebook. Show work:

% Error = _____________

You should be able to achieve 5% error or less on this activity. How does your calculated (observed)

solubility compare to the accepted (A) value? ____________________________________________

How do you account for any excess error? ______________________________________________

Supersaturation

The solution you used in activity 16.8 was saturated. That is, the solvent was "holding" all of thesolute that it could, meaning that all of the solvent molecules were used to surround the solute particlesand keep them apart so that they could not come together to form crystals. It is possible for a solution tobe supersaturated. But what does this mean? How can a solvent hold more solute than it does when thesolution is saturated? Well, the theory goes like this. When the solution is saturated, all of the solventmolecules are bonded to solute particles. Notice the solute–solvent clusters. Although all of the solventmolecules are already used, it is possible for some solute particles to get trapped between thesolute–solvent clusters. Thus, the clusters rather than solvent molecules can serve to keep thesetrapped solute particles apart and, therefore, in solution.

solvent molecule

solute particle


Figure 16.8 Depiction of a Supersaturated Solution

undissolved solute particles

Supersaturated solutions must be very carefully prepared. You should first prepare a saturatedsolution which contains excess solute on the bottom of the container. If the solute is a solid and thesolvent is a liquid, then heating the saturated solution will cause more of the solute to dissolve, becausesolubility increases as temperature increases (for most solids). Then allow the solution to cool slowly. Anyunnecessary agitation may cause the excess solute particles to crystallize. After the solution cools, it willbe supersaturated if some excess solute stays in solution.

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ACTIVITY 16.9 Making a Supersaturated Solution

---> ---> ---> A Teacher Demonstration <--- <--- <---

Your teacher will follow the procedure below. The supersaturated solution you will see containssodium acetate (NaC2H3O2) as the solute and water as the solvent.

Procedure:

1. Obtain from the materials shelf a large stoppered test tube containing a saturated solution of sodiumacetate. You will see some excess solid solute lying on the bottom of the tube below the solution. Set upa hot water bath using a 250 mL beaker about half full of water. Remove the stopper from the tube andplace it in the bath. Allow the water to boil and add water to the beaker if needed to keep it about half full.Stir the solution occasionally, and continue heating until all of the solid solute has dissolved. Carefullyremove the tube from the bath, replace the stopper, and place it upright in your lab drawer until the nextday (or whenever your class meets again). The process of making the supersaturated solution wasendothermic, which means that heat was put into the system.

2. Carefully remove the cool supersaturated solution from your lab drawer. If you see solid solute in thetube, then the solution failed to become supersaturated and you should repeat the activity from thebeginning. If it is supersaturated, obtain a small crystal of solid sodium acetate from the materials shelf.Carefully remove the stopper from the tube and drop the crystal into the solution. Watch what happens!As the excess solute crystallizes out of the solution, note the temperature change of the contents of thetube by touching it with your hand.

How is the temperature of the solution changing? _________________________________________

The heat being evolved is a result of bonds forming between the excess solute particles as they formcrystals. Heat is added to break bonds, so heat is given off when bonds form. The heat in this case isknown as the heat of crystallization. When heat is released by a system, the change is said to beexothermic.

3. Return the stoppered tube of solution to the materials shelf. It can be reused many times.

Endothermic processes absorb heat from the surroundings.

Exothermic processes release heat to the surroundings.

ACTIVITY 16.10 Developing a Solubility Curve for NH4Cl

It was briefly mentioned that the solubility of a substance is dependent on temperature. Earlier inthis chapter, you determined the solubility of NaCl at room temperature. Now you are going to determinethe solubility of a solid (ammonium chloride, NH4Cl) at several different temperatures. (You will need toobtain data form other lab groups which are also doing this activity.) In order to do this, you mustdetermine the mass of the salt, add a known volume of water, and then find the temperature at which thissolution becomes saturated. If a solution is heated until the solid dissolves completely and is then allowedto cool, the solid will begin to crystallize out at the temperature where the solution is saturated. Follow theprocedure below, and record your measurements in Table 16.7.

1. Prepare a hot water bath by heating a 400 or 600 mL beaker about one-half full of tap water. SeeFigure 16.9. To shorten the time required, use a large burner if you have one.

16-16 ©1997, A.J. Girondi

2. Different lab groups in your class will be using different amounts of ammonium chloride. Ask yourinstructor whether your group should use 4.00 g, 4.50 g, 5.00 g, or 5.50 g. Put the NH4Cl into a large (25X 200 mm) test tube.

Figure 16.9Making a Supersaturated Solution

3. Add 10.0 mL of water to the test tube and put it into thehot water bath. Bring the water in the beaker to a boil.

4. Using the thermometer, carefully stir the contents of thetest tube while it is being heated. Continue the heating untilall of the solid has dissolved. (Due to the high concentration,it may look somewhat translucent.)

5. When all of the solid has dissolved, first place athermometer in the solution, wait about 15 seconds, andthen remove the tube from the hot water.

6. While stirring, observe the tube as the solution cools.Record the temperature when crystals start to form, givingthe impression of "snowing" in the tube. You can speed upthe cooling process by allowing tap water to run over theoutside of the tube, or by dipping the tube into a beaker ofcool tap water.) You will know when it is snowing, becausethe solute will be rapidly piling up on the bottom of the tube.

7. Repeat steps 4 through 6 two more times and record thedata. Average the three temperatures. (If one value is notclose to the other two, discard it, and average the two closeones.) If another lab group in your class worked with thesame amount of NH4Cl, obtain their temperatures andaverage them with your own.

8. Obtain average temperature values from the other lab groups in your class which worked with each ofthe three other amounts of ammonium chloride. Use the data to complete Table 16.7.

9. Calculate the solubility (in grams NH4Cl per 100 g H2O) of NH4Cl at each of the four temperatures whichyou recorded. Let's suppose for example, that the 4.00 g sample formed crystals at 21oC. Then thesolubility of NH4Cl at 21oC is 4.00 g NH4Cl / 10.0 mL H2O. Finally, convert this to 40.0 g NH4Cl / 100 g H2Oat 21oC.

10. On the grid which follows Table 16.7, prepare a "solubility curve" of NH4Cl versus temperature in oC.Plot solubility along the "Y" axis and temperature along the "X" axis. Plot four points and draw the bestsmooth curve that you can. Remember, the best curve may not actually pass through each plotted point.Include a title below the graph. Select suitable scales so that you use all or most of the grid.

Conclusion:

What does your graph indicate about the relationship between the solubility of NH4Cl and temperature?__

______________________________________________________________________________

Is this a direct or inverse relationship? ______________________

16-17 ©1997, A.J. Girondi

Table 16.7Solubility versus Temperature

1st Temp 2nd Temp 3rd Temp Average

Mass NH4Cl used: __________g ________ ________ ________ ________

Mass of Sample Crystallization Temperature Solubility (g /100 g H2O)

4.00 g __________oC _________________

4.50 g __________oC _________________

5.00 g __________oC _________________

5.50 g __________oC _________________


This is the end of chapter 16 and the first part of your study of solutions. Read the learningoutcomes below, and be sure you have mastered them before moving on. Take any test or quizzes onchapter 16, and then resume your study of solutions with chapter 17.

_____1. Predict whether a molecule is polar or nonpolar based on its geometry (shape) and its formula.

_____2. Define the relationship between polarity and solubility of solute / solvent combinations.

_____3. Define solute, solvent, solution, saturated solution, supersaturated solution and solubility.

_____4. Calculate the solubility of a solute, given the needed data.

_____5. Explain the phenomena of saturated and supersaturated solutions.

_____6. Describe and explain the effect of temperature on the solubility of most solids in liquid solvents.

16-18 ©1997, A.J. Girondi


Questions:

{1} mixture; {2} alcohol; {3} no; {4} no; {5} yes; {6} yes; {7} 0.8; {8} no; {9} 2.5; {10} no; {11} 1.0; {12} no;

{13} 1.4; {14} yes; {15} It must be nonpolar; {16} NaBr; {17} Ag2S

Problems:

B

H

Br

Br∂ +

∂–

C

Br

Cl Cl

∂ +

∂–Br

1.

Cl2 linear no no noHCl linear yes yes yesBBr3 trigonal planar no yes noAlF3 trigonal planar no no (ionic) noBeI2 linear no yes noH2O bent yes yes yesCF4 tetrahedral no yes noNH3 pyramidal yes yes yesBHBr2 trigonal planar yes yes yesCF2Br2 tetrahedral yes yes yes

2.

(answers are continued on the next page)

16-19 ©1997, A.J. Girondi

∂ +

∂–

C

H H H

∂ +

∂–F

CFH3 tetrahedral polar

LiI Li I∂ + ∂– linear polar

PCl3 P

ClCl Cl

pyramidal polar

GaH3

∂+

∂ –

Ga

H

H

H

trigonal planar nonpolar

CH2Cl2C

H

HCl

∂ +

∂–Cl

tetrahedral polar

H2S S

H H

bent polar

3.

4. 9.20 g Ba(NO3)2 / 100 g H2O

5. 68.5 g NaBr / 100 g H2O

6. 1.3 X 10-16 g Ag2S / 100 g H2O

7. 0.149 L H2O

16-20 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 17

SOLUTIONS(PART 2)

17-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


17-2 ©1997, A.J. Girondi

SECTION 17.1 Molarity

Solutions are classified on the basis of the relative amounts of the solute and solvent present. InChapter 16 solutions were classified as unsaturated, saturated, or supersaturated. In a saturated solution,a maximum amount of solute is dissolved in a solvent. If a comparatively large amount of solute is presentfor a given amount of solvent, the solution is described as being concentrated. If only a small amount ofsolute is present, the solution is described as {1}______________.

There is a definite need to be more precise in specifying the amounts of solutes in solutions,rather than using terms like concentrated or dilute. This is needed because there are an infinite number ofin-between combinations of amounts of solute and solvent. In quantitative work involving solutions, anumber of different units are commonly used to express concentration. In this chapter, we will discussonly the most commonly used method of expressing concentration: molarity. (In Appendix C, molality,which is another method of expressing concentration is defined and used.) The concentration of asolution tells us how much solute it contains relative to the amount of solvent.

The molarity of a solution indicates the numberof moles of solute present in one liter of solution.

Note that this definition refers to one liter of solution, not one liter ofwater. Molarity has units of moles of solute over liters of solution.

Molarity = moles of solute

liters of solution

A capital letter M is the abbreviation for molarity. A 1M solution (onemolar solution) of NaCl contains 1 mole of NaCl dissolved in 1 liter of solution(NaCl and water combined). One mole of NaCl has a mass of 58.4 grams (usingatomic masses from the periodic table). A 1M solution of NaCl is prepared bydissolving 58.4 grams of NaCl in enough water to make 1 liter of solution. Howdo you know how much water is needed to make 1 liter of solution? Chemistsfrequently use volumetric flasks for this purpose. These are flasks that have aspecial mark on their necks indicating the 1 liter point (flasks which measurevolumes other than 1 liter are also available). You simply put the solute in theflask, and add enough water to make one liter of solution.

Figure 17.1Volumetric Flask

You will now be learning how to do two kinds of problems involvingmolarity. The first type requires that you calculate the molarity of a solution,given the needed information. The second type requires that you use themolarity of a solution to calculate some other unknown. Study the exampleson the next page.

Sample Problem: A solution contains 40.2 grams of NaCl dissolved in 350 mL of solution. Calculatethe molarity.

? ? ? ? =

moles of NaCl

liters of solution

The units for molarity are moles solute per liter of solution. Molarity itself is"derived" unit; this means it is actually derived from other simpler units.Molarity is the name of a unit ratio. The term "molarity" itself cannot,therefore, be used in dimensional analysis. So before you set up theproblem to do dimensional analysis, look at the ratio you want to end upwith:

Since we want an answer with information about solute in the numerator and information aboutsolution in the denominator, let's start the problem by putting the information given about solute in thenumerator and the information given about solution in the denominator of a ratio:

17-3 ©1997, A.J. Girondi

40.2 g NaCl

350 mL sol'n X ? ? ? ? X ? ? ? ? =

? moles NaCl

liters sol'n

So, you can see that "g NaCl" has to be changed to "moles NaCl," and "mL solution" has to be changed to"liters of solution." (Solution may be abbreviated as "sol'n.") Finish this problem in the space below.Show the whole fencepost. See if you get the answer given below.

Answer = 2.0 M

As you solve the next two problems, remember that molarity (M) means moles of solute divided byliters of solution. Thus, when all units have canceled, you should be left with "moles solute / literssolution."

Problem 1. Calculate the molarity of a solution made from dissolving 328 grams of calcium nitrate,Ca(NO3)2, in 500. mL of solution.

__________ M

Problem 2. What is the molarity of a solution made by dissolving 2.00 g of Fe2(SO4)3 in 50.0 mL ofsolution?

__________ M

In the second type of problem, you actually use molarity which is part of the information given.Here's an example:

Sample Problem: How many grams of NaCl are needed to prepare 0.46 liters of a 0.25 M solution?

The 0.25 M given in the problem really means 0.25 moles NaCl / 1 L sol'n. Remember that sincemolarity is a derived unit, you can never use "molarity" or "M" in your problem-solving set-up. You must,instead, express molarity as moles solute / liters of solution. We will need to use this ratio in the fencepostsomewhere. We want to end up with "g NaCl," so let's begin with 0.46 L of solution.

0.46 L sol'n X ? ? ? ? ? ? ? ? = ? g NaCl

17-4 ©1997, A.J. Girondi

We can use the molarity ratio to change L sol'n to moles NaCl:

0.46 L sol'n X

0.25 moles NaCl

1 L sol'n X ? ? ? ? ? = 6.7 g NaCl

We still need one more ratio to complete the fencepost. We need the formula mass of NaCl to change"moles NaCl" to "g NaCl." Complete the set-up below and calculate the answer. See if your answeragrees with that given.

0.46 L sol'n X

0.25 moles NaCl

1 L sol'n X = 6.7 g NaCl

Do your answers to the two sample problems above agree with the answers given? ______________.

Let's try one more variation:

Sample Problem: How many mL of 0.500 M solution can be made using 65.0 g of KMnO4?

Let's first rewrite 0.500 M as "0.500 mole KMnO 4 / 1 L sol'n." We are looking for mL of solution solet's begin the fencepost with 65.0 g KMnO4.

Note that three conversion ratios need to be used in the following set-up. Complete the set-upfor the following problem and see if you get the same answer as that given.

65.0 g KMnO 4 X X

1 L sol'n0.500 moles KMnO 4

X = 823 mL sol 'n

Now, try the problems below. Show all work and use complete units on all numbers.

Problem 3. Calculate the number of grams of citric acid, C6H8O7, needed to make 2500. mL of 0.400 Msolution.

_________ g C6H8O7

Problem 4. If you wish to prepare a 0.500 M solution using 133.75 g of CaCl2, how many liters ofsolution can be made?

________ L Solution

17-5 ©1997, A.J. Girondi

Problem 5. How many grams of NaOH would you have to dissolve in 250. mL of solution to make a0.500 M solution?

__________ g NaOH

Remember that molarity (M) means:

moles solute

liters solution

Problem 6. Now, perform the necessary calculations to complete Table 17.1 below. See if you can finda way to use the information given to find that which is not given. Show your calculations in the spacebelow the table.

Table 17.1Review Problems Involving Molarity

Solute Grams NaNO3 Moles NaNO3 Liters Sol'n Molarity

NaNO3 25 __________ __________ ___1.2____

NaNO3 __________ __________ 16 0.023

Calculations:

17-6 ©1997, A.J. Girondi

Problems 7 through 10 below are optional. If you feel you need more practice you can do them on scratchpaper.

Problem 7. Calculate the number of grams of NaCl needed to make 100. mL of a 3.00 M solution.

Problem 8. What would be the molar concentration (M) of a solution containing 3.31 g of Pb(NO3)2 in atotal volume of 500. mL of solution?

Problem 9. How many liters of 3.00 M solution can be made using 210. grams of NH4OH (ammoniumhydroxide)?

Problem 10. If a solution is made by adding 125 grams of BaSO4 to enough water to make 750. mL ofsolution, what is the molarity (M) of this solution?

ACTIVITY 17.2 Making and Using A 0.10 Molar Solution

In this activity, you will be preparing and using a 0.10 M solution of lead nitrate. You will beattempting to experimentally obtain the data needed to determine the mole ratio in the reaction betweenpotassium chromate and lead nitrate:

?? K2CrO4(aq) + ?? Pb(NO3)2(aq) --------> ?? PbCrO4(s) + ?? KNO3(aq)

In the space below, write the balanced equation for this reaction:

_________________________------->___________________________

You will begin this activity by making the 0.10 M solution of Pb(NO3)2. Then you will react it with an excessamount of solid K2CrO4. An excess will be used to insure that all of the Pb(NO3)2 solution is consumed. Ifyou make the 0.10 M Pb(NO3)2 solution accurately, and follow the procedure using good lab techniques,you should obtain the correct result.

Procedure:

1. Obtain a 100 mL volumetric flask and rinse it with distilled water.

2. Determine the precise amount of Pb(NO3)2 in grams needed to make 100.0 mL of 0.10 M Pb(NO3)2 solution. Show your calculations in the space below, and have them verified by your instructor.

3. Obtain the amount of Pb(NO3)2 determined in step 2 above. Caution: Pb(NO3)2 is a toxic substance.

4. Add the Pb(NO3)2 crystals to the flask and fill the flask with distilled water to the mark on its neck.

5. Stopper the flask and invert it again and again until all of the crystals have dissolved.

6. Obtain a clean dry 150 mL beaker and label it with your name. Determine the mass of the labeled beaker. Record the beaker's mass in Table 17.2.

Caution: do not allow K2CrO4 to come into contact with your skin or clothing.

7. With the beaker still on the electronic balance, tare out the mass of the beaker. Add potassium

17-7 ©1997, A.J. Girondi

chromate, K2CrO4, crystals to the beaker until you have between 0.7 and 0.8 gram. You do not need torecord the mass of K2CrO4 since we are using an excessive amount.

8. Add about 40 mL of distilled water to the K2CrO4 crystals in the beaker and stir until they have dissolved.

9. Rinse a 50 mL graduated cylinder with distilled water. Use it to measure out exactly 50.0 mL of your 0.10 M Pb(NO3)2 solution. The remaining Pb(NO3)2 solution should not be poured into the sink. Pour it into the container provided by your instructor.

10. Obtain a ring stand and set up the equipment needed to heat the beaker of K2CrO4 solution. When the K2CrO4 solution in the beaker is hot (do not allow it to boil yet), slowly add the Pb(NO3)2 solution asyou continuously stir the mixture. Note the formation of the yellow precipitate, PbCrO4.

11. Again heat the mixture in the beaker – almost to the boiling point of the liquid – while stirring, then stopheating.

12. Accurately weigh a piece of filter paper. Record its mass in Table 17.2. Fold it into a cone, place it into a funnel, wet the paper with distilled water, and press it snugly against the wall of the funnel using yourthumbs. Mount the funnel properly on a ring stand.

13. Using beaker tongs to handle the hot beaker, separate as much of the the liquid in the beaker as possible from the precipitate by carefully decanting the liquid along a stirring rod into the funnel, while keeping as much of the precipitate as possible in the beaker.

14. Wash the lead chromate precipitate by squirting it with some distilled water (roughly 25 mL) from your wash bottle. Allow any suspended precipitate to settle, and decant the wash water into the funnel. Repeat the washing one more time.

15. Remove the filter paper cone and place it into the labeled beaker which contains the precipitate.

16. Place the beaker and its contents into a warm (not hot) oven to dry until the next day.

17. Determine the mass of the beaker and its dried contents. Record this mass in Table 17.2.

18. Give your precipitate to your instructor for proper disposal. Clean and return all equipment to the proper places.

19. Calculate the mass of PbCrO4 precipitate formed by subtracting the mass of the filter paper and beakerfrom the total mass. Enter the result into Table 17.2.

Table 17.2The Potassium Chromate – Lead Nitrate Reaction

mass of filter paper __________ g

mass of labeled beaker __________ g

mass of beaker + filter paper + precipitate __________ g

mass of solid PbCrO4 (by subtraction) __________ g

17-8 ©1997, A.J. Girondi

20. Convert the mass of PbCrO4 obtained to moles. Show work below.

___________moles PbCrO4

21. In this activity, you used 50.0 mL of a 0.10 M Pb(NO3)2 solution which you prepared. Using this data,calculate the number of moles of Pb(NO3)2 which were actually used.

___________moles Pb(NO3)2

22. Obtain a simple whole number ratio of moles of PbCrO4 to moles of Pb(NO3)2. Do this by dividing thenumber of moles of each substance (steps 20 and 21 above) by the smallest of the two values. Roundthe ratio obtained to whole numbers.

Whole number ratio of Pb(NO3)2 to PbCrO4: ________ to ________

23. Use the whole number ratio obtained, to balance the equation below:

?? K2CrO4(aq) + _____ Pb(NO3)2(aq) --------> _____PbCrO4(s) + ?? KNO3(aq)

Now, add the other coefficients needed to complete the balancing:

_____K2CrO4(aq) + _____Pb(NO3)2(aq) --------> _____PbCrO4(s) + _____KNO3(aq)

Do your results yield a correctly balanced equation? _______________________________________

If not, what do you think was your primary source of error?____________________________________

______________________________________________________________________________

Based on your results, is it your opinion that you correctly prepared the 0.1 M Pb(NO3)2 solution which

was used?__________ Why?_______________________________________________________

17-9 ©1997, A.J. Girondi

SECTION 17.3 How a Solute Affects the Freezing Point of a Solvent

Pure water freezes at 0oC. Ever wonder why it freezes at this temperature? There are attractiveforces which exist between water molecules. These attractive forces are strong enough to make water aliquid rather than a gas at room temperature. Remember that temperature is a measure of the averagekinetic energy (energy of motion) of the molecules. As temperature falls and molecules have less and lesskinetic energy, eventually there comes a point when the attractive forces cause the water molecules toassume a fixed position and the water becomes a solid. This happens at 0oC. However, dissolvingsomething in the water will lower the temperature at which the water freezes. To understand why thishappens, examine Figure 17.2 below. When solute molecules are present in the water, they help toseparate the water molecules and to weaken the forces of attraction which exist between the watermolecules. In addition, there are forces of attraction between the solute molecules and the watermolecules. In order for the water molecules to freeze, they must now lose more kinetic energy (thetemperature must get lower) so that their forces of attraction are strengthened. In this way, the watermolecules can "push" the solvent molecules out of the way and then freeze as pure water. If the numberof solute molecules in the solution is increased, then the temperature of the water must be lowered evenmore in order to make it freeze. Thus, as the concentration of a solution increases, the freezing point ofthe solvent in it decreases.

= solvent molecule= solute molecule

a pure liquid solvent

a water solution with solute molecules found between the solvent molecules

Figure 17.2Pure Solvent vs. Solution

The water in a salt-water solution will not freeze at 0oC. If, however, you were to lower the temperature of asalt water solution sufficiently, eventually the water would freeze. Technically, the solution does notfreeze; only the solvent does.

SECTION 17.4 How a Solute Affects the Boiling Point of a Solvent

You have just learned that a solute lowers the freezing point of a solvent. The effect of the soluteon the boiling point of the solvent is just the opposite. A solute raises the boiling point of a solvent. Tounderstand this phenomenon, we must examine the concept of vapor pressure. Suppose that you havetwo half-full but closed containers. One contains a pure solvent like water, while the other contains asolution such as sugar water. Assume that both containers are at the same temperature. (See Figure17.3.) In both containers some of the water evaporates and, therefore, some water vapor is present in thespace above the liquids. However, as shown in Figure 17.3, the container with the solution contains lessvapor than the one with the pure solvent. Hmmmm. Wonder why.

The molecules in the container of pure solvent are attracted to each other. However, themolecules of water in the solution are not only attracted to each other, but they are also attracted to themolecules of solute. Therefore, the molecules of solvent in the solution require more energy to escape

17-10 ©1997, A.J. Girondi

from the solution and enter the vapor phase. Not all of the solvent molecules in these containers have thesame amount of kinetic energy. Remember, temperature is an average. Since only a small fraction of themolecules in the solution have enough kinetic energy to break these forces of attraction, fewer of themcan enter the vapor phase.

more vapor

pure liquid

less vapor

solution

= solvent molecule= solute molecule

Figure 17.3Effect of Solute on Vaporization of Solvent

(lower pressure)(higher pressure)

Some of the vaporized water molecules will collide with the surface of the liquid and will be recaptured(condensed) into the liquid phase. At a given temperature, an equilibrium will be established in which themolecules are vaporizing and condensing at equal rates. At this point the number of molecules in thevapor phase will be constant. Remember that vapors exert pressure on the walls of their containers. Sincethe space above the solution contains less vapor than that above the pure water, the pressure exerted bythe vapor above the solution will be less than the pressure of the vapor above the pure water. Thus, wecan say that dissolving something in a solvent tends to {2}________________ the vapor pressure of thesolvent in closed systems like those shown in Figure 17.3. Now, how does all this relate to the boilingpoint of the solvent? What is the effect of a solute on the boiling point of a solvent like water? To answerthese questions, let's pursue the concept of vapor pressure a little more.

Suppose we have two half-full containers of pure water. One is at 20oC and the other is at 30oC.The water molecules at 30oC have more kinetic energy than those at 20oC. Therefore, more of the watermolecules at 30oC have enough energy to enter the vapor phase, and there will be more vapor in thecontainer at 30oC. More vapor means more pressure. So, the vapor pressure of water is greater at 30oCthan at 20oC. As temperature increases, so does vapor pressure. This is illustrated in Figure 17.4. Wecan also illustrate this relationship between temperature and vapor pressure by means of a graph. Thegraph is found in Figure 17.5.

O.K., let's review a couple of important things that we just learned. (1) Dissolving a solute in asolvent lowers the vapor pressure of the {3}_______________; (2) Temperature and vapor pressure aredirectly proportional - as one increases, the other {4}_________________. Now, let's put both of theseobservations together. Note that there are two curves plotted on the graph in Figure 17.6. Note that forboth pure water and salt water, vapor pressure increases as temperature goes up; however, also noticethat at any given temperature the vapor pressure of salt water is lower than that of pure water. A liquid in anopen container normally boils when its vapor pressure equals atmospheric pressure. If we assume that wehave an average day and that we are at sea level, then atmospheric pressure is 760 mm Hg. Thus, whenthe vapor pressure of water reaches 760 mm Hg, it will boil. Notice on Figure 17.6 that the vapor pressureof water reaches atmospheric pressure at 100oC. However, since solutes like salt or sugar lower the vaporpressure of a solvent like water, notice that for the salt water solution the vapor pressure of water does notreach atmospheric pressure until the temperature is above 100oC. Thus, dissolving a solute like salt inwater has the effect of {5}_________________ the boiling temperature. Salt water boils at a{6}__________________ temperature than pure water.

17-11 ©1997, A.J. Girondi

= solvent molecule

Figure 17.4Effect of Temperature on Vapor Pressure

less vapor(lower pressure)

more vapor(higher pressure)

20oC 30oC

760

mm HgP

Temp (oC) 100oC

Figure 17.5Vapor Pressure of Water vs. Temperature

20oC 30oC

760

0

pure water

salt water

100 102Temperature (oC)

VaporPressure(mm Hg)

Figure 17.6Vapor Pressure of Pure Water vs. Salt Water

17-12 ©1997, A.J. Girondi

We have learned that dissolving a solute in a solvent lowers the freezing point of the solvent andraises its boiling point. This is why we dissolve antifreeze in the water in the radiators of our cars. Theantifreeze not only helps to prevent winter freeze-up, but it also helps to prevent the water from boilingover in the summer!

ACTIVITY 17.5 The Freezing and Boiling Points of Salt Water

Procedure:

1. Get a bottle of saturated salt (NaCl) solution and a thermometer from the materials shelf. Now, measurethe boiling temperature by heating about 50 to 100 mL of the salt solution in a small beaker. Do notdiscard the saltwater after it boils. It can be used again, so let it cool and then return it to its container. Atwhat temperature did the solution boil vigorously?________oC. This phenomenon is known as boilingpoint elevation.

2. To measure the freezing point of saturated saltwater, get another sample of the solution - about 150mL. Add 3 or 4 crushed ice cubes to the solution. Continuously stir the solution. Record the lowesttemperature attained by the solution. It should level off; if it doesn't, add more crushed ice. This lowesttemperature represents the freezing point of water when it is saturated with NaCl. What is the freezingpoint of the salt solution? __________ Were the predictions you made above correct?________Discard this salt solution (since you diluted it with ice, it will not be reused).

ACTIVITY 17.6 The Relationship Between Boiling Point and Pressure

-> -> -> Teacher Demonstration <- <- <-

In the last section you learned that a liquid boils when its vapor pressure equals the pressureabove the liquid (usually atmospheric pressure). What would happen to the boiling point if we would lowerthe pressure above a liquid? Well, lets take a look at Figure 17.7.

760 mm Hg

0100oCTemperature

VaporPressure(mm Hg)

Figure 17.7Effect of Pressure on the Boiling Point of Water

25oC

24 mm Hg

boiling point is 25oC at 24 mm Hg pressure

normal boiling point is 100oC at atmospheric pressure

17-13 ©1997, A.J. Girondi

Room temperature is usually about 25oC. As shown in Figure 17.7, if we could lower the pressure abovewater to 24 mm Hg, then the vapor pressure of the water would equal the pressure above the water (24mm Hg) at only 25 oC or room temperature. This means that water should boil at room temperature if wecan lower the pressure above it to 24mm Hg!

Let's give it a try. Your teacher will place a small beaker of water under a bell jar on a vacuum pump.When the pump is turned on, the pressure above the water will be reduced until the water boils. Youshould note the temperature of the water before and after the pressure has been reduced.

Thermometer

Pressure Gauge

Vacuum Pump

Glass Bell Jar

Beaker of Water

Figure 17.8Measuring Temperature vs. Pressure

Remember that the water molecules in the beaker contain kinetic energy. Some have more than others. Ifthe pressure exerted by the air molecules is high, then very few molecules of water in the beaker at 25oChave enough kinetic energy to escape into the vapor phase. This is why water does evaporate at roomtemperature, but only very slowly. However, if the pressure of the air above the water is decreased, then amuch higher percentage of the water molecules in the beaker will have enough energy to escape. Theevaporation will be very rapid. We call it boiling. Thus, water can be made to boil by heating it (giving thewater molecules more kinetic energy) or by reducing the pressure above it (see Figure 17.9).

high air pressure

water

bubble containing vapor cannot overcome air pressure

low air pressure

water

bubble containing vapor bursts and vapor escapes

25oC 25oC

Figure 17.9Behavior of Water at High and Low Pressure

17-14 ©1997, A.J. Girondi

SECTION 17.7 Phase Changes in Pure and Impure Substances

Now, let's take a closer look at the effect of a solute on the properties of a solvent. Recall fromChapter 3 that the warming and cooling curves of pure substances have plateaus where the phasechanges are occurring. In a pure substance, phase changes occur at constant (unchanging)temperatures. This is not the case for impure substances. Look at Figures 17.10 and 17.11.

melting

boiling

TempoC

100

0

Time

Figure 17.10Warming Curve of a Pure Substance

TempoC

100

0

Time

Figure 17.11Warming Curve of an Impure Substance

boiling

melting

How are the two warming curves similar to each other?

How are the two warming curves different?

You may wonder why the phase changes (melting and boiling) of a pure substance occur atconstant temperatures. Remember that temperature is a measure of the kinetic energy (energy of motion)

17-15 ©1997, A.J. Girondi

of molecules. As you heat or cool a substance, the molecules are changing their rate of motion. However,during a phase change it is the position of the molecules that is changing, not the rate of motion. Duringboiling, the molecules are moving farther apart. Their {7}_________________ energy (energy ofposition) is changing. During freezing, the molecules are changing position and losing potential energy.Temperature is not affected by changes in {8}___________________ energy; thus, a plateau is seenduring phase changes.

In the winter, rock salt is often placed on icy streets and sidewalks. Propose a reason for this practice,

including some of the principles you have just learned. {9}

SECTION 17.8 Suspensions

Up to this point, all of the solutions that have been described have consisted of solublesubstances dissolved in a solvent. The individual solute molecules in these solutions move apart andbecome uniformly (homogeneously) distributed throughout the solvent molecules. The existence ofindividual solute molecules in the solvent makes this a true solution. If finely divided clay is mixed withwater and allowed to stand, the particles will slowly settle to the bottom. Because the clay particles areinsoluble in the solvent and are much larger than individual molecules, they can reflect light and make themixture appear cloudy. This type of mixture is called a suspension.

Some mixtures of insoluble substances have particles so small that they will not settle out onstanding. Such suspensions are known as colloids or colloidal suspensions. Colloidal suspensions maybe classified using terms which are similar to those which describe true solutions. Instead of solute andsolvent, the suspended particles are called the dispersed phase, and the liquid in which they aredispersed is called the dispersion medium. Collisions between these tiny particles and those of thedispersion medium keep the tiny particles suspended. Nevertheless, the suspended particles in colloidsconsist of small "bundles" of molecules, not individual molecules (as in a true solution).

Figure 17.12Components of a Suspension

particle clusters

dispersion medium

Some mixtures of insoluble substances have particles so small that they will not settle out onstanding. Such suspensions are known as colloids or colloidal suspensions. Colloidal suspensions maybe classified using terms which are similar to those which describe true solutions. Instead of solute and

17-16 ©1997, A.J. Girondi

solvent, the suspended particles are called the dispersed phase, and the liquid in which they aredispersed is called the dispersion medium. Collisions between these tiny particles and those of thedispersion medium keep the tiny particles suspended. Nevertheless, the suspended particles in colloidsconsist of small "bundles" of molecules, not individual molecules (as in a true solution).

Some mixtures of insoluble substances have particles so small that they will not settle out onstanding. Such suspensions are known as colloids or colloidal suspensions. Colloidal suspensions maybe classified using terms which are similar to those which describe true solutions. Instead of solute andsolvent, the suspended particles are called the dispersed phase, and the liquid in which they aredispersed is called the dispersion medium. Collisions between these tiny particles and those of thedispersion medium keep the tiny particles suspended. Nevertheless, the suspended particles in colloidsconsist of small "bundles" of molecules, not individual molecules (as in a true solution).

The term colloid comes from a Greek word meaning "glue." The word was first used tocharacterize the glue-like solutions formed when large organic (carbon-containing) molecules - such asstarch or gelatin - were dispersed in water. The colloidal phenomenon plays an important role in numerousindustrial processes. Colloids are used in the production of rubber products, inks, leather, andphotographic materials. The detergent action of soaps and the action of lubricants all involve theapplication of colloidal principles.

It is relatively easy to distinguish between a colloidal dispersion and an ordinary suspension. If anordinary suspension is shaken, it will become turbid and opaque. The suspension particles settle rapidlyand can be removed by filtration. A colloidal dispersion is generally transparent or translucent and particlesare too small to be removed by filtration. Colloidal particles are extremely small, and they may reflect solittle light that the colloidal suspension may look clear, like a true solution, when viewed with natural light.(True solutions look clear because the individual molecules of solute, and the solute–solvent clusters, aretoo small to reflect light.)

SECTION 17.9 Colloidal Systems

A general classification of colloidal systems depends on the physical state of the dispersed phaserelative to the dispersion medium. The most common systems are known as sols, gels, aerosols, andemulsions . A sol is a colloidal "solution" that exists as a liquid at room temperature, with water as thecontinuous phase and a solid as the discontinuous phase. A gel is a similar colloidal dispersion that is asolid at room temperature. Jello brand dessert is an example of a gel. An aerosol is a colloidal system inwhich either a solid or a liquid is dispersed in a gas. Smoke is an example of an aerosol. Milk is an emulsionconsisting of a liquid (butter fat) dispersed in a liquid (water). A foam consists of a gas which is dispersedthrough a liquid such as whipped cream or a solid such a pumice stone. Table 17.4 lists examples ofvarious types of colloids and the phases of the dispersed phase and the dispersion medium.

Table 17.4Types and Examples of Colloids

Name Dispersed Dispersion Example CommentPhase Medium

gel solid liquid jello solid at room temp. sol solid liquid ink liquid at room temp.aerosol solid gas smoke, dustaerosol liquid gas fog, mist emulsion liquid liquid mayonnaise foam gas liquid whipped cream foam gas solid pumice

17-17 ©1997, A.J. Girondi

ACTIVITY 17.10 A Laser–Aided Look at a Colloidal Suspension


There is a simple test that can be performed to distinguish between true solutions and colloidaldispersions. It has to do with the way in which a beam of light passes through each of these two types ofsolutions. The phenomenon is called the Tyndall effect.

Get the bottles labeled 17.8 from the materials shelf - one contains a true solution, one containsan ordinary suspension, and the third contains a colloidal dispersion. They have been numbered 1, 2,and 3 - but the numbers were randomly assigned and do not identify which is which. You are going toshine a strong light through the side of each bottle. You will need a dark room to do this. Use a laser if youhave one (ask your instructor). Caution: never look directly into a laser beam! The instructor will assist youif you are using a laser.

Any undissolved solid material in the liquids will scatter the light and will be appear as a visiblebeam or cloudiness. The true solution should contain no undissolved solid material, however someimpurities like dust may be present and may cause a little scattering. The colloidal dispersion containssuspended undissolved particles of silver chloride. (You may not be able to see them under roomlighting).

Arrange the bottles so that the light beam travels through the contents of all three bottles. Sincethe particles in the suspension are largest they reflect the most light – followed by the colloidal dispersion.Identify the suspension, colloidal dispersion, and true solution:

Bottle #1 contains the: _________________________________________



Turn off the light source, and return the bottles to the materials shelf.


Problem 11. How many grams of NaOH are needed to make 350. mL of a solution which is 0.150 M?

Problem 12. How many liters of a 2.50 M solution of KCl can be made using 85.0 grams of KCl?

Problem 13. How many moles of MgBr2 are contained in 6.70 L of a solution which is 0.500 M?

Problem 14. What is the molarity (M) of a solution which contains 65.0 grams of FeCl3 in 670. mL ofsolution?

Molality: In Appendix C of your ALICE materials you will find asection entitled "Freezing Point Depression, Boiling PointElevation, and Molality." This may be optional work for you, oryour teacher may choose to make it a required part of yourcourse. If you are told to complete Appendix C, you should do itnow. At this time ask your teacher about Appendix C.

?

17-18 ©1997, A.J. Girondi


Review the learning outcomes below, and check them off if you have mastered them. When theyhave all been checked, arrange to take any quizzes or exams on Chapter 17. Take the quiz on AppendixC too, if applicable. Then move on to Chapter 18.

_____1. Define and calculate concentration in terms of molarity.

_____2. Given two of the following values (molarity, mass, and volume), calculate the third value.

_____3. Given a chemical equation and the molarity or masses of any reactants involved, predict the theoretical yield of products in a chemical reaction.

_____4. Explain the effect of a solute on vapor pressure, melting point, and boiling point of a solvent.

_____5. Distinguish between colloids, true solutions, and suspensions.

_____6. Given the common class name of a colloid, name the dispersed phase and the dispersion medium. Give examples of each type of colloid.

_____7. Explain the principles involved in boiling-point elevation and freezing-point depression.

You will be expected to master the following two outcomes if you are responsible for the material inAppendix C.

_____8. Solve problems in which you are asked to calculate the molality of a solution.

_____9. Solve problems in which you are required to calculate freezing–point depression and boiling–point elevation.

17-19 ©1997, A.J. Girondi


Questions:

{1} dilute; {2} lower; {3} solvent; {4} increases; {5} raising; {6} higher; {7} potential; {8} potential;{9} Salt lowers the melting point of ice

Problems:

1. 4.00 M2. 0.100M3. 192 g4. 2.41 L5. 5.00 g6.

Table 17.1Review Problems Involving Molarity

Solute Grams Solute Moles Solute Liters Sol'n Molarity

NaNO3 25 ___0.29___ ___0.25___ 1.2

NaNO3 ___31.3 ___ ___0.368__ 16.0 0.0230

7. 17.4 g8. 0.02 M9. 2 L10. 0.714 M BaSO4

11. 2.10 g NaOH12. 0.456 L KCl solution13. 3.35 moles MgBr214. 0.598 M FeCl3

17-20 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


“ALICE”

CHAPTER 18

RATES OFCHEMICAL

REACTIONS"Kinetics"

18-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


18-2 ©1997, A.J. Girondi

ACTIVITY 18.1 Comparing Rates of Reactions Between Metals and Acids

You have, thus far, completed various activities in which different types of chemical reactionsoccurred. These chemical reactions involved color changes, temperature changes, precipitate formation,and the evolution of gases. These reactions occurred quickly. Many chemical reactions, however, canproceed rather slowly. In this chapter, you will study the factors that determine the rates of reactions andthe extent to which chemical reactions occur. These two factors are generally called reaction kinetics.

Let's begin by looking at two simple chemical reactions that occur at different rates. Get thematerials labeled 18.1 from the materials shelf. Place about 4 mL of 3 M HCl into two 150 mm Pyrex testtubes. Into one test tube place a small piece of zinc metal. Put a small piece (≈2 cm) of magnesium metalin the other test tube. Answer the following questions about what you just did.

1. How can you tell that a chemical reaction has occurred in the HCl–zinc combination?

{1}________________________________________In the magnesium–HCl combination? {2]_______

______________________________________________________________________________

2. How does the rate of the zinc–HCl reaction compare to the rate of the magnesium–HCl reaction?_____

______________________________________________________________________________

3. Is your answer to the previous question a qualitative or a quantitative statement?{3}____________

Explain: {4}______________________________________________________________________

4. After you have compared the reaction rates, you should test each test tube for the presence of aflammable gas. (Wear safety glasses.) Allow the gas bubbles to accumulate for about 20 to 30 secondswhile you keep your thumb loosely over the mouth of the tube. If needed, put more metal in the tubes.Hold one tube at a 45 degree angle, remove your thumb and very quickly place a flaming wood splint atthe top of the tube. Repeat with the second tube. Identify the gas given off in this reaction: {5}________

SECTION 18.2 Solving Problems Involving Reaction Rates

It is a fairly simple task to quantitatively compare reaction rates. But, is there a way that we mightquantitatively express the rates of the reactions of magnesium and zinc with HCl which you witnessed inactivity 18.1.? Well, actually, there are several ways in which this could be done.

rate =

moles reactant used

time required for change

1. We could weigh the magnesium and record the time requiredfor it to react completely. Rate would then be defined by how fasta reactant is consumed.

rate =

moles produce formed

time required for change

2. Or, we could time the reaction and measure the volume of gasproduced during the experiment. Rate would then be defined byhow fast a product is formed. From this data, the rate of thereaction could then be calculated using the formula at right:

Often it is more convenient for scientists to calculate rate in units of grams/second, orgrams/week, or moles/day, etc. When this is the case, the situation will specify the units you are to usewhen solving a problem.

18-3 ©1997, A.J. Girondi

rate =

moles reactant used

second

If specific units are not requested, you should calculate the rate in "moles reactant used / second."

Sample problem: If 0.048 grams of magnesium, Mg, completely reacted with acid in 20. seconds, whatwould be the rate of this reaction in units of "moles Mg / sec?"

Note that in this problem you are given a quantitative relationship between Mg and time: 0.048 gMg / 20 sec. To change this ratio to "moles Mg / sec" will require a short fencepost. Complete thefencepost below, and calculate an answer. Fill in the blanks.

0.048 g Mg

20. sec X = {6} ___________ mole Mg /sec

If you got a small answer, don't be suspicious. Remember, a mole is a very large number of molecules!Now, let's try another problem.

Sample Problem: In an experiment 0.0070 g of zinc metal completely reacts with acid in 30.0 seconds.Calculate the average rate of this reaction. Complete the fencepost below and do the calculations to see ifyou get the answer given.

0.0070 g Zn X X = 3.6 X 10-6 mol Zn /sec

Next, let's try a problem with a different twist to it.

Sample Problem: It is known that the rate at which magnesium reacts with HCl is 5.0 X 10-2 moles/secand that the reaction takes 5.0 seconds to occur. Calculate the number of moles of magnesium that havereacted.

Complete the fencepost below, and see if you can get the answer that is given.

5.0 X 10-2 mole Mg

1 sec X

1 = 0.25 mole Mg

Problem 1. Calculate the number of grams of Mg that reacted in the problem above. Show your work.

__________ g Mg

Before moving on, it is important that you feel certain about your ability to solve rate problems. Foradditional practice, solve the following problems.

18-4 ©1997, A.J. Girondi

Problem 2. After 2.0 minutes, 0.50 grams of zinc react completely in HCl. Calculate the rate of reactionin grams of zinc per second.

__________ g Zn/sec

Problem 3. Referring to the last problem, calculate the rate of the reaction in moles Zn used persecond.

__________ mole Zn/sec

Problem 4. Given that magnesium reacts with air at a rate of 4.00 X 10-3 moles per 1.00 second, howmany minutes would it take for 4.00 moles of Mg to react?

__________ minutes

Problem 5. Referring to problem 4 above, how many grams of Mg would react after 140. seconds?

__________ g Mg

Problem 6. In 3.0 minutes 0.10 g of CaCO3 will decompose in HCl. Calculate the rate of the reaction inmoles CaCO3 per second.

__________ moles CaCO3/sec

18-5 ©1997, A.J. Girondi

SECTION 18.4 Collision Theory and Activation Energy

Now that you know how to express reaction rates quantitatively, you will do several experimentsthat will enable you to determine the effects of certain factors on the rates of chemical reactions. Theyinclude: (1) nature of the reactants; (2) concentrations; (3) temperature; (4) surface area; and, (5) catalysts.It is usually quite simple to determine the effects of these factors on reaction rates once the basicprinciples involved in chemical reactions are known. An explanation based on the collision theory plays amajor role in this.

The collision theory is based on the assumption that for a chemical reaction to occur, particlesmust first collide with each other. In these collisions, atoms and electrons are rearranged by a reshufflingof chemical bonds that results in the formation of products. A reaction does not necessarily occur everytime there is a collision. First, the reactant molecules must collide in just the right way. According to thecollision theory, the rate of a reaction depends on two factors: (1) the number of collisions per secondbetween the reacting particles, and (2) the fraction of these collisions that are effective. The collidingparticles must collide with sufficient energy to break the reactant's bonds. If sufficient force is not exerted,no reaction will occur. This minimum amount of energy required to rearrange the reactants into products iscalled their activation energy.

Activation energy is easiest to visualize using graphs. Figure 18.1, shows that the products of aparticular reaction possess more energy than the reactants do. There is only a 5 kJ (kilojoule) differencebetween the energy of the reactants and the products. Logically, one would think that adding 5 kJ ofenergy to the reactants would be enough to convert them into products. This is exactly what Figure 18.1shows. Now look at Figure 18.2. This plot is of the same reaction shown in Figure 18.1, but the activationenergy has been added. Recall that the activation energy is the minimum energy needed to rearrangereactants into products. This means that before the reaction in Figure 18.2 can occur, about 12 kJ (not 5kJ) of energy must be added to the reactants.

When reactant molecules collide they form what is called an unstable activated complex whichquickly decomposes to form the products of the reaction. When this decomposition occurs, some of theactivation energy is given off. So while about 12 kJ of activation energy were needed to get the reactionin Figure 18.2 to go, the actual difference in the energy content of the reactants and products is only 5 kJ.

kJ

5

10

15

20

reaction progress

reactants

products

Figure 18.1

kJ

5

10

15

20

reaction progress

reactants

products

Figure 18.2

Activation.Energy

As you will learn in chapter 19, exothermic reactions give off heat energy, while endothermicreactions absorb heat energy. Therefore, in exothermic reactions the products have less energy than thereactants. In endothermic reactions the products have more energy than the reactants. Since theproducts possess 5 kJ more energy than the reactants in Figure 18.2, would this plot represent an

exothermic or an endothermic reaction? {7}_______________________________

18-6 ©1997, A.J. Girondi

Study Figures 18.3 and 18.4. Label the reactants and products, and draw an arrow showing the

activation energy on each curve. How much energy would need to be added to the reactant in Figure

18.3 to get a reaction to occur? {8}________________. How much energy would need to be added to

the reactant in Figure 18.4 to get a reaction to occur?{9}_______________ Is the reaction represented in

Figures 18.3 exothermic or endothermic {10}________________________ Is the reaction represented

in Figures 18.4 exothermic or endothermic? {11}_________________________

kJ

5

10

15

20

reaction progress

Figure 18.3

kJ

5

10

15

20

reaction progress

Figure 18.4

In any reaction system, the molecules do NOT all have equal amounts of energy. That's why wedefine the temperature of a system as a measure of the average kinetic energy of the molecules. As aresult, some molecules may not collide with sufficient energy to produce a reaction. If the activationenergy required for a particular reaction is high, fewer molecules are likely to have enough energy to reactwhen they collide. If a particular reaction has a very high activation energy, would you expect the reactionrate to be fast or slow?{12}___________. Explain why. {13}___________________________________

______________________________________________________________________________

___________________________________________________. Would this same reaction have a

large or a small value for its reaction rate? {14}______________________

The "nature of the reactants" refers to the identity and properties of the reactants. Activationenergy is just one of these properties. Some combinations of reactants have high activation energies,while that for other combinations is low.

Without specific knowledge about the reactants, you can't be certain whether a particular reactionwill occur rapidly or slowly. You may recall that when you were studying equations of single and doublereplacement reactions, you had to predict whether or not a reaction would occur. However, it's possiblethat some of those reactions may occur instantly, while others which are predicted to occur may occurvery, very slowly – perhaps even taking years until they are completed! Educated guesses about reactionrates are possible if some characteristics of the reactants are known. Chemists have developed two rulesconcerning relative rates of reactions:

1. Many reactions that do not involve the breaking of bonds occur rapidly at room temperature.

2. Many reactions in which bonds are broken tend to occur slowly at room temperature.

18-7 ©1997, A.J. Girondi

Use these two general rules to determine which of the reactions below would probably be faster. Writethe word faster next to that equation. Explain why you made your choice.

Reaction 1: H + H -----> H2

Reaction 2: 2 CH4 + 2 Br2 -----> 2 CH3Br + 2 HBr

Explanation: {15}_________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

The two reactions shown above involve either atoms or molecules of substances. It is alsopossible for ions to react with each other. Ions are charged particles consisting of one or more atomswhich can exist as dissolved (aqueous) particles in water solutions or as gases at high temperatures.Reactions between ions tend to occur very rapidly. Two examples of ionic reactions are shown below.Note that the ions have the charges shown:

Ag1+(aq) + Cl1-(aq) ----> AgCl(s)

5 Fe2+(aq) + MnO41-(aq) + 8 H1+(aq) ---> Mn2+(aq) + 4 H2O(l) + 5 Fe3+(aq)

ACTIVITY 18.5 Comparing the Rates of Two Reactions

In this activity you are going to compare the rate at which two compounds react with a solution.The first compound is ferrous sulfate, FeSO4, which is ionicly bonded. The second compound is oxalicacid, H2C2O4, which is covalently bonded. The solution they are going to react with is potassiumpermanganate, KMnO4, which is ionicly bonded. Wear your safety glasses and an apron. Obtain adropper bottle containing a solution of 0.01 M potassium permanganate (KMnO4) which is purple in color.Handle this solution with care since it can stain your skin or clothes. (Be sure to follow proper lab safetyprocedures.) Obtain two 100 or 150 mL beakers and fill each about one-third full of distilled water. Labelthem as beaker 1 and beaker 2. Add a small amount of solid FeSO4 to beaker 1 and a small amount of solidoxalic acid to beaker 2. (Handle oxalic acid carefully. Wash your hands as needed.) Stir each solution for acouple of minutes, rinsing the stirring rod before you move from one solution to the other. Don't beconcerned if some of the solid in either beaker remains undissolved.

Add a drop of the KMnO4 solution to the FeSO4 in beaker 1, stir the solution, and make aqualitative estimate (such as rapid or slow) of the time required for the color to disappear. Repeat thisprocedure by adding another drop of KMnO4 to beaker 1.

Repeat the procedure in the paragraph above, but add the KMnO4 to the oxalic acid solution inbeaker 2. Again, make a qualitative estimate of the time required for the color to disappear compared tothe result for beaker 1. There should be an obvious difference in rate.

Compare the rates of reaction for FeSO4 in beaker 1 and for H2C2O4 in beaker 2? __________________

______________________________________________________________________________

Suggest possible reasons for the difference in rates. (Refer to section 18.4 if you need help.) ________

______________________________________________________________________________

18-8 ©1997, A.J. Girondi

SECTION 18.6 Reaction Mechanisms

There is one additional factor related to the nature of the reactants that must be discussed here.Look at the reactants and products in this reaction:

4 HBr + O2 ----> 2 H2O + 2 Br2

Remember that there must be collisions of sufficient energy between the HBr and O2 molecules beforethe products can be formed. Is it possible that the O2 molecule could collide with the HBr with enoughenergy to become bonded to the HBr, but not with enough energy to actually break the H–Br bond? Ifthis happened, you would have one larger molecule containing all of the reactants:

HBr + O2 ----> HBrO2

Let's take a closer look at what this means. Many chemical reactions involve several intermediate steps.The pathway of intermediate steps through which the reactants pass as they are converted to products iscalled the reaction mechanism. The reaction mechanism for the HBr + O2 reaction involves the threesteps shown below:

Step 1: HBr + O2 ----> HBrO2 <--------- (slow)

Step 2: HBrO2 + HBr ----> 2 HBrO <--------- (fast)

Step 3: 2 HBrO + 2 HBr ----> 2 H2O + 2 Br2 <--------- (fast)

Sum: 4 HBr + O2 + HBrO2 + 2 HBrO ----> HBrO2 + 2 HBrO + 2 H2O + 2 Br2

Net: 4 HBr + O2 ----> 2 H2O + 2 Br2

In step 1, a molecule of HBr combines with an O2 molecule to form a larger molecule of HBrO2. In step 2,the HBrO2 molecule collides with another HBr molecule to form two molecules of HBrO. In step 3, twomore molecules of HBr collide with two HBrO molecules to form the final products which are two moleculesof H2O and two molecules of Br2. Note that the HBrO2 and HBrO molecules are both formed andconsumed during the reaction; therefore, they do not appear as either reactants or products in the overallor net equation. Such substances are sometimes called "intermediates." If you "add up" the threeequations in the reaction mechanism, the result will be the net equation.

Two steps in this reaction mechanism occur extremely fast; however, one step occurs much moreslowly. For that reason, we say that the rate of the reaction depends on the slowest step in themechanism. Just like a relay team of runners, if even one runner is slow, the entire team may beconsidered slow.

No chemical reaction can occur faster than the slowest step in its mechanism. This slow step iscalled the rate-determining step. In the 3-step reaction shown above, which step is the rate-determiningstep? {16}__________ Would the overall reaction be fast or slow?{17}___________________.

Explain: {18}_____________________________________________________________________

______________________________________________________________________________

(Note: whether a reaction has a one-step mechanism or a multi-step mechanism depends on the nature ofthe particular reactants involved.)

18-9 ©1997, A.J. Girondi

The rate of a chemical reaction is also affected by the concentration of the chemicals used.Collision theory proposes that molecules must collide with each other before they can react. Using thisprinciple from collision theory, make a prediction about how increasing the concentration of reactantswould affect the rate at which substances react. Explain both how and why: {19}_______________________________________________________________________________________________________________________________________________________________________________

ACTIVITY 18.7 Solution Concentration and Reaction Rate

Now let's investigate the relationship between concentration and reaction time. For this activityyou will need two solutions from the materials shelf labeled solution 1 and solution 2. You will be timingreactions in this series of experiments. It is important that you read through the entire activity beforebeginning. In this way, you will know what is expected of you. It is always a good idea to do this beforeperforming an experiment.

Procedure:

1. Measure 5 mL of solution 1 and pour it into a 50 mL flask or beaker.

2. Add 5 mL of water (at room temperature) to the same flask. Swirl the container to mix the solution.

3. Carefully measure 5 mL of solution 2.

4. Get a watch ready (use a stopwatch, if available) and then mix solutions 1 and 2. Immediately stir thesolution by swirling. Begin timing the reaction at the moment solution 2 is added to solution 1. Record thetime required for a noticeable reaction to occur in Table 18.1. If you place the flask or beaker on a sheet ofwhite paper, the reaction will be more visible.

What evidence do you see that a chemical reaction has occurred?_____________________________

You have already carried out reaction 1 in Table 18.1. Now prepare the correct combinations shown inTable 18.1 for each of the remaining three reactions. Be sure to use water which is at room temperature.Time each reaction and record your results in the Table.

The last column of Table 18.1 asks that you calculate the concentration of HSO31- ion present ineach of the reaction mixtures. The bisulfite ion, HSO31-, is present in solution 1. To calculate thisconcentration you need to know that the concentration of HSO31- in the original stock solution is 0.072 M.When you add solution 2 and water to the stock HSO31- solution (solution 1), you reduce theconcentration of the HSO31- ion. The total volume in each flask at the end of the reactions is always 15mL.

Study the calculations shown below for determining the concentration of HSO31- in reaction 1.The formula you will use is:

McVc = MdVd Mc = molarity of concentrated sol'n

Vc = volume of concentrated sol'n Md = molarity of dilute sol'n Vd = volume of dilute sol'n

18-10 ©1997, A.J. Girondi

Mc will have a value of 0.072 M HSO31- in each of your calculations. Vc will vary in the four reactions: (5.0mL, 4.0 mL, 3.0 mL,and 2.0 mL). Md is the concentration of the diluted HSO31- solution (the unknown). Vd

will be 15 mL in each reaction.

For reaction 1 the calculation is as follows:

Mc Vc = MdVd so, M d =

McVcVd

and, M d = (0.072) (5.0 mL)

(15 mL) = 0.024 M

The equation above can be used whenever you wish to calculate the new concentration of a solution afteryou have diluted it. Perform the calculations for the remaining three reactions. Show your work below.Enter the results in Table 18.1.

Reaction 2:

__________ M

Reaction 3:

__________ M

Reaction 4:

__________ M

Table 18.1Concentration vs. Reaction Time

Rx No. Volume Sol'n 1 Volume Sol'n 2 Rx Time Molarity of HSO31-

+ water (sec) (mol/L)

1 5 mL + 5 mL 5 mL ______ _________

2 4 mL + 6 mL 5 mL ______ _________

3 3 mL + 7 mL 5 mL ______ _________

4 2 mL + 8 mL 5 mL ______ _________

On the grid which follows, prepare a graph of reaction time versus the concentration of HSO31-.Label each axis, placing the independent variable on the horizontal (x) axis and the dependent variable onthe vertical (y) axis. The independent variable in this example is concentration. The value of thedependent variable depends on the value of the independent variable.

Which variable is dependent here?{20}_______________________

18-11 ©1997, A.J. Girondi

State a general rule that relates the concentration of reactants to the rate of a chemical reaction: {21}_____

______________________________________________________________________________

What effect would diluting solutions with water have on the rate at which they react? {22}____________________________________________________________________________________________

ACTIVITY 18.8 The Effect of Temperature on Reaction Rate

(Note: This activity is best done by all lab groups on the same day. Ask your instructor.)

Another factor that affects the rate of chemical reactions is temperature. Before actuallyconducting this activity, give some thought to what is going to be happening to the kinetic energy(motion) of the particles of the reactants as they are heated.

Scientists use a hypothesis as a means of predicting what is going to happen before they actuallyperform an experiment. This gives them a chance to generate an "educated guess" about the results ofthe experiment.

What effect will temperature have on the rate of a chemical reaction? (State a hypothesis in terms of the

collision theory.) :{23}_______________________________________________________________

______________________________________________________________________________

To test your hypothesis, we will once again be using solutions 1 and 2. To be effective, both ofthese solutions should be at the experimental temperature. This can be done quite easily by preparing awater bath with a known temperature. Your class will be measuring the rate of the reaction at four differenttemperatures. These temperatures should be close to the following: 5oC, 25oC, 40oC, and 50oC. Eachof the time trials will require 3 mL of solution 1, 7 mL of water, and 5 mL of solution 2.

A hot or cold water bath is prepared by filling a 600 mL beaker Figure 18.5 about half-full of water.If there is a need to raise the temperature, use your burner to supply heat. Ice cubes can be used to lower

18-12 ©1997, A.J. Girondi

the temperature. The general setup is shown in Figure 18.5. Be sure that the level of water in the bath ishigher than that of the solutions in the test tubes that will be placed in the bath.

Your lab group will be assigned one temperature at which to time the reaction. Other labs groupswill be assigned different temperatures. Each group should repeat its time trials several times. The timesshould be fairly close. Discard any times which are not close to the others which you obtained, thenaverage the results. If other lab groups in your class are assigned the same temperature as your group,include their times when you calculate your average time.

Procedure:

1. To prepare your solutions, place 5 mL of solution 2 into a large (25 X 200 mm) test tube. Then,measure 3 mL of solution 1 and 7 mL of water into a second large test tube. Do not mix these twosolutions, yet! They can be stoppered and saved. Be sure to label them.

2. Bring your water bath to the assigned temperature for your lab group.When the proper temperature has been achieved, place each unstopperedtest tube in the water bath and give each enough time to reach thetemperature of the water bath. (This will be about 3 to 5 minutes.) Be sure tomaintain a constant and even temperature in the bath while the test tubes arein it. Small pieces of ice or small amounts of heat with lots of stirring will do this.The test tubes, themselves, can be used as stirring rods.

3. After time has elapsed, quickly mix the two solutions by pouring thecontents of one tube into the other tube. To provide good mixing, you shouldimmediately pour the mixture back into the empty tube and quickly place thetube back into the water bath. Record the reaction time. Repeat the reactionat the same temperature until you get two or three values which are close.Average the values and record them in Table 18.2. (As noted above, if othergroups worked at the same temperature as yours, include their values in youraverage.)

thermometer

sol'n 2

sol'n 1 +water

Figure 18.5Heating Solutions

4. Obtain the average time values for the other three temperatures from other lab groups in your class.Enter those values in Table 18.2.

Table 18.2Temperature vs. Reaction Time

Rx No. Relative Experimental Rx Time Temperature Temperature (sec)

1 Cold __________ ______(about 5oC)

2 Room Temp __________ ______(about 25oC)

3 Warm __________ ______(about 40oC)

4 Hot __________ ______(about 50oC)

18-13 ©1997, A.J. Girondi

5. Prepare a graph of temperature vs. reaction time on the grid which follows. Be certain to label each axison your graph. Put the dependent variable on the vertical (y) axis.

What is the dependent variable?{24}_________________________ Does the graph support your

hypothesis concerning temperature and reaction rate?__________ If not, explain: _______________

______________________________________________________________________________

Was the prediction you made in your hypothesis in agreement with the results that you obtained for this

activity? ___________ Experimental error tells us that some of the data we collect during experiments

is not always accurate. A lot of experimental error is really human error. We are often not demanding

enough of ourselves with our laboratory techniques. The end result is data that is not wrong, but not very

accurate, either. List a few sources of experimental error in this activity: _________________________

______________________________________________________________________________

ACTIVITY 18.9 The Effect of Surface Area on Reaction Rate

The next variable which you will study that influences chemical reaction rates is surface area.Surface area has a large effect on rates of reactions in which solids are involved. Chemical reactionsinvolving solids take place only on the surface of the solid. Justify this statement using the collision theoryas your basis. Now, hypothesize about what effect increased surface area has on the rate of a reaction

involving a solid: {25}_______________________________________________________________

______________________________________________________________________________

18-14 ©1997, A.J. Girondi

Part A. Steel Wool

Obtain a sample of steel wool. Tear off two small and approximately equal pieces about the size ofa quarter. Form one piece into a small, marble–sized ball. Separate and spread the fibers of the othersample. Wear your safety glasses! Light a burner and use crucible tongs to hold the small ball of steelwool directly over the flame, allowing the flame to touch the wool. Note the rate at which the oxygen in theair reacts with the iron. Next, do the same thing with the other sample of steel wool. Was the rate ofreaction the same for both samples?__________. If not, reveal which reacted faster and why: ________

______________________________________________________________________________

______________________________________________________________________________

Part B. Iron Filings or Powdered Iron

Next, obtain a bottle of iron filings from the materials shelf. Obtain a "pinch" of the iron filingsbetween your thumb and forefinger, and while holding your hand above the burner flame, sprinkle thefilings into the flame. Watch what happens. Iron filings are very small particles which provide a greatamount of surface area between the iron and the air. When the iron meets the flame it reacts very quicklywith the oxygen in the air. You see "sparkles" in the air!

Many people who own backyard barbecue grills use charcoal briquettes which are pieces ofcarbon, that yield heat as they react with oxygen in the air. Gunpowder also contains carbon that reactswith other chemicals to give off heat. Why must the carbon in the gunpowder be in the form of a powder?______________________________________________________________________________

Part C. A Dust Explosion Using Lycopodium Powder - (Teacher Demonstration)

<--- blow air in

candle

Lid

Paint Can

powder

Figure 18.6Dust Explosion Apparatus

As you witnesses in the demonstration withiron filings or powder, substances that normally burnonly very poorly or slowly burn much more quicklywhen in powdered form. Many dangerous explosionshave occurred at flour mills. We don't normally think ofwheat as being explosive! However, when in thepowdered form it can be very hazardous. Todemonstrate the power of a dust explosion, yourinstructor will use a fine powder which is actually apollen from the Lycopodium plant.

The powder will be placed in a small pile insidean empty paint can. A candle is lit and placed inside thecan. The lid is tightly sealed to the can and a puff of airis blown into the can to suspend the dust. If all goeswell, the lid will be launched toward the ceiling as theexothermic dust explosion causes the gas inside thecan to rapidly expand!

ACTIVITY 18.10 The Effect of a Catalyst on Reaction Rate

The final factor that we will study which affects the rates of chemical reactions is the use ofcatalysts. You experimented briefly with a catalyst called manganese dioxide, MnO2, in an earlier chapter.A catalyst is a substance that can alter the rate of a chemical reaction. Most catalysts are used to speed upthe rate of reactions. You will be performing a "catalyzed" reaction which you saw in a previous chapter.Procedure:

18-15 ©1997, A.J. Girondi

1. Place roughly 10 mL of hydrogen peroxide, H2O2, in each of four clean test tubes (they don't have tobe dry). Even though you cannot see it because it is so slow, there is a chemical reaction occurring inwhich H2O2 decomposes. The equation for this reaction is: 2 H2O2(l) ---> 2 H2O(l) + O2(g) Certain factorscan speed up the rate of this decomposition.

Propose a reason that might explain why H2O2 is normally stored in a dark glass or dark plastic bottle.

{26}_______________________________________ Why is it a good idea to store H2O2 in the

refrigerator?{27}___________________________________________________________________

2. Add a little of the catalyst, MnO2, to one of the tubes of H2O2. Rapid bubbling indicates that the H2O2 israpidly decomposing into water and oxygen gas, and that the catalyst is working. Do you observe rapidbubbling? ______________

3. Try adding a little Fe2O3 (iron (III) oxide) to the second tube and a little common dirt to the third. To the

fourth tube add a little bit of granular aluminum. Which of these substances appear to be good catalysts

for this reaction?_________________________________________________ Which (if any) do not

seem to catalyze this reaction?_______________________________________________________

A catalyst is a substance that can alter a reaction rate, but interestingly enough, it is not consumedin the reaction. The catalyst is not a reactant or a product. For that reason, it does not appear in theequation for the reaction (except that it may be written over or under the arrow (the yield sign) to indicatethat it is being used:

Uncatalyzed reaction: 2 H2O2 ----------------------------------> 2 H2O + O2 (slow)

Catalyzed reaction: 2 H2O2 ----------------------------------> 2 H2O + O2 (fast)catalyst

ACTIVITY 18.11 Catalysts in Matches and in Cigarette Tobacco

--> --> --> A Teacher Demonstration <-- <-- <--

Hopefully, you are getting a bit curious about how a catalyst actually works. It is neither a reactantnor a product, but it must take some part in the reaction, or else it could not affect the reaction rate.

O

O H

H

O O

H

H

Surface Catalyst

Bonds weaken whenH2O2 contacts the catalyst

Figure 18.7Action of Surface Catalysts

A catalyst enters into a reaction by alteringthe reaction mechanism. You will recall thatthe reaction mechanism is the series ofreactions that together comprise the stepsin the overall changes observed. A catalystcan alter the reaction mechanism in a varietyof ways. Some surface catalysts, such asnickel, provide a surface on which one ormore of the reactants can be adsorbed.Once adsorbed onto the surface of thenickel, it is believed that reactant bonds maybe stretched and weakened so much that aless energetic collision is sufficient to causethe reaction.

18-16 ©1997, A.J. Girondi

kJ

5

10

15

20

reaction progressFigure 18.8

Effect of Catalyst onActivation Energy

reactants

products

AB

A = activation energy without catalystB = activation energy with catalyst

Other catalysts - sometimes called cyclic catalysts -may actually combine chemically with a reactant, making itmore reactive. Later on in the reaction, the catalyst ischanged back to its original state, leaving the same mass ofcatalyst at the end of the experiment as there was at thebeginning of it.

In spite of the specific mechanism involved,catalysts perform their function by allowing reactions tooccur without adding as much energy to the reactants.Recall that the activation energy is defined as the minimumamount of energy required to rearrange the reactants intothe products. A catalyst lowers the activation energy (seeFigure 18.8).

Your teacher will follow this procedure:

1. Obtain a book of safety matches and a small piece of very fine sandpaper from the materials shelf. Tearone match from the book and attempt to light it using the sandpaper. What's the result?

The activation energy required to start the reaction is greater than the small amount of heat generated bythe friction between the sandpaper and the match head.

2. Next, try to light another match by rubbing it on the striking pad on the book. Result?

______________________________________________________________________________

The striking pad on a pack of "safety" matches contains a catalyst that mixes with the chemicals on thehead of the match when you strike it. The small amount of heat produced is sufficient to cause a reactionwhen the catalyst is present. This is because the activation energy is lower when the catalyst is present.

Exactly how catalysts do this is an active area of research in science. Why do you think they call these

"safety" matches? ________________________________________________________________

3. Next, obtain two sugar cubes and a container of cigarette ash from the materials shelf. Rub a few sidesof one of the cubes in the ash so that some of the ash adheres to the sides. Place the cubes on a Pyrexwatch glass supported on a ring stand and try to ignite them by flaming them directly with a burner flamewhile holding the burner in your hand.

Describe what happens: ___________________________________________________________

______________________________________________________________________________

Tobacco ash, itself, does not serve as a catalyst. However, the temperature at which cigarettes burn isoften not high enough to insure that cigarettes will continue to burn slowly. To aid the burning process, acatalyst is added to the tobacco when it is processed into cigarettes. The catalytic action prevents thecigarette from going out by lowering the activation energy needed to keep the cigarette burning. Thissame catalyst also aids the burning process with the sugar cube.

18-17 ©1997, A.J. Girondi

A catalyst lowers the activation energy for a reaction by changing the pathway by which the reactionoccurs. (Pathway refers to the intermediate series of reactions which cause reactants to becomeproducts.) Altering the pathway is kind of like finding another way to walk home from school - perhaps aroute that will not require you to walk up a hill! Remember that the rate of a reaction depends on the rate atwhich molecules collide and also the number of collisions which are effective. Raising the temperature will{28}_____________ the rate at which collisions occur, thus speeding up the reaction. Lowering theactivation energy required can increase percentage of collisions which are effective. This is the job of a{29}__________________.

There are also catalysts that can raise the activation energy of a reaction, which serves to decrease thereaction rate. Such substances are called inhibitors .

SECTION 18.12 Some Uses of Catalysts

A catalytic converter is a device that helps to reduce the pollutants emitted by an automobile'sengine. The converter contains a wire screen which is plated with a catalytic metal, probably platinum orpalladium. Engines are not 100% efficient and do not burn gasoline completely into CO2 and H2O.Instead, some unburned hydrocarbons escape through the exhaust system. These are compounds thatcontain both carbon and hydrogen, and they are considered to be pollutants. When the hot butunreacted hydrocarbons pass over the catalyst they react, reducing the hydrocarbon pollution producedby the vehicle. The catalyst in this case is a metal that does not take part in and is not consumed by thereaction. The reaction merely occurs on the surface of the catalyst. How do you think the catalyst affectsthe activation energy needed to allow the burning of the hydrocarbons? {30}_____________________

Only "unleaded fuels" are supposed to be used in cars that have catalytic converters. If fuelscontaining lead are used, the lead can form a coating on the surface of the metal in the converter which willdestroy its catalytic properties.

Catalysts are widely used by industry to increase the rates of many chemical reactions thatotherwise would take place too slowly to be practical. Catalysts are used to increase the yield of high-octane gasoline from petroleum. Catalysts are widely used to increase the rate of formation of ammonia, amajor constituent of fertilizer. Catalysts also occur naturally.

The metal in a catalytic converter functions as a surface catalyst. Nitric oxide (also known asnitrogen monoxide) is an example of a cyclic catalyst. A typical example of catalysis involving NO is itseffect on the decomposition of ozone, O3, in the upper atmosphere of the Earth. Without the presenceof NO, ozone decomposes slowly. In the presence of NO, which serves as a catalyst, the reaction occursrapidly:

2 O3 -----------------> 3 O2 <---- slow

2 O3 -----------------> 3 O2 <---- fast[NO]

The steps involved in the reaction mechanism or pathway for the catalytic decomposition of ozone are asfollows.

Step 1: NO + O3 -----> NO2 + O2 <---- (fast)Step 2: NO2 -----> NO + O <---- (fast)Step 3: O + O3 -----> 2 O2 <---- (fast)

Overall: 2 O3 ---------------> 3 O2 <---- (fast)

NO is consumed in this step

But, NO is reproduced in this step

Overall, the NO is neitherproduced nor consumed. Itis a cyclic catalyst.

[NO]

18-18 ©1997, A.J. Girondi

In the space below, list all of the reactants found in the three steps of this mechanism. (Do not considerthe overall reaction.)

Next, in the space below, list all of the products found in the three steps of this mechanism. (Do notconsider the overall reaction.)

Finally, draw a slash through any particles which appear in both lists, and write the formulas of theremaining particles in the space below.

Remaining reactants:{31}____________________; Remaining products:{32}_____________________

Can you now use the remaining reactants and products to write the overall reaction?_________. Thesum of the steps in any reaction mechanism should "add up to" the overall reaction.

One specific type of catalyst is called an enzyme . Enzymes are complex substances present inbiological systems which function as catalysts for biochemical processes. Pepsin in gastric juice andptyalin in saliva are two examples of enzymes. Ptyalin is the catalyst that accelerates the conversion ofstarch to sugar. Starch will react with water to produce sugar without the presence of the enzyme ptyalin,but it takes weeks for the conversion to occur. We could not survive if our digestive processes occurredthat slowly. We would literally starve to death even as we eat!

SECTION 18.13 Review Questions and Problems

Problem 7. The equation: 2 H2 + O2 ----> 2 H2O describes a reaction in which 0.042 moles of H2O formafter 1.3 hours. Calculate the rate of this reaction in moles of O2 used per minute.

Problem 8. The equation: 2 Al + 3 H2SO4 ----> Al2(SO4)2 + 3 H2 describes a reaction in which 30.0 gramsof Al react with sulfuric acid in 10.0 minutes. Calculate the rate in moles of Al used per hour.

Problem 9. In the reaction: Cu + 2 AgNO3 ----> Cu(NO3)2 + 2 Ag it is found that if 0.082 grams of Cu isconsumed per 1.00 minute, how many moles of Ag will be produced in 1.00 hour?

18-19 ©1997, A.J. Girondi

Problem 10. In the reaction 2 H2 + O2 ----> 2 H2O, it is found that hydrogen gas reacts at a rate of 10.0grams H2 per 1.00 minute. How many minutes will it take for 267 grams of oxygen gas, O2, to react?

Problem 11. On the axes below, sketch an "energy profile curve" (like those found in Figures 18.1 -18.4) for a reaction which has an activation energy of 42 kJ and in which the total energy of the reactants is103 kJ more than the total energy of the products which is 20 kJ. Put a kJ scale on the "y" axis.

kJ

Reaction Progress ---->

Problem 12. Assume that you have 1.25 L of a solution of NaCl which is 0.36 M. If you add enoughwater to this solution to increase the total volume to 2.10 L, what will the molarity (M) of the diluted solutionbe? (See Activity 18.7 if you need help.)

Problem 13. If you have 1.00 L of a 0.65 M solution of HCl which you want to dilute until its molarity isreduced to 0.45 M, how much water will you have to add?

18-20 ©1997, A.J. Girondi


Place a check mark to the left of each learning outcome on the next page after you believe thatyou have mastered it. Arrange to take any chapter exams or quizzes, and then move on to Chapter 19.

_____1. Explain the effects of the following on the overall rate of a chemical reaction: the nature of the reactants, the concentration of the reactants, the temperature of the reactants, and catalysts.

_____2. Explain how the rate-determining step effects the overall rate of a chemical reaction.

_____3. Calculate rates of chemical reactions given the needed information.

_____4. Given the rate of a chemical reaction, calculate the number of grams or moles of product produced after a specified amount of time.

_____5. Given the rate of a reaction and the number of moles produced, calculate the reaction time.

_____6. Define activation energy and explain how a catalyst and an inhibitor affect it.

_____7. Given several chemical equations, hypothesize about which reaction occurs fastest.

_____8. Explain the relationship between the collision theory and rates of chemical reactions.

18-21 ©1997, A.J. Girondi


Questions:

{1} A gas is evolved (given off); {2} A gas is evolved (given off); {3} Qualitative; {4} It is based onobservation rather than measurement; {5} H2; {6} 9.9 X 10-5 mole Mg/sec; {7} endothermic;{8} about 6 kJ needed; {9} about 10 kJ; {10} exothermic; {11} endothermic; {12} slow:{13} not many of the collisions would involved particles with enough energy to react; {14} small;{15} H + H ---> H2 would be faster since no bonds need to be broken; {16} Step 1; {17} slow;{18} If even only one step in a mechanism is slow, the overall reaction will be slow; {19} If substances aremore concentrated there will be more collisions between the reacting particles per unit of time, andtherefore a more rapid reaction; {20} reaction time; {21} More concentrated reactants will react faster thanless concentrated ones; {22} It would reduce the rate of the reaction; {23} At higher temperature there willbe more particle collisions per unit time and a faster rate of reaction; {24} reaction time; {25} Increasedsurface area between reactants will increase reaction rate; {26} Light may act as a catalyst; {27} At coldtemperatures the decomposition of H2O2 would be slower than at room temperature; {28} increase;{29} catalyst; {30} The catalyst lowers the activation energy; {31} 2 O3; {32} 3 O2

Problems:

1. 6.1 g Mg2. 4.2 X 10-3 g Zn/sec3. 6.4 X 10-5 mole Zn/sec4. 16.7 min5. 13.6 g Mg6. 5.6 X 10-6 mole CaCO3/sec7. 0.0027 mole O2/min8. 6.67 mole Al/hr9. 0.155 mole Ag/hr10. 3.37 min11.

kJ

Reaction Progress ---->

20

123

165

12. 0.21 M

13. Must add 0.40 L of water to obtain a final volume of 1.4 L of diluted solution

18-22 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 19

THE ENERGYOF

CHEMICALPROCESSES

19-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


19-2 ©1997, A.J. Girondi

SECTION 19.1 Introduction to Thermodynamics

One of the important principles you learned earlier in this course was the Law of Conservation ofMass. This principle is a result of the fact that atoms cannot be created or destroyed during a chemicalreaction. They can, however, be rearranged from one kind of molecule into another. Since each atom hasa particular mass, it follows that this mass is neither created nor destroyed. Mass is not the only quantitythat is conserved in a chemical reaction. Energy is also conserved. Experiments have shown that energyis neither created nor destroyed in any physical or chemical change. This principle is known as the law ofconservation of energy.

Energy is neither created nor destroyed in ordinary chemical or physicalchanges. This is known as the law of conservation of energy.

The law of conservation of energy is one of a set of laws which are known as the laws of thermodynamics.This set of laws is actually rather complicated, and we will not attempt to cover them in very much detail.Instead, we will merely introduce a few of these laws to you and indicate how they help to explain theenergy changes that accompany chemical and physical changes.

In science, we often use words and descriptions that sound very complicated when firstencountered. The word thermodynamics is a good example of this. However, the word can be translatedinto simpler terms that make it more meaningful. The first part of the word, thermo, means "energy." Thesecond part of the word, dynamics, means "motion." The word thermodynamics means "energy inmotion." Thermodynamics refers to how energy moves around or is converted from one form into anotherduring chemical and physical changes.

The word law sounds very legal, but must not be taken in that sense. When the word law is usedto describe natural principles, scientists are merely trying to describe the "rules" by which nature appearsto operate. It is through experimentation that we "observe" these laws of nature.

The laws of thermodynamics are the result of the work of scientists such as James Maxwell,Ludwig Boltzmann, R.J.E. Clausius, and many other European scientists during the nineteenth century.We will not attempt to explain all of these laws completely, but we will examine the first law as it relates tochemical reactions and other processes.

The reason there is a "0th law" (pronounced "zeroeth" law) is that it was recognized after the otherlaws were formulated, but was considered to be important in describing the other laws. Therefore, it wasnumbered in sequence before them. What are the laws of thermodynamics? When written in scientificterms, they can become quite complicated. The four statements on the next page represent simplifiedversions of these laws.

The Four Laws of Thermodynamics

0th Law: The temperature of a system is a fundamental property of that system which can be measured.(A system is that part of the universe that you are studying, such as the atoms and moleculesinvolved in a chemical reaction.)

1st Law: The total energy of the universe remains constant during any chemical or physical change.

2nd Law: The universe is continually changing from a state of higher organization (order) to a state oflower organization (more disorder).

3rd Law: The lowest possible temperature is zero degrees Kelvin. At this temperature, a system is perfectly ordered or organized – no entropy (disorder) exists.

19-3 ©1997, A.J. Girondi

Problem 1. Below, you will find four simple statements about the laws of thermodynamics. Referring tothe four laws, match the statements below with the correct law by writing the number of the law in theblank. a. _____ Everything in the universe is spreading out.

b. _____ At -273oC all molecular motion ceases.

c. _____ Energy is neither created nor destroyed during a change.

d. _____ It is possible to measure the temperature of any system.

SECTION 19.2 Heat and Temperature

The energy that a system has may be in the form of either potential energy or kinetic energy. You

will recall that potential energy is stored energy which is related to the position of an object. Kinetic energy

is the energy of motion. A boulder sitting on top of a hill possesses {1}_____________ energy. A

boulder that is rolling half-way down a hill has both {2}_________________ and _________________

energy. The motion of a boulder moving at the bottom of a hill represents {3}_______________ energy.

When heat energy is added to a group of molecules or atoms, the temperature will rise. Whenheat energy is removed, the temperature will drop. By measuring temperature changes during a chemicalreaction, we can estimate the change in the energy content of the system. The concepts of heat andtemperature are very important in our study of energy changes, so let's give them a closer look. It isimportant to realize that heat and temperature are not the same thing, but that they are related to eachother. Temperature is frequently defined as a measure of the average kinetic energy of the molecules of asystem. (Heat, on the other hand, is often defined as a measure of the total kinetic energy of a system.)For our purposes, it is sufficient for you to recognize that temperature is a property of a system that can bemeasured, and that this represents the 0th law of thermodynamics.

Problem 2. Look at the data below which describes three gases. Answer the four questions whichfollow as they pertain to the gases.

Compound Temperature Volume Mass Pressure

Gas A 0oC 11.2 L 16 g 1 atm Gas B 25oC 48.9 L 4 g 1 atm Gas C 12oC 11.2 L 22 g 2 atm

a. Which gas has the greatest average kinetic energy? Explain your choice.

b. Which gas has the lowest average kinetic energy? Explain your choice.

c. Notice above that 11.2 L of gas A has a mass of 16 grams. What would be the mass in grams of 1.0mole of gas A? (Remember: at STP (0oC and 1 atm) the volume of 1.0 mole of any gas is 22.4 L.)

16 g Gas A

11.2 L gas A

grams gas A

mole gas AX =

19-4 ©1997, A.J. Girondi

d. Is gas A hydrogen (H2), oxygen (O2), or carbon dioxide (CO2)? Explain your choice. ______________

______________________________________________________________________________

______________________________________________________________________________

The concept that ties the concepts of heat and temperature together is called heat capacity. Allsubstances have definite and measurable heat capacities. This refers to their capacities to gain and holdheat energy without changing their temperature very much. If a substance can absorb a large amount ofheat and have its temperature change by only a small amount, it is said to have a high heat capacity. Onthe other hand, if a small amount of heat added to a substance causes a large increase in temperature, thissubstance is said to have a low heat capacity.Let's look at an example of how heatcapacities differ for different substances.Imagine an automobile that has been sittingin the sun for several hours. You would nothesitate to place your hand on the glass ofthe windshield because, while it will be warm,it is not likely to burn your hand. However,you would hesitate to place your hand on themetal hood because you know it will probablybe too hot to touch without burning yourhand. The windshield and the hood haveboth been exposed to the same amount ofheat energy coming from the sun over thesame period of time, but the metal feelsmuch hotter than the windshield. Glass is asubstance that has a high heat capacity,while metals tend to have very low heatcapacities. As a result, the temperature ofthe glass did not rise as high as did thetemperature of the metal.

Glass Windshield(High Heat Capacity)

Metal Car Body(Low Heat Capacity)

Figure 19.1Heat Capacities of Car Parts

In addition to the definition offered above, temperature can also be considered to be a measure ofthe ability of a substance to transfer its heat to another substance. Heat always flows from substances athigher temperatures to substances at lower temperatures. When the temperatures of both substancesare equal, heat no longer flows. (Remember the meaning of thermodynamics.) With its smaller heatcapacity, the metal part of the car is at a higher temperature and is better able to transfer heat to your handwhen you touch it. This is the reason why the glass felt cooler than the metal parts of the car.

Write two definitions of temperature: {4} ________________________________________________

______________________________________________________________________________

______________________________________________________________________________

Now, let's test your skill at applying the concept of heat capacity to a real life situation. Imagineyour car sitting in the driveway on a very cold winter morning. Assume that all of the parts of the car are inthe same energy environment, absorbing the same amount of energy from the surroundings. Will thefrost melt off of the windshield first, or will it melt from the metal parts of the car first?

Explain your answer. {5}____________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

19-5 ©1997, A.J. Girondi

SECTION 19.3 Specific Heat and Heat Capacity

To make measurements involving any kind of energy (such as heat capacity), a unit is needed. Atraditional unit used to express heat energy is called the calorie. Scientists needed a way to define thisunit and did so by inventing the following definition:

1 calorie is the amount of heat energy needed to raise thetemperature of 1 gram of water by 1 degree Celsius (oC).

This should not be confused with the calorie counts found on food labels. The food calorie is actually1,000 times larger than the scientific calorie, and technically should be called a kilocalorie. When you areon a diet of 1200 calories per day, you are actually eating 1200 kilocalories or 1.2 X 106 calories of energy!

In the International System of Units (the metric system), a more acceptable unit of energymeasurement is the joule. A joule is smaller than a calorie. In fact, 1 calorie = 4.184 joules. Because of itswidespread acceptance, we will use the joule (J) in our energy calculations. Heat capacity depends in parton the mass of a substance.

There are normally two types of heat capacities that are identified for substances. One type ofheat capacity is called specific heat. Specific heat is the number of joules necessary to raise thetemperature of 1 gram of a substance by 1 degree Celsius.

specific heat = Joules

(grams) X (temp change) =

Jg. oC

The other type of heat capacity is called molar heat capacity. This refers to the number of joulesneeded to raise the temperature of one mole of a substance by 1 degree Celsius.

molar heat capacity =

Joules(moles) X (temp change)

= J

mole oC

GO TO APPENDIX D and read the "Swimming Pool Analogy" fora better understanding of specific heat and molar heat capacity.

ACTIVITY 19.4 Determination of the Specific Heats of Three Metals

Specific heat can be measured in the lab with relative ease. However, good lab technique is required. Tomeasure specific heat, a calorimeter must be used. A calorimeter is simply a container made with insulatedwalls. The insulation helps to reduce the heat loss or gain to or from the immediate surroundings. In thisactivity, we will use a foam cup as our calorimeter.

Heat energy always flows from an object at a higher temperature to an object at a lowertemperature. The heat gained by the cooler substance will be equal to the heat lost by the warmersubstance, assuming no heat is lost to the surroundings. In this activity, a measured mass of severalmetals will be heated to a known temperature. The metals will then be placed into a calorimeter containinga known mass of water, also at a premeasured temperature. You probably figured out that the heat will flowfrom the metal to the water until the water's temperature becomes equal to the metal's temperature. Willthe temperature of the water increase or decrease?{6}___________

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heat lost by metal = heat gained by water

Sample ofMetal

water

Figure 19.2Foam Cup Calorimeter

heat lost by hotmetal, is gained bythe water

cardboard lid

From the mass of the metals and that of the water in thecalorimeter, and from the temperature changes, you will be ableto calculate the specific heat of the various metals. Follow theprocedures outlined below to measure the specific heats of thethree metals. Record all of your data in Table 19.1.

Procedure for determining the specific heat of a metal:

1. Prepare a water bath using a 400 mL beaker filled 2 / 3 full ofwater. Bring the water to a boil. While waiting for the water toboil, go on to step 2.

2. You will be using samples of iron, copper, and aluminum. Themetal samples will be in the form of cylinders. Obtain 2 cylindersof each metal and examine them. Be sure that you can identifywhich is which.

3. Measure the total mass of the two cylinders of iron. Repeat forthe copper and aluminum cylinders. Record the masses in Table19.1.

4. Place all of the cylinders in the boiling water bath for 10minutes. Continue on with step 5 while you are waiting for themetal samples to heat.

(The time can be shortened to 5 minutes if necessary.)

5. Obtain 3 foam cups and label the cups 1, 2, and 3. Add exactly 100 mL of water to each cup. Recordthe mass of the water in each cup. Remember, 1 mL water = 1 g.

6. After the metal samples have heated for at least 10 minutes, while keeping the samples in the water,measure the initial temperatures of each of the metals by holding the bulb of a thermometer in the centerthe hot water bath. Record this as the initial temperature of the metals in Table 19.1.

7. Measure and record the initial temperature of the water in calorimeter number 1. Use a secondthermometer to do this.

8. Using your crucible tongs, remove the iron cylinders from the water bath and very quickly place theminto the water in calorimeter 1. With constant stirring (very important), record the highest temperaturereached by the water. (Watch carefully!) The temperature will not rise very much and the change occursvery quickly! Repeat this procedure using the other two metals and calorimeters 2 and 3.

Note: in Table 19.1, for each metal used, the final temperature of the metal will equal the final temperatureof the water.

9. You now have all of the data that is needed to calculate the specific heat of each metal. Before you dothese calculations, clean up your lab area. Dry the metal cylinders and return them to the materials shelf.

Calculations:

Calculating the specific heats of the metals involves the use of two mathematical equations whichare listed below. Keep in mind that the heat lost by the metal is equal to the heat gained by the water.

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Let's call this quantity "q". Equation 1 below is used to calculate "q." The symbol ∆T refers to the changein temperature, and the specific heat of water is 4.184 J / g.oC.

Equation 1: q = (mass in grams of water) (∆T of water in oC) (specific heat of water)

Now substitute your value for "q" from equation 1 (above) into the numerator of equation 2 (below). Enterthe mass of metal used and the temperature change of the metal into the denominator, and solveequation 2 to obtain the specific heat of the metal.

Equation 2: specific heat of metal =

q

(mass of metal in grams) (∆T of metal in oC)

Once again, to calculate the specific heat of a metal first use equation 1 to calculate the heat lost by themetal which is designated as "q" which is the substituted into equation 2. Solving equation 2 yields thespecific heat of the metal:

Use equation 1 to find the value of "q"which is the heat lost by the metal.

Substitute the value of "q" into equation 2.

Solve equation 2 to for the specific heat of the metal.

Calculate the specific heats of the three metals used above. You must use both equations 1 and 2. Showall of your work in the space provided below. Use your data recorded in Table 19.1.

1. Calculation of specific heat of iron:

S.H. iron = __________ J / g.oC2. Calculation of specific heat of aluminum:

S.H. aluminum = __________ J / g.oC

3. Calculation of specific heat of copper:

S.H. copper = __________ J/g.oC

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Data Table 19.1Calculation of the Specific Heat of Three Metals

Metal 1 Metal 2 Metal 3 (Fe) (Al) (Cu)

Mass of metal cylinders (g) ____________ ____________ ____________

Initial temp. of hot metal ____________ ____________ ____________

Final temp. of metal ____________ ____________ ____________

Temp. change (∆T) of metal in oC ____________ ____________ ____________

Mass of H2O used in calorimeter (g) ____________ ____________ ____________

Initial temp. of H2O in calorimeter ____________ ____________ ____________

Final temp. of H2O in calorimeter ____________ ____________ ____________

Temp. change (∆T) of H2O in oC ____________ ____________ ____________

Your results will probably reveal a fair amount of error. Nevertheless, which metal did you find to have the

smallest specific heat?______________ Which metal did you find to have the greatest specific

heat?______________ The accepted values for the specific heats of these three metals, taken from the

Handbook of Chemistry and Physics, are:

copper: 0.389 J

g.o Ciron: 0.444

J

g.o Caluminum: 0.908

J

g.o C

Did your results produce the correct order (smallest, middle, and greatest) of specific heats for the three

metals? _______________________________________________________________________

If samples of these three metals were placed in the sun on a hot summer day, which one would feelhottest to the touch?{7}__________________________ Which of the metals would feel coolest to thetouch? {8}_______________________ The unit for specific heat measurements is J / g.oC. This meansthat if a metal has a high specific heat, it requires {9}___________ energy to cause the temperature of onegram of the metal to rise by 1oC. Therefore, if equal masses of two metals absorb equal amounts ofenergy, the temperature of the one with the higher specific heat will go up {10}_________ than the onewith a lower specific heat value, and the one with the higher specific heat value will feel{11}___________________.

SECTION 19.5 Molar Heat Capacity

When you compare the specific heats of two substances, you are comparing the amount of heatneeded to raise the temperature of equal masses of the two substances by 1oC. Since substances havedifferent densities, equal masses do not contain the same number of molecules or atoms. If you areinterested in comparing the amounts of heat needed to raise the temperature of equal numbers ofmolecules or atoms of two substances by 1oC, you will have to compare their molar heat capacities instead

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of their specific heats. Calculate the molar heat capacities of the three metals in the spaces provided inTable 19.2 below. As an example, the calculations for the molar heat capacity of zinc are shown for you.

Table 19.2Calculating Molar Heat Capacities

Substance Specific Heat Atomic Mass Molar Heat Capacity (J / g.oC) (g / mole) (J / mole oC)

zinc 0.382 65.37 (0.382 J / g.oC)(65.37 g / mole) = 25

iron 0.444 _________

aluminum 0.908 _________

copper 0.389 _________

Note that the specific heats which are based on equal masses (1 gram) of the three metals are different,while the molar heat capacities which are based on equal numbers of atoms (1 mole) are virtually identical.Remember that 1 mole equals 6.02 X 1023 atoms. This indicates that the amount of heat needed to raisethe temperature of an element by 1oC has been found to depend on the number of atoms of the elementpresent, not upon the identity or mass of atoms. This was first discovered by two Frenchmen in 1817,Pierre Dulong and Alexis Petit. Since one mole of every element contains the same number of atoms, afixed amount of heat (about 25 J / mole oC) should then be required to raise the temperature of one moleof a solid element by 1oC. This has come to be known as The Law of Dulong and Petit.

(The molar heat capacity of water is different from that of the metals, but it is a liquid compound, not a solidelement.)

The Law of Dulong and Petit

The amount of heat needed to raise the temperature of an elementby 1oC has been found to depend on the number of atoms of theelement present, not upon the identity or mass of atoms.

This law permitted scientists of the 19th century to begin to approximate the values of atomicmasses of many solid elements. They did this simply by using experimentally determined specific heatvalues of the elements and the molar heat capacity value of 25 J / mole oC.

As an example, let's see if you can determine the atomic mass of element "X" which is found tohave a specific heat of 0.481 J / g.oC. Use the value of 25 J / mole oC for the molar heat capacity andcalculate the atomic mass of this element. Use the calculations in Table 19.2 as a guide. Show your workbelow:

Atomic mass of element "X" = {12}_______ g / mole. What is the identity of element "X"?{13}___________

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SECTION 19.6 Solving Problems Involving Specific Heat and Heat Capacity

Knowing the heat capacity (specific heat) of a substance allows you to calculate how much heatenergy is required to raise the temperature of a known amount of a substance by a specified number ofdegrees. As usual, unit analysis will be useful in solving such problems. Study the sample problembelow.

Sample Problem: How much heat energy is required to raise the temperature of 10.0 grams of waterby 50.0oC? (You must know that the specific heat of water is 4.184 J / g.oC)

Solution: You are asked to find an amount of heat energy, so you want find an answer in joules. Youare given two pieces of information including 10.0 g of water and 50.0oC. You could start the problem witheither of these quantities. We will use dimensional analysis and start with 10.0 g water.

10.0 g H2O X

????

???? X

????

???? = ???? joules

To change information about water into information about energy, you need to include some conversionfactors in the set-up that can relate these two units. That's the job of specific heat. Consider the unit ratiofor specific heat. It relates information about energy (J) to information about water (g) and (oC). So, let'splug it into the fencepost:

10.0 g H 2O X

4.184 J

(1g) (1o C) X

????

???? = ???? joules

Notice that grams of H2O will now cancel, leaving J / oC. We want to end with joules only , so we want oC tocancel. For this purpose, we can use the 50.0oC given in the problem:

10.0 g H 2O X

4.184 J

(1g) (1o C) X

50.0o C

1 = 2090 joules

After including the 50.0oC, all units cancel properly, so we just use a 1 as a denominator under the 50.0oC.The answer turns out to be 2092 joules, which is rounded to three significant figures: 2090 J. That's allthere is to it. Just make sure that units cancel properly, leaving you with the units you want. If we haddecided to start with the 50.0oC instead of the 10.0 g H2O, the fencepost would have ended up lookinglike this:

50.0o C X

4.184 J

(1g) (1o C) X

10.0 g H 2O

1 = 2090 joules

Your knowledge of basic algebra and of dimensional (unit) analysis will allow you to solve a varietyof problems involving specific heat and molar heat capacity. You should not have to memorize anyequations to solve thermodynamics problems, since dimensional analysis allows you to create your ownequations. Here is an example of a variation:

Sample Problem: When 3600. joules of heat energy are added to 140. g of water, how much doesthe temperature go up?

Solution: We are asked for a temperature change, so we want to end with units of oC. Let's start thefencepost with 3600. J:

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Note that the specific heat ratio had to beinverted this time so that joules would cancel.

3600. J X

(1g) (1o C)

4.184 J X

= {14} _ _ _ _ _ _ _ _ _ _ _ oC

Finish the problem above by filling in the blanks in the "fencepost" and calculate an answer. Next, trysome problems on your own. Set up each problem below so that units cancel to give the appropriateanswer. Show the complete set-up for each problem and include all units! Round answers to the correctnumber of significant figures.

Problem 3. How many joules are required to heat 150 mL of water from 20.oC to 84oC? (Note from theexamples that the temperature change is what is important here. Also, recall that 1 g H2O = 1 mL H2O.)

__________ J

Problem 4. How many joules are required to heat 10. g of silicon from 22oC to 44oC? (The specific heatof silicon is 0.76 J / g.oC) Hint: first calculate, ∆T, the change in the temperature.

__________ J

Problem 5. How much will the temperature change when 418.4 J of heat are added to 50. g of water?

__________ oC

Problem 6. A piece of substance "Z" is heated directly and it is found that 5805 Joules of energy areneeded to raise the temperature of the substance from 25.0oC to 35.0oC. The specific heat of substanceZ is known to be 1.28 J/goC. What was the mass in grams of this piece of substance Z? (Note: water is notinvolved in this problem.)

___________ g

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Problem 7. When 2092 J of heat are added to a certain mass of water, the temperature rises by 125oC.What is the mass of the water?

__________ g H2O

Problem 8. A certain substance "Q" has a molar heat capacity of 18.6 J / mole oC. How many joules areneeded to raise the temperature of 2.0 moles of this substance by 200.oC?

__________ J

SECTION 19.7 Entropy

Now let's return to the laws of thermodynamics. The 0th law concerns the concept oftemperature. However, there is more to this law than meets the eye, particularly in relation to the conceptof entropy.

Entropy is a measure of the amount of disorder in a system.

Gases have high entropy (they are highly disordered) because the particles of a gas move around in

random motion. Thus, gases have no definite shape or volume. Solids have low entropy because the

particles of the solid are arranged in a definite order (such as a crystal lattice) and the motion of these

particles is very limited. They have a definite shape and volume.

SOLID LIQUID GAS

lots of order a little order no order

Figure 19.3Degrees of Entropy in Three Phases of Matter

(low entropy) (moderate entropy) (high entropy)

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How do you think the entropy of liquids compares to that of gases and solids?

{15}____________________________________________________________________________

Explain: {16}_____________________________________________________________________

______________________________________________________________________________

As the temperature (kinetic energy) of a system is increased, the particles move more rapidly. This

increased activity can be viewed as an increase in the "disorder" of the system, and, therefore, the

{17}_______________ of the system has increased. On the other hand, when heat energy is removed

from a system, its temperature decreases and the disorder (entropy) tends to decrease. However, phase

changes must also be considered. If a solid turns into a liquid, does entropy increase or

decrease?{18}_____________________. If a gas condenses into a liquid does entropy increase or

decrease ?{19}_____________________.

Problem 9. Identify each situation below as being an example of either increasing or decreasingentropy (circle one):

a. water becoming ice increase decrease

b. a bowl of chili cooling increase decrease

c. paper being shredded increase decrease

d. warming a cup of tea increase decrease

e. heating liquid water to a vapor increase decrease

f. hardening of liquid wax increase decrease

SECTION 19.8 Enthalpy

The 1st law of thermodynamics concerns the conservation of energy in chemical and physicalchanges. By conservation of energy, we mean that when changes occur, energy is neither created nordestroyed. This means that energy does not mysteriously "appear," nor does it mysteriously "disappear"or cease to exist when a change occurs.

You know that a chemical reactionbegins with reactants and ends withproducts. This description is not quitecomplete, because there is also anenergy change associated with areaction. The energy change can beeither endothermic or exothermic.

reactants products

reactants products + heat energy

reactants + heat energy products

Exothermic

Endothermic

In the first case, heat energy is produced in addition to the products. This is normally called an exothermicreaction. An easy way to remember this is to think of the word exit, which means "to leave." In anexothermic reaction, energy leaves the system.

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Problem 10. You must be able to distinguish equations for endothermic reactions from equations forexothermic ones. Test your skills on the equations below. Label each equation as being eitherendothermic or exothermic, based on whether the energy (kilojoules) is written on the reactant or theproduct side.

a. CaCO3 + 177.8 kJ ----> CaO + CO2 __________________________________

b. 2 Ag + Br2 ----> 2 AgBr + 199.2 kJ _______________________

c. 2 SO2 + O2 ----> 186.6 kJ + 2 SO3 _______________________

d. 94.1 kJ + PCl5 ----> PCl3 + Cl2 _______________________

Generally speaking, more chemical reactions are exothermic than endothermic. In both types of reactions,there is an energy difference between the reactants and the products. For an exothermic reaction, thisenergy difference is a result of the excess energy leaving the system and going into the surroundings.

Figure 19.4Movement of Heat During a Reaction

CHEMICALSYSTEM

Surroundings

heat in heat out

During an endothermic reaction, heat energy must enterthe system from the outside. There is never a netincrease or decrease in total energy in the system andthe surroundings. The energy that leaves the systemends up in the surroundings. Thus, energy is neithercreated nor destroyed, but is merely exchangedbetween the system and surroundings. This is the 1stlaw of thermodynamics.

What causes this energy difference between the reactants and the products? It involves morethan just a temperature change. It involves both kinetic and potential energy changes. As you know, heatis a measure of the total kinetic energy of the molecules in a system, while potential energy is related tothe location or the position of molecules. In a larger sense, potential energy has something to do with thearrangement and position of objects.

In chemical compounds, the chemical bonds are the result of attractions between charged parts ofatoms, which is a form of potential energy. Chemical bonds represent an arrangement with a specificamount of potential energy. Potential energy is stored in chemical bonds. During a reaction, chemicalbonds are broken in the reactants and new ones are formed in the products. If the products have lesspotential energy than the reactants, potential energy has been converted into kinetic energy. This makesthe molecules of the products move more rapidly. They ultimately transfer this energy to theirsurroundings, causing the surroundings to become hotter. This situation represents an exothermicprocess.

On the other hand, if the potential energy of the products is greater than that of the reactants (thatis, the chemical bonds in the products have greater potential energy than the chemical bonds of thereactants), heat energy must enter the reactants from the surroundings in order for the reaction to occur.In this case, kinetic energy (or heat energy) is converted into potential energy (in chemical bonds)representing an endothermic process. The temperature in the surroundings decreases in this situation.

There is now a need to quantify all of this. This can be done by referring to the heat content of asystem as its enthalpy, which is a Greek word meaning "warm." Energy changes which involve heat areknown as enthalpy changes. The symbol used to denote the enthalpy of a system is the capital letter H.

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An enthalpy change is denoted by the symbol ∆H. You may pronounce ∆H as either "delta H" or"change in H." For any chemical reaction the enthalpy change can be written:

enthalpy change = ∆H = (Hp – Hr)

In this case, Hp represents the heat content of the products, and Hr represents the heat content of thereactants. Recall that in an exothermic reaction, heat is released as one of the products. This means that

in such a case, Hp must be less than Hr. Steam condensing into water provides an example of such achange:

steam(g) ----> water(l) + heat energyIn exothermic changes, the heat content of the products is less than the heat content of the reactants.

Heat energy must be {20}________________________ the steam to cause it to condense. In this case,the energy contained in the steam is greater than the energy contained in the liquid water. The energychange of an exothermic reaction can be indicated in two ways. First, the energy can be written into theequation as a product as you have already seen:

2 H2 + O2 -----> 2 H2O + 572 kJ

Since the energy is located on the right side of the equation above, this tells us that the energy is givenoff and the reaction is exothermic.

The energy change can be shown separately by writing the ∆H value after the equation:

2 H2 + O2 -----> 2 H2O; ∆H = –572 kJ

When you write it this way, you need some method of indicating whether the energy is given off orabsorbed. To indicate that the energy is given off and that the reaction is exothermic, we make the valuefor ∆H negative.

Now let's look at an example of an endothermic reaction such as the melting of ice.

ice(s) + heat energy -----> water(l)

In endothermic changes, the heatcontent of the products is greaterthan the heat content of thereactants.

Heat energy must be {21}____________________ the ice before it can melt. In this case, is the energy

contained of the ice greater than or less than the energy contained in the liquid water? {22}____________

Because ∆H = Hp – Hr, would ∆H for endothermic reactions always be a positive or a negative number?

{23}____________________

Now look at this chemical reaction: N2 + 2 O2 + 67 kJ -----> 2 NO2

In this chemical reaction, Hp is {24}____________________ than Hr, resulting in an endothermic change.

This equation can also be written: N2 + 2 O2 -----> 2 NO2; ∆H = +67 kJ

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Enthalpy changes are associated with chemical and physical changes. These changes obey the1st law of thermodynamics. You will notice that when a process is exothermic, the heat energy appears onthe right–hand side of the equation with the products.

When the process is endothermic, the heat energy appears on the left–hand side of the equationwith the reactants. Instead of including the energy in the equation, ∆H can be written after the equation.In that case, the sign of ∆H is given a negative sign if the reaction is exothermic and it is given a positivesign if the reaction is endothermic.

Problem 11. Use both methods of indicating the energy change for the chemical reactions defined bythe equations below. Include the energy quantity within the equation

Equation Energy Change

a. 2 H2 + O2 ----> 2 H2O 572 kJ (exo)b. H2 + Cl2 ----> 2 HCl 185 kJ (exo)c. H2 + I2 ----> 2 HI 51.1 kJ (endo)d. 8 H2S ----> 8 H2 + S8 20.1 kJ (endo)

In the left column include the energy quantity within the equation. In the right column use "∆H."

a. ______________________________ ______________________________

b. ______________________________ ______________________________

c. ______________________________ ______________________________

d. ______________________________ ______________________________

ACTIVITY 19.9 How Much Heat Is Required To Melt Ice?

By using a calorimeter, you can measure the amount of heat which is absorbed when ice melts.

Procedure:

1. Obtain a thermometer, a foam cup, and a piece of cardboard large enough to serve as a lid over thecup. Using a graduated cylinder, measure exactly 100. mL of water, and pour the water into the cup.Record the temperature of the water to the nearest 0.5oC.

2. Obtain 2 or 3 ice cubes that have partially melted, so that the temperature of the cubes is 0oC. Removeany excess liquid from the cubes with a paper towel, and add them to the cup. Place the cardboard lidover the cup. The lid should have a small hole through which you should insert the thermometer. Carefully stir the ice–water mixture. While you are providing continuous stirring (don't stop), observe thetemperature every 15 seconds as it goes down. When the temperature becomes constant for twoconsecutive readings, record the final temperature to the nearest 0.5oC.

3. Remove the remaining ice and shake as much water as possible back into the cup from the ice cubes.Do not leave the ice in the water after the temperature has leveled off – remove it immediately. Measureand record the volume of water remaining in the cup. From this data you can determine the volume ofwater that melted from the ice. Record all data in Table 19.3 below.

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Table 19.3Heat Required to Melt Ice

Original Volume of H2O _______mL

Original Temp. of H2O _______oC

Time Elapsed (sec) Temp (oC)

15 _______

30 _______

45 _______

60 _______

75 _______

90 _______

105 _______

120 _______

135 _______

150 _______

165 _______

180 _______

195 _______

210 _______

225 _______

240 _______

Lowest Temp. Reached _______oC

Final Volume of H2O _______mL

Volume H2O from Ice _______mL

Mass of Ice that Melted _______g

(1 mL H2O = 1 g H2O)

water

Figure 19.5Foam Cup Calorimeter

cardboard lid

ice cubes

The 100 mL of water cooled from its original temperature (to) to its final temperature (tf). The

absolute value of this temperature change is represented as |tf - to|. Calculate the change in the heatenergy of this water by using the formula below.

heat energy change = (mass of 100 mL H2O) (∆T) (S.H. of water)

∆T = |tf - to|

S.H. = 4.184 J / g.oC

joules

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Calculation:

Energy change = __________ J

Now calculate the heat needed to melt one gram of ice by dividing the change in energy above (J), by themass of ice that melted (g). Show set-up.

__________ J / g

Now use the heat of fusion per one gram of ice from the calculation above to find the heat required to meltone mole of ice. This is the change in heat energy when 1 mole (18.02 g) of water - in the form of ice -melts. Set up the calculation below to change J / g to J / mole of water.

__________ J / mole

The accepted value from the Handbook of Chemistry and Physics is 6025 J/mole. Calculate thepercentage error involved in your experiment. The formula for %E is in the reference section of yourALICE materials.

___________ %Error

What do you think was the greatest source of error in your experiment?__________________________

______________________________________________________________________________

The ice was absorbing energy as it melted. This energy was not gained as kinetic energy becausethe temperature of the ice remained at 0oC for the entire experiment. Thus, the rate of motion of themolecules was not increasing. If the rate of motion of the molecules was not affected, the energy musthave been gained as potential energy (energy of position). What change in the H2O molecules did thisadditional potential energy cause? {25}__________________________________________________

The amount of heat required to melt a substance is sometimes called its heat of fusion. Thisdoes not mean the same thing as nuclear fusion!

ACTIVITY 19.10 Measuring Enthalpy Changes

The next several experiments are designed to illustrate enthalpy changes. Perform each of thefollowing experiments and answer the questions given in each section.

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Experiment 1: The formation of an aqueous solution of NaOH

1. Place 5 mL of water into a standard 150 mm test tube. Measure the temperature of the water.

T = _________oC

2. Using forceps, put 5 pellets of solid NaOH into the test tube. DO NOT handle the pellets with yourfingers. Wash your hands with plain water if contact is made. Wear safety glasses and an apron. Be sureto quickly replace the lid tightly on the bottle of NaOH. (It absorbs water from the air.) Stir the solution inthe test tube for a minute or two and measure its temperature again. T = __________oC Does the testtube feel warmer or cooler?________________________

In an exothermic process, heat is released by thechemical system and is absorbed by the water whichserves as the surrounding "environment" of thesystem. Thus, the water gets warmer. During anendothermic process heat is absorbed from theenvironment by the chemical system. Since the wateris the environment of the chemical system in this case,an endothermic process would absorb heat from thewater making it cooler. Is the dissolving process forNaOH exothermic or endothermic?______________

chemicalsystem

water getscooler

heat

water environment

Endothermic Reaction In Water

chemicalsystem

water getswarmer

heat

water environment

Exothermic Reaction In Water

Figure 19.6Heat Exchange in Water

Rewrite the equation NaOH(s) -----> Na1+(aq) + OH1-(aq) below and include the term "heat" on the

proper side as a reactant or product:{26}_________________________________________________

Water is not included in the equation above, because dissolving is a physical change, not a chemical one.Therefore, water is not a reactant. In the change above, do the reactants or the products have greaterenthalpy? {27}________________________

Experiment 2: The formation of an aqueous solution of NH4NO3.

1. Place 2 mL of water in a standard 150 mm test tube. Measure the temperature of the water.

T = _________oC

2. Add a small amount (about enough to cover a penny) of solid NH4NO3 to the water. Stir the solution for

a minute or two, and measure the temperature of the solution. T = __________oC

Does the test tube feel warmer or cooler?__________________ How did the temperature

change?________________________________ Was heat released to the water environment or

absorbed from the water environment by the chemical system? ______________________________

Is this process exothermic or endothermic?________________________________

Rewrite the equation NH4NO3(s) -----> NH41+(aq) + NO31-(aq) by adding the term "heat" as a reactant or

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product to the proper side: {28}_______________________________________________________

In the change above, do the reactants or the products have greater enthalpy? {29}__________________

Experiment 3: The reaction of NaOH solution with HCl solution

If you have not already done so, put your safety glasses on. Put 3 mL of 1 M HCl solution into astandard 150 mm test tube. Put 3 mL of 1 M NaOH solution (handle with care!) into a second similar testtube. Wash your hands with plain water if contact is made. Measure the temperature of each solution,rinsing the thermometer between uses.

THCl = __________oC; TNaOH = __________oC

Now mix the two solutions and stir well. Measure the temperature of the reaction mixture.

T = __________oC.

Did the water temperature increase or decrease after mixing?__________________ Is this chemical

reaction exothermic or endothermic?______________________

Rewrite the equation HCl(aq) + NaOH(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq) below by adding the term "heat"

to the proper side as a reactant or product: {30}____________________________________________ Assume that 55 joules of energy were exchanged in experiment 3. Then, write the equation above using∆H to indicate the heat change instead of writing "heat" in the equation. Use the correct sign (+ or –) for∆H.

{31}___________________________________; ∆H = __________J In this reaction, do the reactants or the products have more potential energy? {32}__________________.

ACTIVITY 19.11 (Optional) Heat of Condensation of Water

Earlier in this chapter you determined the heat required to melt a given amount of ice. Whenenergy is added to water at its boiling point, 100oC, the temperature of the water does not increase.Instead, the added energy allows the water molecules to move away from each other and enter the vaporstate. The energy required to vaporize one gram of a substance at its boiling point is known as the heat ofvaporization or the enthalpy of vaporization of the substance. When the vapor cools, the heat released isknown as the heat of condensation (or the enthalpy of condensation). In this activity, we are going to try todetermine the amount of heat released when one gram of water vapor condenses to liquid water.

Materials: foam cup with cardboard lid, 125–mL Erlenmeyer flask, one-holed rubber stopper, glass bend,lab burner, 50-mL graduated cylinder, thermometer, distilled water.

Procedure:

1. Obtain a foam cup and a cardboard lid to cover the top of the cup. The lid should have a small hole in itthrough which a thermometer will fit.

19-21 ©1997, A.J. Girondi

2. Measure the mass of the cup and its lid (together). Enter this as mass "A" in Table 19.4.

3. Place 40.0 mL of distilled water into the foam cup. Measure and record the total mass of the calorimeter,lid, and water as mass "B" in Table 19.4.

cup

H2O

Lid

Figure 19.7Vaporization of Water Apparatus

4. Put some water into the 125-mL Erlenmeyerflask until it is about one-fourth full.

5. Fit the flask with a glass bend and rubberstopper as shown in Figure 19.7.

6. Heat the flask of water on a ring stand with alaboratory burner as shown until it is gentlyboiling. Lower the heat such that the watercontinues to slowly boil.

7. As the water in the flask is getting hot, measureand record the temperature of the water in thefoam cup. (An electronic thermometer canprovide more accuracy - to 0.1oC), if available.)

8. When fog is seen coming out of the glassbend, insert the tip of the glass bend through thehole in the cardboard lid and down into the waterin the calorimeter. Make sure that the glasstubing is actually submerged into the water in thecup. Hold it in this position for 1.5 to 2 minutes.CAUTION: the glass tube and water vapor will behot! Hold the cup into position with beaker tongsto keep your hands away from the hot fog.

9. Remove the glass bend from the cup. Measure the temperature of the water in the cup and record thisas final temperature "E" in Table 19.4.

10. Measure the mass of the cup with its lid, water, and condensed water vapor and record this as mass G.

Analysis and Conclusions:

1. Complete the Table 19.4 by finding the values of C, F, and H. Show work below.

19-22 ©1997, A.J. Girondi

2. The heat released when the water vapor condensed was added to the water in the calorimeter, causingits temperature to increase. This quantity of heat "q" can be found using an equation which you usedearlier in this chapter. You will need to use the initial mass of water (data C), the change in temperature ofthis water (data F), and the specific heat of water. Calculate this amount of heat,"q", in joules for yourexperiment. Show work below.

3. The heat of condensation is the heat released, in joules, when 1.00 gram of water vapor condenses toliquid water. Divide the results of analysis step 2 (data I in Table 19.4) by the mass of steam condensed(data H). Show work below, and record the results.

Table 19.4Heat of Condensation of Water

A. mass of empty calorimeter + lid __________ g

B. mass of calorimeter, lid, and water __________ g

C. mass of water (B – A) __________ g

D. initial temperature of water in calorimeter __________ oC

E. final temperature of water in calorimeter __________ oC

F. change in water temperature (E – D) __________ oC

G. mass of calorimeter, lid, total water __________ g

H. mass of condensed water vapor (G – B) __________ g

I. heat released in condensation process __________ J

J. heat of condensation of water(I ÷ H) __________ J/g

4. The accepted value for the enthalpy of condensation of water is 2260 J/g. How does your experimentalenthalpy of vaporization of water compare with the actual value? _______________________________

Calculate your percentage error:

% Error = __________

19-23 ©1997, A.J. Girondi

5. The amount of heat given off when one gram of water vapor condenses to liquid water is the same asthe amount of heat needed to change one gram of liquid water to a vapor.

water vapor liquid water 2260 J/g

2260 J/g

In Activity 19.9 you experimentally determined the amount of heat required to melt ice. The acceptedvalue for that is 334.7 J/g. The heat required to vaporize liquid water is 2260 J/g. Why do you think thereis a significant difference between the amount of heat needed to melt ice and the amount of heat neededto vaporize liquid water?

{33}____________________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

6. The amount of heat given off when a vapor of ethyl alcohol condenses to a liquid is 855 J/g. For

benzene vapor the value is 395 J/g. How do these values compare with the enthalpy of condensation of

water in J/g? {34}_________________________________________________________________

______________________________________________________________________________

C

C

C

C

C

C

H

H

H

H

H

H

7. The structure of benzene is shown at right. How might you use the

concept of polarity to explain why the amount of heat given off when benzene

vapor condenses to a liquid is less than that given off when an equal mass of

water vapor condenses to a liquid? {35}_____________________________

_________________________________________________________

8. How is the amount of heat given off when a vapor of substance X condenses to its liquid related to the

boiling point of substance X and to the strength of intermolecular forces in the liquid? {36}____________

______________________________________________________________________________

______________________________________________________________________________

9. A burn caused by exposure to hot steam or hot fog at 100oC can be more harmful than a burn caused

by exposure to boiling water at 100oC. Why? {37}__________________________________________

______________________________________________________________________________

19-24 ©1997, A.J. Girondi

ACTIVITY 19.12 Superheated Steam


Once water has been converted into the gaseous state, additional heating cannot change itsphase again. Added heat simply causes the hot gas to increase in temperature. Superheated steam is isgaseous water which is above 100oC. It can cause very severe burns. Your instructor will set upequipment to produce superheated steam. You will see how it can be used to scorch paper!Unbelievable!

Water is first heated to a vapor at 100oC.

Vapor is then heated to temperatures above 100oC.

Superheated Steam

Figure 29.8Making Superheated Steam


12. How many joules of heat are needed to raise the temperature of 56.0 grams water from 45.0oC to52.0oC? (S.H. H2O = 4.184 J/g.oC)

13. If 4.85 KJ of heat are added to 83.0 g of water initially at 67.0oC, what will its final temperature be?

14. The water in a calorimeter absorbs 234 J of heat and its temperature moves from 25.0oC to 38.0oC.What is the mass in grams of water in the calorimeter?

15. Aluminum has a specific heat of 0.908 J/goC. How many kilojoules of heat are released when a 1.5kilogram block of aluminum cools from 135.0oC to 25.0oC?

16. A 85.96 gram block of metal is found to increase in temperature by 12.0oC when it absorbs 458 joulesof heat. Using information given in Section 19.5, identify this metal.

17. It is found that when 46.0 grams of a substance absorbs 672 joules of energy, its temperatureincreases by 45.0oC. What is the value of the specific heat of this substance?

19-25 ©1997, A.J. Girondi


The principles of thermodynamics are important to your understanding of chemistry. Check offthe learning outcomes below after you feel you have mastered them. Arrange to take any tests or quizzeson Chapter 19, and then move on to Chapter 20.

_____1. State, in general terms, the laws of thermodynamics.

_____2. Explain the concepts of specific heat and heat capacity.

_____3. Do calculations involving specific heat and heat capacity.

_____4. Explain transfers of heat involved in calorimetric measurements and be able to do calculations involving these transfers.

_____5. Given a situation, determine whether it is an example of increasing or decreasing entropy.

_____6. Distinguish between exothermic and endothermic reactions.

_____7. Identify an equation as representing an endothermic or exothermic reaction if the equation shows heat as a reactant or product.

_____8. Identify an equation as representing an endothermic or exothermic reaction if the value of ∆H isgiven.

_____9. Explain and distinguish between enthalpy and entropy.

_____10. Identify and use the standard unit of energy in the International System of Units, the joule (J).

19-26 ©1997, A.J. Girondi


Questions:

{1} potential; {2} potential and kinetic; {3} kinetic; {4} measure of average kinetic energy of particles, or ameasure of the ability of a substance to transfer heat; {5} It will melt off of the metal car body first becausemetal has a lower heat capacity and, therefore, its temperature will rise more; {6} increase; {7} The onewith the smallest specific heat value; {8} The one with the largest specific heat value; {9} more; {10} less;{11} cooler; {12} 52 g/mole; {13} Cr; {14} 6.15oC; {15} Liquids have greater entropy than solids but lessthan that of gases; {16} The molecules of a liquid are free to move so that they are more disordered thanthose of a solid - however, they are not able to move farther apart like those of a gas; {17} entropy;{18} increase; {19} decrease; {20} removed from; {21} added to; {22} less than; {23} positive;{24} greater; {25} The molecules moved farther apart during the phase change;{26} NaOH(s) -----> Na1+(aq) + OH1-(aq) + heat; {27} reactants;{28} NH4NO3(s) + heat -----> NH41+(aq) + NO31-(aq); {29} products;{30} HCl(aq) + NaOH(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq) + heat{31} HCl(aq) + NaOH(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq) ; ∆H = – 55 J{32} reactants; {33} Melting requires that bonds between particles be weakened, whereas vaporizationrequires that bonds be broken; {34} The values for ethyl alcohol and benzene are lower than that forwater; {35} The value for benzene is lower because it is nonpolar, while water is polar and polar bonds arestronger; {36} A higher heat of vaporization indicates a higher boiling point and stronger forces ofattraction between the particles; {37} At 100oC hot steam or fog has more potential energy than boilingwater at 100oC.

Problems:

1. a. 2nd; b. 3rd; c. 1st; d. 0th2. a. B - highest temperature; b. A - lowest temperature; c. 32 g;

d. oxygen - it has a molecular mass of 32 g/mole;3. 4.0 X 104 J (rounded to two significant figures)4. 167 J (rounds to 170 J)5. 2.0oC6. 454 g7. 4.00 g H2O8. 7.4 X 103 J (rounded to two significant figures)9. a. decrease; b. decrease; c. increase; d. increase; e. increase; f. decrease10. a. endothermic; b. exothermic; c. exothermic; d. endothermic11. a. 2 H2 + O2 ----> 2 H2O + 572 kJ; 2 H2 + O2 ----> 2 H2O; ∆H = – 572 kJ

b. H2 + Cl2 ----> 2 HCl + 185 kJ; H2 + Cl2 ----> 2 HCl; ∆H = –185 kJc. H2 + I2 + 51.1 kJ ----> 2 HI; H2 + I2 ----> 2 HI; ∆H = 51.1 kJd. 8 H2S + 20.1 kJ ----> 8 H2 + S8; 8 H2S ----> 8 H2 + S8; ∆H = 20.1 kJ

12. 1640 J13. Final Temp = 81.0oC14. 4.30 g15. 150. KJ of heat (rounded to three significant figures)16. iron (S.H. = 0.444 J/g.oC)17. 0.325 J/g.oC

19-27 ©1997, A.J. Girondi

NAME____________________________________ PER____________ DATE DUE____________


"ALICE"

CHAPTER 20

CHEMICAL

EQUILIBRIUM

Keq

20-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


20-2 ©1997, A.J. Girondi

SECTION 20.1 Introduction to Equilibrium Systems

In most of the chemical reactions we have studied so far, it appears as though all of the reactantsare converted to products before a reaction stops. In truth, however, experiments show that theconversion of reactants into products is often incomplete in chemical reactions. This is the case no matterhow long the reaction is allowed to continue.

As a reaction progresses, the concentrations of the reactants decrease, while the concentrationsof the products increase. Eventually, a state is established in which the concentrations of products andreactants no longer change. This is known as the state of equilibrium. You have already studied manyexamples of equilibrium. The most obvious example was probably encountered during your study ofsolutions. At that time, you performed some activities that involved the use of saturated solutions.

Equilibrium is a balance between two equal and opposing processes or forces. When a systemreaches equilibrium, it does not undergo additional change unless the equilibrium is somehow disturbed.A saturated solution is in equilibrium because there is a balance between the two opposing forces ofdissolving and crystallizing. (See Figure 20.1 below.) This is an example of a physical equilibrium sincedissolving and crystallizing are physical changes. An equation can be written to represent what isoccurring in this saturated solution after equilibrium has been reached:

REACTANTS PRODUCTSCaCl2(s) Ca2+(aq) + 2 Cl1-(aq)

The (aq) subscript refers to "aqueous" which means dissolved in water.

Ca2+

Cl1-Cl1-

Saturated CaCl2 Solution

Figure 20.1Solution Equilibrium

The double arrow indicates that the forward reaction is occurringat the SAME RATE as the reverse reaction. Therefore, nooverall change occurs in the system. To an observer, it appearsas though nothing whatsoever is happening in the saturatedCaCl2 solution. Actually, the change shown at right is occurringconstantly in both directions. Note that the equation aboveshows chlorine as 2 Cl1-, not as Cl2. Chloride ions (Cl1-) do notexist in diatomic form. This is because these ions already have astable octet of electrons. Only chlorine atoms form diatomicmolecules: (Cl2). This is also true for the other diatomic elements.

CaCl2(s) <===> Ca2+(aq) + Cl2(aq)WRONG! ---> <--- WRONG!

CaCl2(s) <===> Ca2+(aq) + 2 Cl1-(aq)

The type of change shown in Figures 20.1 and 20.2 is referred to as dynamic equilibrium because,although no changes can be observed in the system, the two opposing changes are happeningconstantly.

Figure 20.2 illustrates two more examples of physical equilibria. Examine what is in each beaker,and note how a double arrow is used in the equilibrium equation for each situation. Note, too, how thestate (solid, liquid, gas, or aqueous) of each substance is represented using subsscripts. Finally, noticethat coefficients are used to balance the equations properly.

20-3 ©1997, A.J. Girondi

Figure 20.2 Equilibrium Systems in Saturated Solutions

Saturated AlCl3 Solution

Al3+ Cl1-

AlCl3(s) <===> Al3+(aq) + 3 Cl1-(aq)

Cl1-

Cl1-

AlCl3

Pb2+

I1-

Saturated PbI2 Solution

I1-

PbI2 (s) <===> Pb2+(aq) + 2 I1-(aq)

PbI2

Problem 1. The three equations below represent equilibrium systems created by dissolving solids inwater. Complete the equilibrium equations below. Include any charges on ions and the state of the ionswhich in this case is "aqueous" or (aq):

a. NaCl(s) <====> _____________________________

b. Fe2(SO4)3(s) <====> _____________________________

c. BaBr2(s) <====> ______________________________

The formula for water is not included in the examples above because dissolving is not a chemical changeand, therefore, water is not consider a reactant. When substances dissolve in water they do notchemically react with it. They merely come part in the water. Later in this chapter you will study reactionsthat involve chemical equilibrium, since the changes are chemical, rather than physical.

There are several criteria that must be satisfied before an equilibrium can be achieved. Let's take a

closer look at some of these. Compare the two systems shown below (beakers A and B). Which beaker

contains a system at equilibrium?{1}_________ What do you think prevents the system in the other

beaker from forming an equilibrium system? {2}____________________________________________

Figure 20.3Closed and Open Systems

H2O(g)

H2O(l)

BEAKER "A"

H2O(g)

H2O(l)

BEAKER "B"

H2O(l) H2O(g)

20-4 ©1997, A.J. Girondi

Could equilibrium ever be achieved in an open system such as that shown in beaker B? {3}________ Why

or why not? {4}___________________________________________________________________

Hopefully, you have arrived at the conclusion that an equilibrium can be established in a systemsuch as this only if the system is closed. In addition, this system must be held under conditions ofconstant temperature, pressure, and volume. If one or more of these factors is altered, the equilibriumchanges as well. (Not all equilibrium systems are affected by pressure or volume, but all equilibriumsystems are affected by temperature. We will discuss this later in the chapter.)

Equilibrium equations give an overall picture of the types of chemical changes that are occurring.The equations do not, however, provide any information about the actual amounts of reactants andproducts present. The equilibrium equation for a saturated sodium chloride solution:

NaCl(s) <====> Na1+(aq) + Cl1-(aq)

merely indicates the identities of the ions and molecules involved in the equilibrium, and the doublearrows show that equilibrium prevails. There is absolutely no way that you can tell what the concentrationsof any of the products or reactants are just by looking at this equation. The reaction container may have afew crystals of solid NaCl present or it may have a large pile of solid salt.

SECTION 20.2 Solving Problems Involving Equilibrium Systems

There is a simple relationship between the concentrations of the products and the reactants.Through much experimental study, scientists were able to come up with a law of chemical equilibrium.Consider the hypothetical equilibrium system: A + B <===> C + D

Brackets such as, [ ], are used to denote molar (M) concentration. So when you see something like[Ag1+], it means "molar concentration of silver ions." This law of chemical equilibrium states that if you findthe product of the concentrations of the products, [C] [D], and divide that result by the product of theconcentrations of the reactants, [A] [B], the result will be a constant value:

constant value =

[C] [D]

[A] [B]

Now, let's look at a real example:

4 NO(g) + 6 H2O(g) <====> 4 NH3(g) + 5 O2(g)

Since this system is at equilibrium, this means that the RATE of the forward reaction (--->) is equal to theRATE of the reverse (<---) reaction. From the equation for the equilibrium system, we can write what iscalled the equilibrium expression for this reaction:

equilibrium expression =

[NH3 ]4 [O2 ]5

[NO]4 [H2O]6

You should notice two things about the expression above. First, note that the numerator contains theproduct of the concentrations of the products, and the denominator is the product of the concentrationsof the reactants. Second, you should see that the coefficients in the equation become exponents in theequilibrium expression. The reasons why we turn coefficients into exponents can be explained usingkinetic–molecular theory and collision theory. However, we will simply accept this, and save theexplanation for a more advanced course. If you know the concentrations of all the reactants and productsin this equilibrium system, you can plug those values into this expression and solve it. The value you will

20-5 ©1997, A.J. Girondi

get is a constant. Why? Well, if you somehow manage to change the concentration of NH3, for example,the concentrations of the other substances in the system will change, too. The net effect will be that theexpression will maintain the same value. Since it is a constant value, it is given the symbol K. Since itinvolves equilibrium, it is often called Keq. (Other authors may designate it as K, Kc, etc.) So,

Keq =

[NH3 ]4 [O 2 ]5

[NO]4 [H2O]6

Problem 2. Write the equilibrium expressions for each of the chemical situations given below.

a. 2 NO2(g) <===> N2O4(g) Keq =

b. N2(g) + 3 H2(g) <===> 2 NH3(g) Keq =

c. Ag1+(aq) + 2 NH3(aq) <===> Ag(NH3)21+(aq) Keq =

d. 2 NO(g) + 2 H2(g) <===> N2(g) + 2 H2O(g) Keq =

Setting up an equilibrium expression is an extremely useful skill after you become aware of whatinformation it can provide. In other words, what good is Keq?

One way to determine the value of Keq for a particular reaction is to allow the reaction to proceed ata given temperature in such a way as to allow the products to accumulate in the reaction container. After aperiod of time, the reaction will reach equilibrium. At this point, it may be possible to experimentallydetermine the concentrations of the reactants and products in the container. The concentration valuesare then substituted into the equilibrium expression which is then solved for Keq.

Let's consider an equilibrium system involving only gases and assume that some method isavailable to determine the concentrations of reactants and products at equilibrium. Equilibrium equationsare frequently accompanied by the temperature at which the reaction was allowed to achieve equilibrium.Since temperature always affects equilibrium, it also affects the value of Keq. For example:

H2(g) + I2(g) <===> 2 HI(g) (at 250oC)

Let's look at how the value of Keq for this system can be determined (at this temperature).

1. Start by writing the correct formula for the Keq of this reaction: Keq =

[HI]2

[H2 ] [I2 ]

20-6 ©1997, A.J. Girondi

2. Below are the experimentally determined equilibrium concentrations of each reactant and product forthe system above. The brackets mean molarity (moles / liter of sol'n).

Temp. [H2] [I2] [HI]

Experiment 1: 250oC 0.00560 0.000590 0.01270

Experiment 2: 250oC 0.00460 0.000970 0.01476

Notice that two experiments were probably performed under slightly different conditions, producingdifferent equilibrium concentrations. Temperature was constant.

3. Substituting this data into the equilibrium expression:

Experiment 1: Keq =

(0.0127)2

(0.0056) (0.00059) = 48.8

Experiment 2: Keq =

(0.01476)2

(0.0046) (0.00097) = 48.8

Note that while the concentrations of reactants and products changed from experiment 1 to experiment 2,the value of Keq did not change.

Complete the following problems. First, write the expression for Keq for each reaction, and thenuse the given data to calculate the value of Keq. Show all work.

Problem 3. Reaction: N2O4(g) <===> 2 NO2(g) (at 520oC)Equilibrium Concentrations: [N2O4] = 0.014; [NO2] = 0.072

Keq = ____________

Problem 4. Reaction: N2(g) + 3 H2(g) <===> 2 NH3(g) (at 583oC)Equilibrium Concentrations: [N2] = 0.20; [H2] = 0.20; [NH3] = 0.016

Keq = ____________

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Problem 5. Reaction: 2 NO(g) + O2(g) <===> 2 NO2(g) (at 500oC)Equilibrium Concentrations: [NO] = 3.49 X 10-4; [O2] = 0.80; [NO2] = 0.25

Keq = ____________

Problem 6. Reaction: NCl3(g) + Cl2(g) <===> NCl5(g) (at 520oC) Equilibrium is established whenthere is 0.0350 mole of NCl3, 0.0200 mole Cl2, and 0.0110 mole of NCl5 in a volume of 2.50 liters. (Hint:

you must first calculate the concentrations in moles per one liter of sol'n.)

Keq = _____________

SECTION 20.3 LeChatelier's Principle

Earlier in this chapter it was mentioned that certain factors can disrupt an equilibrium. If one ormore of these factors is altered, the equilibrium is momentarily upset. These factors are changes in:

1. temperature2. pressure3. volume of the reaction container4. concentration of a reactant and or a product

The exact effect of any change in reaction conditions on equilibrium was studied extensively by HenriLeChatelier, a French chemist. LeChatelier was mainly interested in what occurred when changes in thefour factors listed above were made in a system at equilibrium.

LeChatelier found that changes in these factors could put a "stress" on the system at equilibrium,causing it to move away from the state of equilibrium. This stress causes a change in the rate of either theforward or reverse reaction. After the system moves away from equilibrium, he observed that the amountsof reactants and products adjusted in order to restore the system to equilibrium. The results of hisobservations allowed him to formulate the following law, or principle, known as

LeChatelier's Principle: When a stress is placed on a system at equilibrium, thesystem will adjust to relieve the stress and to restore equilibrium in the system.

The stress changes the RATE of either the forward or the reverse reaction. LeChatelier's principle statesthat when a stress is applied, the RATES of the reactions will adjust so that the RATES once againbecome equal.

20-8 ©1997, A.J. Girondi

As an example, let's look at the equilibrium system: 2 NO(g) + O2(g) <===> 2 NO2(g)

Assume we mix some O2 molecules with an excess (more than needed) of NO molecules:

original equilibrium concentrations

NO O2 NO2

+

(Excess NO present)

length of arrows indicates relativerates of forward and reverse reactions

LeChatelier's principle enables us to predict the direction in which the equilibrium would shift when theconcentration of one or more of the products or reactants is changed. Suppose more oxygen wereadded to the reaction container. We could ask, "What effect would this change have on the rates of theforward and reverse reactions, and would this concentration increase affect the concentrations of theother components in the system?"

You are putting a "stress" on a system in equilibrium whenever you change the concentration ofany of the products or reactants. This means that, as a result of the change, the system is no longer inequilibrium. LeChatelier's principle can be explained using the collision theory. Part of this theory statesthat a collision must occur before a chemical reaction can take place between reactants. Furthermore, thecollisions must produce enough energy and the particles must often collide in just the right way. Not allcollisions result in a reaction. When more oxygen is added, there will be more collisions between the NOand O2 molecules. This will increase the rate of the forward (---->) reaction:

NO O2 NO2

+

(Excess NO present)

some oxygen is added rate of forward reaction is now greater

The system will try to remove the stress by getting rid of the excess oxygen. In this way, an equilibriumcondition can be restored. The only way for the system to get rid of the extra oxygen is to have it react withNO to produce more product, NO2. So, momentarily the system "shifts to the right" and produces moreproduct:

NO O2

+

NO2

(Excess NO present)

more product is now produced

Looking at it another way, the added oxygen molecules temporarily increase the rate of the forward (--->)reaction. However, since there are now more product molecules (NO2), they collide more often whichincreases the rate of the reverse reaction (<-----), and eventually a new equilibrium is established.However, this equilibrium situation is different from the original equilibrium situation because theconcentrations of the reactants and the products are not the same as they were originally. The rates of the

20-9 ©1997, A.J. Girondi

reactions are again equal, but they are not the same as the original rates. Because of the "shift to theright" the system now has more product than it did before more oxygen was added. Notice, too, that theshift caused the concentration of a reactant (NO) to decrease. The stress was relieved in the sense that atleast some of the added oxygen has been removed.

NO O2

+

NO2

(Excess NO present)

new equilibrium rates

new equilibrium concentrations

Since oxygen was added, how do you think the rates of the forward and reverse reactions in the newequilibrium compare to those in the old equilibrium? {5}______________________________________

Using these principles, predict the direction of the shift (forward ---->) or (reverse <----) inequilibrium when the concentration of Cl2 is increased in the system below, and how this will affect theconcentrations of each of the other products and reactants.

4 HCl(g) + O2(g) <===> 2 H2O(g) + 2 Cl2(g)

When the concentration of Cl2 is increased, the equilibrium will shift to the {6}_______________. The

concentration of HCl will {7}____________________ . As the system begins to shift after the addition of

Cl2, the concentration of Cl2 will {8}_________________. The concentration of O2 will

{9}_________________. The concentration of H2O will {10}_________________.

Problem 7. Complete Table 20.1 below. The first blank has been completed as an example.

Table 20.1Direction of Equilibrium Shift

Equation Added Equilibrium Shift Substance to Right or Left

a. CO(g) + 2 H2(g) <===> CH3OH(g) CO ------>

b. PCl5(g) <===> PCl3(g) + Cl2(g) PCl3 ________________

c. N2(g) + 3 H2(g) <===> 2 NH3(g) H2 ________________

d. CO(g) + H2O(g) <===> CO2(g) + H2(g) H2 ________________

e. 4 NO(g) + 6 H2O(g) <===> 4 NH3(g) + 5 O2(g) H2O ________________

f. 2 SO2(g) + O2(g) <===> 2 SO3(g) SO3 ________________

g. 2 NCl3(g) <===> N2(g) + 3 Cl2(g) Cl2 ________________

h. CH3COOH(aq) <===> H1+(aq) + CH3COO1-(aq) CH3COOH ________________

i. C2H6(g) <===> H2(g) + C2H4(g) H2 ________________

20-10 ©1997, A.J. Girondi

ACTIVITY 20.4 Lechatelier's Principle and Changes in Concentration

Now that you are able to predict the direction of an equilibrium shift, you will perform anexperiment that will allow you to apply this skill. The shifting of an equilibrium will be examinedexperimentally by using the reaction for the formation of a "complex ion" with the formula Fe(SCN)2+ bythe combination of Fe3+ ions with an SCN1- ion:

Fe3+(aq) + SCN1-(aq) <===> Fe(SCN)2+(aq)

(pale yellow) (colorless) (reddish)

This is a good example to use because a solution of Fe3+ ions has a faint yellow color, while a solution ofSCN1- ions has no color. When these two solutions are mixed, a deep-red color results due to theformation of the Fe(SCN)2+ ion. In this activity you will examine what happens when certain stresses areplaced on the equilibrium that exists between these three ions. Wear safety glasses.

Before you begin note that: There are two concentrations of Fe(NO3)3 solutions of the materials shelf forthis activity. Be sure to use 0.1M Fe(NO3)3 in step 1 and 0.2M Fe(NO3)3 in step 5, part b.

Procedure:

1. Look at the bottle of 0.1M Fe(NO3)3 solution. This solution contains Fe3+ ions and NO31- ions. Thecolor of this solution is due to the Fe3+ ions. What's the color of the aqueous Fe3+ ions? {11}__________(The NO31- ions are only spectator ions here, so we will ignore them.)

2. Look at the bottle of 0.1M KSCN solution. The KSCN solution contains K1+ and SCN1- ions. Based onthe appearance of this solution, what can you conclude about the color of the K1+ and SCN1- ions?___________________________ (The K1+ ions are only spectator ions here, so we will ignore them.)

3. In a test tube, mix 1 or 2 mL of the Fe(NO3)3 solution with 1 or 2 mL of the KSCN solution. The resultingcolor is due to the product, FeSCN2+. What's the color of the FeSCN2+ ion?{12}________________Look again at the equilibrium equation for this reaction. Note that FeSCN2+ is the product of the reactionbetween Fe3+ and SCN1-.

4. To a 100 or 150 mL beaker, add 1 mL of 0.1M Fe(NO3)3 solution. (Be sure it is the 0.1 M solution.) Next,add 2 mL of 0.1M KSCN solution. Finally, add 75 mL of distilled water. Stir well. This solution containssome reactants (Fe3+ and SCN1-) and some product (FeSCN2+). The reddish color of the product(FeSCN2+) is not very evident at this point, since it is diluted.



5. Now place four 4-mL portions of the solution formed in step 4 into separate clean standard-sized (150mm) test tubes. (Make sure the test tubes all have the same internal diameter.) Label the tubes A throughD and perform the following tests:

a.) Do nothing to tube A. Use it as a control in order to compare its color to the other tubes.

b.) To tube B, add 15 drops of 0.2M Fe(NO3)3 solution and compare the color of test tube B to test tube A.

Has the color gotten lighter or darker?{13}____________________ Does this color change indicate an

increase or a decrease in the concentration of Fe(SCN)2+.{14}_________________________

20-11 ©1997, A.J. Girondi

When you added Fe(NO3)3, you were adding Fe3+ ions to the equilibrium system. This resulted in an

increase in the concentration of Fe3+ ions, [Fe3+], in the system. Explain the effect (on the equilibrium) of

increasing the Fe3+ concentration in the system. {15}_______________________________________

______________________________________________________________________________

c.) To tube C, add several drops of 6M NaOH (dangerous). Handle this solution with great care. If you get

any on you, wash with lots of water. Be sure to wear safety glasses! Mix well. Does the solution's color get

lighter or darker?{16}______________ Does this change indicate an increase or a decrease in the

concentration of Fe(SCN)2+?{17}___________________________

Thus, adding NaOH has the overall effect of DECREASING the concentration of Fe3+ ions in theequilibrium system:



Explain the effect on the equilibrium system which resulted from the addition of NaOH: {18}___________

______________________________________________________________________________

d.) To tube D, add about 1 mL (20 drops) of 0.1M AgNO3 solution. (Avoid getting this solution on yourhands. After several hours, a dark stain can result.) Is the color of this mixture lighter or darker than that oftube A? {19}________________ Does this change indicate an increase or a decrease in theconcentration of Fe(SCN)2+? {20}___________________ When you add AgNO3 to the equilibriumsystem, some of the SCN1- ions react with it. The SCN1- that reacted this way is removed from theequilibrium system. So, adding AgNO3 has the effect of DECREASING the concentration of SCN1- ionsin the equilibrium system. What is the effect of this change on the equilibrium system?{21}________________________________

Problem 8. Using your observations from the activities above, complete Table 20.2 for the Fe(SCN)2+

equilibrium system.

Table 20.2Shifts in the Fe(SCN)2+ Equilibrium System

Test Substance Added Effect Shift Forward (--->) Toward Products, or Shift Reverse (<---) Toward Reactants

a. nothing none none

b. Fe(NO3)3 increases Fe3+ ___________________

c. NaOH decreases Fe3+ ___________________

d. AgNO3 decreases SCN1- ___________________

SECTION 20.5 LeChatelier's Principle and Changes in Pressure

Now let's "shift" our discussion to the effects of pressure changes on equilibrium systems. Forreactions which occur in liquid solutions, a change in pressure will not affect the equilibrium to any great

20-12 ©1997, A.J. Girondi

extent because the volume of the liquid solution will not change very much even if extreme pressure isplaced on the system. However, when a gas is involved, changing pressure will have an effect on theequilibrium. Consider the equilibrium system below:

N2O4(g) <===> 2 NO2(g)

If a container full of these two gases is at equilibrium and the pressure is changed, a stress is placed on thesystem. If we increase the pressure on the system, it will adjust itself to reduce the pressure. This is inaccord with LeChatelier's principle. Since the pressure of the system is directly proportional to thenumber of gas molecules present, the only way to reduce the pressure (at constant temperature) is toreduce the total number of molecules in the system. This can occur if two NO2 molecules combine to formone N2O4 molecule. Therefore, to relieve the strain caused by increasing the pressure, the equilibrium willshift to the left toward the reactant which is N2O4.

Increasing the pressure on a gaseous system atequilibrium causes the equilibrium to shift to the side withthe fewest number of molecules.

On the other hand, if the pressure is decreased, more gas molecules must be formed to bring thesystem's pressure back to equilibrium. In this case, some N2O4 molecules will decompose into two NO2

molecules. This will increase the number of gas molecules present, thereby increasing the pressure inthe system.

Decreasing the pressure on a gaseous system atequilibrium causes the equilibrium to shift to the side withthe larger number of molecules.

Consider a container in which the reactionshown at right has come to equilibrium: N2(g) + 3 H2(g) <===> 2 NH3(g)

In which direction will the equilibrium shift if the pressure on the system above is

decreased? {22}_____________ Explain: {23}____________________________________________

______________________________________________________________________________

How should the pressure be changed on the system above in order to produce a larger amount of

ammonia, NH3? {24}____________________ Explain: {25}_________________________________

______________________________________________________________________________

Shown at right is another reaction which you studiedpreviously involving the formation of HI gas by the reaction:

H2(g) + I2(g) <===> 2 HI(g)

How will the equilibrium in the reaction above be affected by an increase in pressure? {26}_____________

Explain: {27}_____________________________________________________________________

In part a of Problem 9 below, the equation shows 1 molecule of N2O4 on the left side and 2molecules of NO2 on the right side. Therefore, a decrease in pressure would cause a shift of theequilibrium to the right (toward more molecules). In part b of Problem 9, the equation reveals 2 moleculeson the left side of the arrow and only 1 molecule on the right side. Indicate the direction of shift when thepressure decreases in the space provided. Then, complete Problem 9.

20-13 ©1997, A.J. Girondi

Problem 9. Complete Table 20.3 below. Indicate the direction in which the equilibrium will shift if thepressure is changed in the manner indicated.

Table 20.3Pressure Changes and Equilibrium Shifts

Equilibrium Equations Pressure Shifts

a. N2O4(g) <===> 2 NO2(g) decreased ---->b. PCl3(g) + Cl2(g) <===> PCl5(g) decreased _____c. 2 SO3(g) <===> 2 SO2(g) + O2(g) decreased _____d. 2 CO(g) + O2(g) <===> 2 CO2(g) decreased _____e. N2(g) + O2(g) <===> 2 NO(g) increased _____f. 2 H2(g) + O2(g) <===> 2 H2O(g) decreased _____g. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g) decreased _____h. 2 N2O(g) <===> 2 N2(g) + O2(g) increased _____i. 2 HBr(g) <===> H2(g) + Br2(g) increased _____j. CH4(g) + 2 O2(g) <===> CO2(g) + 2 H2O(g) increased _____

SECTION 20.6 LeChatelier's Principle and Changes in Volume

LeChatelier's principle can also be used to predict the effect of changes in the volume of thereaction container on equilibrium. Equilibrium shifts caused by volume changes are similar to thosecaused by pressure changes. When the volume of a particular reaction container is reduced, themolecules get crowded together and collide more frequently. This stress can be relieved by decreasingthe number of molecules present. Look again at the nitrogen dioxide equilibrium:

2 NO(g) + O2(g) <===> 2 NO2(g)

A decrease in the volume of the reaction container can be compensated for by forming fewer molecules.The result is that the equilibrium above shifts to the right to form more molecules of NO2. In the process,three molecules of reactants will become two molecules of product. In general terms, this relationship canbe stated as follows:

For reactions in which there is a change in the number ofgas molecules, a decrease in the volume favors the reactionthat produces fewer molecules. An increase in volumefavors the reaction that produces the larger number ofmolecules.

If the forward and reverse reactions of an equilibrium systemproduce the same number of molecules, then changes in volumehave no effect on the system. Consider the system shown at right:

H2(g) + Cl2(g) <===> 2 HCl(g)

Note that the forward reaction (--->) produces two molecules of HCl, while the reverse reaction (<---) alsoproduces two molecules – one H2 and one Cl2. (This is also true for pressure changes.)

20-14 ©1997, A.J. Girondi

Problem 10. Based on this generalization, predict whether equilibrium shifts toward the products (--->)or the reactants (<---) in each example below when the volume of the reaction container is decreased:

a. PCl5(g) <===> PCl3(g) + Cl2(g) _____________________________

b. N2(g) + 3 H2(g) <===> 2 NH3(g) _____________________________

c. 2 CO(g) + O2(g) <===> 2 CO2(g) _____________________________

Problem 11. In which direction will the systems below shift if the volume of the reaction container isincreased:

a. H2(g) + I2(g) <===> 2 HI(g) _____________________________

b. CO(g) + 2 H2(g) <===> CH3OH(g) _____________________________

c. C3H8(g) + 5 O2(g) <===> 3 CO2(g) + 4 H2O(g) _____________________________

Problem 12. Study the equations in Table 20.4. Determine the direction of equilibrium shift. Answerby writing either "forward" or "reverse," or by using arrows: ---> or <---. These equations are a little morecomplicated, because they involve liquids and solids in addition to gases. Pressure changes do not havemuch, if any, effect on liquids and solids, so you should only consider molecules of gases in an equilibriumsystem when effects of pressure changes are being evaluated. Therefore, substances which are liquidsor solids with subscripts (l) or (s) in the equations should be ignored in Table 20.4.

Remember: As you complete Table 20.4, ignore any substances below which are solids or pure liquids.

Table 20.4Equilibrium Shifts and Pressure Changes

Pressure ShiftEquilibrium Equation Change Direction

a. CO(g) + H2O(l) <===> CO2(g) + H2(g) increase ___________

b. 4 FeS(s) + 7 O2(g) <===> 2 Fe2O3(s) + 4 SO2(g) increase ___________

c. 4 NH3(g) + 5 O2(g) <===> 4 NO(g) + 6 H2O(l) decrease ___________

d. C(s) + O2(g) <===> CO2(g) decrease ___________

When an equilibrium system involving gases is present in a closed container, changes in volumecause changes in pressure. If the size of the container is decreased (volume is decreased) the pressuregoes up. For example, imagine that you are squeezing a balloon which contains gases into a smallervolume. To relieve the increased pressure, the system will shift in the direction of fewer gas atoms ormolecules. If the size of the container is increased, the system will shift in the direction which will providemore gas atoms or molecules which can occupy the added volume of space. Changes in volume – likechanges in pressure – do not affect solids or pure liquids.

20-15 ©1997, A.J. Girondi

Problem 13. Complete Table 20.5 below. Reminder: Ignore any substances which are solids (s) orpure liquids (l).

Table 20.5Equilibrium Shifts and Volume Changes

Volume ShiftEquilibrium Equation Change Direction

a. CO(g) + H2O(l) <===> CO2(g) + H2(g) increase ___________

b. 4 FeS(s) + 7 O2(g) <===> 2 Fe2O3(s) + 4 SO2(g) increase ___________

c. 4 NH3(g) + 5 O2(g) <===> 4 NO(g) + 6 H2O(l) decrease ___________

d. C(s) + O2(g) <===> CO2(g) decrease ___________

SECTION 20.7 Equilibrium Systems Involving Solids and/or Liquids

Many of the equilibrium equations that you have seen so far in this chapter have reactants andproducts which are either gases (g) or water (aqueous) solutions (aq). Changes in volume or pressurehave an effect on gases. Changes in concentration have an effect on both gases or aqueous solutions.However, changes in volume, pressure, or concentration in equilibrium systems do not affect solids orpure liquids. They are not variables that play a role in determining the value of Keq. We will save an in-depth discussion of the reasons for this for a future chemistry course! Solids and pure liquids are neverincluded in Keq expressions. Examine the following examples.

CaO(s) + CO 2(g) <===> CaCO 3(s) Keq =

1

[CO2 ]

H2O(l) + HF(g) <===> H 3O 1+

(aq) + F 1-(aq) Keq =

[H3O1+ ] [F1- ]

[HF]

Problem 14. Write the equilibrium (Keq) expressions for the following systems. Do not includesubstances which are solids (s) or pure liquids (l).

a. BaCO3(s) <===> BaO(s) + CO2(g) Keq =

b. HCN(aq) + H2O(l) <===> H3O1+(aq) + CN1-(aq) Keq =

c. CuSO4•3H2O(s) + 2 H2O(g) <===> CuSO4•5H2O(s) Keq =

20-16 ©1997, A.J. Girondi

SECTION 20.8 LeChatelier's Principle and Changes in Temperature

Changes in temperature will also create a stress on a system at equilibrium. In this case, the effectis more complicated than that for stresses caused by concentration and pressure changes. This isbecause Keq is temperature dependent, meaning that its value will be numerically different at differenttemperatures for a given reaction. Thus, heating or cooling an equilibrium system will result in a shifting ofthe equilibrium forward (to the right) or reverse (to the left), depending on which of the two reactions isexothermic and which is endothermic.

Let's first consider a reaction that is exothermic (as most reactions tend to be). Such a reactioncan be written as: reactants <===> products + heat energy

Suppose the temperature of the system is increased. This involves adding heat energy to the system.

This is like increasing the amount of a product of the reaction. (Since heat appears on the right side of the

equation, it can be considered a product.) You might think of this as placing a stress on the right side of

the equation. To relieve this stress, the equilibrium will shift to the left. The concentrations of products will

then {28}_____________ , and the concentrations of reactants will {29}_______________. .

Lowering the temperature of an exothermic reaction that has come to equilibrium should have the

opposite effect. Lowering the temperature amounts to removing heat energy from the system. This

stress will result in the equilibrium shifting to the right. This will {30}______________the concentration of

products and {31}________________ the concentration of reactants.

Next, let's consider an endothermic reaction that can be written as:

reactants + heat energy ----> products

The heat energy can be regarded as part of the reaction, in this case as one of the reactants. Adding heat

energy by increasing the temperature of this system at equilibrium amounts to placing a stress on the left

side of the equation. To relieve the stress, the concentration of reactants will decrease by forming more

{32}___________. (The system shifts to the right.) Decreasing the temperature of this endothermic

reaction will cause the equilibrium to shift to the {33}_____________.

In summary, you know that in an equilibrium system one of the reactions is exothermic and the

other is {34}______________. Raising the temperature (which amounts to adding heat) will cause an

increase in the rate of both reactions. In terms of collision theory, why would this be true? {35}_________

______________________________________________________________________________

However, when heat is added, the rate of the endothermic reaction (the reaction which uses heat) will

generally increase more than the rate of the exothermic reaction (which gives off heat). In other words,

adding heat causes a shift in favor of the endothermic reaction. Removing heat, causes a shift in favor of

the exothermic reaction.

20-17 ©1997, A.J. Girondi

Problem 15. Complete Table 20.6 by indicating the direction of the equilibrium shift when thetemperature is changed as indicated.

Table 20.6The Effect of Temperature Changes on Equilibrium Systems

Temperature Shift Equilibrium Equation Change Direction

a. H2(g) + Cl2(g) <===> 2 HCl(g) + 44184 kJ decrease __________b. 50.2 kJ + H2(g) + I2(g) <===> 2 HI(g) decrease __________c. CH4(g) + 2 O2(g) <===> CO(g) + 2 H2O(l) + 887 kJ increase __________d. C(s) + O2(g) <===> CO2(g) + 393 kJ decrease __________e. N2(g) + 3 H2(g) <===> 2 NH3(g) + 46 kJ increase __________f. 2376 kJ + 8 SO2(g) <===> S8(s) + 16 O2(g) decrease __________g. 75.3 kJ + CH4(g) <===> C(s) + 2 H2(g) increase __________

ACTIVITY 20.9 Testing LeChatelier's Principle With Cobalt Ions

This next activity will allow you to study the effects of concentration and temperature changes onan equilibrium system that exists between two different cobalt complexes. In water solutions, the Co2+ ionis pink. The pink color is due to the existence of a complex ion with the formula: Co(H2O)62+. This is theform in which cobalt normally exists in water. When Cl1- ions are also present in high concentrations, theCo(H2O)62+ is converted to Co(H2O)4Cl2, which is blue:

Co(H2O)62+(aq) + 2 Cl1-(aq) <===> Co(H2O)4Cl2(aq) + 2 H2O(l)

pink blue

The two colored Co2+ species can be converted to one another by appropriate changes in theconcentration of Cl1- ion or of water and by changes in temperature. Follow the procedure below, and besure to wear your glasses.

1. Mark two 50 mL Erlenmeyer flasks or 50 mL beakers "1" and "2."

2. Prepare the following two solutions:

Solution 1: Dissolve 0.50 g of CoCl2•6H2O in 10 mL of 6M HCl (hydrochloric acid). Handle the HCl withcare. The high concentration of Cl1- in the HCl pushes the equilibrium to the right and most of the cobalt inthis mixture is in the form of Co(H2O)4Cl2. Stir the solution until the solid is completely dissolved. What isthe color of Co(H2O)4Cl2 in solution 1? {36}______________

Solution 2: Dissolve 0.50 g of CoCl2•6H2O in 15 mL of water. The high concentration of H2O in thismixture pushes the equilibrium to the left and most of the cobalt in this mixture is in the form ofCo(H2O)62+. Stir the solution until the solid is completely dissolved. What is the color of Co(H2O)62+ insolution 2? {37}_______________

Adding HCl (and therefore Cl1-) to this system causes it to shift to the {38}____________ and the color

turns more {39}___________. Adding H2O to this system causes it to shift to the {40}____________and

the color turns more {41}________________.

20-18 ©1997, A.J. Girondi

2. Add 5 mL of water (or more if necessary) to solution 1 until a color change occurs. Now what is thecolor of the solution? {42}_______________ Heat the flask of solution 1 on a hotplate or with a burner untila color change occurs. What is the color of the heated solution 1?{43}______________

3. What do you think will happen to the color of the solution if it is cooled? {44}__________________Now place the flask of solution 1 into an ice water bath. Allow the flask to remain in the ice water bath until achange occurs. Was your prediction correct?{45}_________________

4. Keeping in mind that solution 2 contains CoCl2•6H2O, what two things could you do to solution 2 to getit to form more Co(H2O)4Cl2? {46}______________________________________________________Now do both of these two things – simultaneously – to a 15 mL portion of solution 2 in a 50 mL flask orbeaker. Describe the result: {47}______________________________________________________Did you manage to make more Co(H2O)4Cl2 in solution 2? {48}_________________________________

How do you know? {49}_____________________________________________________________

5. Based on your results, is the forward (--->) reaction for this cobalt system endothermic or exothermic?

{50}________________. Explain how you know: {51}_______________________________________

______________________________________________________________________________

If you have a little time left try this. Add some solid CoCl2 to a small volume of water in a test tube and shaketo dissolve. Pour the solution onto a piece of filter paper. Note that it is red in color. Now hold the wetfilter paper with your crucible tongs and warm it gently over the flame of a lab burner. Be careful not toignite the paper as it drys. Note the color change as you evaporate the water out of the system. Wet thepaper withplain water to restore the red color.

SECTION 20.10 A Review of LeChatelier's Principle

1. State LeChatelier's principle: {52}___________________________________________________

______________________________________________________________________________

2. In which direction will an equilibrium system shift if the concentration of one of the products isdecreased (at constant T and P)? {53}__________________________________________________

3. In which direction will an equilibrium shift if a reaction has more gas molecules on the left (reactant side)than on the right (product side) and if the pressure of the system is increased? {54}__________________

4. Suppose for a hypothetical equilibrium system such as A(g) <===> B(g), the forward reaction (from leftto right) is exothermic. In which direction (forward --->) or (<--- reverse) will the equilibrium shift if thetemperature is increased? {55}___________________

5. Will an equilibrium reaction shift forward (--->) or reverse (<---) if the concentration of one of the

reactants is decreased (at constant T and P)? {56}__________________

6. For a reaction involving equal numbers of gas molecules on both sides of the equation, will the

equilibrium shift forward or reverse if the pressure is decreased? {57}____________________________

7. For an endothermic reaction, will the equilibrium shift toward products or reactants if the temperature is

increased? {58}_______________________________

20-19 ©1997, A.J. Girondi

SECTION 20.11 Using Keq Values to Make Predictions

Equilibrium constants (Keq) are quite useful to chemists because they provide a clue about theamount of product that can be produced in a given chemical reaction. Normally, a chemical reaction iscarried out because an experimenter wants to obtain and use the product. Ideally, one would like to get a100% yield, which means that all of the reactants would be converted into products. However, 100%yields are not always possible. Instead, a system may go to equilibrium resulting in less than a 100% yield.

We are able to predict the extent to which reactants will be converted into products based on thesize of Keq. Remember that Keq is related to a ratio involving products over reactants. Look at the Keq

values calculated below:

Keq =

100

2 = 50; Keq =

10

0.001 = 1 X 10 4 ; Keq =

50

0.02 = 2.5 X 103

1. Suppose the three Keq values above represent very similar reactions. Which Keq value represents thereaction which produced the most product?{59}_________________

2. Which Keq value represents the reaction which produced the least product?{60}__________________

3. Explain how Keq values can be used to determine which of a series of similar reactions will produce the

most product? {61}________________________________________________________________

4. For the gaseous reaction A + B <===> C, a chemist is interested in getting as large a yield of the

product C as possible. He varies the reaction temperature which is the one variable that can change the

Keq value of a system. At 300oC he experimentally calculates that Keq for the system is 26.2. At 10oC he

finds that Keq = 0.012. To maximize the amount of C produced, should the chemist heat the reaction

container or cool it?{62}_______________ Explain: {63}_____________________________________

______________________________________________________________________________

Problem 16. For the equilibrium system: 2 SO2(g) + O2(g) <===> 2 SO3(g) the value of K eq at roomtemperature is 30.0. Predict the concentration of SO3 gas in the system when it is at equilibrium if theother concentrations are: [SO2] = 0.20M and [O2] = 0.30M.

__________M

20-20 ©1997, A.J. Girondi


Problem 17. A five-liter flask contains the system: CO(g) + Cl2(g) <===> COCl2(g). The flask contains1.50 moles of CO, 1.00 mole of Cl2, and 4.00 moles of COCl2. (These are all gases.) Calculate the value ofthe equilibrium constant for this system. (Remember values used must be in moles per 1.00 liter.)

________________

Problem 18. In a 1-liter flask the following system is at equilibrium: C(s) + H2O(g) <===> CO(g) + H2(g).The amounts of substances present in the 1-liter flask are 0.16 mole of C, 0.58 mole of H2O, 0.15 mole ofCO, and 0.15 mole of H2. Calculate the value of Keq for this system. (Note that C is a solid while the othersubstances are in the gas phase.)

________________

There is a supplementary discussion of another type of equilibrium constant known as thesolubility product constant, Ksp, in Appendix E of your ALICE materials. Ask your teacher if you shouldstudy that Appendix, or if you should end Chapter 20 here.

Go To Appendix E???(Ask the Instructor)

20-21 ©1997, A.J. Girondi


Equilibrium is an extremely important topic in chemistry and in all of the sciences. It will be veryuseful to you in the upcoming chapter on acids and bases. Look over the learning outcomes and makesure that you have mastered each of them. Check them off once you are satisfied. Arrange to take anyquizzes or exams on Chapter 20, and then move on to Chapter 21.

_____1. Define equilibrium and state the general characteristics of a system in equilibrium.

_____2. Write equilibrium expressions for chemical systems involving solids, liquids, and gases.

_____3. Calculate Keq values given the equilibrium concentrations of the products and reactants.

_____4. State LeChatelier's principle in general terms.

_____5. Use LeChatelier's principle to predict the direction an equilibrium will shift if there is a change in pressure, concentrations, temperature, or volume of the reaction container.

_____6. Given the Keq values for two or more similar systems in equilibrium, predict which system contains the greater concentration of products.

The following learning outcomes are to be included only if you studied the material concerning Ksp inAppendix E.

_____7. Determine the identity of unknown chemicals by testing and comparing them with a set of knownchemicals.

_____8. Given the solubility of a substance, calculate its Ksp value.

_____9. Given the Ksp value and the concentration of one ion, calculate the concentration of the other ion.

20-22 ©1997, A.J. Girondi


Questions:

{1} beaker A; {2} molecules are escaping; {3} no; {4} molecules that escape cannot return to liquid phase;{5} they will be greater; {6} left; {7} increase; {8} decrease; {9} increase; {10} decrease; {11} amber(depends on your color vision); {12} deep red (depends on your color vision); {13} darker; {14} increase;{15} causes more collisions between Fe3+ ions and SCN1- ions and shifts system toward right (products);{16} lighter; {17} decrease; {18} system shifted toward the left (toward reactants); {19} lighter;{20} decrease; {21} system shifts toward the left (toward reactants}; {22} shift to left toward reactants;{23} decreased pressure is a stress, so system shifts to the left to form more molecules to help increasethe pressure; {24} increase the pressure; {25} increasing the pressure will create a stress which thesystem will try to relieve by forming fewer molecules (shift to the right) which will help lower the pressure;{26} no effect; {27} since both sides of equation have same number of molecules, shifting would notchange pressure; {28} decrease; {29} increase; {30} increase; {31} decrease; {32} products;{33} left toward reactants; {34} endothermic; {35} more collisions between particles occur at highertemperatures; {36} blue (depends on your color vision); {37} pink (depends on your color vision);{38} right; {39} blue; {40} left; {41} pink; {42} pink; {43} blue; {44} will turn pink; {45} I hope so!{46} heat it and add more HCl; {47} solution should shift toward blue; {48} yes; {49} because of thechange in color; {50} endothermic; {51} because adding heat speeds up an endothermic reaction morethan it speeds up an exothermic one; {52} when a stress is placed on a system at equilibrium, the systemwill adjust to relieve the stress and to restore equilibrium in the system; {53} shifts to the right towardproducts; {54} shifts to the right toward products; {55} shift to the left toward reactants; {56} shifts to theleft (reverse) toward reactants; {57} neither, it will not shift either way; {58} shift to the right towardproducts; {59} 1X 104; {60} 50; {61} bigger Keq value means more products; {62} heat it; {63} since Keq isgreater at the higher temperature, the forward reaction which forms C is endothermic

Problems:

1. a. NaCl(s) <===> Na1+(aq) + Cl1-(aq); b. Fe2(SO4)3(s) <===> 2 Fe3+(aq) + 3 SO42-(aq)

c. BaBr2(s) <====> Ba2+(aq) + 2 Br1-(aq)

2. a. Keq = [N2O4] / [NO2]2; b. Keq = [NH3]2 / [N2] [H2]3; c. Keq = [Ag(NH3)21+] / [Ag1+] [NH3]2

d. Keq = [N2] [H2O]2 / [NO]2[H2]2

3. 0.37

4. 0.16

5. 6.4 X 105

6. 39.3

7. a. --->; b. <---; c. --->; d. <---; e. --->; f. <---; g. <---; h. --->; i. <---

8. a. none; b. --->; c. <---; d. <---

9. a. --->; b. <---; c. --->; d. <---; e. no effect; f. <---; g. --->; h. <---; i. no effect; j. no effect

10. a. <---; b. --->; c. --->

11. a. no effect; b. <---; c. --->

12. a. <---; b. --->; c. <---; d. no effect

13. a. --->; b. <---; c. --->; d. no effect

14. a. Keq = [CO2]; b. Keq = [H3O1+] [CN1-] / [HCN]; c. Keq = 1 / [H2O]2

15. a. --->; b. <---; c. <---; d. --->; e. <---; f. <---; g. --->

20-23 ©1997, A.J. Girondi

16. 0.60

17. 13.3

18. 3.9 X 10-2

20-24 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


"ALICE"

CHAPTER 21

ACIDSAND

BASESBehavior In Water

21-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


21-2 ©1997, A.J. Girondi

SECTION 21.1 Acids and Bases - The Arrhenius Definitions

In preceding chapters you have studied substances such as gases, salts, metals, and so forth.These labels describe classes of chemical substances which have certain properties in common. Thischapter will introduce you to two additional classes of substances known as acids and bases.

Many years ago it was found that when certain substances were placed into water, the resultingsolutions had a sour taste. When the early alchemists discovered this sour taste, they called thesesubstances acids. The alchemists were fond of using Latin names and phrases when describing theirwork. The Latin word meaning "sour" is acidus. The alchemists discovered that these acid solutions weresometimes capable of dissolving compounds that would not dissolve in plain water. You are probablyfamiliar with the sour taste of vinegar. Vinegar contains acetic acid. The Latin word for vinegar is acetum,reflecting its sour taste.

Alchemists also worked with another kind of substance prepared from the ashes of dried plants.When these ashes were placed in water, some of the components in the ashes dissolved in the water. Itwas found that when these "ash solutions" were mixed with acid solutions, the sour taste of the aciddisappeared. These substances derived from ashes were named bases. Early chemists found that abase could neutralize, or cancel out, the properties of an acid.

Chemical descriptions of acids and bases have developed and improved during the past 300years. The actual ions and molecules present in acids and bases have been identified. It was found thatthe common ion present in acid solutions was the positively charged hydrogen ion, H1+, while the mostcommon ion found in solutions of bases was the negatively-charged OH1- ion. What is the name of thision? {1}_________________________

In modern chemistry there are three ways to define acids and bases. In this chapter, we willexamine two of these ways. Please keep in mind that in this chapter we will be discussing how acidsbehave when you put them into pure water, and how bases behave when you put them into pure water.In the next chapter, you will study how acids and bases react with each other.

The first and simplest definition of acids was provided by the Swedish chemist, Arrhenius, in1887. He defined an acid as a compound that contains hydrogen and which would produce hydrogenions (H1+) when you dissolve it in water. For example, when hydrogen chloride gas (HCl) is dissolved inwater it breaks up or "dissociates" into hydrogen ions and chloride ions:

HCl(g) -----> H1+(aq) + Cl1-(aq)

This solution which contains hydrogen and chloride ions is called hydrochloric acid. It is the acid which aidsdigestion in your stomach. Its acidic properties are due to the presence of the hydrogen ions, H1+.

We will generally show hydrogenas the first element in the formulaof an acid.

HCl

In some acids, hydrogens whichare part of a polyatomic ion do notdissociate into hydrogen ions.

HC2H3O2

acetate ion

In the case of acetic acid (the acid in vinegar) the first H in the formula represents what we call an "acidic"hydrogen, which is an H that can form an H1+ ion when the molecule is put into water. The three H's whichare part of the acetate ion (C2H3O21-) are not "acidic" H's:

HC2H3O2(aq) <===> H1+(aq) + C2H3O21-(aq)

21-3 ©1997, A.J. Girondi

The Arrhenius definition of bases is the simplest one. According to Arrhenius, a base is acompound that contains the hydroxide ion, and produces hydroxide ions (OH1-) when it is dissolved inwater. When solid sodium hydroxide is dissolved in water, it breaks up or "dissociates" into sodium ionsand hydroxide ions:

NaOH(s) -----> Na1+(aq) + OH1-(aq)

A water solution of NaOH has the properties of a base because of the presence of the hydroxide ions.Sometimes basic solutions are described as being "alkaline."

The Arrhenius Definitions:

An acid produces H1+ in water solutionA base produces OH1- in water solution

General Properties of Acidic Solutions

they taste sour they neutralize bases they affect chemical indicators

they are electrolytes, meaning they conduct electricity

Problem 1. Complete the following equations, showing how the acids listed dissociate in water to formhydrogen ions and a negatively-charged ion (anion).

a. HBr(aq) __________________________________________________

b. HI(aq) __________________________________________________

c. HClO4(aq) __________________________________________________

they taste bitter they neutralize acids they affect chemical indicators

they are electrolytes, meaning they conduct electricity they feel slippery

General Properties of Basic Solutions

Problem 2. Complete the following equations, showing how the bases listed dissociate in water to form OH1- ions.

a. KOH(s) __________________________________________________

b. LiOH(s) __________________________________________________

c. CsOH(s) __________________________________________________

21-4 ©1997, A.J. Girondi

SECTION 21.2 The Bronsted–Lowry Definitions

Arrhenius's definitions were generally accepted by chemists of his time. They explained manyunanswered questions about acids and bases. His definitions of acids and bases are still widely used bychemists who work with aqueous (water) solutions.

However, eventually compounds were discovered that had the properties of bases, but which didnot contain the OH1- ion. This meant that a better definition of bases was needed. The Arrheniusdefinition was no longer adequate. A definition was needed which could explain why some compounds,other than those containing hydroxide, had "basic" properties. The term "alkaline" is sometimes used todescribe substances which have "basic" properties. To solve this problem, a second definition of acidsand bases was suggested by Thomas M. Lowry and Johannes N. Bronsted. It will be helpful for you torealize that a hydrogen ion (H1+) can also be called a proton, since all that is left of a hydrogen atom whichhas lost its electron is a proton.

Hydrogen Ion = H1+ = a proton

H

A common hydrogen atomconsists of one proton in thenucleus and one electron.

If a hydrogen atom loses itselectron, it becomes a hydrogenion - just a proton.

H1+

Bronsted and Lowry proposed that an acid be defined as a molecule or ion that can give away or donate aproton (H1+) to some other particle. A base was defined as a substance that can combine with or accept aproton (H1+) from some other particle. According to the Bronsted–Lowry concept, an acid became knownas a proton donor, and a base as a proton acceptor.

The Bronsted–Lowry Definitions

An acid donates a proton (H1+)A base accepts a proton (H1+)

Note that the Arrhenius definition of an acid and the Bronsted–Lowry definition of an acid are very similar.Both definitions refer to the formation of H1+: HCl ----> H1+ + Cl1-.

However, the Arrhenius definition of a base refers only to substances which can provide OH1- ions insolution, whereas, the Bronsted–Lowry definition of a base refers to any particle which can accept ahydrogen ion (proton), H1+. Perhaps the following example will help. Arrhenius would consider NaOH tobe a base because it forms OH1- when you put it into water solution:

NaOH(s) ----> Na1+(aq) + OH1-(aq)

Bronsted and Lowry would consider NaOH to be a base because a solution of it contains the OH1- ionwhich is a proton acceptor:

H1+(aq) + OH1-(aq) ----> HOH(l)

When OH1- accepts a proton its forms a molecule of water, HOH.

Now remember, when Arrhenius defined a base he was thinking only of OH1-. However, otherparticles such as the fluoride ion, F1-, can also act a proton acceptors. For example:

H1+(aq) + F1-(aq) ----> HF(aq)

21-5 ©1997, A.J. Girondi

Solutions which contain the fluoride ion have "basic" properties similar to solutions which contain thehydroxide ion, OH1-. Other particles which can function as bases (proton-acceptors) of this kind includeboth molecules and ions like C2H3O21-, NH3, CO32-, and many others. Therefore, many more particles canbe classified as bases according to Bronsted–Lowry than according to Arrhenius. Here are a few moreequations which illustrate how these bases function as proton-acceptors:

H1+ (aq) + C2H3O21-(aq) ----> HC2H3O2(aq)

H1+(aq) + NH3(aq) ----> NH41+(aq)

H1+(aq) + CO32-(aq) ----> HCO31-(aq)

What is it about particles like OH1-, F1-, C2H3O21-, NH3 , and CO32- that allows them to function as protonacceptors? If you look at the electron dot structure of these ions, you will see that they have one or moreunshared pairs of electrons:

O1-

F1-

hydroxide ion (a base) fluoride ion (a base)

HThe OH1- ion has 3 unshared pairs ofelectrons, while the F1- ion has 4 unsharedpairs.

The H1+ ion is seeking the stable helium configuration (1s2). It can achieve that configuration by sharing apair of electrons with another particle. So, protons (H1+) tend to bond to particles which have an unsharedpair of electrons in their valence shells. Many such particles can act as proton–acceptors which are basesaccording to the {2} ______________________definition. Check out the electron-dot structures ofOH1- and the F1- ions below:

H1+ + HO1-

H O H

H1+ + HF1-

F

If you examine the electronic structures of other bases like NH3 molecules or acetate ions, C2H3O21-, youwill see that they also have unshared pairs of electrons:

H

NH HH1+ +

H

NH HH

1+

C CO

OH

H

H

H1+ + C CO

OH

H

H

H

1–

O

HH

You might ask, "Why do acids lose H1+ when they are put into water?" Well, thereason is that water molecules take them! You see, since water has twounshared pairs of electrons on the oxygen atom, it too can accept protons andfunction as a base. Note the electron-dot structure of water shown at right.

21-6 ©1997, A.J. Girondi

To illustrate how water can act as a proton-acceptor, consider the equation below showing the reaction or"dissociation" of HCl in water:

HCl(g) + H2O(l) ----> H3O1+(aq) + Cl1-(aq)

The shorthand way of writing this equation is: HCl(g) -----> H1+(aq) + Cl1-(aq)

(Note that HCl is a gas and therefore is accompanied by the (g) subscript before it is put into water. Most of the other acids you will see will be written using the (aq) subscript.)

When you include water in the equation, you must represent the hydrogenion as H3O1+ instead of as H1+. This is actually a more accuraterepresentation of what happens. The H3O1+ particle is known as thehydronium ion. It is a water molecule which is bonded to a proton (H1+). Itselectron-dot structure is shown at right.

O HH

H

1+

The hydronium ion forms when water acts as a base and accepts a proton from an acid molecule like HCl.

O HH

H1+

H Cl + O HH

Cl+ 1-

H1+

Equations representing what happens when you put an acid molecule in water can, therefore, be writtenin two ways. You can choose to show the water and the hydronium ion (the more accurate way to do it), oryou can choose not to show them (the shorthand way of writing it). Below are the equations representingwhat happens when you put nitric acid in water:

HNO3(aq) + H2O(l) ----> H3O1+(aq) + NO31-(aq)

or

HNO3(aq) ----> H1+(aq) + NO31-(aq)

According to the Bronsted-Lowry definition, water can act as a base since it can accept protons. (Later you will learn that water can also give away a proton and, therefore, act as an acid.)

Problem 3. For each of the acids below, write two forms of the equation which represents whathappens when you put them into water.

a. HBr _____________________________________

_____________________________________

b. HClO4 _____________________________________

_____________________________________

Not all particles with unshared pairs of electrons make good bases (proton acceptors) in water. Forexample, the chloride ion has the necessary unshared pairs of electrons, but it is not a basic ion. It isdescribed as being neutral. It is NOT a proton acceptor:

ClH1+ +1-

no reaction

Some other neutral ions include Br1-, I1-, NO31-, SO42-, and ClO41-.

21-7 ©1997, A.J. Girondi

SECTION 21.3 Strong Acids Versus Weak Acids

A. Strong Acids

The reason that solutions of acids and bases are electrolytes is because they contain ions. Theions move about and carry the electric charge through the solution. The equation below demonstrates thereaction of hydrochloric acid in water. Note that the ions produced are "aqueous" meaning in watersolution. Acids are hydrogen compounds that form water solutions which contain ions, one of which isthe hydrogen ion.

For example: HI(aq) ----> H1+(aq) + I1-(aq) or HI(aq) + H2O(aq) ----> H3O1+(aq) + I1-(aq)

There are only six strong acids, but there are many, many weak acids. The six strong acids are

HBr, HCl, HI, HNO3, H2SO4, HClO4. You should memorize the six strong acids. You may be wonderingwhat makes an acid strong or weak. Acids such as HCl and HNO3, are strong acids because theydissociate completely to form ions when they are put into water. In other words, all of the molecules of astrong acid will dissociate into ions when you put the acid into water solution. We say that they are "100percent dissociated in water." Solutions of strong acids, therefore, contain a high concentration ofhydrogen ions. Perchloric acid, HClO4, is an example:

HClO4(aq) ----> H1+(aq) + ClO41-(aq)

OR

HClO4(aq) + H2O(l) ----> H3O1+(aq) + ClO41-(aq)

THE SIX STRONG ACIDS

HIhydroiodic acid

HClhydrochloric acid

HBrhydrobromic acid

H2SO4

sulfuric acidHNO3

nitric acid

HClO4

perchloric acid

B. Weak Acids

Acetic acid, HC2H3O2, is classified as being weak. Because of the nature of the bonding betweenthe acidic hydrogen and the acetate ion, a molecule of acetic acid does not dissociate very much in water.As a result, most molecules of weak acids remain in the form of molecules when they are put into water.Many of the ions which form when the weak acid molecules dissociate will recombine to form the originalmolecules. Thus, the concentration of hydrogen ions is lower than it would have been if the acid had beenstrong. Only a small percentage of molecules of a weak acid will be dissociated at any given point in time.An equilibrium is established in which the equilibrium is strongly favored toward the reactants (<---):

21-8 ©1997, A.J. Girondi

HC2H3O2(aq) + H2O(l) <====> H3O1+(aq) + C2H3O21-(aq)

double arrow indicates equilibrium low concentration of hydronium ions

Or, using the shortcut (but less accurate) representation:

HC2H3O2(aq) <====> H1+(aq) + C2H3O21-(aq)

The use of the double-headed arrow indicates that the acid is weak and exists mostly in the form ofmolecules (HC2H3O2) rather than as ions. Equilibrium is characteristic of {3}__________ acids in water.When equilibrium is established, the system contains mostly reactants and, therefore, not many ions. Thisfact provides us with a method for determining the strength of acids. The greater the concentration ofions in an acid solution, the better the solution will conduct electricity. Since stronger acids dissociate intoions much, much better than weak ones do, the stronger acids are {4}___________ conductors ofelectricity. You will use this property to determine the relative strengths of some acids in the next activity.

There are many weak acids. However, at this point you will be expected to memorize the namesand formulas of only the three listed below.

SOME WEAK ACIDS

HC2H3O2

Acetic AcidHF

Hydrofluoric AcidH3PO4

Phosphoric Acid

Weak acids form an equilibrium system in water in which the reverse reaction is favored. Why?Each of the two reactions involved in such an equilibrium system is an acid-base reaction. In the systemshown below, HF is a weak acid. In the forward reaction (---->) the acid is HF and the base is{5}__________. That is, the HF is donating a proton, and the H2O is accepting it to form the products onthe right. In the reverse reaction (<----) the acid is H3O1+ and the base is {6}_________. That is, the H3O1+

is donating a proton, and the F1- is accepting it to form the products on the left:

HF + H2O <====> H3O1+ + F1-

acid base acid base

It turns out that H3O1+ is a stronger proton donor (acid) than HF and F1- is a stronger proton acceptor(base) than H2O. Therefore, the reverse reaction (<----) is better than the forward (---->) one. Whenequilibrium is established, there will be much more HF and H2O in the solution than H3O1+ and F1-.Solutions of weak acids, therefore, have a relatively low concentration of H3O1+ ions.

Problem 4. Complete the following equations, showing how the weak acids listed below dissociate inwater to form ions. The anions contained in these acids are ClO31- and OCl1-. Be sure to use a double-headed arrow in the equation. Include H2O in the equations.

a. HClO3(aq) ______________________________________________

b. HOCl(aq) ______________________________________________

21-9 ©1997, A.J. Girondi

ACTIVITY 21.4 Comparing the Conductivity of Strong and Weak Acids

There is a fairly simple method that can be used to determine the extent to which an acid or basedissociates. This method involves measuring the conductivity of a solution. When ions are present in asolution, it is possible for the solution to conduct electricity. Other things being equal, the greater theconcentration of ions in a solution, the greater the electrical current that will pass through the solution.

The number of ions which form when an acid or base is added to water depends on the degree ofdissociation. Since stronger acids and bases dissociate much more than weaker ones do, they are muchbetter conductors of electric current. The greater the electrical conductivity of an acid or base solution,the stronger it is.

Your teacher will give you specific instructions about how to use the conductivity device. Use it tomeasure the conductivity of the three acids listed in Table 21.1. Record your observations in the spacesprovided in the table. If the apparatus has a meter, record the meter reading in the table. If it has a lightbulb, record the strengths of the acids as high, medium, or low depending on the brightness of the bulb.If the bulb does not light, this is probably because the ion concentration is too low to allow enough currentto flow to light the bulb. It does not necessarily mean that there are no ions in the solution. (Hint: one ofthe acids should be rated "high;" one should be rated "medium;" and one should be rated "low." Youdetermine which is which.

Table 21.1Conductivity of Acid Solutions

Acid Formula Acid Name Conductivity

0.01M HCl hydrochloric _________

0.01M HC2H3O2 acetic _________

0.01M H8C6O7 citric _________

Based on the results, what conclusions can you draw about the relative strengths of these three acids?

{7}____________________________________________________________________________

Even though all three acid solutions have the same concentration, they do not conduct the same amount

of electric current and are not equally strong. Why not? {8}___________________________________

______________________________________________________________________________

______________________________________________________________________________

SECTION 21.5 The Dissociation Constant of a Weak Acid, Ka

As you have already learned, weak acids form an equilibrium system when they are put into water.There is both a forward and a reverse reaction. We can write equilibrium expressions for weak acidsystems. Let's consider the acetic acid equilibrium system:

HC2H3O2(aq) + H2O(l) <====> H3O1+(aq) + C2H3O21-(aq)

You should recall that an equilibrium expression consists of the product of the molar concentrations of the

21-10 ©1997, A.J. Girondi

products divided by the product of the molar concentrations of the reactants. Furthermore, you may recallthat solids and pure liquids are not included in equilibrium expressions. Water is a pure liquid in thesystem above. The equilibrium constant for a weak acid is called the dissociation constant and is giventhe symbol, K a. The Ka expression for the acetic acid system is given below. Compare the Ka expressionto the equation shown above.

Ka =

[H3O1+ ] [C 2H3O21− ]

[HC2H3O2 ]

Problem 5. The two equilibrium equations below involve weak acids. Write the Ka expression for eachsystem.

a. HNO2(aq) + H2O(l) <===> H3O1+(aq) + NO21-(aq) Ka =

b. HCN(aq) + H2O(l) <===> H3O1+(aq) + CN1-(aq) Ka =

Look at the expressions for Ka which you wrote above. Note that as the concentration of hydronium ions,H3O1+, increases, so does the value of Ka. Weak acids vary considerably in strength. Some are muchstronger than others, although none of them approach the strength of the six "official" strong acids.Thus, the weak acids which are "strongest" have larger Ka values. Indeed, by comparing Ka values, youcan determine which of any given set of acids is strongest. Three weak acids are listed in Table 21.2.Which is strongest?{9}______________________ Weakest?{10}______________________

Table 21.2Ka Values of Selected Weak Acids

Acid Dissociation Constant

hypochlorous acid, HClO Ka = 3.2 X 10-8

formic acid, HCOOH Ka = 1.8 X 10-4

phosphoric acid, H3PO4 Ka = 7.1 X 10-3

Reference sources such as textbooks and The Handbook of Chemistry and Physics contain tables of Ka

values of many weak acids.

SECTION 21.6 The Behavior of Strong and Weak Bases In Water

A. Strong Bases

Bases are also divided into groups that are strong and weak. The strong bases are the hydroxidecompounds of most of the Group 1A and 2A metals. Examples include LiOH, NaOH, KOH, Ca(OH)2, etc.All other bases are considered to be weak. Strong bases are completely dissociated in water.

Strong base: NaOH(s) ---------> Na1+(aq) + OH1-(aq)

21-11 ©1997, A.J. Girondi

Strong bases form solutions which contain lots of ions. The base (NaOH in this case) simply breaks apartin water. They are considered to be strong bases because 100% of the compound dissociates intoindividual ions when it dissolves. The hydroxides of family 1A metals, like NaOH, are potent bases for tworeasons: (1) they are very soluble in water, and (2) they are 100 percent dissociated in solution. As aresult, strong bases of family 1A can produce lots of OH1- ions in solution. The high concentration ofOH1- ions in strong bases makes them dangerous.

The strong bases are the hydroxidecompounds of the solid metals infamilies1A and 2A of the periodic table.(Except for beryllium and magnesium)

Using a periodic table, you should be able to write the formulas of the strong bases from memory.

The hydroxide compounds of the family 2A metals, like Ca(OH)2, are considered to be strongbecause, like family 1A hydroxides, they too are 100% dissociated in solution. However, the family 2Ahydroxides are not nearly as soluble as those of family 1A. As a result, while they are strong bases, theyare not as potent because they do not form solutions with high concentrations of OH1- (since they are notvery soluble). Only a small amount of Ca(OH)2 will dissolve before its solution becomes saturated. In fact,Ca(OH)2 is called "lime" and is mild enough to be used to neutralize acids in lawn and garden soils.Mg(OH)2 is used in some stomach antacids like "milk of magnesia." You wouldn't want to use NaOH forthat purpose! NaOH is the active ingredient in many drain cleaners!

Strong base: Ca(OH)2(s) ----> Ca2+(aq) + 2 OH1-(aq)

[Ca(OH)2 is 100% dissociated, but not much will dissolve.]

Water is not included in equations which show the dissociation of strong bases. Strong and weak acidsactually react with water so H2O can be included in the equation. Strong bases do not react with water;they just come apart (dissociate).

Problem 6. Write equations showing the complete dissociation of the following strong bases in water.

a. RbOH(s) ----> _____________________________

b. Ba(OH)2(s) ----> _____________________________

B. Weak Bases

Weak bases are in some ways similar to weak acids. For example, they form ions when youdissolve them in water. Furthermore, they form equilibrium systems in water, because they dissociateonly slightly (similar to weak acids). Some are polar covalent molecules which react with water to form ions.The most common example of a weak base is ammonia, NH3.

Ammonia: NH3(g) + HOH(l) <===> NH41+(aq) + OH1-(aq) weak weak base acid

Note: Remember that water can also be written as HOH. We will use this form of the formula in equations dealing with

weak bases, because it makes it easier to see what is happening.

21-12 ©1997, A.J. Girondi

Some weak bases are ions such as the fluoride ion, F1-:

Weak base: F1-(aq) + HOH(l) <===> HF(aq) + OH1-(aq)

base acid acid base

The electron dot notation of the fluoride ion, F1-,

molecule is shown at right. What is the feature of this

molecule that allows it to function as a base?

{11}______________________________________

________________________________________

F1-

Notice that water acts as an acid in the last two equations. In previous sections, you saw water acting as abase. Water can act as an acid because it contains hydrogen which can form ions (H1+), and it containsoxygen which has unshared pairs of electrons which allow it to act as a Bronsted–Lowry base.Substances such as water which can act as either acids or bases are said to be amphoteric.

HHO

WATER

Can act as an acid bygiving away hydrogenin the form of H1+

Can act as a base byattracting H1+ to one of itsunshared electron pairs.

There is an important difference between weak bases and strong bases. Both produce hydroxideions, OH1-, when you dissolve them in water; however, with strong bases the OH1- comes directly fromthe strong base, like NaOH: NaOH(s) ----> Na1+(aq) + OH1-(aq)

Note that the formula for a weak base like ammonia, NH3, does not contain hydroxide. The most commonand important weak bases are not hydroxide compounds. Yet, when you dissolve them in water,hydroxide ions are formed. So, where do these hydroxide ions come from?

NH3(g) + HOH(l) <===> NH41+(aq) + OH1-(aq)The formula of this weak base doesNOT contain hydroxide (OH1-).

In a solution of a weak base, the OH1- ions come from the water!

Whenever you write an equation showing the reaction of a weak base with water, you should alwaysinclude the water in the equation (as is done in the equation above). If you write the equation withoutincluding the water, it will not be balanced:

Wrong -----> NH3(g) <===> NH41+(aq) + OH1-(aq)

Right -----> NH3(g) + HOH(l) <===> NH41+(aq) + OH1-(aq)

21-13 ©1997, A.J. Girondi

Problem 7. Write balanced equations showing the behavior of the following weak bases in water. Keepin mind that the particle of weak base will accept a H1+ ion from water.

a. BrO1-(aq) BrO1-(aq) + HOH(l) <===> ________________________

b. S2-(aq) S2-(aq) + HOH(l) <===> __________________________

Problem 8. Write balanced equations which illustrate the behavior of the substances listed below whenthey are added to water. Refer back to the examples in previous sections of this chapter. Be sure toinclude charges on ions.

a. HBr (a strong acid) ___________________________________________________

b. LiOH (a strong base) ___________________________________________________

c. HNO2 (a weak acid) ___________________________________________________

d. CN1- (a weak base) ___________________________________________________

(Check to be sure that you showed charges on any ions in the products.)

There are two reactions involved in an equilibrium system. In the acetic acid system, note that inthe forward reaction (--->) we can classify one substance as an acid and one as a base, and we can do thesame thing for the reverse reaction (<---):

HC2H3O2 + H2O(l) <====> C2H3O21-(aq) + H3O1+(aq) acid base base acid

We are making use of the Bronsted–Lowry definitions of acid and base here.

What molecule is the proton donor in the forward (--->) reaction? {12}_______________ What ion is the

proton donor in the reverse reaction?{13}_____________________

Problem 9. In the three equations below, label each substance as an acid or a base:

a. HF + H2O <===> H3O1+ + F1-

_______ _______ _______ _______

b. HCO31- + HBr <===> H2CO3 + Br1-

_______ _______ _______ _______

c. HCN + NH3 <===> CN1- + NH41+

_______ _______ _______ _______

At equilibrium, the system contains mostly reactants. This was also the case for the weak acids. In otherwords, the weak bases do not form a high concentration of OH1- ions. Even though weak bases do notcontain OH1-, it is still the OH1- ion that makes the solution basic. Notice that the OH1- ions which areproduced in a solution of a weak base are formed indirectly. By that we mean that the OH1- ions do notcome from the base itself, but they come from the water! Examine the following equation.

21-14 ©1997, A.J. Girondi

NO21-(aq) + HOH(l) <===> HNO2(aq) + OH1-(aq)

base acid acid base

In solutions of weak bases, the OH1- ions come from the water!

As you can see, some negative ions (anions) like NO21- can acts as weak bases in water:

In solutions of strong bases like NaOH, the OH1- ion comes from the base itself. Water is not shown in theequation which represents a strong base in water:

NaOH ----> Na1+ + OH1-

Explain the difference in the source of the OH1- ions in solutions of strong bases versus solutions of weak

bases: {14}______________________________________________________________________

______________________________________________________________________________

Since weak bases involve equilibrium systems, an equilibrium expression can be written. For thesystem below:

NO21-(aq) + HOH(l) <===> HNO2(aq) + OH1-(aq)

The equilibrium expression =

[HNO2 ] [OH1- ]

[NO21− ]

Water is omitted since it is a pure liquid. This expression is equal to a constant which is given the symbolKb. It is known as the dissociation constant of a weak base:

Kb =

[HNO2 ] [OH1- ]

[NO21−]

Notice that Kb gets larger as the concentration of OH1- increases.

You probably already realize that the larger the Kb value, the stronger the base. A very common andimportant weak base is ammonia, NH3. Ammonia is very useful in the production of explosives andfertilizers. When it is dissolved in water, the solution is usually called "ammonium hydroxide." When youlook at the reaction, you can guess why:

NH3(g) + HOH(l) <===> NH41+(aq) + OH1-(aq)

base acid acid base

ammonium ionhydroxide ion

waterammonia

The electron dot structure of ammonia is shown in the space atthe right. What feature makes it a base (even though it's aweak one)? {15}___________________________________

N H

H

H

21-15 ©1997, A.J. Girondi

The products of the reaction are an ammonium ion and a hydroxide ion. However, since this is a weak

base , a better name for the solution would be "ammonia water." Explain why: {16}__________________

______________________________________________________________________________

______________________________________________________________________________

NH3(g) + HOH(l) <===> NH41+(aq) + OH1-(aq)

99 1

When 100 of these are added to water

Number of particles present at equilibrium

In fact, if you put 100 ammonia molecules inwater, only one would be dissociated atequilibrium. Thus, for every 100 ammoniamolecules put into a solution, only one OH1-

ion would be formed at equilibrium.

Problem 10. The equations below involve weak bases in water. Write the Kb expression for eachequation.

a. CN1-(aq) + H2O(l) <===> HCN(aq) + OH1-(aq) Kb =

b. NH3(aq) + H2O(l) <===> NH41+(aq) + OH1-(aq) Kb =

Three weak bases are listed below along with their Kb values. Which is strongest?{17}________________

Weakest?{18}___________________

Base Dissociation Constant

ammonia, NH3 Kb = 1.8 X 10-5

fluoride ion, F1- Kb = 1.4 X 10-11

sulfite ion, SO32- Kb = 1.8 X 10-7

It is not easy to identify a weak base just by looking at its formula. You would have to draw itselectron-dot structure to see if it contains an unshared pair of valence electrons, and you would have tostudy how it reacts with water. Some particles like the chloride ion, Cl1-, do not act as bases even thoughthey do have unshared pairs of valence electrons. One simple way of explaining this is to say that if theCl1- ion were to accept a proton, it would form a molecule of HCl. However, HCl is a strong acid, and strongacids are completely dissociated in water solutions. Therefore, HCl cannot form. Particles such as Cl1-

which would form a molecule of a strong acid by accepting a proton cannot, therefore, serve as bases.They are neutral particles. The anions (negative ions) which are found in the six strong acids are neutralanions. As mentioned earlier in this chapter, they include Cl1-, Br1-, I1-, SO42-, NO31-, and ClO41-.

Keep in mind that the reactions we are considering here all occur in water. Therefore,let's say thatthe Cl1- ion is trying to accept a proton from water:

Cl1-(aq) + H2O(l) ----> HCl(aq) + OH1-(l) <----- This reaction WILL NOT happen!

21-16 ©1997, A.J. Girondi

Why won't the reaction shown in the equation above happen? Well, let's just say that the reaction wouldresult in the formation of a strong acid (HCl in this example). Strong acid molecules like HCl do not exist inwater because they are completely dissociated. Therefore, the reaction does not occur.

ACTIVITY 21.7 Comparing the Conductivity of Strong and Weak Bases

This activity is like Activity 21.4 in which you determined the relative strengths of a few acids. Themethod involves measuring the conductivity of a solution. When ions are present in a solution, it ispossible for the solution to conduct electricity. Other things being equal, the greater the concentration ofions in a solution, the greater the electrical current that will pass through the solution.

The number of ions which form when an acid or base is added to water depends on the degree ofdissociation. Since stronger acids and bases dissociate much more than weaker ones do, they are muchbetter conductors of electric current. The greater the electrical conductivity of an acid or base solution,the stronger it is.

Because of your experience in Activity 21.4, you should already know how to use the conductivitydevice. Use it to measure the conductivity of the two bases listed in Table 21.3. Record yourobservations in the spaces provided in the table. If the apparatus has a meter, record the meter readingin the table. If it has a light bulb, record the strengths of the acids as high, medium, or low depending onthe brightness of the bulb. If the bulb does not light, this is probably because the ion concentration is toolow to allow enough current to flow to light the bulb. It does not necessarily mean that there are no ions inthe solution. (Hint: one of the acids should be rated "high;" one should be rated "low." You determinewhich is which.

Table 21.3Conductivity of Basic Solutions

Base Formula Base Name Conductivity

0.1M NaOH sodium hydroxide _________

0.1M NH3 ammonia _________

Based on the results, what conclusions can you draw about the relative strengths of these two bases?

{19}____________________________________________________________________________

Even though both base solutions are 0.1M, they do not conduct the same amount of electric current and

are not equally strong. Why not? {20}__________________________________________________

______________________________________________________________________________

SECTION 21.8 Polyprotic Acids

There are only six acids which are generally recognized as being strong (100% dissociated). Theyinclude HCl, HNO3, H2SO4, HBr, HI, and HClO4. (You should have memorized these by now!) Thenumber of hydrogens in the formula has nothing to do with the acid's strength. Thus, H2SO4 is not twiceas strong as the other strong acids. Phosphoric acid, H3PO4, is weak even though it contains threehydrogens! Remember, it is the percentage of dissociation that determines an acid's strength. (In otherwords, how well the acid molecule reacts with water.) That is, the strength is determined by how many H1+

ions actually form when you put the acid particle in water, not by how many H's are in the acid's formula. If

21-17 ©1997, A.J. Girondi

an acid can provide only one H1+ ion, it is said to be monoprotic. The general term for acids which can yieldmore than one H1+ ion is "polyprotic."

If an acid can provide two H1+ ions, it is "diprotic." Sulfuric acid is strong because it ionizescompletely as shown in the equation below:

H2SO4(aq) + H2O(l) ----> H3O1+(aq) + HSO41-(aq)

Notice that only one hydrogen formed, even though H2SO4 contains two hydrogens. The secondhydrogen is now part of the HSO41- ion, and it is very hard for a second H1+ to break away since theHSO41- ion, with its negative charge, strongly "holds on" to it. As shown in the equation below, a few domanage to break away, and as a result, the second hydrogen makes a small, very insignificant contributionto the total number of H1+ ions in a solution of sulfuric acid. The HSO41- ion is a relatively weak acid:

HSO41-(aq) + H2O(l) <===> H3O1+(aq) + SO42-(aq)This equilibrium reaction does not produce much H3O1+.

Therefore, when describing the behavior of sulfuric acid in water, it is sufficient for our purposes to saythat it undergoes only one significant step:

H2SO4(aq) + H2O(l) ----> H3O1+(aq) + HSO41-(aq)

Now, what we have just discussed is the behavior of sulfuric acid in pure water. However, sulfuric acid isdiprotic and it can give up two hydrogen ions when it reacts with a strong base instead of with water. Youwill learn more about this when you study acid-base reactions in the next chapter.

Acetic Acid CH3COOH or HC2H3O2

C CO

OH

H

H

H

Not all acids with more than one "H" are polyprotic. As mentioned earlierin this chapter, acetic acid has the formula HC2H3O2. Only the firsthydrogen in the formula can dissociate - at all - to form H1+. The otherthree hydrogens are part of the acetate ion (a polyatomic ion), C2H3O21-,and they are bonded differently in the molecule. As a result, they cannotform H1+ ions. Therefore, it is only a monoprotic acid. Sometimes, thosewho work primarily with carbon compounds (organic chemists) write theformulas for acids differently. For example, instead of writing acetic acidas HC2H3O2, they may write CH3COOH. The hydrogen that forms H1+

now appears at the end of the formula instead of the beginning. Thisalternate way of writing formulas gives more of a clue as to the structure ofthe molecule. Note the structure of acetic acid shown at right.

In the acetic acid molecule, it is the hydrogen which is bonded to oxygen which can dissociate inwater as an H1+ ion. The H–O bond is polar, and so is the water molecule. Thus there is an attractionbetween them, and water can act as a base and "pull" a few of those particular H's off of the acetic acidmolecule. The C–H bonds are not polar, so those H's are not pulled away by water at all. Acetic acid hasonly a 1-step dissociation. The following equation showing electron-dot structures may be helpful:

21-18 ©1997, A.J. Girondi

C CO

OH

H

H1–

C CO

OH

H

H

H

polar bond

nonpolar bond polar water molecule

O

H

H+ H O

H

H1+

+

Acetic Acid Water Hydronium Ion Acetate ion

One final word about the hydronium ion. As you know, the hydronium ion forms when water(which has 2 unshared pairs of electrons) acts as a base and accepts a proton. This is happening in theequation shown above. Since the hydronium ion, H3O1+, ion still contains one unshared pair ofelectrons, students often ask why H3O1+ cannot accept a second proton to form an ion with the formulaH4O2+. The proposed equation for this reaction is shown below.

H

H O

H

H1+

H O

H

H2+

+ H1+This reaction does NOT occur.

Suggest a reason why the hydronium ion cannot act as a base by accepting a proton. {21} ____________

______________________________________________________________________________

______________________________________________________________________________

21-19 ©1997, A.J. Girondi

SECTION 21.9 LEARNING OUTCOMES

This is the end of Chapter 21. Check the learning outcomes below and arrange to take the examon Chapter 21. Then, go on to Chapter 22 which is a continuation of the subject of acids and bases.

_____1. List the general properties of acids and bases.

_____2. Compare and contrast the definitions of acids and bases according to the Arrhenius and the Bronsted–Lowry models.

_____3. Write balanced equations showing the dissociation of strong and weak acids and bases in water.

_____4. Given a list of acids, classify them as strong or weak.

_____5. Given a list of bases, classify them as strong or weak.

_____6. Explain why some acids are strong, while others are weak.

_____7. Explain why some bases are strong, while others are weak.

21-20 ©1997, A.J. Girondi


Questions:

{1} hydroxide; {2} Bronsted-Lowry; {3} weak; {4} better; {5} H2O; {6} F1-; {7} strongest is HCl and weakest is acetic; {8} They do not all dissociate into ions to the same extent. HCl dissociates most, while acetic dissociates least; {9} phosphoric; {10} hypochlorous; {11} An unshared pair of electrons on the fluoride ion; {12} HC2H3O2; {13} H3O1+; {14} In solutions of strong bases, the OH1- comes from the base itself, whereas, in solutions of weak bases, the OH1- comes from the water that the weak base reacts with; {15} An unshared pair of electrons on the nitrogen atom; {16} The reverse reaction is better than the forward reaction, so there is more ammonia and water present at equilibrium; {17} NH3; {18} F1-; {19} Sodium hydroxide is strong but ammonia is weak; {20} The 0.1M NaOH dissociates to a high degree to form a lot of OH1- ions, but 0.1M ammonia does not; {21} The positively-charged H3O1+ ion will repel the positively-charged H1+ ion too much;

Problems:

1. a. HBr(aq) ----> H1+(aq) + Br1-(aq)

b. HI(aq) ----> H1+(aq) + I1-(aq)

c. HClO4(aq) ----> H1+(aq) + ClO41-(aq)

2. a. KOH(s) ----> K1+(aq) + OH1-(aq)

b. LiOH(s) ----> Li1+(aq) + OH1-(aq)

c. CsOH(s) ----> Cs1+(aq) + OH1-(aq)

3. a. HBr(aq) ----> H1+(aq) + Br1-(aq)

HBr(aq) + H2O(l) ----> H3O1+(aq) + Br1-(aq)

b. HClO4(aq) ----> H1+(aq) + ClO41-(aq)

HClO4(aq) + H2O(l) ----> H3O1+(aq) + ClO41-(aq)

4. a. HClO3(aq) + H2O(l) <===> H3O1+(aq) + ClO31-(aq)

b. HOCl(aq) + H2O(l) <===> H3O1+(aq) + OCl1-(aq)

5. a. Ka =

[H3O1+ ] [NO21− ]

[HNO2 ]b. Ka =

[H3O1+ ] [CN1− ]

[HCN]

6. a. RbOH(s) ----> Rb1+(aq) + OH1-(aq)

b. Ba(OH)2(s) ----> Ba2+(aq) + 2 OH1-(aq)

7. a. BrO1-(aq) + HOH(l) <===> HBrO(aq) + OH1-(aq)

b. S2-(aq) + HOH(l) <===> HS1- + OH1-(aq)

8. a. HBr(aq) + H2O(l) ----> H3O1+(aq) + Br1-(aq)

b. LiOH(s) ----> Li1+(aq) + OH1-(aq)

c. HNO2(aq) + HOH(l) <===> NO21-(aq) + H3O1+(aq) d. CN1-(aq) + HOH(l) <===> HCN(aq) + OH1-(aq

9. a. acid, base, acid, baseb. base, acid, acid, basec. acid, base, base, acid

10 . a. Kb =

[HCN] [OH1- ]

[CN1- ]b. Kb =

[NH41+ ] [OH1- ]

[NH3 ]

21-21 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


"ALICE"

CHAPTER 22

ACID – BASEREACTIONS

Acid–Base EquationsThe pH Scale

Chemical IndicatorsAcid–Base Titrations

22-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]


22-2 ©1997, A.J. Girondi

SECTION 22.1 Acid – Base Reactions

When a water solution of an acid reacts with a water solution of a base, the products are generallywater and a salt compound. This type of reaction is called neutralization:

ACID + BASE ----> WATER + A SALT

HCl + NaOH ----> HOH + NaCl acid base water salt

(Salts are compounds consisting of a metal combined with a nonmetal. NaCl is only one of many salts. The particular

salt which forms in a neutralization reaction depends on which acid and base are used.)

Problem 1. Complete the formula equations below which illustrate acid–base "neutralization" reactions.The hydrogen in the acids combines with the hydroxide in the base to form water in every case. The otherproduct is a salt. First, use oxidation numbers to write the correct formulas for the salts. Then, balanceeach equation.

a. _____HCl(aq) + _____LiOH(aq) ----> __________________________________

b. _____HNO3(aq) + _____KOH(aq) ----> __________________________________

c. _____HBr(aq) + _____NaOH(aq) ----> __________________________________

When HCl and NaOH are placed in water they dissociate into ions as shown below.

HCl(g) ----> H1+(aq) + Cl1-(aq)

NaOH(s) ----> Na1+(aq) + OH1-(aq)

When the two solutions are mixed, the positive hydrogen ions and negative hydroxide ions then attracteach other, and water is formed:

H1+(aq) + Cl1-(aq) + Na1+(aq) + OH1-(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq)

Notice that in the equation above, the salt, NaCl, does not actually form. This is because NaCl is soluble inwater, so it remains dissociated as aqueous sodium and chloride ions. (Soluble salts are dissociated whenin solution.) If the water in the system were evaporated, then the Na1+ and Cl1- ions would combine toform solid salt, NaCl.

The equation above in which dissociated reactants and products are shown as a mix of individualions is called the ionic equation. Notice that the sodium and chloride ions appear on both sides of theequation in identical form. They are not reacting in any way, so we call them spectator ions. If we dropthem out of the equation, the particles that remain are called the reacting species. That ionic equation isrepeated below. Draw a slash through the "spectator ions" on both sides of this equation now.

22-3 ©1997, A.J. Girondi

Ionic Equation:

H1+(aq) + Cl1-(aq) + Na1+(aq) + OH1-(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq)

An equation that includes only the reacting species is called a net ionic equation. Write the net ionicequation for this reaction below:

Net ionic equation: {1}_____________________________________________________________

By using the appropriate subscripts to indicate phases and complete formulas for the compounds, the"formula" equation can also be written as a traditional double replacement equation:

HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)

Let's review the three methods for writing the equation for the reaction between HCl and NaOH:

Formula Equation: HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)

Ionic Equation: H1+(aq) + Cl1-(aq) + Na1+(aq) + OH1-(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq)

Net Ionic Equation: H1+(aq) + OH1-(aq) ----> HOH(l)

Following the examples given in this chapter, write the balanced formula, ionic, and net ionicequations for the strong acid–strong base combinations in problems 2 and 3. Use oxidation numbers towrite the correct formulas for the products. Be sure that each equation is balanced. (If no spectator ionsare present, the ionic and net ionic equations are identical.)

Problem 2. Complete the equations for the reaction between HBr and KOH.

a. formula: _____HBr(aq) + _____KOH(aq) ---> _________________________

b. ionic: _______________________________________________________

c. net ionic: _______________________________________________________

What are the spectator ions in the reaction in problem 2 above? {2}________________________

Problem 3. Complete the equations for the reaction between HI and LiOH.

formula: _____HI(aq) + _____LiOH(aq) ----> __________________________

ionic: _______________________________________________________

net ionic: _______________________________________________________

Some acids contain more than one hydrogen "acidic" H and some bases contain more than onehydroxide ion. More than one molecule of water is formed in reactions between such acids and bases.You will gain some practice with such acids and bases in problem 4 below.

22-4 ©1997, A.J. Girondi

Problem 4. Complete the formula equations below which illustrate acid–base "neutralization" reactions.Be sure to balance each equation after you have written the correct formulas for the products (usingoxidation numbers). The first one has been completed for you as an example:

a. _____H2SO4(aq) + _____Ca(OH)2(aq) ----> ______2 HOH(l) + CaSO4(aq)______

b. _____H3PO4(aq) + _____Ba(OH)2(aq) ----> _____________________________

c. _____H2CO3(aq) + _____CsOH(aq) ----> _____________________________

d. _____HClO4(aq) + _____Sr(OH)2(aq) ----> _____________________________

The three forms of equations (formula, ionic, and net ionic) can be written for reactions of acidswhich have more than one acidic H and for bases which contain more than one OH1- ion. For example, inpart a of problem 4 we wrote the formula equation for the reaction between H2SO4 and Ca(OH)2. The ionicequation for that reaction is written as:

2 H1+(aq) + SO42-(aq) + Ca2+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + Ca2+(aq) +SO42-(aq)

If we drop out the spectator ions, the result is the net ionic form of the equation:

2 H1+(aq) + 2 OH1-(aq) ----> 2 HOH(l)

Since a balanced equation should always have the lowest ratio of coefficients, we reduce the above to:

H1+(aq) + OH1-(aq) ----> HOH(l)

Problem 5. Write the three forms of the equation for the reaction between H2SO4 and LiOH:

a. formula: ___________________________________________________

b. ionic: ___________________________________________________

c. net ionic: ___________________________________________________

Problem 6. Write the three forms of the equation for the reaction between H3PO4 and NaOH:

a. formula: ___________________________________________________

b. ionic: ___________________________________________________

c. net ionic: ___________________________________________________

Problem 7. Write the three forms of the equation for the reaction between HBr and Ba(OH)2:

a. formula: ___________________________________________________

b. ionic: ___________________________________________________

c. net ionic: ___________________________________________________

22-5 ©1997, A.J. Girondi

ACTIVITY 22.2 The Production of a Salt by an Acid–Base Reaction

In this activity you will be producing solid NaCl by reacting an HCl solution with a solution of NaOH.Follow the procedure carefully.

The reaction equation is: HCl(aq) + NaOH(aq) ----> NaCl(aq) + HOH(l)

Procedure:

1. Weigh a clean dry evaporating dish and watch glass together to the nearest 0.01 g. Record the mass inTable 22.1.

2. Obtain 10.0 mL of 1M HCl solution and pour it into the evaporating dish.

3. Obtain 10.0 mL of 1M NaOH solution and carefully add it to the HCl solution in the evaporating dish. Caution: keep NaOH solution off of skin and clothes.

4. Cover the dish with the watch glass and, using a laboratory burner, heat the contents of the evaporating dish to a boiling. Boil the mixture gently until it is dry.

5. When the dish has cooled, weigh the dish with contents (NaCl) and watch glass to the nearest 0.01 g.

Table 22.1Production of Salt

1. Mass of dish and watch glass _________g

2. Mass of dish, watch glass, and NaCl _________g

3. Mass of NaCl formed (2 - 1) _________g

Calculations:

1. Starting with 10.0 mL of 1.0 M NaOH solution, calculate the theoretical amount of NaCl which shouldhave been produced in this reaction. Do this by finishing the partially completed set-up below.

10.0 mL NaOH X 1 L NaOH

1000 mL NaOH X

1 L NaOH X

1 mole NaOH X

g NaCl

1 mole NaCl

= _______ g NaCl

2. Using the theoretical mass of NaCl above as the accepted value, and the mass of NaCl produced in your experiment as the observed value, calculate your percentage error.

Error = __________%

22-6 ©1997, A.J. Girondi

SECTION 22.3 The Dissociation Constant Of Water, Kw

Pure water is capable of conducting a very small amount of electrical current. It must, therefore,contain a very small concentration of ions. Pure water contains ions because polar water molecules canreact with themselves! Under the proper conditions, it is possible for the partially negatively charged(oxygen) end of one water molecule to pull a hydrogen ion away from another water molecule:

H2O + H2O <====> H3O1+ + OH1-

The result of this reaction is the formation of a hydronium ion and a hydroxide ion. The reversereaction is very good, since H3O1+ is an excellent acid and OH1- is an excellent base. So an equilibrium isestablished, with far more reactants than products present. In fact, only about two water molecules in onebillion are in the form of ions at any point in time! The equation for the reaction is:

H2O(l) + H2O(l) <===> H3O1+(aq) + OH1-(aq)

Since there is an equilibrium, we can write an equilibrium expression. There will be no denominator, sincethe reactant, H2O, is a pure liquid: Kw = [H3O1+] [OH1-]

The equilibrium constant for this system is given the symbol, Kw, and is called the dissociation constant ofwater. Experiments have been done to determine the concentrations of H3O1+ and OH1- in water. It wasfound that in pure water:

[H3O1+] = 1 X 10-7 and [OH1-] = 1 X 10-7

Therefore, the value of Kw is calculated:

Kw = [H3O1+] [OH1-] = (1 X 10-7)2 = 1 X 10-14

Kw = 1 X 10-14

Since water contains some hydrogen ions and some hydroxide ions, all water solutions must alsocontain at least a small concentration of each of these ions. This includes all of the solutions of acids andbases that we discussed in Chapter 21. Water solutions of acids contain both H1+ and OH1- ions, andwater solutions of bases contain both H1+ and OH1- ions. Pure water is considered neutral because theconcentrations of H1+ and OH1- are equal at 1 X 10-7 M. Acid solutions have hydrogen ion concentrationsthat are greater than their hydroxide ion concentrations. Basic solutions have hydroxide ionconcentrations that are greater than their hydrogen ion concentrations. However, the product[H1+][OH1-] for all water solutions must always equal 1 X 10-14.

Water: [H1+] = [OH1-] = 1 X 10-7 M Acids: [H1+] > [OH1-]Bases: [H1+] < [OH1-]

For Acids, Bases, and All Other Water Solutions: [H1+] [OH1-] = 1 X 10-14 (Always!)

22-7 ©1997, A.J. Girondi

Examine Table 22.2 below. It shows the relationship between [H1+], [OH1-], and Kw.

Table 22.2The Relationship Between [H1+], [OH1-], and Kw in Water Solutions

[H3O1+] [OH1-] Kw

Strong Acids -----> 1 X 100 1 X 10-14 1 X 10-14

1 X 10-1 1 X 10-13 1 X 10-14

1 X 10-2 1 X 10-12 1 X 10-14

1 X 10-3 1 X 10-11 1 X 10-14

1 X 10-4 1 X 10-10 1 X 10-14

Weak Acids ----> 1 X 10-5 1 X 10-9 1 X 10-14

1 X 10-6 1 X 10-8 1 X 10-14

Neutral -----> 1 X 10-7 1 X 10-7 1 X 10-14

1 X 10-8 1 X 10-6 1 X 10-14

Weak Bases ----> 1 X 10-9 1 X 10-5 1 X 10-14

1 X 10-10 1 X 10-4 1 X 10-14

1 X 10-11 1 X 10-3 1 X 10-14

1 X 10-12 1 X 10-2 1 X 10-14

1 X 10-13 1 X 10-1 1 X 10-14

Strong Bases ----> 1 X 10-14 1 X 100 1 X 10-14

Now let's use this information and the concept of Kw to solve some problems. We'll begin with anexample:

Sample Problem: For a particular solution, [OH1-] = 1.0 X 10-5. Calculate [H1+] for this solution, andidentify it as acidic, basic, or neutral.

You are given [OH1-] and asked to find [H1+]. An equation that includes both of these quantities is theexpression for Kw: Kw = [H1+] [OH1-]. Substitute the given value for [OH1-] and for Kw into the equation,and solve for [H1+]:

Kw = [H1+][OH1- ] [H1+ ] = Kw

[OH1- ][H1+ ] =

1 X 10 -14

1 X 10 -5

thus, [H1+ ] = 1 X 10-9

We have found that [H1+] = 1 X 10-9, while [OH1-] = 1 X 10-5 for this solution. Since [OH1-] > [H1+], thesolution is basic.

Table 22.2 reveals that as [H1+] gets larger, [OH1-] gets smaller. Kw remains constant. Now try theproblems below. Be sure to enter exponential numbers into your calculator in the correct way, using theexponent key, either [EXP] or [EE]. Show your work.

22-8 ©1997, A.J. Girondi

Problem 8. A solution has [H1+] = 1.0 X 10-4. Calculate [OH1-] for this solution, and identify it as acidic,basic, or neutral. Show work.

[OH1-] = _____________ Solution is _____________

Problem 9. A solution has [OH1-] = 4.8 X 10-6. Calculate [H1+] for this solution, and identify it as acidic,basic, or neutral. Show work.

[H1+] = _____________ Solution is ______________

Problem 10. A solution has [H1+] = 8.3 X 10-2. Calculate [OH1-] for this solution, and identify it as acidic,basic, or neutral. Show work.

[OH1-] = ____________ Solution is ______________

SECTION 22.4 The pH Scale

As you can see in the problems above, in chemistry we often work with small concentrations ofH1+ and OH1- ions. The quantities in the measurements are usually negative exponential numbers.These are not fun to work with! So, in order to characterize a solution as acidic, basic, or neutral withouthaving to use negative exponential numbers, scientists devised what is known as the pH scale. The smallletter "p" in chemistry is frequently used as an abbreviation meaning "negative logarithm of." The H standsfor hydrogen ion concentration: [H1+]. Thus, pH is a shorthand way of saying "the negative logarithm ofthe hydrogen ion concentration of a solution." Or, pH = - log [H1+]. Now you must be wondering why thisconcept was ever devised! Well, follow along with this explanation.

Logarithm is simply another word for exponent. Common logarithms are powers of 10. Thelogarithm of 100 is 2. Why? Because 100 = 102. If someone asks you, "What is the log of 100?" Whatthey are really asking is, "To what power must you raise 10 to get 100?" The answer, of course, is 2. It iseasy to find "logs" using a scientific calculator. Simply enter the number and press the log key.

22-9 ©1997, A.J. Girondi

Problem 11. Find the logs of the numbers listed below.

a. log 1000 = ______________ d. log 450 = ______________

b. log 10 = ________ _____ e. log 8932 = __ ___________

c. log 10000 = ______________ f. log 4.5 X 103 = ______________

To calculate the negative logarithm of a number, follow the procedure for finding the log on yourcalculator, and then change the sign of your answer using the change sign key (+/-).

Problem 12. Do the problems below.

a. - log 679 = ________________ c. - log 0.034 = _________________

b. - log (4.60 X 102)= ________________ d. - log (8.2 X 10-3) = _________________

Let's find the negative log of [H1+] for water: - log(1 X 10-7) = 7. You can see that by finding the negativelog of a negative exponential number, you end up with a positive integer. It is much easier to work withpositive integers than with negative exponents. So, scientists decided that it would be better to describethe acidic or basic nature of solutions using positive integers than by using the actual values of [H1+] or[OH1-] for the solution. Instead of talking about the hydrogen ion concentration of a solution, we can talkabout its pH. If we take the values in Table 22.2, and find the negative logs of them, we can express themall as positive integers. [H1+] becomes pH, [OH1-] becomes pOH, and Kw becomes pKw. This hasbeen done for you in Table 22.3.

Table 22.3The Relationship Between pH, pOH,

and pKw in Water Solutions

pH pOH pKw

Strong Acids ---> 0 14 141 13 142 12 143 11 144 10 14

Weak Acids ---> 5 9 146 8 14

Neutral Solutions ---> 7 7 148 6 14

Weak Bases ---> 9 5 1410 4 1411 3 1412 2 1413 1 14

Strong Bases ---> 14 0 14

Notice that the sum of pH and pOH of a solution is always equal to 14

22-10 ©1997, A.J. Girondi

Problem 13. Find the pH of solutions with the following [H1+]:

a. 4.60 X 10-3 __________________ c. 0.00680 __________________

b. 8.80 X 10-9 __________________ d. 3.40 X 10-7 __________________

To do the next two problems, you will need to use the formulas for Kw and for pH:

Problem 14. What is the pH of a solution with [OH1-] = 5.20 X 10-5?

pH = __________

Problem 15. What is the pOH of a solution with [H1+] = 1.20 X 10-10?

pOH = __________

As pH {3} creases, the strength of an acid solution decreases. Acids get stronger as their [H1+]

{4} creases. A neutral solution has a pH of {5}___ _ . The stronger the acid, the {6}____________

the pH. The stronger the base, the {7}_______________ the pH.

<--- stronger acids stronger bases --->0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Figure 22.1 The pH Scale

The pH scale was first devised by the Danish biochemist, Sven Sorenson in 1909. It is an open-ended scale which means that it is possible for solutions to have pH values which are less than 0 or greaterthan 14. However, the vast majority of solutions fall into this range. The pH scale of 0 to 14 includes anextremely large range of [H1+] concentrations. Remember that pH values are logs, which means they arepowers of ten. When pH = 1, [H1+] = 1 X 10-1. When pH = 7, [H1+] = 1 X 10-7. This difference between pH= 1 and pH = 7 represents a difference of 6 powers of 10 which is 1,000,000! Let's try the problem below.

A solution with pH = 2 has how many times more H1+ ions per liter than one with pH = 7? {8}___

How many times more hydrogen ions are in a solution of pH 1 compared to a solution of pH 12?

Answer: {9}______________________________________

Another type of problem which is a little trickier than what you have done involves finding [H1+] for a solution when you know its pH. For example, if the pH of a solution is 8.5, what is its [H1+]? The correct answer is 3.2 X 10-9 M. This kind of problem involves the use of inverse log which is also known as antilog. Problems of this type are discussed in Appendix F of your ALICE materials. Ask your teacher if he/she

22-11 ©1997, A.J. Girondi

wishes to include this material in this chapter. If so, read Appendix F and complete the problems there. Otherwise, move on to Activity 22.3.

GO TO APPENDIX F -----> ??? (CHECK WITH YOUR INSTRUCTOR.)

ACTIVITY 22.5 Acid-Base Indicator Solutions

There are several ways to determine the pH of solutions. One method involves the use ofchemical indicators. These are substances which change color when added to acidic or basic solutions.Bromthymol blue is an example. Many of these indicators are actually very, very weak acids which are plantpigments. Let's abbreviate the chemical formula of bromthymol blue as "HIn." Since it is a very weak acid,it exists in water in equilibrium:

HIn(aq) + H2O(l) <===> H3O1+(aq) + In1-(aq) yellow blue

If we add few drops of this indicator solution to an acid (which has an excess of H3O1+), the H3O1+ ions in

the acid would push the equilibrium to the (left / right) {10}_____________. This would cause the color to

shift more to (yellow / blue) {11} . If we add some bromthymol blue to a basic solution (which

has an excess of OH1- ions), the OH1- ions in the base will react with the H3O1+ in the indicator, producing

water. The result is a lower [H3O1+], which will shift the bromthymol blue equilibrium to the (left/right)

{12}_____________. The color would then shift more to (yellow/blue) {13} . Notice that to

serve as an indicator, the molecule (HIn) form of the weak acid must have a different color than the anion

form (In1-).

Bromthymol blue has a pH interval of 6.0 to 8.0. This means that it is yellow in a solution with a pHof less than 6.0, while it is blue in a solution with a pH above 8.0. Between these two pH values theindicator is changing its color. So, if you add bromthymol blue to a solution and the result is a yellow color,you can predict that the pH is less than 6.0. If it turns blue, you would conclude that the pH is above 8.0. Ifit turns green, then the pH is somewhere in the middle of the pH interval, around 7. There are many acid-base indicators, and they have different pH intervals. A few are listed in Table 22.3.

Table 22.4A Selection of Acid-Base Indicators

Indicator Acid Color Base Color pH Interval

bromthymol blue yellow blue 6.0 - 8.0litmus red blue 5.5 - 8.0methyl orange red yellow 3.1 - 4.4methyl violet yellow blue 0.0 - 1.6phenolphthalein colorless pink 8.2 - 10.0alizarin yellow yellow red 10.1 - 12.0

In this activity you will use several acid-base indicators in an attempt to estimate the pH of someselected solutions.

1. Obtain a dropper bottle of any solution with a pH < 3.1 and a bottle of any solution with a pH > 4.4. (Theexact pH values are not important.)

22-12 ©1997, A.J. Girondi

2. Place ten drops of these solutions into separate wells of a dropping plate. Add a drop of methyl orangeindicator to each solution. Note the colors. Do not discard these mixtures.

3. Using two clean wells of the same dropping plate, repeat the procedure above using any two solutionswith pH < 6.0 and pH > 8.0, and adding a drop of bromthymol blue to each. Note the colors. Do notdiscard the mixtures.

4. Repeat the procedure once more using any two solutions with pH < 8.2 and pH > 10.0, and addingphenolphthalein indicator solution. Note the colors. Do not discard the mixtures. Compare your results tothe colors and pH intervals listed in Table 22.4.

5. Add ten drops of white vinegar to each of three clean wells of a second dropping plate. Add a drop ofmethyl orange to one of the wells of vinegar, a drop of bromthymol blue to the second, and a drop ofphenolphthalein to the third. Compare the colors to those of the indicators in the solutions of known pH.Record your observations in Table 22.5 below, and estimate the pH of the vinegar. Your estimates mightbe something like these: < 4.4; > 8.2; or maybe, < 2.3.

6. Repeat the procedure in step 5 with the following solutions: household ammonia, a solution oflaboratory detergent, a colorless soft drink solution (like 7-up), and tap water. Record all observations andpH estimates in Table 22.5 below.

Table 22.5pH of Common Household Substances

Color in Color in Color inSubstance Methyl Orange Bromthymol Blue Phenolphthalein pH

vinegar ___________ _____________ _____________ ______

ammonia ___________

detergent ___________

soft drink ___________

A more accurate method for determining pH involves the use of an instrument called a pH meter.It is a sensitive device that must be used with care, and the proper procedure must be followed. Yourteacher will assist you in the use of the pH meter as you measure the pH of each of the substances inTable 22.5 above. Record your results below:

Vinegar: __________ Ammonia: __________ Detergent: __________ Soft Drink: __________

Do all of the estimates of pH which you made using the indicators agree with the more specific results

obtained using the pH meter? If not, which ones did not agree?_____________________

. They all should be in agreement, unless there was

some error in the procedure or equipment.

22-13 ©1997, A.J. Girondi

SECTION 22.6 Acid–Base Titrations

At this point we are ready to determine the outcome when acidic and basic solutions are addedtogether. Earlier in this chapter you completed an exercise in which reactions between acids and basesproduced salts and water. These were examples of neutralization reactions. Some degree ofneutralization occurs whenever an acid is mixed with a base. Water is always one of the products formed insuch a reaction. Consider the neutralization reaction that occurs when HCl and NaOH solutions are mixed.One way that we wrote this reaction was: HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq) According to thisequation, 1 mole of HCl is required to neutralize 1 mole of NaOH. Laboratory experiments often involveneutralization reactions. Suppose you knew the molar concentration (M) of one of the solutions (the acidor the base) and you wanted to determine the concentration of the other solution. You could determinethe unknown concentration by mixing the acid and base together until the neutralization was complete.Then, the unknown concentration can be calculated using the volumes of the acid and base solutionswhich were consumed. Such an experimental procedure is called a titration. Let's try some examples:

Sample Problem: What is the molarity of a solution of HCl if 48 mL of 0.25 M NaOH solution arerequired to neutralize 35 mL of HCl?

HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)

We will solve this problem by setting up a "fencepost" and by using dimensional (unit) analysis. Since weare looking for the molarity of HCl, we will want to be left with units of moles HCl divided by liters of HClsolution: moles HCl / L sol'n. Although we could start the fencepost with any of the information given,let's go with 48 mL NaOH:

48 mL NaOH X 1 L NaOH

1000 mL NaOH X

0.25 mole NaOH

1 L NaOH X

1 mole HCl

1 mole NaOH X

1

35 mL HCl X

1000 mL HCl

1 L HCl

= 0.34 mole HCl

1 L HCl or, 0.34 M HCl

Sample Problem: What is the molarity (M) of a solution of KOH if 45 mL of 0.20 M H2SO4 are required toneutralize 34 mL of KOH solution? H2SO4 + 2 KOH ----> 2 HOH + K2SO4

45mL H 2SO4 X 1 L H2SO4

1000 mL H 2SO4 X

0.20 mole H 2SO4

1 L H2SO4 X

2 moles KOH

1 mole H2SO4 X

1

34 mL KOH ...

X 1000 mL KOH

1 L KOH =

0.53 mole KOH

1 L KOH or, 0.53 M KOH

Sample Problem: A beaker contains 0.11 L of 0.35 M Ca(OH)2 solution. What volume (in L) of 0.50 MH3PO4 solution will be required to neutralize it? 2 H3PO4 + 3 Ca(OH)2 ----> 6 HOH + Ca3(PO4)2

0.11 L Ca(OH)2 X

0.35 mole Ca(OH)2

1 L Ca(OH)2 X

2 moles H3PO4

3 moles Ca(OH)2 X

1 L H3PO4

0.50 mole H 3PO4 = 0.051 L H 3PO4

Study the three sample problems above carefully, and then do the problems which follow. Show your set-up in each case. All measurements should have units. Complete a balanced equation before startingeach problem.

22-14 ©1997, A.J. Girondi

Problem 16. If 24.60 mL of 0.18 M HNO3 is titrated with 22 mL of KOH solution. What is the

concentration (molarity) of the base? HNO3 + KOH ---> _________ _______

__________ M KOH

Problem 17. If 40.0 mL of H3PO4 are neutralized by 22 mL of 0.60 M NaOH, What is the concentration

of the acid? H3PO4 + 3 NaOH ---> __________________________

_________ M H3PO4

Problem 18. How many mL of 2.0 M H2SO4 will be needed to neutralize 40.0 mL of 0.80 M Ca(OH)2?

H2SO4 + Ca(OH)2 ---> CaSO4 + ______________________

__________ mL H2SO4

Problem 19. If 16 mL of 0.15 M H3PO4 are needed to neutralize a 0.10 M solution of Mg(OH)2, what is

the volume (in mL) of the base solution? 2 H3PO4 + 3 Mg(OH)2 ---> ________________________

__________ mL Mg(OH)2

22-15 ©1997, A.J. Girondi

ACTIVITY 22.7 A Titration of Vinegar

You are now going to put what you have learned topractical use. You are going to determine the percentage ofacetic acid in vinegar. You will perform a titration by adding abase to the acetic acid until neutralization occurs. One of themost important parts of a titration, is knowing when the acid andbase have been completely neutralized. We will get thisinformation by using phenolphthalein indicator solution.

When doing titrations, volumes of solutions must bevery carefully measured. A graduated cylinder is normally usedto measure volume. But, in titrations, we use a piece ofglassware that is even more accurate. It is called a buret (alsospelled burette). The volume of solution in the buret shouldbe read the same way that you read a graduated cylinder.Read the bottom of the meniscus.

Get the materials labeled 22.7 from the materials shelf.Follow the procedure below. Wear safety glasses and anapron.

Figure 22.2Reading a Meniscus

Procedure:

1. Rinse two burets with some distilled water. Mount the clean burets on a ringstand with a double buret holder as shown in Figure 22.3.

2. Place funnels in the mouths of the burets, and fill one of the burets withwater. Practice using the burets by measuring out volumes of 5 and 10 mL.Drain any remaining water.

3. Pour a few mL of 0.50 M NaOH into one buret to rinse it. Drain the solutionout, and then fill that buret with the 0.5 M NaOH solution. Remove the air in thetip of the buret below the stopcock by draining some of the solution. Recordthe starting level and the concentration of NaOH (0.50 M) in Table 22.6.

4. For the other buret, follow the procedure in step 3 but use white vinegarinstead of NaOH solution.

5. Carefully measure three 10 mL volumes of vinegar from the buret into three125 mL Erlenmeyer flasks. Label your flasks 1, 2, and 3. Add 5 to 10 drops ofphenolphthalein indicator to each flask.

Figure 22.3 Burets and Holder

6. Place one flask on a small piece of white paper under the buret containing NaOH, and add a few mL ofthe NaOH solution from the buret while you swirl the solution in the flask. The pink color of the indicatormay appear briefly, but swirl the solution until the color disappears. (Use a magnetic stirrer if one isavailable in your lab. It will make stirring easier.)

7. Continue to add NaOH solution to the flask, dropwise, with thorough mixing. We will use this firsttitration to get a rough idea of about how much NaOH is required to neutralize the vinegar. Therefore, youcan allow the drop rate to go rather rapidly. When you reach the point where a slight pink color persists(does not disappear), you have reached the end point of the titration. Record the level of base left in theburet in Table 22.6.

22-16 ©1997, A.J. Girondi

Table 22.6Titration of Vinegar

Sample 1 Sample 2 Sample 3

1. Initial buret reading (mL)

2. Final buret reading (mL)

3. mL of base added

4. Volume of vinegar (mL)

5. Concentration (M) of NaOH 0.50 0.50 0.50

8. Repeat the titration for the other two samples of vinegar. In each case, fill the NaOH buret, record theinitial level, run the titration, and record the final level of base. The second two titrations should go morequickly, because you already know the approximate final level of base in the buret. Run an amount ofbase into the second flask that is 1 or 2 mL less than that required in the first titration. Then complete thetitration by carefully adding the last amount of base dropwise with good mixing. A good end point willleave the mixture in the flask with only a faint pink color.

9. Rinse both burets with water when you are finished. Record all data in Table 22.6.

Calculations:

1. Calculate the concentration (M) of acetic acid (HC2H3O2) in the vinegar. Follow the examples in section22.6. (For the volume of NaOH, use an average of the two closest volumes from your 3 titrations.)

HC2H3O2 + NaOH ---> HOH + NaC2H3O2

__________ M

2. Using atomic masses from the periodic table, calculate the molecular mass of acetic acid, HC2H3O2.

__________ grams / mole

3. Calculate the mass in grams of acetic acid in a liter of vinegar. (Your answer to calculation one is in molesper liter. Change that value to grams per liter using the molecular mass you found in calculation 2.)

__________ g HC2H3O2 / L

22-17 ©1997, A.J. Girondi

4. Assume that a liter of vinegar has a mass of 1000. grams. Calculate the mass percent of acetic acid invinegar:

mass of acid per litermass of 1 liter

X 100 = % acetic acid in vinegar

__________% HC2H3O2

5. Check the bottle of vinegar from which your samples came. What is the percent of acid in the vinegar

according to the label? _____________ %

6. Comment on how your result compares to that on the label of the vinegar bottle.

7. Calculate your percent error. The formula is in the reference section of your ALICE materials.

__________ % Error


The following problems are optional, and can be done on a separate sheet of paper if you feel thatyou need additional practice.

Problem 20. Write and balance the three forms of the equation for the reaction between HClO3(aq) andCa(OH)2(aq).

Problem 21. If the [H3O1+] of a solution is 8.6 X 10-12, find the [OH1-] of the same solution.

Problem 22. If the [H3O1+] of a solution is 5.2 X 10-9, calculate the pH and pOH and identify the solutionas an acid or a base.

Problem 23. If the [OH1-] of a solution is 3.5 X 10-4, calculate the pH and pOH and identify the solutionas an acid or a base.

Problem 24. What volume (in mL) of a 0.22M HCl solution are required to neutralize 64 mL of a 0.12Msolution of Ba(OH)2?

Problem 25. During a titration procedure, it is found that 38 mL of a 2.5M solution of KOH are neededto neutralize 57 mL of a solution of H2SO4. What is the molarity (M) of the H2SO4 solution?

22-18 ©1997, A.J. Girondi


Understanding acid-base chemistry is very important, because many chemical reactions involve acidsand bases. Check the learning outcomes below. When you feel you have mastered them, arrange to takeany quizzes or exams on Chapter 22. Then, move on to Chapter 23.

_____1. Given an acid and a base, properly complete and balance the equation for the reaction that occurs between them in the formula, ionic, and net ionic forms.

_____2. Explain pH and classify a solution as acid, base, or neutral given its pH, pOH, [H1+], or [OH1-].

_____3. Describe the relationship between the conductivity of an acid solution, the size of its Ka, and its pH.

_____4. Given [H1+], calculate [OH1-] and pH; Given [OH1-], calculate pH and [H1+].

_____5. Explain the proper procedure for conducting a titration.

_____6. Calculate concentrations or volumes in titration problems.

The following learning outcome pertains to Appendix F:

_____7. Determine the [H1+] or [OH1-] of a solution, given its pH or pOH.

22-19 ©1997, A.J. Girondi


Questions:

{1} H1+(aq) + OH1-(aq) ----> HOH(l) ; {2} Br1- and K1+; {3} in; {4} de; {5} 7.0; {6} lower; {7} higher;

{8} 1 X 105; {9} 1 X 1011; {10} left; {11} yellow; {12} right; {13} blue

Problems:1. a. HCl + LiOH ---> HOH + LiCl

b. HNO3 + KOH ---> HOH + KNO3

c. HBr + NaOH ---> HOH + NaBr2. a. HBr + KOH ---> HOH + KBr

b. H1+(aq) + Br1-(aq) + K1+(aq) + OH1-(aq) ---> HOH(l) + Br1-(aq) + K1+(aq)

c. H1+(aq) + OH1-(aq) ---> HOH(l)

3. a. HI(aq) + LiOH(aq) ---> HOH(l) + LiI(aq)

b. H1+(aq) + I1-(aq) + Li1+(aq) + OH1-(aq) ---> HOH(l) + I1-(aq) + Li1+(aq)

c. H1+(aq) + OH1-(aq) ---> HOH(l)

4. a. H2SO4(aq) + Ca(OH)2(aq) ----> 2 HOH(l) + CaSO4(aq)

b. 2 H3PO4(aq) + 3 Ba(OH)2(aq) ----> 6 HOH(l) + Ba3(PO4)2(s)

c. H2CO3(aq) + 2 CsOH(aq) ----> 2 HOH(l) + Cs2CO3(aq)

d. 2 HClO4(aq) + Sr(OH)2(aq) ----> 2 HOH(l) + Sr(ClO4)2(aq)

5. a. H2SO4(aq) + 2 LiOH(aq) ----> 2 HOH(l) + Li2SO4(aq)

b. 2 H1+(aq) + SO42-(aq) + 2 Li1+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + 2 Li1+(aq) + SO42-(aq)

c. 2 H1+(aq) + 2 OH1-(aq) ---> 2 HOH(l) which reduces to H1+(aq) + OH1-(aq) ---> HOH(l)

6. a. H3PO4(aq) + 3 NaOH(aq) ----> 3 HOH(l) + Na3PO4(aq)

b. 3 H1+(aq) + PO43-(aq) + 3 Na1+(aq) + 3 OH1-(aq) ----> 3 HOH(l) + PO43-(aq) + 3 Na1+(aq)


7. a. 2 HBr(aq) + Ba(OH)2(aq) ----> 2 HOH(l) + BaBr2(aq)

b. 2 H1+(aq) + 2 Br1-(aq) + Ba2+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + 2 Br1-(aq) + Ba2+(aq)


8. 1.0 X 10-10; acid9. 2.1 X 10-9; base10. 1.2 X 10-13; acid11. a. 3; b. 1; c. 4; d. 2.7; e. 3.951; f. 3.712. a. -2.83; b. -2.66; c. 1.5; d. 2.113. a. 2.34; b. 8.06; c. 2.17; d. 6.4714. pH = 9.7215. pOH = 4.0816. 0.20 M KOH17. 0.11 M H3PO4

18. 16 mL H2SO4

19. 36 mL Mg(OH)220. 2 HClO3(aq) + Ca(OH)2(aq) ----> 2 HOH(l) + Ca(ClO3)2(aq)

2 H1+(aq) + 2 ClO31-(aq) + Ca2+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + Ca2+(aq) + 2 ClO31-(aq)

2 H1+(aq) + 2 OH1-(aq) ----> 2 HOH(l) which reduces to H1+(aq) + OH1-(aq) ----> HOH(l)

21. 1.16 X 10-3

22. pH = 8.28; pOH = 5.72; base23. pH = 10.5; pOH = 3.46; base24. 69.8 mL HCl25. 0.83M H2SO4

22-20 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


CHAPTER 23

OXIDATIONAND

REDUCTION(Part 1)

23-1 ©1997, A.J. Girondi

SECTION 23.1 Introduction To Oxidation–Reduction Reactions

In earlier chapters, we studied chemical reactions labeling them as combination, decomposition,single replacement, and double replacement. In this chapter, we will study yet another group calledoxidation-reduction reactions. We can greatly simplify the classification of reactions by grouping them allinto two broad classes. These two classes are: (1) reactions in which there is no electron transfer from onesubstance to another, and (2) reactions in which electrons transfer from one substance to another. Allchemical reactions fit into one or the other of these two categories.

In previous chapters we made use of oxidation numbers, but we did not define them. Now it istime to do so.

An oxidation number is a signed number which is assigned to an atom or ionaccording to a set of rules. It represents the "oxidation state" of the atom or ion.

The "oxidation state" of an atom or ion changes when it loses or gains electrons. You have previouslyused oxidation numbers to write the correct formulas for compounds and polyatomic ions. You may recallthat there is a table of common oxidation numbers in the reference section of your ALICE materials. Therules used to assign oxidation numbers to atoms and ions are listed below. You should know them wellenough to use them from memory during tests and quizzes.

Rules for Assigning Oxidation Numbers

1. The oxidation number of an atom of a free element is zero. Elements are free if they are not combinedwith other elements. If atoms of an element are combined with themselves, they are still considered to befree. For example, a free atom of Ag has an oxidation number of zero. In addition, atoms in molecules likeH2, Cl2, N2, O2, F2, Br2, I2, P4, S8, etc., have oxidation numbers of zero.

2. The oxidation number of a monatomic ion is equal to its charge. A monatomic ion is one that formedfrom only one atom. Ex: Ag1+

3. The algebraic sum of the oxidation numbers of the atoms in the formula of a compound is zero.

4. In compounds, the oxidation number of hydrogen is +1. (There is one exception. In compounds knownas hydrides, it can be -1. Sodium hydride is NaH.)

5. In compounds, the oxidation number of oxygen is -2. (Exceptions: oxygen is -1 in peroxidecompounds like H2O2, and +2 when it combines with fluorine in OF2.)

6. In combinations of nonmetal atoms, the oxidation number of the less electronegative element ispositive and of the more electronegative element is negative. For example, in NO2, N = +4, and O = -2.The element with the positive oxidation number is written first in the formula of a compound, such as inNO2. An exception is ammonia, NH3, in which nitrogen (-3) is written before hydrogen (+1).7. The algebraic sum of the oxidation numbers of the atoms in the formula of a polyatomic ion is equal tothe charge on the polyatomic ion. Example: In Cr2O72-, each chromium atom is +6, while each oxygenatom is -2. Note that (+6 X 2) + (-2 X 7) = -2.

Because of your work in previous chapters, you are already somewhat familiar with rules 3 through 7.Now, rules 1 and 2 will also be needed in order to understand redox reactions.

Reactions in which no electrons are transferred usually involve the separation and rejoining ofatoms or ions. For example, silver nitrate (AgNO3) and sodium chloride (NaCl) exist as ions in solution.

AgNO3(s) ----> Ag1+(aq) + NO31-(aq)

23-3 ©1997, A.J. Girondi

NaCl(s) ----> Na1+(aq) + Cl1-(aq)

When added together, the positive Ag1+ ions combine with the negative Cl1- ions to form a precipitate ofAgCl. The Na1+ and NO31- ions are spectator ions which will combine only if the water is removed from thesystem.

Formula Equation: AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)

Ionic Equation: Ag1+(aq) + NO31-(aq) + Na1+(aq) + Cl1-(aq) ----> AgCl(s) + NO31-(aq) + Na1+(aq)

Net Ionic Equation: Ag1+(aq) + Cl1-(aq) ----> AgCl(s)

Ag has an oxidation number of {1}______ when combined with NO31- and with Cl1-. The oxidation numberremains the same before and after the reaction. When the oxidation number does not change, there hasbeen no transfer of electrons. You can see from the equations above that there has been neither anincrease nor a decrease in oxidation number. This is true for the silver ions and for the other ions involvedin the reaction.

In this chapter, however, we want to concentrate on the second type of reaction in whichelectrons are transferred from one particle to another. Such oxidation-reduction equations are oftencalled "redox" equations for short. You have seen some redox reactions in previous chapters, althoughyou probably did not realize it. For example, single replacement reactions are also redox redox reactions.We will now study some of the "whys" and "hows" of these reactions. Let's begin with some definitions.

Oxidation is defined as the LOSS of electrons by a substance.

Reduction is defined as the GAIN of electrons by a substance.

An example of a typical redox equation is: 2 Mg(s) + O2(g) ----> 2 MgO(s)

The reactants, Mg and O2, are neutral particles. The product, MgO, is made of equal numbers of Mg2+ andO2- ions. These oppositely charged ions attract each other, and the result in an ionically bondedcompound. We can gain a better understanding of the reaction between Mg and O2 by breaking it into twoseparate parts, something that can be done to any redox equation. The separate parts are called half-equations or half-reactions. In the first half-equation below, magnesium is shown losing two electrons(e-). Remember that each electron has a charge of {2}_______. Magnesium is undergoing oxidation.Oxygen, on the other hand, is shown gaining two electrons per atom (4 electrons total) and is, therefore,undergoing reduction.

In oxidation half-equations, the electrons are always written on the product side (right side) toshow that they are being lost:

Oxidation half-equation: Mg ----> Mg2+ + 2 e-

In the half-equation above, each magnesium atom is losing {3}________ electrons.In reduction half-equations, the electrons are written on the reactant side to show that they are beinggained.

Reduction half-equation: O2 + 4 e- ----> 2 O2-

In the half-equation above, a total of {4}_______ electrons are being gained. Since two oxygen atoms areinvolved, each oxygen atom is gaining {5}________ electrons.

23-4 ©1997, A.J. Girondi

In the overall equation, note that the oxidation number of the magnesium atom is changing from 0 to{6}________. Also notice that the oxidation number of each of the two oxygen atoms is changing from 0to {7}________.

2 Mg(s) + O2(g) ----> 2 MgO(s)

0 0 +2 -2

These changes in the oxidation numbers of magnesium and oxygen can be explained by the loss andgain of electrons as shown in the half-equations.

You must learn to easily identify whether half-equations represent oxidations or reductions. Oneway to remember the definitions is to think of electrons going Out in Oxidation, and electrons Returningin Reduction.

Problem 1. Complete the exercises below by labeling each of the half-equations as oxidations or asreductions.

Half-Equations Oxidation or Reduction

a. Cu1+ + e- ---> Cu ______________________

b. F2 + 2e- ---> 2 F1- ______________________

c. Cr3+ ---> Cr6+ + 3e- ______________________

d. Bi3+ ---> Bi5+ + 2e- ______________________

e. N5+ + 2e- ---> N3+ ______________________

f. 2 Cl1- ---> Cl2 + 2e- ______________________

SECTION 23.2 Balancing Redox Half-equations

The total charge on each side of a redox equation must be equal. This is also true for half-equations. Notice that the sum of the oxidation numbers is zero on both sides of this equation takenfrom section 23.1:

2 Mg(s) + O2(g) ----> 2 MgO(s)

0 0 +2 -2

The total charge on each side of a redox half-equation must also be equal. In each of the half-equations shown previously, this is also true. For example:

Cr3+ ----> Cr6+ + 3e-

+3 +6 -3

You can see that the total charge on both sides of the half-reaction is {8}________.

23-5 ©1997, A.J. Girondi

This is an important rule about all redox equations:

In any redox equation or half-equation, the total charge of the reactants mustequal the total charge of the products.

This rule makes it possible for you to determine exactly how many electrons are gained or lost in any half-equation. If you were told that manganese changes from a charge of +7 to +4, you could write a half-equation for this change using the previous examples as your guide. For example, examine the half-equation below: Mn7+ ----> Mn4+

The charges are not balanced (equal) on the reactant and product sides. By adding electrons (negativecharges) to the reactant (left) side, the charges can be balanced:

Mn7+ + 3 e- ----> Mn4+

Each side of the half-equation now has a total charge of {9}__________.

You have already learned that a chemical equation must be balanced according to mass, whichmeans that it must have the same number of atoms or ions on both sides. This is because of the Law ofConservation of Mass which states that matter cannot be created or destroyed in ordinary chemicalreactions. Now you have also learned that charge cannot be created or destroyed, either. This is the Lawof Conservation of Charge.

A redox reaction or half-reaction must be balanced in two ways: first, according to {10}___________, and,

second, according to {11}______________.

Problem 2. Complete the exercises below by adding the correct number of electrons needed tobalance the charges in each half-equation. Then, label each half-equation as an oxidation or reduction.Review the first two rules for assigning oxidation numbers before you begin.

Half-Reactions Oxidation or Reduction

a. Na ---> Na1+ + _____e- ______________________

b. Cl2 + _____e- ---> 2 Cl1- ______________________

c. 2 H1+ + _____e- ---> H2 ______________________

d. 2 Al ---> 2 Al3+ + _____e- ______________________

e. 3 F2 + _____e- ---> 6 F1- ______________________

f. Fe2+ ---> Fe3+ + _____e- ______________________

g. Mn7+ + _____e- ---> Mn2+ ______________________

h. 6 Cl1- ---> 3 Cl2 + _____e- ______________________

i. Cr6+ + _____e- ---> Cr3+ ______________________

j. Cl1- ---> Cl5+ + _____e- ______________________

23-6 ©1997, A.J. Girondi

Problem 3. Write balanced half-equations for each situation presented below. You will need to decideon which side to place the electrons. Make sure your half- equations are balanced according to mass andcharge. Then label each change as oxidation or reduction.

Half-Equation Balanced Half-Equation Ox or Red?

a. Cu2+ ---> Cu1+ __________________________ ____________________

b. I7+ ---> I1+ __________________________ ____________________

c. 3 Br2 ---> 6 Br1- __________________________ ____________________

d. Sn2+ ---> Sn4+ __________________________ ____________________

e. P5+ ---> P3+ __________________________ ____________________

f. Cu2+ ---> Cu __________________________ ____________________

g. 2 Br2 ---> 4 Br1- __________________________ ____________________

h. 6 N3- ---> 3 N2 __________________________ ____________________

SECTION 23.3 Adding Redox Half-Equations Together

Oxidation and reduction are processes that must occur together. This is true because it isimpossible to create or destroy electrons in any chemical change. Electrons can only be transferred fromone substance to another. The electrons that are lost in oxidation are gained in a reduction reaction. Thisis why all oxidation-reduction reactions are the sum of two half-reactions. Let's consider the case in whichsodium reacts with fluorine:

Sodium atoms tend to lose {12}_______ electron(s) to become a stable ion: Na ----> Na1+ + e-

Fluorine atoms tend to gain one electron to become stable. Since fluorine is diatomic, F2 would gain atotal of {13}________ electrons:

F2 + 2e- ----> 2 F1-

After the diatomic molecule gains electrons, is it still diatomic? {14}________ It has been changed into twoindependent stable ions. Since F2 needs two electrons and sodium gives up only one electron per atom,two sodium atoms will have to react with one F2 molecule, so we multiply through the oxidation half-equation by two: 2 (Na ----> Na1+ + e-) OR 2 Na ----> 2 Na1+ + 2 e-

and then we add the two half-equations together:

2 Na ----> 2 Na1+ + 2e- <-- oxidation half-equationF2 + 2e- ----> 2 F1- <-- reduction half-equation

2 Na + F2 + 2e- ----> 2 Na1+ + 2 F1- + 2e- <-- overall redox equation

Since there are two electrons on each side of the net equation, they can be dropped to yield the finalredox equation:

2 Na(s) + F2(g) ----> 2 NaF(s)

23-7 ©1997, A.J. Girondi

It's fairly simple to write redox equations from two half-equations. Let's try to write the redoxequation for the reaction between Cu atoms and N3+ ions. The two half-reactions are:

Cu ---> Cu2+ and 2 N3+ ---> N2

The 2 is needed in front of N3+ in the second half-reaction because nitrogen is diatomic as a free element,and the equation must be balanced by mass. While the two half-equations are balanced according tomass, you can see that they are not balanced according to {15}_______________. To balance thecharge, we add electrons to the proper side of each half-reaction:

Cu ---> Cu2+ + 2e- <---- oxidation 2 N3+ + 6e- ---> N2 <---- reduction

Now, since the number of electrons lost in the oxidation must equal the number of electrons gained in thereduction, we multiply through the oxidation half-equation by {16}________: 3 Cu ----> 3 Cu2+ + 6 e-

(When trying to balance some redox equations, you may have to multiply through the reduction half-equation, or you may have to multiply through both half-equations.)

Then, we add the two half-reactions together:

3 Cu ----> 3 Cu2+ + 6e- <-- oxidation half-equation2 N3+ + 6e- ----> N2 <-- reduction half-equation

2 N3+ + 3 Cu + 6e- ----> N2 + 3 Cu2+ + 6e- <-- overall redox reaction

Since there are {17}________ electrons on both sides of the equation, they should be eliminated to givethe overall redox equation: 2 N3+ + 3 Cu ----> N2 + 3 Cu2+

How can you tell that the equation above is balanced according to mass?

{18}____________________________________________________________________________

Notice that in the equation above, {19}________ electrons are lost in the oxidation and {20}________electrons are gained in the reduction. Electrons must always be balanced in this way in a redox equation.

Using the same procedure as above, write the overall redox equation for each of the half-equationcombinations given below. Identify each half-equation as oxidation or reduction. (Some elements inthese equations may have oxidation numbers that do not appear on your list of common oxidationnumbers.)

Problem 4. Li ----> Li1+ and O2 ----> 2 O2-

23-8 ©1997, A.J. Girondi

Problem 5. Fe ----> Fe2+ and Hg2+ ----> Hg

Problem 6. Cr3+ ----> Cr2+ and Al ----> Al3+

Problem 7. F2 ----> 2 F1- and Zn ----> Zn2+

SECTION 23.4 Changes In Oxidation Numbers

So far, the equations we have examined have been pretty simple. All of them involve either singleelements or single ions. In such cases, it is pretty easy to determine which elements or ions are losingelectrons and which are gaining electrons. Many redox equations are more complicated than this. Someinvolve polyatomic ions like PO43- or NO31-. Before we can attempt to understand redox equations thatinclude such ions, we must know how to determine the oxidation number of each atom in a polyatomic ion.This is not new to you. You worked with this concept back in Chapter 14. For example, what is theoxidation number of the Mn atom in the ion MnO41-? The total of the oxidation numbers of the atoms inthis ion must must equal the charge on the ion, which in this case is -1. Rule 5 in section 23.1 tells us thatthe oxidation number of oxygen is -2. Since there are four of them, the total charge from oxygen is -8. Ifthe total is to be -1, the oxidation number of the one manganese atom present must be +7.

Let's try another one. What is the oxidation number of each chromium atom in the Cr2O72- ion?The seven oxygen atoms have a total charge of -14. The total charge must add up to -2, since that is thecharge on the whole ion. The total positive charge from the chromium atoms must, therefore, be +12.

Since there are two Cr atoms present, each one must have an oxidation number of {21}_________.

You may want to reread the rules in section 23.1 and review the work you did with oxidationnumbers in Chapter 14 before completing problem 8.

23-9 ©1997, A.J. Girondi

Problem 8. Oxidation Numbers of Elements in Compounds or Polyatomic Ions

Compound or Ion Determine Ox. No. of: Oxidation No. is:

a. HAsO2 As __________

b. HBr Br __________

c. KI I __________

d. MnO2 Mn __________

e. MnO41- Mn __________

f. H3PO4 P __________

g. HCO31- C __________

h. CO32- C __________

i. ClO41- Cl __________

j. ClO21- Cl __________

k. SeO42- Se __________

l. Cr2O72- Cr __________

We can now use oxidation numbers to determine which atoms are being oxidized (losingelectrons) and which are being reduced (gaining electrons). In addition to losing and gaining electrons,another definition of oxidation and reduction can be given in terms of changes in oxidation number:

When an atom or ion undergoes oxidation, its oxidation number goes up (becomes morepositive or less negative). When an atom or ion undergoes reduction, its oxidationnumber goes down (becomes more negative or less positive.)

Oxidation involves an increase in oxidation number, while {22}__________________ involves a decreasein oxidation number.

According to our other definition, the half-equation below would be labeled as an oxidationbecause electrons are being lost. Note that the oxidation number of Fe increases from 0 to +3, so it alsofits our new definition of oxidation:

Fe ----> Fe3+ + 3e- oxidation

A similar relationship occurs when reduction occurs. In the half-equation shown below, Br2 gains 2electrons to become 2 Br1-. Electrons are gained, so this is reduction. The oxidation number of brominegoes from 0 to -1. A decrease in oxidation number also indicates that a reduction has occurred:

Br2 + 2 e- ----> 2 Br1- reduction

Use these additional definitions of oxidation and reduction to determine whether iron, Fe, is beingoxidized or reduced in each of the following processes.

23-10 ©1997, A.J. Girondi

Problem 9. Determine the oxidation number of iron in each case. Write the oxidation number of ironabove each Fe shown, whether it is alone or combined with another element.

Process Ox or Red

a. Fe2O3 becomes Fe __________________

b. FeO becomes Fe2O3 __________________

c. FeF2 becomes Fe __________________

d. FeO becomes FeCl3 __________________

SECTION 23.5 Oxidizing Agents And Reducing Agents

When discussing oxidation-reduction reactions, the terms oxidizing agent and reducing agent areoften used. The substance that undergoes a decrease in its oxidation number is called the oxidizingagent. The substance whose oxidation number increases is called the reducing agent. The oxidizingagent does the oxidizing, and, as a result, gets reduced. The reducing agent does the reducing, and, asa result, gets oxidized. This can be a little confusing at first, so let's organize the data in another way.

Oxidizing Agent Reducing Agent

1. its oxidation number decreases 1. its oxidation number increases2. it does the oxidizing 2. it does the reducing3. it gets reduced (gains e-) 3. it gets oxidized (loses e-)

It makes sense if you think about it. The oxidizing agent does the oxidizing, meaning that it causessomething else to lose electrons. In the process the oxidizing agent gains those electrons; thus, theoxidizing agent gets reduced. On the other hand, the reducing agent has to give away electrons in orderto do the reducing. Therefore, the reducing agent ends up losing electrons and being oxidized.Complete problem 10 by identifying the reducing agent and oxidizing agent in each equation. As youcomplete the table, be sure to indicate whether you are answering with the atom of an element or the ion(the first two are done for you). Be sure to include the charge when answering with ions.

Problem 10. Identifying Oxidizing and Reducing Agents

Reducing Agent Oxidizing Agent

a. 2 Na + Ni2+ ----> 2 Na1+ + Ni Na Ni2+

b. Hg2+ + 2 Fe2+ ----> Hg + 2 Fe3+ Fe2+ Hg2+

c. 2 Al + 3 Pb2+ ----> 2 Al3+ + 3 Pb ______________ _______________

d. Mg2+ + 2 Li ----> Mg + 2 Li1+ ______________ _______________

e. O2 + 4 K ----> 2 K2O ______________ _______________

f. Fe + Ni2+ ----> Fe2+ + Ni ______________ _______________

g. Cu + 2 Ag1+ ----> Cu2+ + 2 Ag ______________ _______________

h. Cd + 2 H1+ ----> Cd2+ + H2 ______________ _______________

i. 3 Co2+ + 2 Al ----> 3 Co + 2 Al3+ ______________ _______________

23-11 ©1997, A.J. Girondi

ACTIVITY 23.6 Observing Redox Reactions

So far you have worked with redox equations. Now it is time to actually carry out several redoxreactions in the laboratory. You will be able to see changes in substances being oxidized and reduced,and you will be attempting to identify the substances being oxidized and reduced. Be sure to wear safetyglasses and an apron.

Procedure - Part A:

1. Place 20 mL of 0.2 M CuSO4 solution into a clean 125 mL Erlenmeyer flask. This solution containsCu2+ ions. We can ignore the SO42- ions, because they do not get involved in the reaction.

2. Put a small amount of powdered zinc metal into the Cu2+ solution. Allow the reaction to proceed for atleast five minutes before recording any observations. Swirl the mixture in the flask every minute or so.

What happens to the zinc metal after 5 minutes? {23}________________________________________

______________________________________________________________________________

What happens to the solution's color?{24}________________________________________________

______________________________________________________________________________

3. Atoms of zinc metal, Zn, have been converted into colorless Zn2+ ions which go into solution. The blueCu2+ ions present in the CuSO4 solution have been converted into solid copper atoms, Cu, which appearon the bottom of the flask. Write a redox equation that describes the reaction between Zn and Cu2+:

{25}____________________________________________________________________________

4. Complete the items below. Be sure to indicate whether your answer refers to an atom or an ion of anelement.

substance oxidized:{26}______________ oxidizing agent: {27}______________

substance reduced: {28}______________ reducing agent: {29}______________

Procedure - Part B:

1. Obtain a solution of 3.0 M HCl. Place 20 mL of the HCl into a 125 mL Erlenmeyer flask. This solutioncontains H1+ ions. Place the beaker under the fume hood and add a small (2 cm) piece of magnesiummetal to the solution. Observe what happens. What evidence was there that a chemical reaction

occurred? {30}_____________________________________________________________________

2. Magnesium atoms, Mg(s), react with the hydrogen ions, H1+, in the HCl solution to form hydrogen gas,H2(g), and magnesium ions, Mg2+(aq), which go into solution. Write a redox equation for the reaction that

occurred. {31}_____________________________________________________________________

3. Complete the items below. Be sure to indicate whether your answer refers to an atom or an ion of anelement.

substance oxidized: {32}______________ oxidizing agent: {33}______________

substance reduced: {34}______________ reducing agent: {35}______________

23-12 ©1997, A.J. Girondi

Procedure - Part C:

1. Add a small piece of copper metal, Cu, to 20 mL of 0.1 M HCl. Do you observe any reaction?_______

2. Based on your observation, is the following statement true or false {36}_____________?

"H1+ ions will oxidize Mg metal much more readily than they oxidize Cu metal."

From these results you see that oxidizing and reducing agents are not always strong enough for a redoxreaction to occur spontaneously. H1+ was able to oxidize Mg, but it was not strong enough to oxidize thecopper. Cu + HCl ----> no reaction OR Cu + H1+ ----> no reaction

If this reaction had occurred, it would have fallen into the category of single replacement. Singlereplacement reactions are also redox reactions. Remember when you used the activity series in chapter6? (A copy of the activity series can also be found in your reference notebook.) If you check it, you willfind that copper is below hydrogen on the activity series, meaning no reaction will occur. You will also notein the series that Mg is located above both copper and hydrogen, which explains why the reactions inparts A and B of this activity did occur.

SECTION 23.7 Balancing Redox Equations

In Chapters 6 and 7 you learned how to balance chemical equations. The equations you workedwith were classified into four groups - combination, decomposition, single replacement, and doublereplacement. You learned how to balance them by inspection. This means that you looked from side toside in each equation changing the coefficients, in your attempt to make the number of atoms of eachelement on the left side equal to the number of atoms of each element on the right side.

Many redox equations are difficult to balance by inspection. For example:

?? H2S + ?? HNO3 ----> ?? S + ?? NO + ?? H2O

Balancing such equations can be made easier by using oxidation numbers. This fact is the basis for amethod of balancing redox equations known as the electron-transfer method. It is a much moresystematic method of balancing equations than is balancing by inspection. The steps involved in theelectron-transfer method will be illustrated using the equation above. (For simplicity's sake, sulfur isrepresented as S in this example, not S8)

Step 1: Assign oxidation numbers to each element in the equation. (Follow the rules in section 23.1)+1 -2 +1 -2 0 +2 -2 +1 -2

\ / \ / | \ / \ / H2S + HNO3 ----> S + NO + H2O

| +5

23-13 ©1997, A.J. Girondi

Step 2: Identify the elements that have been oxidized and reduced, and then write the half-equations(balanced according to mass and charge).

Sulfur's oxidation number increases from -2 to 0, so it has been oxidized. Nitrogen's oxidationnumber decreases from +5 to +2, so it has been reduced. Make sure both half-equations are balancedaccording to mass and according to charge.

oxidation: S2- ----> S + 2e-

reduction: N5+ + 3e- ----> N2+

Step 3: Conserve electrons by multiplying each half-equation by a coefficient that makes the number ofelectrons lost in the oxidation equal to the number of electrons gained in the reduction.

3 (S2- ----> S + 2e-) = 3 S2- ----> 3 S + 6e-

2 (N5+ + 3e- ----> N2+) = 2 N5+ + 6e- ----> 2 N2+

Step 4: Add the two half-reactions. (The electrons should cancel out.)

oxidation: 3 S2- ----> 3 S + 6e-

reduction: 2 N5+ + 6e- ----> 2 N2+

overall: 3 S2- + 2 N5+ ----> 3 S + 2 N2+

Step 5: Place the coefficients in front the proper substances in the original equation.

3 H2S + 2 HNO3 ----> 3 S + 2 NO + ?? H2O

Step 6: Add any other coefficients or make any changes needed to balance the equation. (In ourexample, you need a coefficient in front of the H2O. A 4 will balance it.)

3 H2S + 2 HNO3 ----> 3 S + 2 NO + 4 H2O DONE!!

Here's another example. Let's follow the six steps to balance this equation:

I2 + HNO3 ----> HIO3 + NO2 + H2O

Step 1: 0 +1 -2 +1 -2 +4 -2 +1 -2| \ / \ / \ / \ /

I2 + HNO3 ----> HIO3 + NO2 + H2O | |

+5 +5

23-14 ©1997, A.J. Girondi

Step 2: oxidation: I2 ----> 2 I5+ + 10e-

reduction: N5+ + e- ----> N4+

Notice that if gases are diatomic in the equation to be balanced, they are also written in diatomic form in thehalf-reactions. This is why the oxidation half-reaction above is I2 ---> 2 I5+ + 10e- rather thanI ---> I5+ + 5e-. A coefficient 2 is needed in front of I5+ to balance mass, and then 10e- are needed tobalance charge.Step 3: 1 (I2 ----> 2 I5+ + 10e-) = I2 ----> 2 I5+ + 10e-

1 0 (N5+ + e- ----> N4+) = 10 N5+ + 10e- ----> 10 N4+

Step 4: oxidation: I2 ----> 2 I5+ + 10e-

reduction: 10 N5+ + 10e- ----> 10 N4+

overall: I2 + 10 N5+ ----> 2 I5+ + 10 N4+

Step 5: I2 + 10 HNO3 ----> 2 HIO3 + 10 NO2 + ?? H2O

Step 6: I2 + 10 HNO3 ----> 2 HIO3 + 10 NO2 + 4 H2O DONE!!

Problem 11. Use the six steps of the electron-transfer method to balance the redox equations below.Again, for simplicity, sulfur is represented as S instead of S8. (Some elements in these equations mayhave oxidation numbers that do not appear on your list of common oxidation numbers.)

a. HNO3 + H2S -----> H2O + NO + S

b. S + HNO3 -----> SO2 + NO + H2O

c. HIO3 + NO2 + H2O ----> HNO3 + I2

23-15 ©1997, A.J. Girondi

d. Al + H1+ + SO42- ----> Al3+ + SO2 + H2O

(In order to balance more complex equations, you may need to change one or more coefficients evenafter you have completed the 6 steps. This will be necessary in equation e below.

e. HCl + KMnO4 ----> KCl + MnCl2 + Cl2 + H2O

The equations below are optional. You may balance them on a separate page if you feel that you needmore practice.

f. HNO3 + KI ----> KNO3 + I2 + NO + H2O

g. Fe2+ + MnO41- + H1+ ----> Mn2+ + Fe3+ + H2O

h. Sn2+ + Ce4+ ----> Sn4+ + Ce3+

j. Sn + H1+ + NO31- ----> SnO2 + NO2 + H2O

j. NaI + H2SO4 ----> H2S + I2 + Na2SO4 + H2O (This one can be fun!)

23-16 ©1997, A.J. Girondi


This is the end of Chapter 23. The subject of oxidation-reduction is continued in Chapter 24.Review the learning outcomes below. Arrange to take any quizzes or exams on Chapter 23, and thenmove on to Chapter 24.

_____1. Distinguish between oxidation and reduction.

_____2. Identify the oxidizing and reducing agents in a redox equation.

_____3. Assign oxidation numbers to all elements in any compound or polyatomic ion in a redox equation using the rules for assigning oxidation numbers.

_____4. Determine the number of electrons transferred in redox half- equations and overall equations.

_____5. Write balanced net redox equations given two half-equations.

_____6. Balance redox equations by the electron-transfer method.

23-17 ©1997, A.J. Girondi


Questions:

{1} +1; {2} -1; {3} 2; {4} 4; {5} 2; {6} +2; {7} -2; {8} +3; {9} +4; {10} mass; {11} charge; {12} one; {13} 2;{14} no; {15} charge; {16} three; {17} six;{18} It has the same number of atoms of each element on both sides of the equation;{19} six; {20} six; {21} +6; {22} reduction; {23} It gets oxidized (Zn(s) ---> Zn2+(aq) + 2e-;{24} It fades as the blue Cu2+ ions are reduced to Cu atoms; {25} Zn(s) + Cu2+(aq) ---> Zn2+(aq) + Cu(s)

{26} Zn; {27} Cu2+; {28} Cu2+; {29} Zn; {30} A gas is evolved (given off);{31} Mg(s) + 2 H1+(aq) ---> Mg2+(aq) + H2(g); {32} Mg; {33} H1+; {34} H1+; {35} Mg; {36} true

Problems:

1. a. red; b. red; c. ox; d. ox; e. red; f. ox

2. a. Na ---> Na1+ + __1___e- oxb. Cl2 + ___2__e- ---> 2 Cl1- redc. 2 H1+ + __2___e- ---> H2 redd. 2 Al ---> 2 Al3+ + __6___e- oxe. 3 F2 + __6___e- ---> 6 F1- redf. Fe2+ ---> Fe3+ + ___1__e- oxg. Mn7+ + __5___e- ---> Mn2+ redh. 6 Cl1- ---> 3 Cl2 + __6___e- oxi. Cr6+ + __3___e- ---> Cr3+ redj. Cl1- ---> Cl5+ + ___6__e- ox

3. a. Cu2+ ---> Cu1+ Cu2+ + 1e- ---> Cu1+ redb. I7+ ---> I1+ I7+ + 6e- ---> I1+ red c. 3 Br2 ---> 6 Br1- 3 Br2 + 6e- ---> 6 Br1- redd. Sn2+ ---> Sn4+ Sn2+ ---> Sn4+ + 2e- oxe. P5+ ---> P3+ P5+ + 2e- ---> P3+ redf. Cu2+ ---> Cu Cu2+ + 2e- ---> Cu redg. 2 Br2 ---> 4 Br1- 2 Br2 + 4e- ---> 4 Br1- redh. 6 N3- ---> 3 N2 6 N3- ---> 3 N2 + 18 e- ox

4. ox: 4(Li ---> Li1+ + e-) 5. ox: Fe ---> Fe2+ + 2e-

red: O2 + 4e- ---> 2 O2- red: Hg2+ + 2e- ---> Hgoverall: 4 Li + O2 ---> 4 Li1+ + 2 O2- overall: Fe + Hg2+ ---> Fe2+ + Hg

6. ox: Al ---> Al3+ + 3e- 7. ox: Zn ---> Zn2+ + 2e-

red: 3(Cr3+ + e- ---> Cr2+) red: F2 + 2e- ---> 2 F1-

overall: 3 Cr3+ + Al ---> 3 Cr2+ + Al3+ overall: F2 + Zn ---> 2 F1- + Zn2+

8. a. 3+; b. 1-; c. 1-; d. 4+; e. 7+; f. 5+; g. 4+; h. 4+; i. 7+; j. 3+; k. 6+; l. 6+

9. a. red; b. ox; c. red; d. ox

10. c. Al, Pb2+; d. Li, Mg2+; e. K, O2; f. Fe, Ni2+; g. Cu, Ag1+; h. Cd, H1+; i. Al, Co2+

23-18 ©1997, A.J. Girondi

11. a.ox: 3 (S2- ---> S + 2e-)red: 2 (3e- + N5+ ---> N2+)overall: 3 S2- + 2 N5+ ---> 3 S + 2 N2+

final: 2 HNO3 + 3 H2S ---> 4 H2O + 2 NO + 3 S

11. b.ox: 3 (S ---> S4+ + 4e-)red: 4(3e - + N5+ ---> N2+)overall: 3 S + 4 N5+ ---> 3 S4+ + 4 N2+

final: 3 S + 4 HNO3 ---> 3 SO2 + 4 NO + 2 H2O

11. c.ox: 10(N4+ ---> N5+ + e-)red: 1(10e- + 2 I5+ ---> I2)overall: 10 N4+ + 2 I5+ ---> 10 N5+ + I2

final: 2 HIO3 + 10 NO2 + 4 H2O ---> 10 HNO3 + I2

11. d.ox: 2(Al ---> Al3+ + 3e-)red: 3(2e - + S6+ ---> S4+)

overall: 2 Al + 3 S6+ ---> 2 Al3+ + 3 S4+

final: 2 Al + 12 H1+ + 3 SO42- ---> 2 Al3+ + 3 SO2 + 6 H2O

11. e.ox: 5(2 Cl1- ---> Cl2 + 2e-)red: 2(5e - + Mn7+ ---> Mn2+)overall: 10 Cl1- + 2 Mn7+ ---> 5 Cl2 + 2 Mn2+

final: 16 HCl + 2 KMnO4 ---> 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H2O

11. "f" through "j" - answers not given to the optional ones. You figure them out!

23-19 ©1997, A.J. Girondi

NAME PER DATE DUE_________________


CHAPTER 24

OXIDATIONAND

REDUCTION(Part 2)

Corrosion Electrochemical Cells

24-1 ©1997, A.J. Girondi

ACTIVITY 24.1 Redox Reactions in a Petri Dish

Corrosion is a type of oxidation in which an uncombined metal reacts with an oxidizing agent toform compounds. Corrosion involves an oxidation–reduction reaction.

Rusting is a very common type of corrosion. In rusting, iron atoms lose electrons (usually 3) tobecome Fe3+ ions. These iron (ferric) ions then react with oxygen and water to form iron (III) hydroxide,Fe(OH)3, which eventually reacts with more oxygen in the air to form the reddish–brown compound wecall rust (Fe2O3). Oxygen molecules are reduced (gain electrons). The half–equations for these reactionsare written below.

oxidation: Fe ---> Fe3+ + 3e-

reduction: O2 + 2 H2O + 4e- ---> 4 OH1-

You can see, then, that for iron to corrode, both water and an oxidizing agent like oxygen are needed.Theoretically, you could prevent this kind of corrosion of iron or steel objects by keeping them in anenvironment free of water or oxidizing agents. In this activity you will observe the corrosion (oxidation) ofsome metals. The metals will be placed into a "gel" made by mixing agar with water. The oxidizing agentwill be the dissolved oxygen (O2) contained in the water in the gel.

Procedure: (Watch your time. About 40 minutes may be needed.)

1. Get 2 petri dishes and 4 iron nails. Obtain a strip of bare zinc wire or ribbon 4 to 5 cm in length, and a 5cm piece of bare copper wire.

2. Prepare 100 mL of agar-agar solution by obtaining 1.0 g of powdered agar-agar. Heat 100 mL ofdistilled water to a boiling in a 250 mL beaker. Stop heating, and then slowly add the agar-agar to the hotwater with constant stirring. Next, add 10 drops of 0.1 M K3Fe(CN)6 and 10 drops of phenolphthaleinsolution to the agar solution and stir again.

3. While the agar solution is cooling, set aside the nails for dish 1 which include straight nail "A" and a nail"B" bent to a very sharp 90 degree angle. Use pliers to bend nail B, sharply.

4. Next, prepare the nails for dish 2. Wrap nail "C" with the zinc metal wire and nail "D" with the copper wireas shown in Figure 24.1. The wire coils should not touch each other, and should extend the entire lengthof the nails.

Agar

nail wrapped with copper wire nail wrapped with zinc wire

Figure 24.1 Redox in Petri Dishes

plain nails

24-3 ©1997, A.J. Girondi

5. Using forceps or tongs, dip the 4 nails in a beaker of 1,1,1- trichloroethane (or another degreasingagent) to remove any oil or grease. Dry the nails with a towel and try not to touch them more thannecessary after the degreasing.

6. Place nails A and B in dish 1, and nails C and D in dish 2. Do not allow the nails to touch each other orthe edge of the dishes.

7. While keeping the nails well separated, slowly pour the agar-agar solution into the dishes to cover thenails. Do not cover or disturb the dishes until the solution forms a solid gel.

8. After the agar has solidified, cover the dishes and turn them upside down. If time allows, you can watchfor any changes which may be occurring in the dishes. The results will be observed after 24 hours havepassed. Before leaving class, place the dishes (upside down) in your lab drawer.

9. After the 24 hours have passed, observe the dishes and answer the questions below. Empty thecontents of the dishes into a paper towel, and clean and dry the dishes. Return them to the materialsshelf.

Discussion: The chemicals you added to the gel will cause the color pink to appear wherever reductionhas occurred and blue or green to appear wherever oxidation (corrosion) of iron has occurred. Whiteareas indicate sites when zinc has been oxidized. Metals will be oxidized, while the dissolved oxygen, O2,in the water in the agar solution will be reduced.

Questions:

1. List the locations on the nails where oxidation occurred.

When nails are produced, the metal is drawn out into a wire. The wire is coated with a substance whichinhibits oxidation. The wire is then cut into nails.

Why do you think the oxidation occurred at the ends of the nails?

2. List the locations in the dishes where reduction occurred.

Why do you think it occurred in those areas?

3. Any white areas around the zinc wire represent substances formed by the oxidation of zinc. Explainwhy there was no oxidation of the iron in the nail which was wrapped with zinc, while there was oxidation ofthe iron nail which was wrapped with copper wire.

24-4 ©1997, A.J. Girondi

4. Why do you think that zinc blocks are attached to the iron hulls of naval vessels?

5. Steel which is given a coating of zinc is said to be galvanized. List 3 or more possible uses forgalvanized steel, since it does not corrode.

SECTION 24.2 Introduction to Electrochemistry

In the remainder of this chapter, we will be studying the role of oxidizing and reducing agents in afield of study known as electrochemistry. Electrochemistry is the study of chemical reactions that involveelectric currents. Some redox reactions can be made to produce an electric current. Other redoxreactions will not occur unless an electric current is passed through the reactants. When a chemicalreaction occurs, there is usually a net increase or decrease in potential energy. In most cases, this energytakes the form of heat given off or absorbed from the surroundings. Occasionally, however, the change inpotential energy appears as electrical energy. It is possible to produce an electric current from anoxidation-reduction reaction if it occurs spontaneously.

In the previous chapter, you carried out an experiment with copper and zinc. This redox reactioninvolved energy changes; but, the experiment did not require that you measure or use any of this energy.By changing the set-up of the experiment, we can use the energy difference between the reactants andthe products to produce an electric current. The secret is to allow the oxidation and reduction half-reactions to occur in containers that are connected in such a way as to complete an electrical circuit. Thiscan be done by organizing your equipment into an arrangement called a voltaic cell. A spontaneous redoxreaction produces an electric current when it occurs in a voltaic cell.

Now let's take a detailed look at a typical voltaic cell. Just as redox reactions have two parts (twohalf–reactions), a voltaic cell is also made of two parts (two half–cells). Oxidation occurs in one half–cell,and reduction occurs in the other half-cell. Look at the diagram of the voltaic cell in Figure 24.2 below.

Cu

Cu Zn

VOLTMETER

CuSO4(aq)

ZnSO4(aq)

Cathode Anode

Figure 24.2 A Voltaic Cell

Porous Cup

24-5 ©1997, A.J. Girondi

At first glance, a voltaic cell it may seem complicated; but, if you study it carefully, you will see that it is not.There are several ways to organize a voltaic cell. The one pictured above is only one example.

First, look at the basic parts of the cell. Every voltaic cell must have two electrodes. An electrodeis a surface, usually (but not always) made of a metal where oxidation or reduction occurs. The electrodewhere oxidation takes place is called the anode. The electrode where reduction takes place is called thecathode. In Figure 24.2, a piece of zinc metal serves as the anode, while the cathode is a copper strip.Notice the use of the word "where" in this paragraph. Sometimes it is useful to think of the anode andcathode as places rather than as things. For example, the surface of the anode is the place whereoxidation occurs. Depending on the reaction, the anode material itself may or may not be involved in thereaction. In the cell pictured in Figure 24.2, the zinc anode actually is a reactant in the reaction, while thecopper cathode is just the place where reduction of Cu2+ ions in the solution will occur. The two half–reactions in this cell are:

Zn ---> Zn2+ + 2e- oxidation

Cu2+ + 2e- ---> Cu reduction

Notice that the beaker contains a solution of ZnSO4 which would be in the form of dissociated Zn2+ andSO42- ions. The porous cup contains a solution of CuSO4 in the form of dissociated Cu2+ and SO42- ions.

The electrodes are attached to a voltmeter which measures the force with which electrons aremoved through a circuit. They could just as well have been attached to a motor, a light bulb, or some otherdevice that makes use of electricity. In a setup like this one, we are not really putting the electricity that isproduced to a useful purpose. We are simply measuring the voltage produced.

Everything in the cell has now been located except the electrons. The wires connected to theelectrodes allow the electrons to move around the system, so we will have to follow them. Follow the pathof the electrons through the voltaic cell as you read.

Pathway of electrons:

1. Begin on the surface of the zinc electrode. The oxidation-reaction, Zn ---> Zn2+ + 2e-, occurs here. A

solid zinc atom loses two electrons and becomes a Zn2+ ion which becomes part of the solution in the

beaker. If you allow this oxidation to continue for a time, what would happen to the mass of the zinc

electrode? {1} ________________________________

2. The electrons move up the wire and pass through the voltmeter. You know that electrons are passingthrough the voltmeter if some voltage registers on the meter.

3. The electrons leave the voltmeter and continue through the wire and into the copper electrode. Theymove to the surface of the copper electrode where they meet the Cu2+ ions in the solution in the cup. ACu2+ ion will pick up two electrons to become a copper atom (the reduction reaction): Cu2+ + 2e- ---> Cu.The copper atom that forms stays on the surface of the copper electrode as a "plating."

This process happens every time a zinc atom loses electrons which pass through the circuit. If this

reaction continues for a time, what will happen to the mass of the copper electrode?

{2}____________________________________________________________________________

Theoretically, the oxidation and reduction reactions should continue for as long as the zinc keeps

giving up its electrons to the Cu2+ ions. The reaction would stop when the half-cells have consumed

either all of the zinc electrode or all of the Cu2+ ions.24-6 ©1997, A.J. Girondi

You may be wondering why the porous cup is needed. Porous means that certain things canpass through the walls of the cup. In this case, the SO42- ions can seep through its walls. If this did nothappen in the cell, there would be an accumulation of positive charges in the zinc half–cell as a result ofzinc ions being produced and added to the solution in the beaker:

Zn ---> Zn2+ + 2e-

In addition, there would be an accumulation of negative charges in the copper half–cell as a result of Cu2+

ions being reduced and removed from the copper solution in the porous cup:

Cu2+ + 2e- ---> Cu

The movement of the SO42- ions from the copper solution to the zinc solution through the walls of the cupmanages to keep both solutions electrically neutral.

Electricity is nothing more than the movement of charge through a conductor such as a wire or asolution. In a wire the charge is carried by a flow of electrons, while charge is carried by moving ions insolutions. The redox reaction is producing electricity which we can measure and use. In physics we learnthat a force is needed to move something or to do work. The force behind the flow of the electrons (whichthe voltmeter measures) can be used by us to do useful work. Actually this same redox reaction wouldoccur if you simply drop a piece of zinc metal into a solution of Cu2+ ions. However, the electrons wouldflow from the zinc to the Cu2+ ions inside the solution. When that happens, we cannot "tap into the flow"of the electrons so that we can make use of the force. A voltaic cell arrangement causes the electrons toflow outside the cell where we can harness some of their energy.

ACTIVITY 24.3 Construction and Testing of Voltaic Cells

1. Look back at Figure 24.2 and assemble the materials needed to construct a voltaic cell.

2. Clean one metal strip of copper and one of zinc, if needed, with a piece of steel wool or fine sandpaper.

3. Place the Cu electrode in the porous cup which should be about 2 / 3 full of CuSO4 solution. The zincelectrode should be placed in a 250 mL beaker which should be about 1 / 2 to 2 / 3 full of ZnSO4 solution.

4. Connect the electrodes to a voltmeter so that you get a positive voltage. If you are using a multimeterwhich has numerous scales, check with your teacher for the correct setting. If the meter needle goes thewrong way, or if you get a negative reading with a digital meter, reverse the wires. You should read thevoltage immediately after you connect the meter.

What reading do you get from the voltmeter? V

The half–reaction occurring at the anode is: Zn ----> Zn2+ + 2e-

Is this an oxidation or a reduction? {3} .

The half–reaction occurring at the cathode is: Cu2+ + 2e- ----> Cu


24-7 ©1997, A.J. Girondi

The net reaction in this voltaic cell is: Zn + Cu2+ ----> Zn2+ + Cu

Are the electrons moving from the zinc metal to the copper ions or from the copper ions to the zinc metal?

{5} __ .

What is the reducing agent in your cell? {6} .

What is the oxidizing agent in your cell? {7} .

Does the number of SO42- ions in the entire cell increase, decrease, or stay the same during the reaction?

{8} Why? {9} .

.

5. While you still have all of your materials, measure the voltage produced in a cell where a different redoxreaction is occurring. Pour the ZnSO4 solution out of the cell into the original container, because it can bereused.

6. Rinse the cell with water, and replace the zinc solution with a solution of lead nitrate, Pb(NO3)2.

7. Replace the zinc electrode with a lead strip (clean it if needed).

8. Keep the Cu electrode and CuSO4 solution the same. Connect the voltmeter and read the voltage.

The measured voltage is: V

9. The Pb(NO3)2 solution can also be reused, so pour it back into the original container.

10. The solution in the porous cup can be discarded.

11. Rinse the porous cup with water, and return all materials to the materials shelf.

The half-equation for the reaction occurring at the anode is: Pb ----> Pb2+ + 2e-


The half-equation for the reaction occurring at the cathode is: Cu2+ + 2e- ----> Cu


The overall equation for the redox reaction in this voltaic cell is: Pb + Cu2+ ----> Pb2+ + Cu

Are the electrons moving from the lead metal to the copper ions or from the copper ions to the lead metal?

{12}___________________________________________________________________________.

What is the reducing agent in your cell? {13} .

What is the oxidizing agent in your cell? {14} .

24-8 ©1997, A.J. Girondi

SECTION 24.4 Predicting Voltages Of Voltaic Cells

It is possible to perform a large number of experiments in which many different electrodes andsolutions are used. Some combinations of electrodes and solutions produce large voltages, while othercombinations produce no voltage at all. Measurements such as these have already been done bychemists. They have measured the contribution of each half–reaction to the voltage of the cell. Anarbitrary value of zero has been assigned to the hydrogen reduction half–reaction:

2 H1+ + 2e- ---> H2 Half-Cell Voltage (Eo) = 0.00

Scientists have combined many other half-reactions, one at a time, with hydrogen and measured the cellvoltage. These measured values are called half–cell potentials and are represented by the symbol Eo.The potential of a half-cell representing a reduction is called a reduction potential and is given the symbol,Eored. The potential of a half-reaction representing an oxidation, is called an oxidation potential and isgiven the symbol, Eoox.

The overall cell potential or cell voltage is equal to the sum of Eored and Eoox, and is given thesymbol Eocell or simply Eo.

Eored + Eoox = Eo

A listing of many of these values is shown in Table 24.1. They are all listed as reduction potentials.

The greater the value of a reduction potential, the greater the ability of a substance to be reduced (gainelectrons).

Table 24.1 is extremely useful. It allows you to predict the amount of voltage which will be produced byany combination of half-reactions.

Let's say you wish to determine the voltage produced by the combination of magnesium andcesium in a voltaic cell. Begin by writing the two half-equations and their Eored values, taken from Table24.1.

Cs1+ + 1e- ----> Cs Eored = –2.92 volts Mg2+ + 2e- ----> Mg Eored = –2.37 volts

The half-equations are both written as reduction reactions. You must decide which reaction to write as anoxidation. The substance with the smallest Eored value always undergoes oxidation. The substance withthe largest Eored value undergoes reduction.

To convert the Cs half-reaction into oxidation, you merely reverse the reaction and the sign infront of the Eored value which converts it into an Eoox value. Obtain the net reaction in the same way youdid earlier in this chapter:

2 Cs ----> 2 Cs1+ + 2e- Eoox = +2.92 V oxidation

Mg2+ + 2e- ----> Mg Eored = –2.37 V reduction

2 Cs + Mg2+ ---> 2 Cs2+ + Mg Eo = +0.55 V overall reaction

Notice that the oxidation equation has been multiplied by two so that the number of electrons lost in the oxidation half–equation equals the number of electrons gained in the reduction half–equation. However, you see that the Eoox value for that reaction was NOT multiplied by two.

24-9 ©1997, A.J. Girondi

Table 24.1Standard Half-Cell Reduction Potentials

Reduction---> <---Oxidation E ored (Volts)

Li1+ + e- <====> Li -3.00Rb1+ + e- <====> Rb -2.92 K1+ + e- <====> K -2.92 Cs1+ + e- <====> Cs -2.92 Ba2+ + 2e- <====> Ba -2.90 Sr2+ + 2e- <====> Sr -2.89 Ca2+ + 2e- <====> Ca -2.87

Na1+ + e- <====> Na -2.71 Mg2+ + 2e- <====> Mg -2.37 Al3+ + 3e- <====> Al -1.66 Mn2+ + 2e- <====> Mn -1.18 Zn2+ + 2e- <====> Zn -0.76 Cr3+ + 3e- <====> Cr -0.74 Cd2+ + 2e- <====> Cd -0.40 Fe2+ + 2e- <====> Fe -0.44 Cr3+ + e- <====> Cr2+ -0.41 StrongerCo2+ + 2e- <====> Co -0.28 ReducingNi2+ + 2e- <====> Ni -0.25 AgentsSn2+ + 2e- <====> Sn -0.14 Pb2+ + 2e- <====> Pb -0.13

------------- 2 H1+ + 2e - <====> H2(g) 0.00 ------------- Sn4+ + 2e- <====> Sn2+ +0.15 Cu2+ + e- <====> Cu1+ +0.15

Stronger Cu2+ + 2e- <====> Cu +0.34 Oxidizing Cu1+ + e- <====> Cu +0.52 Agents I2 + 2e- <====> 2 I1- +0.53

Fe3+ + e- <====> Fe2+ +0.77 Hg2+ + 2e- <====> Hg(l) +0.78 Hg22+ + 2e- <====> 2 Hg(l) +0.79 Ag1+ + e- <====> Ag +0.80Br2(l) + 2e- <====> 2 Br1- +1.06Cl2(g) + 2e- <====> 2 Cl1- +1.36Au3+ + 3e- <====> Au +1.50F2(g) + 2e- <====> 2 F1- +2.87

Oxidizing Reducing Agents Agent

Use the same procedure to write the overall reaction, and calculate the cell voltages for the pairs ofelements listed in the problems which follow. Show the proper half-equations and Eo values.

24-10 ©1997, A.J. Girondi

Problem 1. Ni and Al

oxidation: Eoox = V

reduction: Eored = V

overall: Eo = V

Problem 2. Zn and Cl2

oxidation: Eoox = V


overall: Eo = V

Problem 3. Co and K

oxidation: Eoox = V


overall: Eo = V

Problem 4. Mn and Al

oxidation: Eoox = V


overall: Eo = V

Problem 5. Li and Au

oxidation: Eoox = V


overall: Eo = V

Problem 6. F2 and Ag

oxidation: Eoox = V


overall: Eo = V

24-11 ©1997, A.J. Girondi

Problem 7. Mg and Au

oxidation: Eoox = V


overall: Eo = V

ACTIVITY 24.5 BATTERIES

As was mentioned earlier, there are a number of types of voltaic cells. These include thehydrogen/oxygen fuel cells used in space technology, and dry cells used in devices such as flashlights.We often misuse the term battery. A battery is actually two or more voltaic cells joined together. Forexample, the lead battery in your car most likely contains six cells linked together in an arrangement knownas a "series." Each cell produces two volts, giving the battery a total of twelve volts. A standard nine voltbattery such as you might use in a radio or a smoke detector is actually six 1.5 volt cells joined together (6 X1.5 = 9). So, the nine volt battery really is a battery, but a 1.5 V dry cell is not. Dry cells get their name fromthe fact that they do not contain solutions like the voltaic cell did that you put together. Instead, theycontain a moist paste. A typical carbon–zinc dry cell contains a carbon rod in the center which issurrounded by a dark paste. This whole arrangement is enclosed in a zinc metal case. The zinc metal isthe anode and the carbon rod is the cathode. The zinc gets oxidized, while manganese dioxide, MnO2, inthe paste get reduced at the surface of the carbon rod. The half- reactions are:

Zn(s) ----> Zn2+(aq) + 2e- Eoox = +0.76 V

2 MnO2(s) + 2 NH41+(aq) + 2e- ---> Mn2O3(s) + 2 NH3(aq) + H2O(l) Eored = ???

On the line below, write the overall redox equation that occurs in a dry cell:

{15} Eo = +1.50 V

Problem 8. Since you know that a carbon-zinc dry cell produces 1.50 V, use the information given tocalculate the half-cell potential for the reduction half-reaction shown above. Show the calculation.

Eored = _________ V

Your teacher has some different types of dry cells which have been cut in half for you to examine.In the space below, sketch cross–sections of a 1.5 V carbon-zinc cell and a 9 V carbon–zinc battery.Label the anodes and cathodes.

1.5 V Cell 9 V Battery

24-12 ©1997, A.J. Girondi

SECTION 24.6 Spontaneous and Nonspontaneous Redox Reactions

You have learned how to calculate the standard voltage (Eo) of a number of redox reactions. By"standard" we mean that any solutions in the cell are 1 M and any gases involved would be at 1 atm ofpressure. If you go on to study more advanced chemistry, you will learn that changes in these conditionscan change the voltage of a cell.

Most of the reactions we have worked with so far have resulted in positive cell voltages. Ingeneral, when the standard voltage is calculated for a spontaneous redox reaction, the cell voltage will bea positive value. A spontaneous reaction happens on its own without us having to do anything to make ithappen.

The opposite is also true. If we calculate a negative standard voltage, then the redox reaction willbe nonspontaneous. Reactions that are nonspontaneous require additional energy to proceed.

If Eo is positive, the reaction is spontaneous. If Eo is negative, the reaction is nonspontaneous.You may recall an activity you performed in Chapter 23. You placed a piece of copper metal in an acid (H1+)solution and nothing happened:

Cu(s) + H1+(aq) ----> NO REACTION

If something would have happened, it would have had to have been:

Cu(s) + 2 H1+(aq) ----> Cu2+(aq) + H2(g)

However, this reaction does not occur. Let's figure out why it doesn't.

Problem 9. The half-equations for this redox reaction are given below. Look up the half–cell reductionpotential for the first one using Table 24.1 or Reference Table 11, and enter the data in the blank to theright. Since the half–reaction involving Cu is an oxidation, use the same table to find the oxidationpotential for this half–reaction. Enter the value in the blank to the right. Add the two half–cell potentials tofind Eo for the reaction, and enter the result in the proper blank below.

2 H1+ + 2e- ----> H2 Eored = __________ V

Cu ----> Cu2+ + 2e- Eoox = __________ V

Cu + 2 H1+ ----> Cu2+ + H2 Eo = __________ V

What evidence exists to support the conclusion that the reaction represented by the equation above, is

nonspontaneous: {16} _____________________________________________________________

Problem 10. Indicate whether the reactions below would be spontaneous or nonspontaneous.

a. Fe + 2 H1+ ----> Fe2+ + H2 spontaneous nonspontaneous

b. Ba + Pb2+ ----> Ba2+ + Pb spontaneous nonspontaneous

c. 2 Ag + Cu2+ ----> 2 Ag1+ + Cu spontaneous nonspontaneous

24-13 ©1997, A.J. Girondi

Problem 11. An interesting nonspontaneous reaction involves the decomposition of water intohydrogen and oxygen gases. We can make this reaction happen by adding energy in the form ofelectricity. The half–equations are listed on the next page. It is interesting because water is the reactant inboth of the half–reactions. In other words, some water molecules get oxidized and some get reduced asthe reaction proceeds. The half–cell potentials are given. Calculate the cell voltage, and enter the valuein the space provided.

oxidation: 2 H2O ----> O2 + 4 H1+ + 4 e- Eoox = –1.23 V

reduction: 2(2 H2O + 2 e- ----> H2 + 2 OH1-) Eored = –0.83 V

sum: 6 H2O + 4e- ----> O2 + 2 H2 + 4 H2O + 4e-

If we subtract some H2O and some electrons from both sides we get:

net: 2 H2O ----> O2 + 2 H2 Eo = __________ V

Referring to the Eo value for the reaction, explain why energy is needed to decompose water:

{17}___________________________________________________________________________

ACTIVITY 24.7 The Electrolysis of Water

The teacher will have set up equipment to demonstrate the decomposition of water usingelectricity. Fill in the blanks below as the teacher operates the equipment for you.

1. Turn on the power supply and slowly increase the voltage until you see gases being given off at theelectrodes. Note the voltage as soon as evidence of a reaction is seen. Repeat this procedure severaltimes, by turning the power down so that the voltage returns to zero, and then turning it back up again.When electricity is used to decompose molecules in this way, the process is known as electrolysis.Although your result will be just a rough estimate, you should note a relationship between the cell voltageyou calculated in Problem 11 above, and the amount of voltage needed to cause this reaction to occur. Inorder to make this reaction happen, electricity must be supplied in excess of how many volts?{18}______V

The equipment pictured In Figure 24.3 is an example of an electrolytic cell. The reaction in anelectrolytic cell is nonspontaneous, so we must add energy to make the cell function. Electrolytic cellshave (positive/negative) {19}_________________________ cell voltages (potentials).

2. Increase the voltage just high enough to allow the reaction to proceed at a good rate. Allow it to runlong enough to permit you to determine the ratio of the volumes of the two gases being produced.

What does the ratio appear to be? ______to ______. Based on the balanced equation for this reaction,

does the observed ratio appear to be in agreement with the balanced equation for this reaction? .

Explain: {20}_____________________________________________________________________

.

24-14 ©1997, A.J. Girondi

VOLTMETER

POWER SUPPLY

gasgas

water + acid

Figure 24.3 Electrolysis Apparatus

ACTIVITY 24.8 Electroplating

Electrolytic cells are used to electroplate metals. For example, copper plating is done using asolution which contains Cu2+ ions. By running a current through the solution the Cu2+ ions are reducedto Cu and the Cu metal "sticks" to the object being plated which is one of the electrodes in the cell.

Should the object being plated be the anode in the cell or the cathode? {21}_________________________________

In this activity you will be electroplating a coin with a layer of copper metal. Nickels, dimes or quarters seemto produce the best results.

Copper anode

Solution of Cu2+ ions

Object to be plated

Cu2+

Power Supply

Figure 24.4 Copper Plating

Procedure:

1. Have your teacher advise and assist you as youcomplete this activity. Obtain a coin and clean it well bywashing it with soap and water. Place a few mL of 1 MHCl into a small beaker or flask and rinse the coin in it fora few minutes.

2. Decant the HCl into the sink without dumping thecoin. Rinse the coin with water. Decant the water.

3. Obtain a battery or power supply and two wire leadswith alligator clips. Attach one clip to a strip of coppermetal and the other clip to your coin. Try to cover aslittle of the coin as possible when you attach the clip.

24-15 ©1997, A.J. Girondi

4. Place the copper strip and the coin into a small (100 or 150 mL beaker) as shown in figure 24.4 whichillustrates silver plating.

5. Obtain the bottle of copper sulfate electroplating solution from the materials shelf, and pour enough ofthe solution into the beaker to completely submerse the coin. Handle the solution – which also containssulfuric acid – with care.

6. Connect the battery or power supply to the wire leads making sure that the coin is connected to thecathode (the negative terminal). If you are using a power supply, connect the leads in the same mannerand adjust the voltage to between 1 and 3 V. There may be a mark on the dial to help you with this. If not,use a voltmeter. Your instructor will assist you. DO NOT allow the copper strip to touch the coin. This willcause a short circuit.

7. Allow the current to flow for a few minutes. Pull the coin out of the solution every minute or so to checkon the progress of the plating. Change the position of the clip and immerse the coin again. After a fewminutes, the coin should be covered with a thin layer of copper.

8. Remove the coin and rinse with water. If polishing compound is available, put some on the coin and rubit between your fingers or with a soft moist cloth. Buff the coin to a nice shine with a dry rag.

9. Pour the plating solution back into the bottle from which it came. It can be reused many times. Rinsethe copper strip and the beaker with water. Return all equipment to the materials shelf.

The electrons gained by the copper ions in the solution turned them into copper atoms. This reactionoccurred at the surface of the coin. These electrons came from the copper metal in the copper strip.Reaction at the coin (reduction): Cu2+(aq) + 2e- ----> Cu(s)

Reaction at the copper strip (oxidation): Cu(s) ----> Cu2+(aq) + 2e-

Even though the plating process removes Cu2+ ions from the solution, the concentration of Cu2+ ions inthe plating solution remains constant.

How is this possible? {22}___________________________________________________________

What do you think is happening to the mass of the coin as it is being plated? {23}___________________

What do you think is happening to the mass of the copper strip as the plating of the coin goes on?

{24} Why? {25} __________________________________________________

.


This concludes the discussion of redox processes. As you have seen, redox reactions are veryimportant, indeed. Review the learning outcomes below. When you feel you have mastered them,arrange to take any quizzes or exams on Chapter 24, and move on to Chapter 25.

1. Explain how redox reactions and "cells" can be used to produce electrical currents.

2. Trace the pathway of electrons through a voltaic cell.

3. Explain the difference between voltaic and electrolytic cells.

4. Use a table of oxidation potentials to determine the overall voltage of a redox reaction.

24-16 ©1997, A.J. Girondi


Questions:

{1} It would decrease; {2} It will increase; {3} oxidation; {4} reduction; {5} from zinc metal to copper ions;{6} zinc metal; {7} copper ions, Cu2+; {8} Stays the same; {9} they are not involved in the reaction;{10} oxidation; {11} reduction; {12} from lead metal to copper ions; {13} lead metal; {14} copper ions;{15} Zn(s) + 2 MnO2(s) + 2 NH41+(aq) ----> Zn2+(aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l)

{16} The cell voltage has a negative value; {17} Energy is needed because the reaction is nonspontaneous (has a negative cell voltage); {18} +2.06 V; {19} negative; {20} The ratio of H2 to O2 in the equation is 2:1, the same as the volume ratio of H2 to O2 when water is decomposed; {21} cathode;{22} The piece of Cu metal oxidizes to form Cu2+ ions to replace those Cu2+ ions which are plated out of the solution; {23} It is increasing in mass; {24} It is decreasing in mass; {25} Because it is being oxidized.

Problems:

1. ox: 2(Al ----> Al3+ + 3e-) Eoox = +1.66 V red: 3(Ni2+ + 2e- ----> Ni) Eored = –0.25 V overall: 3 Ni2+ + 2 Al ----> 3 Ni + 2 Al3+ Eo = +1.41 V

2. ox: Zn ----> Zn2+ + 2e- Eoox = +0.76 Vred: Cl2 + 2e- ----> 2 Cl1- Eored = +1.36 Voverall: Cl2 + Zn ----> 2 Cl1- + Zn2+ Eo = +2.12 V

3. ox: 2(K ----> K1+ + e-) Eoox = +2.92 Vred: Co2+ + 2e- ----> Co Eored = –0.28 Voverall: Co2+ + 2 K ----> Co + 2 K1+ Eo = +2.64 V

4. ox: 2(Al ----> Al3+ + 3e-) Eoox = +1.66 Vred: 3(Mn2+ + 2e- ----> Mn) Eored = –1.18 Voverall: 2 Al + 3 Mn2+ ----> 2 Al3+ + 3 Mn Eo = +0.48 V

5. ox: 3(Li ----> Li1+ + e-) Eoox = +3.00 Vred: Au3+ + 3e- ----> Au Eored = +1.50 Voverall: 3 Li + Au3+ ----> 3 Li1+ + Au Eo = +4.50 V

6. ox: 2(Ag ----> Ag1+ + e-) Eoox = –0.80 Vred: F2 + 2e- ----> 2 F1- Eored = +2.87 Voverall: 2 Ag + F2 ----> 2 Ag1+ + 2 F1- Eo = +2.07 V

7. ox: 3 (Mg ----> Mg2+ + 2e-) Eoox = +2.37 Vred: 2(Au3+ + 3e- ----> Au) Eored = +1.50 Voverall: 3 Mg + 2 Au3+ ----> 3 Mg2+ + 2 Au Eo = +3.87 V

24-17 ©1997, A.J. Girondi

8. 0.74 VcarboncathodesZn anode

carboncathode

zinc anodes1.5 V Cell

9 V Battery

9. Eoox = –0.34 VEored = 0 VEo = –0.34 V

10. a. spontaneous; b. spontaneous; c. nonspontaneous

11. –2.06 V

24-18 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


CHAPTER 25

INTRODUCTIONTO ORGANICCOMPOUNDS

(Part 1)

25-1 ©1997, A.J. Girondi

SECTION 25.1 Introduction to Carbon Compounds

All substances can be classified as being either organic or inorganic. So far, our study ofchemistry has dealt mainly with inorganic compounds. Originally, organic substances were considered tobe those carbon compounds that were extracted from living things, while inorganic ones werecompounds that did not originate in living systems. An organic compound is defined as a substance thatcontains the element carbon. However, some compounds that contain carbon are considered to beinorganic. A better definition may be that organic compounds have a carbon base, that carbon is the"backbone" of the compounds.

Organic chemistry plays a very important role in our daily lives. Many of the clothes we wear aremade of rayon, dacron, nylon and orlon. These are all synthetic (man-made) organic compounds. Plasticsof all sorts are synthetic organic compounds, too. Petroleum is a naturally occurring organic substance,but synthetic rubber and plastics are two of the by-products of petroleum.

A large number of modern chemical materials have been developed from by-products ofpetroleum. In addition to these items, other materials such as sulfa drugs, penicillin, cortisone, perfumes,detergents, vitamins, pesticides, anesthetics, and many of the more modern antibiotics are among thecontributions made to society through a study of organic chemistry.

Throughout the 18th century, early chemists unsuccessfully tried to synthesize organicsubstances, starting with inorganic materials in their laboratories. Their failures gave rise to the "vital forcetheory" which stated that organic compounds could only be produced by a "vital force" which wasresponsible for life itself. This conclusion was closely tied to religious beliefs at the time. However, in1828, the German chemist, Friedrich Wohler, succeeded in synthesizing an organic compound known asurea, starting with two inorganic compounds. Thereafter, many other organic compounds weresynthesized in the same way in laboratories around the world. By 1850, the "vital force theory" wasdiscredited. From that time on, organic and inorganic chemistry were recognized as two major fields of thescience. There are over 90,000 known inorganic compounds. However, there are well over one millionknown organic compounds, and many more are being synthesized by chemists every year!

Why are there so many organic compounds? Well, carbon atoms can attach themselves to eachother in wide variety of ways. They can join together to form short or long chains, and they can form ringsof many kinds, as well:

C–C–C–C– C–C

C

C

C–C–C–C–C

CC–C– C

CC C

C C C

C C C C

C

C

C

Carbon Chains Carbon Rings

The chains and rings can have branches and cross-links with atoms of other elements (mainly hydrogen)attached to the carbon atoms. Different arrangements of carbon atoms correspond to differentcompounds, and each compound has its own characteristic properties.

We are going to approach the subject of organic chemistry in terms of organic nomenclature.Nomenclature involves the naming of compounds. We will restrict ourselves to the simpler organiccompounds, because the more complex ones can get really complicated. You will be given a set of rulesto follow as you name compounds. These rules must be followed very carefully. Success in learningorganic nomenclature will involve some memorization on your part, but it will rely mainly on a logicalapproach to the problems presented.

The second most abundant element found in organic compounds is hydrogen. This chapter willdeal exclusively with compounds composed of only carbon and hydrogen. These are called

25-3 ©1997, A.J. Girondi

hydrocarbons. These two elements can combine in countless ways. The structures of somehydrocarbons are shown below. The lines between the atomic symbols represent bonds. There arethree types of carbon to carbon bonds:

CH C H

H H

HH

C CH

HC C HH

single bond double bond triple bond

In each case you will note that carbon has a total of four bonds. This is because carbonhas four valence electrons. There are only a few carbon compounds in which carbondoes not have four bonds. One example is carbon monoxide. In this chapter,however, we will deal only with organic compounds in which the carbon atoms havefour bonds. After we have studied the hydrocarbons, Chapters 26 and 27 willintroduce you to the names and structures of organic compounds which contain otherelements in addition to carbon and hydrogen.

C Ocarbon

monoxide

Section 25.2 The Alkanes

The alkane family represents the simplest of the hydrocarbons. The general formula for thecompounds in this family is CnH2n+2, where "n" equals the number of carbon atoms in the molecule. Forexample, if you substitute a 1 into this formula you will get CH4. Substitute a 2 and you will get C2H6.These are the first two members of the family. The compounds in the alkane family are often calledsaturated compounds, which means that the molecules contain only single bonds between the carbonatoms.

Naming alkanes is fairly simple. The prefix in the name of each compound indicates the number ofcarbon atoms present. All alkanes have a suffix of -ane. A list of alkane prefixes is shown in Problem 1which has been partially completed for you. To make writing formulas or drawing structures easier, thehydrogens on the carbons are not always shown (note the structures on page 25-3); however, you shouldassume that enough hydrogen atoms are present to give each carbon atom 4 bonds.

Problem 1. Give the name and molecular formula for each compound below. Use the formula CnH2n+2

to determine the formula, and add the suffix "ane" to the prefixes to obtain the names.

Prefix No. of Carbons Name Molecular Formula

a. meth- 1 ___methane__ ____CH4___

b. eth- 2 ____________ __________

c. prop- 3 ____________ ____C3H8___

d. but- 4 ____________ __________

e. pent- 5 ___pentane___ __________

f. hex- 6 ____________ __________

g. hept- 7 ____________ __________

25-4 ©1997, A.J. Girondi

h. oct- 8 ____________ __________

i. non- 9 ____________ __________

j. dec- 10 ____________ ___C10H22__

In problem 1, you were writing molecular formulas. The kinds of formulas seen at the top of page25-4 are known as structural formulas. Writing structural formulas for organic compounds can become verycumbersome when all of the chemical bonds are included in the drawings. To remedy this problem,chemists have developed a shorthand method of writing structural formulas that involves condensing thestructures. In this shorthand method, the carbon atoms are still written separately (separated by hyphens),but the hydrogens which are bound to carbons are not. Instead, the hydrogens are written to the right ofthe carbon atoms to which they are bonded. This method of representing organic compounds is knownas the condensed structural formula. Study the examples of condensed structural formulas below.

C–C–C–C–H

H H H H

H H H H

H

C–H

H

H

H

Compound Structural Formula Condensed Structural Formula

methane

butane

CH4

CH3-CH2-CH2-CH3

Molecular Formula

CH4

C4H10

Problem 2. Complete the exercise below.

Compound Name Molecular Formula Condensed Structural Formula

a. methane ______CH4______ _____________CH4_____________

b. ethane _______________ _____________________________

c. propane _______________ _____________________________

d. butane _____C4H10______ ________CH3-CH2-CH2-CH3________

e. pentane _______________ _____________________________

f. hexane _______________ _____________________________

g. heptane _______________ _____________________________

h. octane _______________ _____________________________

i. nonane _______________ _____________________________

j. decane _______________ _____________________________

25-5 ©1997, A.J. Girondi

Section 25.3 Alkyl Groups

Carbon chains are not rigid structures. They can bend and flex freely. When we say that an alkanehas a "straight" chain, we don't really mean straight. We mean that it is a continuous chain, rather than abranched chain. The two structures below both contain six carbon atoms. The one on the left is"straight," while the one on the right is branched.

CH3

CH2 CH2 CH2

CH2 CH3

CH3 CH2 CH2 CH CH3

CH3

This is one continuous chain of carbon atoms.

This is a branched chain of carbon atoms.

Now that you have mastered the straight-chain (or should we say "continuous" chain) alkanes, it istime to try something more challenging. Most alkanes exist as "branched" molecules such as the oneshown below. The longest continuous chain of carbon atoms in the molecule below is 7 (enclosed bybox). Therefore, the parent compound here is heptane. (Remember, the longest continuous chain is notnecessarily straight!)

CH3 CH2 CH2 CH CH3

CH3

CH

CH2

CH3CH2

The longest continuous chain contains 7 carbon's.

Having identified the parent compound, we must next identify the side chains. These side chains arecommonly called alkyl groups. Alkyl groups are attached to the longest continuous chain. When writtenalone, they are usually shown with a free-bonding site represented by a dash (like this: –CH3). Thisbonding site represents a spot where a hydrogen atom has been removed. Thus, the general formula forthe alkyl groups is CnH2n+1. The free bonding site is what allows the alkyl group to bond to the parentcompound. Alkyl groups are named with the same prefixes as the alkanes themselves. The suffix ischanged from "ane" to "yl." Complete Problem 3 below by entering the formulas and condensedstructural formulas of the first six alkyl groups.

Problem 3. Complete the exercise below.

Name of Alkyl group Condensed Structural Formula

a. ___methyl _____________–CH3____________

b. __________ _____________________________

c. __________ _____________________________

d. ___butyl___ ________–CH2–CH2–CH2–CH3_____

e. __________ _____________________________

f. __________ _____________________________

25-6 ©1997, A.J. Girondi

Depending on where the hydrogen atom is removed, the bonding site on some alkyl groups can changeposition. This would change the way in which the alkyl group bonds to the parent compound. Forexample, note the two alkyl groups shown below. Both are composed of three-carbon chains, but thebonding site differs:

CH2 CH2 CH3 CH3 CH CH3

propyl isopropyl

The compound on the left below has a propyl group attached to the parent compound which is octane.The compound on the right has an isopropyl group attached to the parent compound (heptane). Notethat all carbons in the molecules have four bonds.

CH3 CH2 CH2 CH CH2 CH3CH2 CH2

CH2

CH2

CH3

CH3 CH2 CH2 CH CH2 CH3CH2 CH2

CHCH3 CH3

Propyl group attached to an 8-carbon chain Isopropyl group attached to an 8-carbon chain

The carbon atoms on the end of the chain are called terminal carbons. When the bonding site of an alkylgroup occurs on a terminal carbon, the alkyl group is said to be "normal" and its name is sometimespreceded by the letter n. Thus, the propyl group above could also be called n-propyl (pronounced"normal propyl"). We will consider the use of this "n" prefix as optional. The other structure with thebonding site on the center carbon is called isopropyl .

SECTION 25.4 IUPAC Rules for Naming Alkanes

A system for naming organic compounds has been developed by the International Union of Pureand Applied Chemists (IUPAC). The system is accepted and used throughout the world. There is also amethod by which many organic compounds are given "common" names, but we will use only the IUPACsystem in this chapter. We will consider the rules one at a time and apply them to some practice problems.

RULE 1: Locate the longest continuous chain of carbon atoms. This will give you the name of the"parent" compound.

For example, if the longest chain contains four carbons, the parent compound is butane. The longestchains in the following two molecules are enclosed by a box:

CH3 CH2 CH2CH CH2 CH3CH2

CH2

CH2

CH3

CH

CH2 CH3CH2 CH2

longest continuous chain = 11 carbons

CH3 CH2 CH CH2 CH3CH2CH

CH2 CH3

CH2 CH3

longest continuous chain = 8 carbons

25-7 ©1997, A.J. Girondi

Problem 4. Draw a box around the longest continuous chain of carbon atoms in the structures below,and name the parent compound for each one.

CH3 CH2 CHCH2 CH3

CH3

a. CH3 CH

CH3

CH3

b. c. CH3 CH2CHCH2 CH3

CH2 CH2 CH3

d. CH3 CH CH2 CH3

CH2 CH3

CH

CH2 CH3

e. CH3 CH CH2 CH3

CH2 CH3

CH3 CH2C CH3

CH3

f.

CH2

CH3

a. parent: __________________________ d. parent: __________________________

b. parent: __________________________ e. parent: __________________________

c. parent: __________________________ f. parent: __________________________

RULE 2: The name of the parent compound is modified by noting what alkyl groups are attached to thechain. Number the longest chain so that the alkyl group(s) will be on the lowest numbered carbons.

Note in the molecules shown below, that the longest chain should be numbered from right to leftin order to give the carbon which is bonded to the methyl group the lowest possible number:

CH3 CHCH2 CH3

CH3

1 2 3 4

Incorrect Numbering

CH3 CHCH2 CH3

CH3

4 3 2 1

Correct Numbering

The correct name of this compound is 2-methylbutane. The "2-" indicates that the methyl group isattached to the second carbon in the longest chain. Note that the name of the alkyl group is added to thatof the parent compound (butane) to form one word, and that hyphens are used to separate numbers fromalphabetical parts of the name.

Problem 5. For the following compounds, draw a box around the longest continuous carbon chain andname each molecule. The name of the molecule in part "b" is given to help you.

CH2CH3 CH CH2 CH3

CH3

a. Name: ___________________________________

CH2CH3 CHCH2 CH3b. Name: __3-ethylhexane______________________CH2

CH3 CH2

25-8 ©1997, A.J. Girondi

CH2CH3 CHCH2c.

Name: ____________________________________

CH2 CH3CH2 CH2

CH3CH2 CH2

Name: ____________________________________

CH2CH3 CHCH2d. CH2 CH2

CH3CH3 CH

CH3CH2 CH2

RULE 3: When the same alkyl group occurs more than once in a molecule, the numbers of the carbons towhich they are attached are all included in the name. The number of the carbon is repeated as many timesas the group appears. The number of repeating alkyl groups is indicated in the name by the use of Greekprefixes for 2, 3, 4, 5, etc. (di, tri, tetra, penta, etc.).

To better understand rule 3, study the following examples.

CH3 CH CH2 CH3CH

CH3

CH3

is called 2,3-dimethylpentane

Note that numbers used in the name are separated from each other by commas, and note that thenumbers are separated from the rest of the name with a hyphen.

CH2CH3 is called 3,3-diethylhexaneCH2 C CH3CH2

CH2 CH3

CH2 CH3

Problem 6. Name the four molecules whose structures are drawn below.

a. CH3CH3 C

CH3

CH3

b. CH3CH3 C

CH2 CH3

CH3 CH2

CH2CH3 CH2 C CH3CH2

CH2 CH3

CH2 CH3

CH2 CH2c.

CH3 CH CH2 CH3CH2

CH CH3

CH CH3CH3

d.a.

b.

c.

d.

25-9 ©1997, A.J. Girondi

RULE 4: If there are two or more different kinds of alkyl groups attached to the parent chain, name them inalphabetical order.

CH3 CH CH CH3CH2

CH3

CH3CH2

For example:is called 3-ethyl-2-methylpentane

It is NOT called 2-methyl-3-ethylpentane

However, when you are determining the alphabetical order, do not consider any Greek prefixes that arebeing used. For example:

CH3 C CH2 CH CH2 CH2 CH3

CH3

CH3 CH2 CH3is called 4-ethyl-2,2-dimethylheptane

It is NOT called 2,2-dimethyl-4-ethylheptane

Problem 7. Name the four molecules drawn below.

CH2 CH3

CH3 CH2 CH CH2 CH2 CH3CH2 CH CH2

CH3 CH2 CH3

a.

CH3 CH2 CH2 CH2 CH3CH2CHCH2b. C

CH3

CH3 CH2 CH3CH2

CH3

c. CH3 CH2 CH CH CH2 CH CH3

CH CH3CH3

CH2 CH3

d. CH3 CH CH2 CH CH2 CH2 CH CH3

CH3 CH3

RULE 5: To put the finishing touches on the name of an alkane, keep the following points in mind: (a)hyphens are used to separate numbers from names of substituents; (b) numbers are separated from eachother by commas; (c) the last alkyl group to be named is prefixed to the name of the parent alkane, formingone word; and (d) the suffix "-ane" indicates that the molecule is an alkane.

ACTIVITY 25.5 Using Molecular Models

The structure of alkanes is more understandable if you see them in three dimensions. We will usemolecular model kits for this purpose. Obtain a box containing a molecular model kit and determine whichparts represent carbon atoms, hydrogen atoms, carbon to carbon bonds, and carbon to hydrogen bonds.When you have done this, assemble models of the six molecules drawn in Problem 4. Pick up one of your

25-10 ©1997, A.J. Girondi

models and rotate one section of the model while holding the other. Do you see how rotation is possiblearound a single bond?_____________. Holding the model with both hands, bend and flex it a bit. Notethe bond angles between the carbons themselves and between the carbons and the hydrogens. Do yousee why these molecules are not really "straight" chains? ______________.

Because free rotation is possible around a single bond, what can you conclude about the 2 molecules

shown below?{1}____________________________ If you named these two molecules, what would

you discover?{2}________________________ What is the name?{3}___________________________

CH2 CH CH3CHCH3

CH3 CH3

CH2 CH CH3CHCH3

CH3

CH3

SECTION 25.6 Cyclic Alkanes

The compounds we have studied so far have been either "straight" or "branched" chains. Carbonatoms can also form rings which result in the formation of cyclic alkane molecules with the general formula,CnH2n. Naming the cyclic alkanes is not difficult, but the rules do differ a bit from those used to name thestraight and branched chained compounds.

The name of a cyclic molecule requires the addition of the prefix "cyclo" to the name of thehydrocarbon. Note the two condensed structural formulas below.

CH2 CH2

CH2

CH2 CH2

CH2 CH2

cyclopropane cyclobutane

To make cyclic compounds easier to draw, a shorthand notation is used in which the hydrogens andcarbons which are part of the ring are not represented at all. The rings are represented by lines, and acarbon atom is assumed to be present at each angle in the ring. The proper number of hydrogen atoms isassumed to be attached to each carbon.

For example:

cyclopropane cyclobutane cyclopentane cyclohexaneC3H6 C4H8 C5H10 C6H12

{4}__________________________ Name this compound

25-11 ©1997, A.J. Girondi

Like the "straight-chained" compounds, cyclic molecules can also contain alkyl side chains. Thesame general rules for alkane nomenclature apply to the cyclics, except that all positions in a ring areequivalent, so a number is not needed to indicate the position of the alkyl group if there is only one alkylgroup on the ring. For example:

CH3 This is called methylcyclohexane

(It is NOT called 1-methylcyclohexane)

The carbon on which the alkyl group is located is automatically assumed to be number 1.

Problem 8. Name the cyclic molecules below.

CH2CH3CH2 CH3

CH2 CH3CH2

a._____________________ b._____________________ c._____________________

If there are two or more substituents on a ring, numbers must be used to indicate their positions.One of the substituents is always assigned position number 1, and starting at position 1, the chain isnumbered either clockwise or counterclockwise so as to give the other substituents on the ring thesmallest possible numbers. For example:

CH2 CH3

CH3

This is called 1-ethyl-2-methylcyclopentane

CH3

CH3

This is called 1,2-dimethylcyclopentane

(It is NOT called 1,5-dimethylcyclopentane)

CH3

CH2CH3

This is called 1-ethyl-4-methylcyclohexane

(You may have wanted to call it 4-ethyl-1-methylcyclohexane,but we chose to assign the number 1 position to ethyl since itcomes first, alphabetically, and since we get the samenumbers,1 and 4, either way.)

CH3

CH3

CH2CH3 This is called 4-ethyl-1,2-dimethylcyclopentane

25-12 ©1997, A.J. Girondi

In the last example, we assign position 1 to the carbon in the lower right corner and number the ringcounterclockwise. This gives the lowest possible set of numbers for the three substitutents on the ring.

CH2 CH3CH3

CH3

CH3

This is called 3-ethyl-1,1,2-trimethylcyclobutane

(We numbered clockwise this time)

In the molecule drawn above, if we assigned position #1 to the carbon which is bonded to the ethyl group,we would have had to number counterclockwise and name the molecule: 1-ethyl-2,3,3-trimethylbutane.This was avoided because it resulted in higher numbers.

The three structures drawn below are identical. Write the name: {5}_____________________________

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

Problem 9. Name the cyclic alkanes shown below:

CH2 CH3

CH2 CH3

CH3a.

CH3

CH3

b.

c.CH2 CH3

CH3CH3

CH2 CH3

d. CH2 CH3

e. CH3

CH3

CH3CH

CH3

CH3

CH3

f. g. CH2 CH2 CH3

CH3

a. __________________________________ e. __________________________________

b. __________________________________ f. __________________________________

c. __________________________________ g. __________________________________

d. __________________________________

25-13 ©1997, A.J. Girondi

ACTIVITY 25.7 Models of Cyclic Alkanes

Using a molecular model kit, construct the four cyclic molecules drawn below. The models giveyou some idea of what these cyclic compounds look like in three dimensions. You will also see the effectsof the bond angles on the shapes of the molecules. Be sure to include all needed hydrogen atoms, evenif they are not shown on the drawings.

cyclopropane cyclobutane cyclopentane cyclohexaneC3H6 C4H8 C5H10 C6H12

Do any of these cyclic compounds have what you might consider to be flat rings? If so, which one(s)?

{6}____________________________________________________________________________

Here is a summary of the rules used to name alkanes:

RULE 1: Locate the longest continuous chain of carbon atoms. This will give you the name of the"parent" compound.

RULE 2: The name of the parent compound is modified by noting what alkyl groups are attached to thechain. Number the longest chain so that the alkyl group(s) will be on the lowest numbered carbons.

RULE 3: When the same alkyl group occurs more than once in a molecule, the numbers of the carbons towhich they are attached are all included in the name. The number of the carbon is repeated as many timesas the group appears. The number of repeating alkyl groups is indicated in the name by the use of Greekprefixes for 2, 3, 4, 5, etc. (di, tri, tetra, penta, etc.).

RULE 4: If there are two or more different kinds of alkyl groups attached to the parent chain, name them inalphabetical order.

RULE 5: The put the finishing touches on the name of an alkane, keep the following points in mind: (a)hyphens are used to separate numbers from names of substitutents; (b) numbers are separated fromeach other by commas; (c) the last alkyl group to be named is prefixed to the name of the parent alkane,forming one word; and (d) the suffix "-ane" indicates that the molecule is an alkane.

SECTION 25.8 Naming Alkenes

Now that you are an expert on alkanes, let's take a look at the alkene functional group. Afunctional group is a feature of a class of compounds that is responsible for its characteristic properties.The functional group of the alkanes is the single bond. The functional group of the alkenes is the doublebond. Alkenes contain at least one double bond which exists between a pair of carbon atoms. Thegeneral formula for the straight-chained alkenes is CnH2n. The suffix to be used in the names of alkenes is"-ene." The rules for naming alkenes are the same as those for alkanes with a few additional restrictions.

25-14 ©1997, A.J. Girondi

Additional Rules for the Nomenclature of Alkenes:

RULE 1: The chain chosen as the parent chain must contain the carbon–carbon double bond (C=C).

RULE 2: The parent chain must be numbered to give the carbon-carbon double bond the lowest possiblenumber.

RULE 3: The name of the alkene must contain a number to indicate the position of the double bond.

Note the example below. The longest carbon chain alkene is numbered correctly, giving the double bondthe lowest possible number.

CH2CH3 CH CH C CH3

CH2 CH3

CH3

12

3

4567 As we number the carbons, the first carbon involved in thedouble bond is #3, so the parent chain is called 3-heptene.Methyl groups are located on carbons #3 and #5.

3,5-dimethyl-3-heptene

A number is not used to locate the double bond in chains which are shorter than four carbons. Twoexamples are below.

CH2 CH2 This is called ethene, not 1-ethene

CH CH2CH3 This is called propene, not 1-propene

Why is it that these two molecules do not require the use of the number? {7}______________________

______________________________________________________________________________

Problem 10. Name the alkenes below. After you have located the longest chain containing the doublebond, be sure to number the chain so that the double bond gets the lowest possible number.

a. CH3 – CH2 – CH = CH2 ___________________________________________________________

b. CH3 – CH = CH – CH3 ___________________________________________________________

c. CH3 – CH2 – CH = CH – CH3 ___________________________________________________________

d. CH3 – CH2 – CH = CH – CH2 – CH3 ___________________________________________________________

e. CH2 = CH2 ___________________________________________________________

f. CH3 – CH = CH2 ___________________________________________________________

CH2 CH CHCHCH3

CH3

CH3

CH2g.

25-15 ©1997, A.J. Girondi

CH CH3CH3 C

CH2 CH3

h.

CHCH3i. CH CH2

CH3

CH3CCHCH3j. CH CH2

CH2 CH3

CH2 CH3

SECTION 25.9 Naming Cycloalkenes

Cycloalkenes are named similarly to straight chained alkenes. The carbons in the ring that containthe double bond are always assigned the #1 and #2 positions, so numbers are used only to locate thepositions of substitutents attached to the ring - not to locate the position of the double bond. The generalformula for cyclic alkenes in CnH2n-2. Study the examples below.

CH3

CH3

CH3

cyclobutene 3-methylcyclohexene 3,4-dimethylcyclopentene

Problem 11. Name the following cycloalkenes.

a. CH2 CH3

b.

CH3

CH3

CH3 CH2

c.

CH3

CH3

25-16 ©1997, A.J. Girondi

d.

CH CH3CH3

e.

CH2 CH2 CH3

CH3

CH3

f.

CH2 CH3

CH2 CH2 CH2 CH3

SECTION 25.10 Naming Alkynes

The functional group of the compounds known as the alkynes is a triple bond. The generalformula for straight-chained alkynes is CnH2n-2. Alkynes are named in much the same way as the alkenes,except that their names end with the suffix "-yne", signifying the triple bond. Once again, the triple bondmust be located within the parent chain, and it should be assigned the lowest possible number.

Additional Rules for the Nomenclature of Alkynes:

RULE 1: The chain chosen as the parent chain must contain the carbon- carbon triple bond.

RULE 2: The parent chain must be numbered to give the carbon-carbon triple bond the lowest possible number.

RULE 3: The name of the alkyne must contain a number to indicate the position of the triple bond.

As was the case with the alkenes, no number is used to locate the triple bond if the parent chain is shorterthan four carbons:

CH CH

ethyne

CH C CH3

propyne

CH C CH2 CH3

1-butyne

C CH3CH3 C

2-butyne

CCH3 C CH2 C CH3

CH3

1 2 3 4 5 6

For the example at right, the correct name is 5-methyl-2-hexyne

25-17 ©1997, A.J. Girondi

Problem 12. Name the alkynes drawn below. Be sure to number the parent chain so as to give thetriple bond the lowest possible number.

a. CH C – CH2 – CH2 – CH3 __________________________________________

b. CH3 – CH2 – CH2 – C C – CH3 __________________________________________

c. CH3 – CH2 – C C – CH3 __________________________________________

d. CH3 – CH2 – CH2 – C CH __________________________________________

e. CH3 – C C – CH2 – CH2 – CH2 – CH3 _______________________________________________________________

f. CH3 CH C CH

CH3

g. CH3 C CHCH2 CH CH

CH3

CH2 CH3

CH2 CH3

h.

CH2

CH C C CH3

CH3

i. CH3 CH2CHC CH3C CH

CH3

CH2 CH3CH2

Table 25.1Summary of General Formulas forAlkanes, Alkenes, and Alkynes

Class of Compound General Formula

Straight-chained alkanes CnH2n+2

Cycloalkanes CnH2n

Alkenes CnH2n

Cycloalkenes CnH2n-2

Alkynes CnH2n-2

25-18 ©1997, A.J. Girondi


Problem 13. The names of the compounds listed below are NOT correct. Using the incorrect name,draw the structural formula in the work area. Then write the correct name of each compound on the lineprovided.

Incorrect Name Correct Name Work Area

a. 4,4-dimethylhexane ___________________________

b. 2-n-propylpentane ___________________________

c. 1,1-diethylbutane ___________________________

d. 1,4-dimethylcyclobutane ___________________________

e. 3-methyl-2-butene ___________________________

f. 5-ethylcyclopentene ___________________________

g. 2-n-propyl-1-propene ___________________________

h. 2-isopropyl-3-heptene ___________________________

i. 2,2-dimethyl-3-butyne ___________________________

j. 5-octyne ___________________________

25-19 ©1997, A.J. Girondi

Problem 14. Write condensed structural formulas for the following:

Name Condensed Structural Formula

a. 4-isopropyloctane

b. 3,4-dimethyl-4-n-propylheptane

c. 1,1-dimethylcyclobutane

d. 3-ethyl-3-heptene

e. 3-ethyl-2-methyl-1-hexene

f. 3-octene

g. 3,3-dimethyl-1-butyne

h. 4,4-dimethyl-2-pentyne

i. 3-n-butyl-2-ethylcyclohexene

j. 3,4-diethyl-4,6-dimethylnonane

25-20 ©1997, A.J. Girondi


Before leaving this chapter, read through the learning outcomes listed below. Place a checkbefore each outcome when you feel you have mastered it. When you have completed this task, arrangeto take any quizzes or exams on this chapter, and move on to Chapter 26.

_____1. Distinguish between organic and inorganic compounds.

_____2. Distinguish between alkanes, alkenes, and alkynes.

_____3. Determine the number of carbon atoms in the longest chain of any alkane, alkene, or alkyne.

_____4. Use the IUPAC system to name alkanes, alkenes, and alkynes, given their condensed structural formulas.

_____5. Given the IUPAC names, be able to draw condensed structural formulas for alkanes, alkenes, and alkynes.

25-21 ©1997, A.J. Girondi


Questions:

{1} They are identical; {2} They would have the same name; {3} 2,4-dimethylpentane; {4} cyclooctane;{5} 1,1,2-trimethylcyclobutane; {6} cyclopropane and cyclobutane;{7} the double bond can only be in the #1 position

Problems:

1.a. meth- 1 methane CH4

b. eth- 2 ethane C2H6

c. prop- 3 propane C3H8 d. but- 4 butane C4H10

e. pent- 5 pentane C5H12

f. hex- 6 hexane C6H14

g. hept- 7 heptane C7H16

h. oct- 8 octane C8H18

i. non- 9 nonane C9H20

j. dec- 10 decane C10H22

2.a. methane CH4 CH4

b. ethane C2H6 CH3–CH3

c. propane C3H8 CH3–CH2–CH3

d. butane C4H10 CH3–CH2–CH2–CH3

e. pentane C5H12 CH3–CH2–CH2–CH2–CH3

f. hexane C6H14 CH3–CH2–CH2–CH2–CH2–CH3

g. heptane C7H16 CH3–CH2–CH2–CH2–CH2–CH2–CH3

h. octane C8H18 CH3–CH2–CH2–CH2–CH2–CH2–CH2–CH3

i. nonane C9H20 CH3–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH3

j. decane C10H22 CH3–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH3

3.a. methyl –CH3

b. ethyl –CH2–CH3

c. propyl –CH2–CH2–CH3

d. butyl –CH2–CH2–CH2–CH3

e. pentyl –CH2–CH2–CH2–CH2–CH3

f. hexyl –CH2–CH2–CH2–CH2–CH2–CH3

4. a. pentane; b. propane; c. hexane; d. heptane; e. pentane; f. pentane

5. a. 2-methylpentane; b. 3-ethylhexane; c. 4-propyloctane; d. 4-isopropylnonane

6. a. 2,2-dimethylpropane; b. 2,2-dimethylpentane; c. 4,4-diethyloctane; d. 2,3,4-trimethylheptane

7. a. 4-ethyl-6-methylnonane; b. 6-propyl-3,3-dimethylnonane; c. 4-ethyl-5-isopropyl-2-methylheptane;d. 4-ethyl-2,7-dimethyloctane

25-22 ©1997, A.J. Girondi

8. a. ethylcyclobutane; b. ethylcyclopropane; c. propylcyclopentane (or n-propylcyclopentane)

9. a. 1,3-diethyl-5-methylcyclohexane; b. 1,2-dimethylcyclopropane;c. 1-ethyl-2,3-dimethylcyclopropane; d. 1,2-diethylcyclopentane; e. 1,3,5-trimethylcyclohexane;f. 1-isopropyl-3-methylcyclobutane; g. 1-methyl-2-propylcyclooctane

10. a. 1-butene; b. 2-butene; c. 2-pentene; d. 3-hexene; e. ethene; f. propene;g. 3,5-dimethyl-1-hexene; h. 3-methyl-2-pentene; i. 3-methyl-1-butene; j. 4,4-diethyl-2-hexene

11. a. 4-ethylcyclopentene; b. 6-ethyl-3,3-dimethylcyclohexene; c. 1,3-dimethylcyclobutene;d. 3-isopropylcyclopropene; e. 3,5-dimethyl-6-propylcyclooctene; f. 2-butyl-3-ethylcyclobutene

12. a. 1-pentyne; b. 2-hexyne; c. 2-pentyne; d. 1-pentyne; e. 2-heptyne; f. 3-methyl-1-butyne;g. 4-ethyl-3-methyl1-hexyne; h. 3,3-dimethyl-1-hexyne; i. 6-methyl-4-propyl-2-heptyne

13.

a. 3,3-dimethylhexane CH3 CH2 CH2 C CH2 CH3

CH3

CH3

b. 4-methylheptane CH3 CH2 CH2 CH2 CH3CH CH2

CH3

c. 3-ethylhexane CH3 CH2 CH CH2 CH3CH2

CH2 CH3

d. 1,2-dimethylcyclobutane

CH3

CH3

e. 2-methyl-2-butene CH3CH3 CH C

CH3

f. 3-ethylcyclopentene CH2CH3

g. 2-methyl-1-pentene CH2 C CH2 CH3CH2

CH3

h. 2,3-dimethyl-4-octene CH3 CH CH CH CH CH2 CH2 CH3

CH3

CH3

25-23 ©1997, A.J. Girondi

i. 3,3-dimethyl-1-butyne CH3 C C CH

CH3

CH3

j. 3-octyne C C CH2 CH3CH2CH2CH2CH3

14.

CHCH3 CH2CH2 CH2 CH2 CH3CH2

CH CH3CH3

a.

CH3 CH2 CH2 CH3CH2b.

CH2 CH2 CH3

CH C

CH3 CH3

c.CH3

CH3

CH3CH2CH2CH3 CH2 C CH

CH2 CH3

d.

CH3CH2 CH2CHe. C

CH3

CH2 CH3

CH2

CH3f. CH2 CH CH CH2 CH2 CH2 CH3

g. C C CH3

CH3

CH3

CH

h. C C CH3

CH3

CH3

CCH3

25-24 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


CHAPTER 26


(Part 2)

26–1 ©1997, A.J. Girondi

SECTION 26.1 Alcohols

Alcohols are molecules in which an alkyl group is attached to a hydroxy group (–OH). Thehydroxy group is responsible for the characteristic properties of alcohols so we refer to it as the functionalgroup for alcohols. There are three different methods for naming alcohols, but we will use only the IUPACsystem. The rules that you used for naming alkanes and alkenes (in Chapter 25) are similar to those usedfor the alcohols. The modified rules are listed below.

Additional Rules for the Nomenclature of Alcohols:

RULE 1: Locate the longest continuous chain of carbon atoms which contains the "hydroxy" (–OH) group. This chain will serve to identify the parent compound.

RULE 2: Number the chain so as to give the carbon atom which is bonded to the –OH group the lowest possible number.

RULE 3: A number is included before the name of the parent compound to indicate the position of the –OH group.

RULE 4: The suffix "ol" is added to the name to indicate that the molecule is an alcohol.

Study the examples below. Note that the number indicating the position of the –OH group is not used if

the chain is shorter than 3 carbons. Why? {1}_____________________________________________

______________________________________________________________________________

CH3OH

CH3CH2OH

CH3CH2CH2OH

CH3CHOHCH3

OH

CH3 OH

CH2 OHCH3

CH2 OHCH2CH3

CH

OH

CH3 CH3

C

C C

C C

OH

HH

H

H

H

H

HH H

methanol

ethanol

1–propanol

2–propanol

cyclopentanol

Name Formula Condensed Structural Formula

26–3 ©1997, A.J. Girondi

In addition, you see in the example above that the position of the –OH ("hydroxy") group is not included in

the names of cyclic alcohols, either. Why not? (Remember that this is also the case for the double bond

in cyclic alkenes. {2}________________________________________________________________

(The hydroxy group,–OH, should not be confused with the hydroxide ion, OH1–. The hydroxy group hasthe same formula, but it is not an ion.)

Problem 1. Name the alcohols given below.

a. CH3–CH2–CH–CH2–CH3

OH

b. CH3–CH–CH2–CH3

OH

c. CH3–CH–CH–CH3

OH

CH3

d. CH3–CH–CH2–CH–CH–CH3

OH

CH3

CH3

e. CH3–CH2–CH–CH2–CH2–CH2–OH

CH2 CH3

f.

OH

g.OH

CH3

CH3

h. CH3 OH

i.

OH

CH2 CH3CH3

26–4 ©1997, A.J. Girondi

j. CH3–CH–CH3

OH

This compound is commonly called "rubbing alcohol." Give its IUPAC name.

Problem 2. Draw the condensed structural formulas for the following.

a. 4,4–dimethyl–2–hexanol b. cyclopropanol

c. 2,3–diethylcyclohexanol d. 3,4–diethyl–2–heptanol

Section 26.2 Ethers

Ethers are compounds which contain an oxygen atom bonded to two carbon atoms within thecarbon chain. The functional group is the C–O–C arrangement found within the chain. When you look atan ether molecule, you will see an alkyl group on each side of the oxygen. For example,CH3–CH2–O–CH3 has an ethyl group on the left of the oxygen atom and a methyl group on the right. The"common name" for this molecule is methyl ethyl ether. Although common names are still frequently usedfor ethers, we will stick to our "game plan" and use the IUPAC system.

CH3–CH2–O–CH3

Parent compound is "ethyl"

Functional group is "methoxy"

In the IUPAC system, the larger of the two alkyl groups attached to the oxygen is considered to bethe parent compound. For the ether mentioned in the last paragraph above, the parent compound wouldbe ethane. The smaller alkyl group and the oxygen atom are considered to be a substituent group on theparent compound. The –O–CH3 group is the substituent and it is called "methoxy." So the name of thatether is methoxyethane. If the substituent had been CH3–CH2–O–, it would have been called "ethoxy."Collectively these functional groups of the ethers are known as alkoxy groups. Only one modified ruleneeds to be mentioned here regarding the nomenclature of ethers.

26–5 ©1997, A.J. Girondi

Additional Rule for the Nomenclature of Ethers:

RULE: For ethers with parent chains that contain 3 or more carbon atoms, a number is included to indicate the position of the alkoxy group.

Study the examples below.

CH3–O–CH3 CH3–CH2–O–CH2–CH3 CH3–O–CH2–CH2–CH3 CH3–O–CH–CH3

CH3

methoxymethane ethoxyethane 1–methoxypropane 2–methoxypropane

Problem 3. Name the following ethers:

a. CH3–O–CH2–CH2–CH2–CH3 ___________________________________

b. CH3–CH2–CH2–O–CH2–CH2–CH3 ___________________________________

c. CH3–CH2–O–CH–CH2–CH2–CH3 _____________________________________________________

CH3

Draw condensed structures for the following ethers:

d. methoxycyclohexane e. 3–methoxycyclopentene

f. 4–ethoxynonane g. 2–isopropoxybutane

Section 26.3 Aldehydes and Ketones

The next two organic functional groups we will study are those of the aldehydes and ketones.Aldehydes and ketones contain a carbonyl group, which consists of an oxygen atom which isdouble–bonded to a carbon atom. There are two kinds of carbonyl groups involved here. In aldehydes, atleast one hydrogen is attached to the carbonyl carbon, while in ketones, two carbon atoms are alwaysattached to the carbonyl carbon.

C

O

C

O

H C

O

CC

carbonyl group aldehyde group ketone group

26–6 ©1997, A.J. Girondi

It is helpful to note that in an aldehyde the carbonyl carbon is always a terminal carbon, which means italways occurs at one end of the carbon chain. In ketones, the carbonyl carbon is never a terminal carbon.The nomenclature of aldehydes requires a few rule modifications:

Additional Rules for the Nomenclature of Aldehydes:

RULE 1: The longest continuous chain containing the aldehyde group is considered to be the parent compound.

RULE 2: The carbonyl carbon is part of the parent chain and is always considered to be in the #1 position.

RULE 3: The suffix "al" is added to the name of the parent compound to indicate that the compound is analdehyde.

Note the examples of aldehydes shown below. You see that no number is needed to indicate theposition of the functional group since it is always at position #1.

CH3–CH2 O

H

C CH3–CH–CH2–CH2 O

H

C

CH3

CH2–CH2–CH2–CH–CH2–CH2–CH–CH2–CH3C

O

H

CH2–CH3 CH3

propanal 4–methylpentanal

5–ethyl–8–methyldecanal

The nomenclature of ketones also requires a few rule modifications.

Additional Rules for the Nomenclature of Ketones:

RULE 1: The longest continuous chain containing the ketone group is considered to be the parentcompound.

RULE 2: A number is included before the name of the parent compound to indicate the position of theketone group. The chain is always numbered so that the carbonyl carbon has the lowestpossible number.

RULE 3: The suffix "one" is added to the name of the parent compound to indicate that the compound is a ketone.

For example:

C

O

CH3CH3 C

O

CH2CH2

CH3

CH3 CH3–CH–CH2–CH2–CH2–C O

CH3CH3

2–propanone 3–pentanone 6–methyl–2–heptanone

26–7 ©1997, A.J. Girondi

Why would it be impossible for a ketone to have a name like 3–methyl–1–hexanone? {3}_____________

_____________________________________________________________________________

Problem 4. Name the molecules shown below.

C

O

CH3CH2CH3a.

C

O

HCH2CHCH3

CH3

b.

CH3–CH2–CH–CH–CH3

CH3

O

H

C

c.

CH3–CH–CH–CH2– OC

CH3

CH3

CH2–CH3

d.

OC

CH3–CH2–CH2

CH2–CH2–CH–CH3

CH3e.

Section 26.4 Organic Acids

Organic acids are molecules that contain a carboxyl group (sometimes called a carboxylic acidgroup). This functional group consists of a carbon which is doubled bonded to an oxygen atom, as wasthe case with aldehydes and ketones. However, in an acid a hydroxy group (–OH) is also bonded to thatsame carbon. Be careful not to confuse organic acids with alcohols, aldehydes, or ketones. As was thecase with aldehydes, this functional group always occurs on a terminal carbon of the parent chain.Therefore, a number is not used in the name to locate the carboxyl group.

Additional Rules for the Nomenclature of Carboxylic Acids:

RULE 1: The longest continuous chain containing the carboxyl group is considered to be the parent compound.

RULE 2: The carboxyl carbon is part of the parent chain and is always considered to be in the #1 position.

RULE 3: The suffix "oic" is added to the name of the parent compound, and the word "acid" is added to the name.

26–8 ©1997, A.J. Girondi

For example:

C

OH

H

O

C

OH

O

CH3 CH3–CH–CH2–CH2 C

OH

OCH3

methanoic acid ethanoic acid 4–methylpentanoic acid

Acids also have common names. For example, ethanoic acid is also called acetic acid or "vinegar." We willwork only with the IUPAC names.

As you attempt to name the carboxylic acids, note that the carboxyl group is written in shorthand as–COOH in the condensed structural formulas.

Problem 5. Name the organic acids below.

b. CH3–CH2–CH2–CH2–CH2–COOH __________________________________

c.__________________________________

d.__________________________________

e. __________________________________

f. __________________________________

CH3–CH2–CH2

CH3–CH2–CH–CH2–COOH

CH3–CH–CH2–CH2–CH–CH2–CH2–COOH

CH3 CH3

CH–CH2–COOH

CH3

CH3

CH3–C–CH2–CH2–COOH

CH2–CH2–CH2–CH3

CH2–CH2–CH2–CH3

a. CH3–CH2–CH–CH2–CH2–COOH __________________________________

CH2 – CH3

Section 26.5 Esters

C

O

O R

Esters are organic compounds which are very common in nature. Forexample, fats and oils are esters. Esters are also responsible for many of the odorsand flavors of fruits. Oil of wintergreen and aspirin are esters. Esters can beconsidered to be derivatives of carboxylic acids. The functional group of esters lookssimilar to the carboxyl group of acids, except that the hydrogen atom on the hydroxygroup is replaced with an organic group such as an alkyl group. The letter "R" in thestructure at right represents some organic group (methyl, ethyl, etc.).

26–9 ©1997, A.J. Girondi

carboxyl group general ester group sample ester group

C

O

O R

C

O

O CH3

C

O

O H

Esters are named by first naming the "R" group followed by the name of the acid portion. The suffix of theacid derivative is then changed from "–ic" to "–ate." For example, in the leftmost structure below, theparent acid is ethanoic acid. The "R" group is methyl, so the name of the ester is methyl ethanoate. In thecenter structure, the parent acid is butanoic, while the "R" group is ethyl, so the ester is named ethylbutanoate. Notice that the names of esters consist of two words, while the names of most of the previoustypes of compounds you have studied consisted of only one word.

C

O

O– CH3

CH3 – C

O

O– CH2 – CH3

CH3 – CH2 – CH2 – C

O

O– CH2 – CH3

H –

methyl ethanoate ethyl butanoate ethyl methanoate (pineapples) (artificial rum flavor)

Artificial flavors of strawberry, apple, raspberry, cherry, etc., are made from esters.

Additional Rules for the Nomenclature of Esters:

RULE 1: Determine the name of the "R" group.

RULE 2: Place the name of the "R" group in front of the name of the parent acid, forming two words.

RULE 3: Determine the name of the parent acid, and change its suffix from "–ic" to "–ate." Drop the word "acid."

Problem 6. Name the esters below.

C

O

O – CH2 – CH2 – CH3

CH3 – CH2 – CH2 –a. C

O

O– CH3

CH3 – CH2 – CH2 – CH2 –b.

C

O

O – CH2 – CH2 – CH2 – CH3

CH3 – CH2 – c. C

O

O– CH – CH3

CH3 –

CH3

d.

26–10 ©1997, A.J. Girondi

C

O

O– CH2 – CH2 – CH3

CH3 – CH2 – CH2 – CH2 –e. C

O

O– CH2 – CH2 – CH2 – CH2 – CH3

H –f.

Section 26.6 Amines

Amines are organic compounds which are related to ammonia (NH3). All amines have the elementnitrogen in them. There are three basic kinds of amines:

1. In primary amines one hydrogen atom in ammonia has been replaced by an alkyl group.

2. In secondary amines two hydrogen atoms in ammonia have been replaced by two alkyl groups.

3. In tertiary amines all three hydrogen atoms in ammonia have been replaced by three alkyl groups. Examine the examples below:

CH3 N H

H

CH3 CH2 N

H

CH2 CH3 N CH2 CH3CH3

CH CH3CH3

A Primary Amine A Secondary Amine A Tertiary Amine

According to the IUPAC system, primary amines are named by treating the –NH2 (amino) group inthe molecule as a substituent group on the longest (parent) chain of carbon atoms. For example, theprimary amine shown above is called aminomethane . Two more examples are shown below.

CH3 – CH2 – CH – CH2 – CH2 – CH3

N HH CH3 – CH – CH2 – CH – CH2 – CH2 – CH3

CH3 NH2

3–aminohexane 4–amino–2–methylheptane (a primary amine) (a primary amine)

Secondary and tertiary amines are named according to a "common" naming system. Primaryamines can have either IUPAC or common names. Amines are the only organic compounds for which wewill learn common names. In the common system, amines are named by adding the names of the alkylgroup(s) attached to the nitrogen atom to the word "amine." In the past, the alkyl groups were named inorder of size (smallest first) instead of in alphabetical order is normally done in the IUPAC system.However, today we follow the IUPAC rules and name the alkyl groups in alphabetical order. For example,the name of the secondary amine shown above is diethylamine. The name of the tertiary amine above isethylisopropylmethylamine. Study the examples below. Note that the primary amine can have two names.

26–11 ©1997, A.J. Girondi

CH3 N CH3

CH3

NH

CH2

CH3

CH2 CH3

NH

CH3CH2 CH2 CH2 CH2

H

trimethylamine methylpropylamine pentylamine (common) 1–aminopentane (IUPAC)

a tertiary amine a secondary amine a primary amine

Additional Rules for the Nomenclature of Amines:

RULE 1: In primary amines only, the IUPAC system treats the NH2 (amino) group as a substituent group on the parent chain.

RULE 2: When using the common naming system, the names of the alkyl groups which are attached to the nitrogen atom are listed in alphabetical order and are attached to the suffix "amine" to form one word. Greek prefixes are used if specific alkyl groups occur more than once in a molecule. Name the amines below. Where two lines are present, give two names.

Problem 7. Name the amines below. Where two lines are present, give two names.

CH2CH3 N CH3

CH3a.

CH2CH3 N CH3

Hb.

c. CH3 – CH2 – CH2 – CH – CH2 – CH – CH3

CH3 NH2

CH2CH3 CH2 CH2 CH2

NH CH3

d.

CH3 CH2 N CH2 CH3

CH2 CH3e.

CH3 – CH – CH2 – CH2 – CH2 – CH2 – CH2 – NH2

CH2 CH3f.

g.CH3 CH3

NH2

CH

26–12 ©1997, A.J. Girondi

h.

NH2

Section 26.7 Amides

You are already familiar with the carboxyl group which is the functional group of a carboxylic acid. Ifyou replace the hydroxy group (–OH) in the carboxyl group with an amino group (–NH2), you get thefunctional group of a class of organic compounds known as primary amides.

C

O

O H

C

O

NH2

carboxyl group amide group

There are three classes of amides just as there were for amines, but we will consider only primaryamides, and we will name them according to the IUPAC system. Amides are considered to be derivativesof carboxylic acids, which means they are formed from acids. Thus, the amides are named as derivatives ofacids. To name an amide, simply identify the name of the organic acid from which the amide was derived,and change the "–oic" suffix in the acid's name to "–amide." The examples of amides shown below werederived from ethanoic, propanoic, and butanoic acids.

C

O

CH3

NH2

C

O

CH2

NH2

CH3 C

O

CH2

NH2

CHCH3

CH3

ethanamide propanamide 3–methylbutanamide

Additional Rules for the Nomenclature of Amides:

RULE 1: Identify the carboxylic acid from which the amide was derived and change the suffix of the acidname from "–oic" to "–amide," and drop the word acid.

RULE 2: Add the names of any alkyl groups to the name of the parent compound, forming one word.

Problem 8. Name the amides shown below. Note that the amide functional group is written inshorthand as CONH2.

a. HCONH2 b. CH3–CH2–CH2–CH2–CONH2

__________________________________ ____________________________________

26–13 ©1997, A.J. Girondi

CH3–CH2–CH–CH2–CH2–COHN2

CH2–CH3 CH3–C–CH2–CONH2

CH3

CH3

__________________________________ ____________________________________

c. d.

CH3–CH2–CH–CH2–CH–CH2–CONH2

CH2–CH3

CH3

CH3–CH–CH2–C–CH2–CONH2

CH3 CH3

CH3

e. f.

__________________________________ ____________________________________

Section 26.8 Halogenated Hydrocarbons

The last group of compounds we are going to discuss includes some that are of great importanceand interest today. Included are the chlorofluorocarbons that are used in refrigeration and air conditioningsystems and which are thought to be involved in the depletion of ozone in the upper atmosphere.

This class of organic compounds is known as the halogenated hydrocarbons. In addition to theiruse in refrigerants they are used as solvents, aerosol sprays, antiseptics, dry cleaning fluids, insecticides,herbicides, and anesthetics. Most of these compounds are synthetic (human– made).

In these compounds, the functional group is a single atom of a halogen such as fluorine, chlorine,bromine, or iodine. In the IUPAC system, the halogen atoms are considered to be substituents on theparent chain. The "–ine" suffix of the halogen's name is dropped and the letter "o" is added before beingadded to the name of the parent compound. For example, fluorine becomes "fluoro," chlorine becomes"chloro", bromine becomes "bromo," and iodine becomes "iodo." Note the examples below.

H–C–I

H

H

H–C–C–Cl

H H

HH

H–C–C–C–Br

H H

HH

H

H

H–C–C–C–C–C–C–C–C–H

H F

HH

H

H

H I H H H

F H H H H

iodomethane chloromethane 1–bromopropane 2,4–difluoro–5–iodooctane

CH3–I CH3–CH2–Cl CH3–CH2–CH2–Br CH3–CHF–CH2–CHF–CHI–CH2–CH2–CH3

Numbers are not used to indicate the position of a single halogen atom substituent unless the parentcarbon chain is longer than 2 atoms; however, if more than one halogen atom substituent is present, thennumbers are needed on a two–carbon chain, too! Study the following examples.

H–C–C–Cl

H

HH

Cl

Cl–C–C–Cl

H H

HH

H–C–H

Br

F1,1–dichloroethane 1,2–dichloroethane bromofluoromethane

26–14 ©1997, A.J. Girondi

Additional Rules for the Nomenclature of Halogenated Hydrocarbons:

RULE 1: Drop the "–ine" suffix from the name of the halogen atom(s) and add a suffix consisting of the letter "o".

RULE 2: Add the altered name(s) of the halogen atom(s) to that of the parent compound.

Problem 9. Name the halogenated compounds below.

a. CH3 – CH2 – CH – CH – CH3

Cl

Cl

b. F – C – F

H

H

c. CH3 – CH – CH – CH – CH2 – CH3

I

F Br

d. CH3 – CH – CH = CH2

Br

e. CH2 – CH2 – CH2 – Br

CH2 – CH2 – CH2 – CH3

f. Cl – C – Cl

F

F

Problem 10. Write condensed structural formulas (such as those shown above) for the following.

a. tetrafluoromethane

b. 1,1,1–trichloroethane

26–15 ©1997, A.J. Girondi

c. chlorocyclopentane

d. 1,3–difluoro–2–iodocyclohexane

e. 3,4–dibromo–6–methyl–1–heptyne

f. 3–chlorocyclopentene

g. 2,3–dichlorocyclobutene

Section 26.9 A Review of Organic Nomenclature

The remainder of this chapter consists of a review of nomenclature of the various classes oforganic compounds which you have studied.

Problem 11. Some of the names of the six compounds listed below are incorrect. If the name iscorrect, respond with "O.K." If the name is incorrect, provide the correct name.

a. 3–chloropentane ___________________________________

b. 1,1–dimethyl–1–propanol ___________________________________

c. 2,2,3–trimethyl–4–bromoheptane ___________________________________

d. 4–methyl–4–hexanol ___________________________________

e. 2,2–dimethyl–3–chloro–3–butanol ___________________________________

f. 1–ethyl–2–ethanol ___________________________________

26–16 ©1997, A.J. Girondi

Problem 12. Draw condensed structural formulas for the compounds named below.

a. 1,3,5–tribromocyclohexane b. 2,3–dichlorobutane

c. 2–ethyl–3–methyl–1–pentanol d. 1–ethoxypropane

e. 2–iodo–3–isopropylcyclohexanol f. 3,3–dimethylbutanal

g. 2–methoxy–3–heptanone h. 3–pentanone

i. 3,4–diethylhexanal j. 2,4–difluorohexanoic acid

k. 2–hydroxybutanoic acid l. ethyl ethanoate

m. n–propyl octanoate n. 4–bromo–3–chloroheptane

26–17 ©1997, A.J. Girondi

o. ethylmethylamine p. isopropyldimethylamine

q. propanamide r. 3–methylbutanamide

s. 4–chloro–2–pentanone t. 2,3,4–triiodopentanoic acid

Problem 13. Give another name for each of the following:

a. ethylamine _____________________________________________

b. isopropylamine _____________________________________________

Section 26.11 Learning Outcomes

Before leaving this chapter, read through the learning outcomes listed below. Place a checkbefore each outcome when you feel you have mastered it. When you have completed this task, arrangeto take any quizzes or exams on this chapter.

_____1. Given their names or condensed structural formulas, distinguish between alcohols, ethers, aldehydes, ketones, organic acids, esters, amines, amides, and halogenated compounds.

_____2. Given their names, draw condensed structural formulas for the classes of compounds given in outcome 1 above.

_____3. Given their condensed structural formulas, give the IUPAC names of molecules belonging to the classes of compounds listed in outcome 1 above.

_____4. Given their condensed structural formulas, give the common names of secondary and tertiary amines.

26–18 ©1997, A.J. Girondi

Section 26.12 Answers to Questions and Problems

Questions:

{1} Only one position is possible for the –OH group; {2} Whatever position the –OH group occupies isautomatically #1 if the compound is named as an alcohol; {3} Such a compound would be an aldehyde, nota ketone

Problems:

1. a. 3-pentanol; b. 2-butanol; c. 3-methyl-2-butanol; d. 4,5-dimethyl-2-hexanol;e. 4-ethyl-1-hexanol; f. cyclobutanol; g. 2-methylcyclohexanol; h. 3,4-dimethylcyclopentanol;i. 2-ethyl-3-methylcyclopropanol; j. 2-propanol

CH3 – CH – CH2 – C – CH2 – CH3

OH CH3

CH3

OH

CH2 – CH3

CH2 – CH3

OH

CH3 – CH – CH – CH – CH2 – CH2 – CH3

CH2 – CH3

CH2 – CH3

OH

2. a. b.

c. d.

3. a. 1-methoxybutane; b. 1-propoxypropane; c. 2-ethoxypentane

– O – CH3 – O – CH3

CH3 – CH2 – CH2 – CH – CH2 – CH2 – CH2 – CH2 – CH3

O

CH2 – CH3 CH3 – CH – CH2 – CH3

O

CH3 – CH – CH3

d. e.

f. g.

4. a. 2-butanone; b. 3-methylbutanal; c. 2,3-dimethylpentanal; d. 5,6-dimethyl-3-heptanone;e. 7-methyl-4-octanone

5. a. 4-ethylhexanoic acid; b. hexanoic acid; c. 3-ethylhexanoic acid; d. 4,7-dimethyloctanoic acid;e. 3-methylbutanoic acid; f. 4-n-butyl-4-methyloctanoic acid (The "-n-" is optional.)

6. a. n-propyl butanoate (the n is optional here and in parts c, e, and f); b. methyl pentanoatec. n-butyl propanoate; d. isopropyl ethanoate; e. n-propyl pentanoate; f. n-pentyl methanoate

7. a. ethyldimethylamine; b. ethylmethylamine; c. 2-amino-4-methylheptane;d. methylpentylamine; e. triethylamine; f. 1-amino-6-methyloctane;g. isopropylamine (or 2-aminopropane); h. cyclobutylamine (or aminocylcobutane)

8. a. methanamide; b. pentanamide; c. 4-ethylhexanamide; d. 3,3-dimethylbutanamidee. 5-ethyl-3-methylheptanamide; f. 3,3,5-trimethylhexanamide

26–19 ©1997, A.J. Girondi

9. a. 2,3-dichloropentane; b. difluoromethane; c. 4-bromo-2-fluoro-3-iodohexane;d. 3-bromo-1-butene; e. 1-bromoheptane; f. dichlorodifluoromethane

F – C – F

F

F

Cl – C – C – H

Cl

Cl H

H

Cl

F

F

I

CH C – CH – CH – CH2 – CH – CH3

Br

Br

CH3

Cl

Cl

Cl

10. a. b. c. d.

e. f. g.

11. a. OK; b. 2-methyl-2-butanol; c. 4-bromo-2,2,3-trimethylheptane; d. 3-methyl-3-hexanol;e. 2-chloro-3,3-dimethyl-2-butanol; f. 1-butanol

Br Br

Br

CH3 – CH – CH – CH3

Cl

Cl

CH3 – CH2 – CH – CH – CH2 – OH

CH3

CH2 – CH3

CH3 – CH2 – CH2 – O – CH2 – CH3 C

O

HCH3 – C – CH2 –

CH3

CH3

CH3 – CH – C – CH2 – CH2 – CH2 – CH3

O

O – CH3

CH3 – CH2 – C – CH2 – CH3

O

C

O

HCH3 – CH2 – CH – CH – CH2 –

CH2 – CH3

CH2 – CH3

C

O

HCH3 – CH2 – CH – CH2 – CH –

F

F

12. a. b. c.

d.

OH

I

CH – CH3

CH3

e. f.

g. h.

i. j.

12. k. CH3–CH2 –CH C

OH

OOH

C

O – CH2 – CH3

O

CH3

C

O – CH2 – CH2 – CH3

O

CH3 – (CH2)6 CH3 – CH2 – CH – CH – CH2 – CH2 – CH3

Cl

Br

l.

m. n.

26–20 ©1997, A.J. Girondi

C

NH2

O

CH3 – CH2 H – N – CH3

CH2 – CH3

H – C – N – CH3

CH3

CH3

H3C

C

NH2

O

CH3 – CH – CH2

CH3

CH3 – CH – CH2 – C – CH3

Cl O

CH3 – CH – CH – CH C

OH

O

I I

I

12. o. p. q.

r. s.

t.

13. a. aminoethane; b. 2-aminopropane

26–21 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


CHAPTER 27


(Part 3)

27-1 ©1997, A.J. Girondi

SECTION 27.1 Aromatic Compounds

An important class of carbon compounds exists which was not discussed in Chapters 25 or 26.This class of compounds was discovered during the early years of the study of organic chemistry.Because these compounds had sweet smelling odors they came to be known as the aromatics, althoughsome of them have no such odor.

These compounds are still called aromatics, although the name no longer applies to any odor.Instead, compounds are now classified as aromatic if they contain a particular structure.

The alkanes, alkenes, and alkynes discussed in Chapter 25 are categorized according to whetherthey contained single, double, or triple bonds. Alcohols, ethers, aldehydes and the other classes ofcompounds discussed in Chapter 26 are categorized according to the presence of a specific functionalgroup. The parent compound found in all aromatic molecules is benzene.

Benzene was discovered in 1825 by Michael Faraday who was at the time analyzing "illuminatinggas." Benzene, itself, is the simplest member of the aromatic family. About ten years after its discovery, itsformula was determined to be C6H6. However, the problem of determining the structure of benzenepersisted for several decades. The molecule contains only one hydrogen atom for each carbon atom.Many different structures were proposed, some of which are shown below.

H–C C – C = C – C = C – H

H

H

H

H

H – C = C – C C – C = C – H

H

H H

H

H – C C – C C – C – C – H

H H

HH

H – C – C C – C C – C – H

H H

HH

H–C C – C = C = C – C – H

H H

H

H

H – C – C C – C = C = C – H

H

HH

H

Structure 1 Structure 2 Structure 3

Structure 4 Structure 5 Structure 6

Benzene was found to be a rather stable molecule. Eventually, chemical testing revealed thatnone of these six "straight-chained" structures could explain the properties of benzene. Finally, in 1865,the chemical structure of benzene was predicted by the German chemist, Kekule. As the story goes,Kekule fell asleep one night sitting in front of a fire and had a dream about chains of carbon atomsbehaving like twisting snakes. Suddenly, one of the snakes bit its own tail forming a ring, giving Kekulethe inspiration to spend the rest of the night devising a ring structure for benzene. As a result, Kekule isremembered by his famous quote, "Let us learn to dream gentlemen, and then perhaps we shall learn thetruth."

H

H

H

H

H

H

The structure which Kekule proposed was a ring consisting of sixcarbon atoms joined by alternating single and double bonds. Eachcarbon atom was also bonded to one hydrogen atom. The stability of themolecule needed to be explained, as did the fact that when it reacts in aone-to-one mole ratio with bromine, benzene forms only one di-substitutedproduct (containing two bromine atoms), when two different productswould be predicted as shown below. Benzene

27-3 ©1997, A.J. Girondi

H

H

H

H

H

H

H

Br

H

H

H

H

H

H

H+ Br2 OR

Br Br

Br

This evidence suggested that all of the bonds in benzene were equivalent. To solve this problem,Kekule suggested that the double bonds and single bonds in the ring shifted back and forth, and thisshifting made all of the bonds sites equivalent:

H

H

H

H

H

H

However, benzene does not react like the alkenes which really do have double bonds. Today, we use thetheory of resonance to describe the structure of benzene. This theory suggests that benzene does notcontain single bonds or double bonds. Instead, the bonds between the carbon atoms are all identical.They are called resonance hybrids meaning that they have properties which are between single anddouble bonds, such as the length of the bonds. The length of carbon to carbon single bonds and carbonto carbon double bonds are 1.54 Angstroms and 1.34 Angstroms, respectively. The length of the carbonto carbon resonance hybrid bonds in the benzene ring is 1.39 Angstroms. According to this theory eachcarbon on the molecule would be identical. Therefore, only one product would be expected whenbenzene reacts with bromine. You may still see benzene represented with alternating single and doublebonds, but it is understood that they are actually identical resonance hybrids. Some chemists use othersymbols to represent benzene, including the structure shown below with the circle inside the ring. (Thehydrogen atoms present on the ring are usually not written. They are assumed to be there, unless theyare replaced by an atom of another element.)

OR

Benzene Benzene

Section 25.2 Naming Aromatic Compounds

Aromatic compounds have both common names and IUPAC names. In Chapters 25 and 26 wedid not include common names. However, because they are still widely used with aromatic compounds,we will include them in this chapter. When a hydrogen atom on the benzene ring is replaced by someother "substituent," the compound can be named as a derivative of benzene. In other words, benzene isconsidered to be the parent compound. Common substituents include:

bromo –Br nitro –NO2

chloro –Cl amino –NH2

fluoro –F hydroxy –OHiodo –I methyl –CH3

ethyl –C2H5

27-4 ©1997, A.J. Girondi

Naming the compounds in this way, gives the IUPAC names:

methylbenzene ethylbenzene isopropylbenzene

CH3 CH2 – CH3 CH3 – CH – CH3

Cl Br NO2 NH2

chlorobenzene bromobenzene nitrobenzene aminobenzene

Note how the IUPAC names always end with "benzene." When the ring contains only one substituent, itis not necessary to include numbers in the name, because all of the carbons are equivalent.

When the ring contains more than one substituent, the names become more complex. When thering has two substituents, the prefixes ortho, meta, and para are used to describe the positions of thesubstituents. Ortho refers to adjacent positions, while meta describes two positions separated by onecarbon atom. Para positions are located across from each other on the ring. The prefixes can beabbreviated as o-, m-, and p- as in the examples below. The substituents appear in the names inalphabetical order:

Br

Cl

OH

CH3

Cl

NO2

o-bromochlorobenzene m-hydroxymethylbenzene p-chloronitrobenzene

Prefixes are used when more than one identical substituent occurs on the ring. Note the use of the prefix"di" in the name of the structure below.

Prefixes are used when more than oneidentical substituent occurs on the ring. Notethe use of the prefix "di" in the name of thestructure shown at right.

The structure at left has only twosubstituents. Note that in the IUPAC namethe substituents are in alphabetical order(chloro before hydroxy).

Cl

Cl

p-dichlorobenzene(paradichlorobenzene)

Cl

OH

m-chlorohydroxybenzene

27-5 ©1997, A.J. Girondi

Problem 1. Give IUPAC names for the structures shown below. Write your answers in the spaceprovided.

I

Cl

NH2

OH

Cl

F

F

a. b.

c. d.

a.__________________________________

b.__________________________________

c.__________________________________

d.__________________________________

Structural isomers are compounds that have the same formula but different structures. Threestructural isomers of dibromobenzene are possible. They are shown below with their IUPAC names.

Br

Br

Br

Br

Br

Bro-dibromobenzene

(orthodibromobenzene)m-dibromobenzene

(metadibromobenzene)p-dibromobenzene

(paradibromobenzene)

When the ring has three or more substituents, they arelocated by numbering the ring. The carbons in the ringare numbered so as to give the substituents the lowestpossible numbers (as was the case with the cycliccompounds you studied in previously). The compoundshown at right is called 1,2,4–trichlorobenzene.

Cl

Cl

Cl

Prefixes like di, tri, etc., are not considered whenputting names of substituents in alphabetical order.Consider the name of the structure at right. It is called1-fluoro-2,3-dimethylbenzene. Note that "fluoro" comesbefore "methyl." The prefix "di" is ignored.

CH3

F

CH3

Note that the structure at right is 1-bromo-2-chloro-4-hydroxybenzene. It is not called 1-hydroxy-3-chloro-4-bromobenzene,because numbering the ring that waywould give put higher numbers in the name.

Br

Cl OH

27-6 ©1997, A.J. Girondi

When numbering a ring in either of two ways givesthe same set of numbers, then give the lowest number tothe substituent which appears first in the name (seestructure at right). It is called 1-bromo-2-chloro-3-iodobenzene. It is not called 1-iodo-2-chloro-3-bromobenzene.

I

Br

Cl

Problem 2. Name the compounds shown below using the IUPAC system:

Cl

CH3 CH2 – CH3

Br

F

OH

a. b. c.a.

b.

c.

d. e. f.

Cl CH3

CH3

Cl

Br

Cl

NO2

NO2

F

F

d.

e.

f.

Problem 3. Draw structures which satisfy each of the following IUPAC names:

a. 1,2-dichloro-4-methylbenzene b. 1,3,5-trimethylbenzene

c. o-diiodobenzene d. m-hydroxyiodobenzene

27-7 ©1997, A.J. Girondi

e. p-diethylbenzene f. 1,3-dibromo-5-chlorobenzene

Problem 4. The following IUPAC names are incorrect. Draw the structure that satisfies the name, andthen write the correct name:

a. 2,3-dichlorobenzene b. 1-chloro-5,6-dibromobenzene

_____________________________ _____________________________

c. 2-hydroxy-4,6-difluorobenzene d. 1-chloro-2-amino-5-chlorobenzene

_____________________________ _____________________________

Section 27.3 The Common Names of Aromatic (Benzene) Compounds

Remember, in the IUPAC system compounds are named as derivatives of benzene and,therefore, IUPAC names end with "benzene." Note the common names of the compounds shown below.The IUPAC names are also given.

CH3 OH NH2

Common Name:IUPAC Name:

toluenemethylbenzene

phenolhydroxybenzene

anilineaminobenzene

27-8 ©1997, A.J. Girondi

We will now consider the common names of some benzene compounds that have twosubstituents on them. Note, however, that they are named as derivatives of toluene, phenol, or aniline,rather than benzene.

CH3

Cl

OH

Br

NH2

NH2

o-chlorotoluene m-bromophenol p-aminoaniline

Problem 5. Give the IUPAC names of the three structures shown above.

a.______________________________________________

b.______________________________________________

c.______________________________________________

When the two substituents on the benzene ring are methyl groups, the compounds are known asxylenes in the common naming system. There are three forms of xylene:

CH3

CH3

CH3

CH3

CH3

CH3

o-xylene m-xylene p-xylene

Since xylenes are a special case, they should not be named as derivatives of toluene. For example,o-xylene should not be named o-methyltoluene. The common naming system, like the IUPAC system,makes use of the numbering of the ring for structures with three or more substituents. Note the commonnames of the structures below:

CH3

NH2

I

Cl

OH

Cl

NH2

Br

Common Names: 2-amino-4-iodotoluene 2,5-dichlorophenol m-bromoaniline

27-9 ©1997, A.J. Girondi

Problem 6. Give the IUPAC names of the three structures above.

a.___________________________________________

b.___________________________________________

c.___________________________________________

TNT is an abbreviation that comes from the commonname for an explosive. Its structure is shown at right.Its common name is 2,4,6-trinitrotoluene. What is itsIUPAC name?

{1}_______________________________________________________

NO2NO2

NO2

CH3

CH3

Br

Br

Examine the structure at right. Its common name is 2,3-dibromotoluene. Note that "methyl" is not part of its commonname because the methyl group is part of the toluene structure.Since it is part of the parent compound in the common name(toluene), the methyl group is numbered as being on the number1 carbon in the ring. That's why we do not call it 1,2-dibromotoluene.

Problem 7. Name the following compounds twice, using the common and IUPAC systems.

F

OH

F

a.

CH2 – CH3

NH2b.

Common: _________________________________________

IUPAC:___________________________________________

Common: _________________________________________

IUPAC:___________________________________________

CH3c.

CH3

Common: _________________________________________

IUPAC:___________________________________________

27-10 ©1997, A.J. Girondi

CH3d.

Bre.

Common: _________________________________________

IUPAC:___________________________________________

Common: _________________________________________

IUPAC:___________________________________________

f. Common: _________________________________________

IUPAC:___________________________________________

Cl

Cl

OH

I

Cl

NH2

Problem 8. The following common names are incorrect. Draw the structure that satisfies the namegiven, and then write the correct common name:

a. 4,5-dichlorophenol

_____________________________________

b. 3-bromotoluene

_____________________________________

c. 4,5-dibromoaniline

_____________________________________

d. 4-methyltoluene

_____________________________________

Section 27.4 Condensed Ring Structures

There are some common aromatic structures which are not composed of benzene rings withsubstituents on them. These compounds are known as condensed ring structures. They look somewhatlike a number of benzene rings which have been bonded together. Three examples of condensed ringstructures which are extracted from coal tar include:

27-11 ©1997, A.J. Girondi

naphthalene anthracene phenanthrene

Like p-dichlorobenzene, naphthalene is often used to make moth balls. Anthracene is used inthe manufacture of dyes, and steroids are compounds which are based on the structure of phenanthrene.Since aromatic chemistry developed in a rather haphazard way for many years, many of these compoundswere given common names (such as those above) which are still used today. These compounds do notactually exist in coal tar; instead, they are formed when coal tar is heated (distilled). It is believed that somecondensed ring structures are formed whenever organic molecules are heated to a high temperature.This includes the burning of tobacco in cigarettes. The bad news is that many of these compounds havebeen shown to produce cancer. Workers at plants where coal tar is distilled have had higher than normalrates of skin cancer. Needless to say, condensed ring structures are also suspect in the development oflung cancer.

Section 27.5 Learning Outcomes

This is the end of your study of organic nomenclature in the ALICE program. You should knowthat organic chemistry is a very large field of study. New organic compounds and products are being madeevery year. Students of organic chemistry not only learn about the classification and nomenclature ofcompounds. This is actually only a very small part of what they study. Some of the most importantresearch being done in science today, involves the study of reactions that organic compounds undergo.This field is very important in agriculture, medicine, manufacturing, and many other fields of humanendeavor. Be sure that you have mastered each of the learning outcomes below.

_____1. Distinguish between organic structures which belong to the aromatic class of compounds and those which do not.

_____2. Draw the structure of simple aromatic molecules given their IUPAC names.

_____3. Write the IUPAC names of simple aromatic molecules given their structures.

_____4. Draw the structures of simple aromatic molecules given their IUPAC names.

_____5. Draw the structures of simple aromatic molecules given their common names.

_____6. Recognize the structure of condensed ring structures.

27-12 ©1997, A.J. Girondi

Section 27.6 Answers to Questions and Problems

Questions:

{1} 1-methyl-2,4,6-trinitrobenzene

Problems:

1. a. iodobenzene; b. m-aminochlorobenzene; c. p-chlorohydroxybenzene; d. o-difluorobenzene

2. a. m-chloromethylbenzene; b. p-bromoethylbenzene; c. m-fluorohydroxybenzene;d. 4-chloro-1,2-dimethylbenzene; e. 1-bromo-2,4-dichlorobenzene;f. 1,2-difluoro-4,5-dinitrobenzene

a. b. c.Cl

Cl

CH3

CH3CH3

CH3

I

I

I

OH

CH2 – CH3

CH2 – CH3

Cl

Br

Br

d. e. f.

3.

a. Cl

Cl

d.

4.

1,2-dichlorobenzene

Cl

Br

Br

b.

1,2-dibromo-3-chlorobenzene

OHF

F

c.

1,3-difluoro-5-hydroxybenzene

NH2

Cl

Cl

1-amino-2,4-dichlorobenzene

5. a. o-chloromethylbenzeneb. m-bromohydroxybenzenec. p-diaminobenzene

27-13 ©1997, A.J. Girondi

6. a. 2-amino-4-iodo-1-methylbenzeneb. 1,4-dichloro-2-hydroxybenzenec. m-aminobromobenzene

7. a. 3,5-difluorophenol1,3-difluoro-5-hydroxybenzene

b. m-ethyltoluene1-amino-3-ethylbenzene

c. m-xylenem-dimethylbenzene

d. 2,5-dichlorotoluene1,4-dichloro-2-methylbenzene

e. p-bromophenolp-bromohydroxybenzene

f. 2-chloro-4-iodoaniline1-amino-2-chloro-4-iodobenzene

a. OH

Cl

d.

8.

3,4-dichlorophenol

CH3

Br

b.

o-bromotoluene

c.

2,3-dibromoaniline

Cl

CH3

p-xylene

CH3

Br

Br

NH2

27-14 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


CHAPTER 28

NUCLEARCHEMISTRY

(Part 1)

28-1 ©1997, A.J. Girondi

SECTION 28.1 Nuclear Notation and Isotopes

Nuclear chemistry involves changes that occur in the nucleus of an atom. These changes in anucleus often result in the release of great amounts of energy – much greater than the amount of energyreleased in any chemical reactions. You will recall that chemical reactions involve the formation andbreaking of bonds between atoms. In addition to the release of energy, certain types of particles areemitted from a nucleus during nuclear reactions. Before going on, there are a few basic facts which youshould know:

1. Most of the mass of an atom is found in the nucleus. This is a result of the relatively close "packing" of the protons and neutrons in it.

2. All protons carry a positive charge.

3. Because they all carry a positive charge, the protons in a nucleus repel each other with a strongforce; yet, the nucleus of a stable atom does not fall apart.

You may recall that it is the number of protons in the nucleus of an atom (the atomic number) thatdetermines what element the nucleus represents. A nuclear change sometimes involves a change in thenumber of protons in the nucleus. When this happens, a nucleus of one element is changed into anucleus of a different element. This is called a transmutation. In a previous chapter, you wereintroduced to nuclear notation. Let's review it now to refresh your memory. The general form for nuclearnotation can be represented by the expression shown below:

XA

Zsymbol of the element

mass number(sum of protons and neutrons)

atomic number(number of protons)

What would the expression A minus Z, or A - Z, represent? {1}_________________________________

If the value of Z changes, will X change?{2}_______________ If Z changes by a value of 2, what will

happen to the value of A? {3}_________________________________________________________

You learned previously that atoms of an element can exist in different forms known as isotopes.Isotopes are atoms of an element that contain different numbers of neutrons. Therefore, isotopes havedifferent masses and different mass numbers – although they have the same atomic number. Someelements have many isotopes, while others have only a few. In addition, some isotopes of elements arenaturally-occurring while others are man-made. Some isotopes are unstable, meaning that theydecompose or break apart on their own. Many elements possess both stable and unstable isotopes.Unstable isotopes are said to be radioactive. They give off energy and/or nuclear particles when theydecompose. In Table 28.1, the mass numbers of isotopes of some selected elements are shown.

If ALL of the isotopes of an element happen to be radioactive, then the element itself iscategorized as being radioactive. With this in mind, which of the selected elements listed in Table 28.1should be categorized as radioactive? {4}____________________________ How many radioactive isotopes doescarbon (C) have?{5}_________ How many nonradioactive isotopes does nitrogen (N) have?{6}_________ How many man-made radioactive isotopes does helium (He) have?{7}_________ All

elements on the periodic table with an atomic number of 84 or greater are radioactive. These elements areshown as they occur on the periodic table in Table 28.2.

28-3 ©1997, A.J. Girondi

Table 28.1 Isotopes of Some Selected Elements

In this table, mass numbers of naturally-occurring nonradioactive isotopes are given in plain type; massnumbers of naturally-occurring radioactive isotopes are double-underlined; mass numbers of any otherisotopes are single-underlined. Naturally-occurring isotopes are listed in their order of abundance. All otherisotopes are listed in order of decreasing half-life which is discussed later in this chapter.

Element Mass numbers of isotopes H 1, 2, 3 He 4, 3, 6, 8 Be 9, 10, 7, 11, 6 B 11, 10, 8 C 12, 13, 14, 11, 10, 15, 16, 9 S 32, 34, 33, 36, 35, 38, 37, 31, 30, 29 N 14, 15, 13, 16, 17, 18

Ca 40, 44, 42, 48, 43, 46, 41, 45, 47, 49, 50, 39, 38, 37

Sn 120, 118, 116, 119, 117, 124, 122, 112, 114, 115, 126, 123, 113, 125, 121, 110, 127, 128, 111, 109, 108, 129, 131, 130, 132

U 238, 235, 236, 234, 233, 232, 230, 237, 231, 240, 229, 239, 228, 227 Lr 260, 256, 255, 254, 257, 256, 252, 251, 258

The simplest element, hydrogen, has three isotopes. The most common form ofhydrogen (protium) has one proton and no neutrons in its nucleus. Its atomic number is 1,and its mass number is 1. The nuclear notation for protium is shown at right. In natureapproximately 99.985% of all hydrogen atoms are protium.

H1

1

The remaining 0.015% of hydrogen consists of deuterium atoms. Also known as heavy

hydrogen, deuterium differs from protium in that it has one neutron in the nucleus in addition to one

proton.Using the letter D instead of H as the symbol, write the nuclear notation for deuterium:

{8}_______________ Protium and deuterium are both stable, naturally-occurring isotopes. Water (H2O)

molecules which contain deuterium instead of protium are known as "heavy water" which is sometimes

represented as D2O. About two water molecules in every billion are "heavy." A third form of hydrogen is

man-made and is radioactive. It is known as tritium, and it is a common by-product of the nuclear reactions

that occur in a nuclear power plant. Tritium has two neutrons in its nucleus. Using the letter T instead of H

as the symbol, write the nuclear notation for tritium. {9}________________

28-4 ©1997, A.J. Girondi

1A

2A 3A 4A 5A 6A 7A

8A

84 Po

85 At

86 Rn

87 Fr

88 Ra

89 Ac

104Unq

105 Unp

90 Th

91 Pa

92 U

93 Np

94 Pu

95 Am

96Cm

97 Bk

98 Cf

99 Es

100 Fm

101 Md

102 No

103 Lr

106 Unh

107 Uns

108 Uno

109 Une

110 Uun

111 Uuu

Table 28.2The Radioactive Elements

(All of their isotopes are radioactive)

43 Tc

61 Pm

It is common to identify which particular isotope of an element is being discussed by writing themass number after the name of the element with a dash in between. For example, protium is hydrogen-1,while deuterium is hydrogen-2. Following this method, how would tritium be written?{10}_______________ What is meant by mass number? {11}___________________________________________________

SECTION 28.2 Four Types of Nuclear Reactions

The equation at right represents a nuclear change. We will refer toit as a nuclear equation. More specifically, it depicts the change of an atomof carbon-14 into an atom of nitrogen-14:

C -----> N + e14

6

14

7

0

-1

0

-1e

p1

+1

n1

0

electron:

proton:

neutron:

Nuclear equations often include a special type of notation to represent subatomicparticles such as electrons, protons, and neutrons. This notation looks similar tonuclear notation which represents a nucleus, but it is not the same. Thenotations describing an electron, a proton, and a neutron are shown below. Notethat the superscripts represent the mass numbers of each particle. The massnumber of an electron is zero. However, the subscripts represent the charge onthe particle. Note that neutrons have no charge, so the subscript for them is zero.Thus, the difference between nuclear notation and the notation for thesesubatomic particles lies in the meaning of the subscript.

28-5 ©1997, A.J. Girondi

14

6C p

1

+1

NUCLEAR NOTATION SUBATOMIC PARTICLE NOTATION

mass number

atomic number

mass number

charge

According to the data in Table 28.1, is carbon-14 a radioactive isotope?{12}_________ How aboutnitrogen-14?{13}_________ Note that if the atomic number changes during a nuclear reaction, theidentity of the resulting element changes, too. In the equation shown below, a nucleus of carbonbecomes a nucleus of nitrogen as the atomic number changes from 6 to 7. An electron is also given off asa product. But hey! If the atomic number changes from 6 to 7, this means that one addition proton is nowpresent. Where did it come from? Hmmmm.

C -----> N + e14

6

14

7

0

-1

Electrons are sometimes called beta particles (pronounced "bay-ta"). So, the giving off of anelectron in a nuclear reaction is called a beta emission. In order for carbon-14 to change to nitrogen-14,there was an increase in the number of {14}_________________ in the nucleus. When a neutrondecomposes, the products are a proton and an electron. The new proton causes the atomic number toincrease by one, and the electron is given off. When the decomposition of a neutron produces a proton,the mass number remains unchanged. Since one element is changed into another in this reaction, thisparticular type of nuclear reaction is called a {15}____________________________.

There are four types of nuclear reactions that release energy:

1. Natural Radioactive Decay

Natural radioactive decay refers to the ability of a nucleus to decompose (decay) and give offenergy spontaneously (without any external stimulation). As a result, the number of {16}______________(atomic number) in the nucleus may increase or decrease, depending on the type of radioactive decay.The equation below in which carbon-14 is converted to nitrogen-14 represents a natural radioactivedecay.

C -----> N + e14

6

14

7

0

-1

2. Artificial Transmutation

During artificial transmutation, a nucleus changes its identity as a result of some externalstimulation created by man. For example, an external particle such as a neutron could be used to bombardthe nucleus, causing it to decompose. This kind of nuclear disintegration results in the formation of anartificial (man-made) isotope of the element. The equation below shows the conversion of naturalnonradioactive cobalt-59 to radioactive cobalt-60 by a process known as slow neutron bombardment.

n1

0Co +

59

27Co

60

27---->

Notice that since a neutron is being added to the nucleus, the mass number of the nucleus increases byone, from 59 to 60. The atomic number remains unchanged since the number of {17}______________________ isunchanged. Since the atomic number remains unchanged, the identity of the nucleus (cobalt) remainsthe same. What we have done here is to change one isotope of cobalt into a different isotope of cobalt.

28-6 ©1997, A.J. Girondi

3. Fission

In fission, a nucleus with a large mass splits into two nuclei with smaller masses. To cause fission,man bombards certain nuclei with special particles. The fission process is used to generate heat in nuclearpower plants, and is the kind of reaction which occurs during the explosion of an atomic bomb. Let's seewhere this energy comes from. Look at the equation below which represents the fission of uranium- 235.Find the total of the mass numbers of the two particles on the left side of the equation: {18}__________.

n1

0n

1

0U +

235

92----> Ba +

138

56Kr +

95

363 + energy

Next, find the total of the mass numbers of the five particles on the right side: {19}__________. How dothese totals compare? {20}______________________________ As a result, you would think that massthe amount of matter) is conserved (neither created nor destroyed). However, this is a bit misleading.Keep in mind that the mass number is the total number of the protons and neutrons in the nucleus, nottheir exact total mass. Remember that masses of atoms and subatomic particles are expressed in very tinyunits called atomic mass units (amu). The mass of an atom of U-235 is actually a little greater than 235 amu,and the masses of Ba-138 and Kr-95 are actually a little less than 138 and 95, respectively. Therefore, inthe equation above, there is a small loss of mass which appears as a great amount of energy. In otherwords, some mass is converted into energy. An atomic bomb gives off a tremendous amount of heatbecause some mass is converted into energy. A tiny amount of mass can produce a tremendous amountof energy. When the uranium nucleus splits into smaller nuclei, the energy which was needed to hold thewhole thing together in the first place is no longer needed. This is the energy which is given off.

4. Fusion

When fusion occurs, the nuclei of two lower mass elements are combined to form a nucleus with agreater mass representing a different element. Exceedingly high temperatures are needed to causefusion to occur, since the two nuclei repel each other due to their similar positive charges. Fusionreactions are the source of the sun's energy where hydrogen nuclei combine to form helium nuclei. Theequation below shows the fusion of 2 deuterium nuclei to form one helium-4 nucleus (also called an alphaparticle).

H2

1H

2

1+ -------> He

4

2+ energy

Fusion reactions were used in weapons such as the hydrogen bomb. Scientists are experimenting withfusion reactions in devices known as breeder reactors which may someday replace fission reactors innuclear power plants. Fusion, like fission, results in a loss of mass which is converted into a great amountof energy. However, fusion releases much more energy per gram of fuel than fission does.

Problem 1. Let's practice writing nuclear notation. Keep in mind that the superscript is the massnumber (sum of protons and neutrons) and the subscript is the atomic number (number of protons) if theparticle is a nucleus. If the particle is a subatomic particle (proton, electron, or neutron,) then the subscriptis the charge on the particle. Write the nuclear notation for each of the following:

a. an isotope of carbon (C) which contains 6 protons and 8 neutronsb. an isotope of helium (He) which contains 2 protons and 4 neutrons c. an isotope of uranium (U) which contains 92 protons and has a mass number of 233d. an isotope of tin (Sn) which contains 50 protons and 60 neutrons

a.____________ b.____________ c.____________ d.____________

28-7 ©1997, A.J. Girondi

Now, let's try working with some nuclear equations. Keep in mind that in a balanced nuclearequation, the total of the superscripts of all particles must be equal on both sides of the equation. Thesum of the subscripts of all particles must also be equal on both sides. For example, consider theequation below.

Ra226

88Rn

222

86-----> + He

4

2

In this example, an isotope of radium (Ra) decomposes into an isotope of radon (Rn), and thisdecomposition is accompanied by the emission of a helium nucleus which is also called an alpha particle.What is the sum of the superscripts on the right side of the equation?{21}_______________ How does thiscompare with the superscript on the left side?{22}_____________________ What is the sum of thesubscripts on the right side?{23}__________________ How does this compare to the subscript on the left side?{24}___________________________________ Is this nuclear equation balanced?{25}_________________

Problem 2. Complete the following transmutation reactions, indicating in each case, the nuclearnotation of the element formed. What element is formed in the first equation below? Well, if you check itout, the atomic number of the missing particle will have to be 6. What element has an atomic number of 6?{26}________________________ Therefore, what element symbol will the missing particle have?{27}________________

Be 9

4He

4

2+ -----> + n

1

0a.

Si 28

4D

2

1+ -----> + n

1

0b.

Al 27

13+ -----> +c. n

1

0He

4

2

D2

1Mn

55

25+ -----> +d. n

1

02

Na 24

11-----> +e.

0

-1e

Complete the following equations indicating in nuclear notation, in each case, what particle - if any - wasejected. Answers may include:

electron: 0

-1e proton: p

1

+1neutron: n

1

0alpha particle: He

4

2

N 14

7+ -----> +f. n

1

0B

11

5

B10

5D2

1Be

9

4+ -----> +g.

28-8 ©1997, A.J. Girondi

He4

2+ -----> +h. Al

27

13P

30

15

U 239

92-----> Np +

239

93i.

The radioactive elements with atomic numbers 84 through 92 (up to and including uranium) havesome naturally-occurring radioactive isotopes. The elements beyond uranium (with atomic numbersgreater than 92) do not have any naturally occurring isotopes. These elements beyond uranium areknown as the transuranium elements . They are all synthetic elements since all of their isotopes are man-made. Most of the radioactive elements (with atomic numbers 84 and above) are too unstable to beassigned an atomic mass (atomic weight). If you look at a periodic table, you will notice that the atomicmasses of these elements are given in parentheses. (Check this out on a periodic table now.) Thenumber in the parentheses represents the atomic mass of the single most stable isotope. You will recallthat atomic mass is defined as the average mass of the various naturally occurring isotopes of an elementin the proportions in which they occur in nature. The radioactive isotopes of elements with atomicnumbers 84 and above are constantly decomposing. These isotopes have different half-lives, whichmeans that they are decomposing at different rates. Use this information to explain why these elements

cannot have an atomic mass as defined above: {28}________________________________________

______________________________________________________________________________

Most elements with atomic numbers smaller than 84 are stable because NONE of their naturally-occurring isotopes are radioactive. There are some exceptions to this rule. For example, K-40 and Ca-46are radioactive. Most of the elements below atomic number 84 are stable enough to be assigned anatomic mass. (Elements #43 and #61, Technetium and Promethium, are exceptions.) Man-maderadioactive isotopes have been synthesized for many of these elements, but synthetic isotopes are notincluded in the calculation of atomic masses since they are not found in nature.

Section 28.3 Early Studies of Radioactivity

In1896, a French scientist by the name of Henri Becquerel accidentally discovered naturalradioactivity while conducting experiments with a uranium compound called potassium uranyl. In one ofhis experiments, Becquerel wrapped a photographic plate in black, lightproof paper and placed some ofthe uranium compound on top of the covered plate. He then placed this arrangement in the sunlight.Although the sunlight could not pass through the lightproof paper, the plate became exposed in the areaof the uranium compound, as indicated by a dark area on the photograph. Becquerel thought thatperhaps energy from the sun had been changed into some more penetrating form which was able to passthrough the paper. He then attempted to repeat the experiment, but cloudy weather prevented him fromdoing so at that time. He decided to store his second set-up in a closed drawer. Later, on a sunny day,Becquerel repeated the experiment using a fresh photographic plate instead of the one he had stored inthe closed drawer. He then developed both of the photographic plates. Since the stored plate had notbeen exposed to sunlight, Becquerel expected the developed photograph to be blank or almost blank.Instead, he found that it had a dark area like that of the fresh plate which had been exposed to sunlight.Becquerel reasoned that the uranium compound must have emitted some type of energy on its ownwithout the stimulation of sunlight. This ability of a nucleus to emit energy spontaneously (withoutexternal stimulation) is called natural radioactivity. Uranium ore exhibits natural radioactivity with thegreatest amount of energy coming from its most abundant naturally-occurring isotope, U-238.

28-9 ©1997, A.J. Girondi

Becquerel also discovered that as the energy is emitted from a radioactive nucleus and passesthrough molecules of oxygen and nitrogen in the air, it causes these molecules to lose electrons, formingpositively charged ions. As a result, the air becomes ionized. The fact that radioactive nuclei can ionizegases is a principle used in the construction of equipment which can detect the presence of radioactivity.You probably have a smoke detector in your home. The most common form of smoke detector contains asmall sample of a radioactive element (probably americium). The radiation emitted is capable of ionizingsmall particles in the air. When enough particles are present (as during a fire), the ions which are producedallow an electric current to form and the alarm goes off.

An electroscope is a device which can detect and store an electric charge. See Figure 28.1. Asimple electroscope can be constructed by attaching two pieces of thin metal foil to a metal rod. Thisapparatus is then sealed inside a glass container such as a jar. When the electroscope is in its normal"uncharged" state, the two pieces of metal foil will hang beside each other. To convert the electroscopeto its "charged" state, we have to supply it with an excess of electrons. How do you do this? Well, thereare many ways. Even by combing your hair and then touching the comb to the metal rod on theelectroscope will do it. The electrons on the comb (which came from your hair) will flow into the rod andinto the two pieces of metal foil. At that point, both pieces of foil would carry a negative charge and theywould repel each other. The greater the amount of charge they hold, the more they repel each other. So,an electroscope is a crude device for detecting and measuring an electrical charge. The air around the foilin the electroscope acts as an insulator, helping to prevent the electroscope from losing its stored chargeright away. It is much harder for electrons to flow through air than through metal. If you touch the metal rodon the electroscope with any substance which is a good "acceptor" or conductor of electrons (such as apiece of metal), the excess electrons will flow out of the electroscope which will then lose its charge.

discharged weakly charged highly charged

Figure 28.1An Electroscope

When nuclear radiation ionizes the air forming positively-charged particles, these positive particlescan draw negatively-charged electrons away from an electroscope in which they might be stored. It ispossible to measure the rate at which radioactive emissions occur by measuring the rate at which anelectroscope loses its charge. Marie Sklodowska, a student of Becquerel, used an electroscope to studythe radioactivity of uranium and its various ores. She found that one uranium ore, pitchblende, gave off

28-10 ©1997, A.J. Girondi

much more radioactivity than even pure uranium. After her marriage to the physicist Pierre Curie, theyboth studied the radioactivity of pitchblende. The Curies discovered that the increased radioactivity ofpitchblende was due to the presence of two elements in the ore. Madame Curie called the first radioactiveelement which they discovered in the ore "polonium" after her native land, Poland. Find polonium (Po) onthe periodic table. What is its atomic number?{29}_______________ On the periodic table, the mass number ofpolonium is (210). What is so special about Po-210 and why is this mass number given in parentheses?It took the Curies four years to complete the processing of the ore from which they extracted only 0.1 gramof the second radioactive element, radium, in the form of radium chloride. Radium (Ra) has what atomicnumber on the periodic table?{30}___________ Its mass number is given as (226). Both polonium andradium were found to be more radioactive than uranium. Although the use of the electroscope allowedthe Curies to measure the rates at which radiation was emitted, it did not provide any indication as to thenature of the radiation. In other words, it did not indicate whether the radiation consisted of energy, orparticles, or both.

In 1903, Ernest Rutherford performed an experiment which provided some new informationabout the properties of radiation. He placed a piece of pitchblende into a hole drilled deep into a block oflead. (See Figure 28.2) Most of the radiation emitted by the pitchblende was absorbed by the lead. Onlythe radiation that was traveling in a straight line through the hole could escape. A photographic plate waspositioned in the path of the escaping radiation. When the plate was developed, a small single spotappeared where it was struck by the radiation.

Next, Rutherford placed the poles of a U-shaped magnet at right angles to the stream of radiation.This forced the radiation to pass through a magnetic field. Since a magnetic field deflects oppositelycharged particles in opposite directions, it was possible to determine the charge of any particles in theradiation. Streams of radiation which do not contain particles would not be affected by the magnetic field.When the magnetic field was used, three distinct spots were produced. (See figure 26.2.) The threespots indicated that the magnet had separated the radiation into three distinct streams. Two streams weredeflected in opposite directions, whereas one stream was not deflected at all. How many of these threestreams contained particles?{31}__________ Why were the two affected streams deflected in oppositedirections?{32}___________________________________________________________________

The two deflected streams are called alpha (∝) and beta (ß) radiation in Figure 28.2. The unaffectedstream was called gamma (∂) radiation. What must be true about the stream of gamma radiation that wasnot deflected? {33}_________________________________________________________________________________________________

radiation

single spot formed

photographic plate

Lead

pitchblende

Figure 28.2Rutherford's Study of Radiation from Pitchblende

radiation

photographic plate

Lead

pitchblende

(–) (+)

magnet

3 spots formed

28-11 ©1997, A.J. Girondi

He4

2Nuclear Notation for Helium-4

or for an Alpha Particle

The particles which were deflected only slightly in adirection indicating a positive charge were called alpha particles.

The Greek symbol for alpha is: ∝. The fact that they were onlyslightly deflected indicated that they had a relatively large masscompared to beta particles. In later experiments, it was shownthat alpha particles were actually bundles composed two protonsand two neutrons. They have the same structure as heliumnuclei. You can say that the term alpha particle is another namefor a helium nucleus. Alpha particles are, therefore, designatedby the same nuclear notation as is the most common isotope ofhelium which is helium-4. Alpha particles travel at 10,000 to20,000 miles per second, but can be stopped by a sheet ofpaper. They have a great ability to cause ionization by knockingelectrons loose from atoms or molecules through which theypass.

The very low mass particles were deflected much more than the alpha particles and in theopposite direction. Apparently, they were negatively charged. Rutherford called them beta particles. TheGreek symbol for beta is: ß . They were later shown to be electrons which travel at a rate of up to 100,000miles per second! Their ability to penetrate matter when they strike it is much greater than that of alphaparticles; nevertheless, they still cannot penetrate more than a few inches of solid material. Beta particlescause much less ionization than alpha particles.

The radiation emitted between the alpha and beta streams was not deflected at all by the magneticfield and, therefore, carries no electric charge. This stream was called gamma radiation. The Greek symbolfor gamma is: ∂. Gamma rays are similar to x-rays, but are higher in energy. Their penetrating power ismuch greater than either alpha or beta radiation, and they can penetrate almost one foot of solid lead!Gamma rays travel at the speed of light (186,000 miles per second). They cause practically no ionization atall when they interact with atoms or molecules. Table 28.3 summarizes some of the informationpresented about the three forms of radioactivity. Complete the column headed "Penetrating Power" byinserting the terms high, low, and moderate in the proper slots. Next, complete the column headed"Ionizing Power" by inserting the terms high, moderate, and almost none in the proper slots.

Table 28.3The Three Forms of Natural Radioactivity

Penetrating IonizingDecay Product Symbol Charge Power Power

alpha particle {34}_________ {37}_________

beta particle {35}_________ {38}_________

gamma rays none none {36}_________ {39}_________

He4

2

0

-1e

+2

–1

In general, a radioactive isotope of an element emits alpha particles or beta particles, but not both. The

emission of gamma rays generally accompanies both alpha emissions and beta emissions. Which of the

three kinds of radioactive emissions is needed in order for a transmutation to occur? {40}______________

Explain: {41}_____________________________________________________________________

______________________________________________________________________________

28-12 ©1997, A.J. Girondi

Name three radioactive elements found in pitchblende: {42}___________________________________

SECTION 28.4 Methods of Detecting Radiation

Electroscopes

Radioactivity has an effect on matter as it passes through it. We can, therefore, study radioactivityby recording and measuring these effects. You already know that nuclear emissions can exposephotographic plates and can ionize gases. Some measuring devices make use of the fact that gases willconduct electricity when they become ionized as a result of exposure to radiation. For example, theelectrical charge stored in an electroscope can be lost when the air inside and around the electroscopebecomes ionized. See Figure 28.3 below.

incoming radiation ionizes the air

chargelost

ions of airinside heremolecules

of air

charged foil strips

Figure 28.3Effect of Radiation on Stored

Charge

Ionization chambers

In an ionization chamber, radiation passes through a gas. The radiation causes the gas particles tobe split into pairs of ions which are then collected on the surfaces of oppositely charged electrodes. Thenumber of pairs of ions produced can be measured. An example of a measuring instrument using thisprinciple is the self-reading dosimeter. With such a device, radiation can be measured in units calledRoentgens. This may sound a bit complicated, but a Roentgen is the amount of gamma radiation requiredto produce 1.61 X 1012 pairs of ions when it is absorbed by 1 gram of air.

Geiger Counter

A Geiger counter (more accurately known as a Geiger–Mueller counter) consists of a sealed tubecontaining argon gas at a low pressure. One end of the tube contains a thin glass window. There are twoelectrodes in the tube (see Figure 28.4). The negative electrode is a metal cylinder located just inside thetube. The positive electrode is a wire which runs down the center of the cylindrical tube. A high voltageexists between these electrodes, but electric current does not flow, since the uncharged (un-ionized)argon gas atoms cannot carry the current from one electrode to the other. When radiation enters throughthe thin window, it ionizes some of the argon atoms, forming argon ions and free electrons. The argonions become conductors of electric current between the electrodes. The electrical impulses are then sentinto an amplifier. From there they may be sent to a counter or to an amplifier to be converted into soundsor flashes of light.

28-13 ©1997, A.J. Girondi

1000 Volts

To amplifier or counter

negative electrode

positive electrode

argon gas

thin glasswindow

incomingradiation

Figure 28.4Geiger Counter

Photographing Particle Trails

Figure 28.5Particle Trails in a Cloud Chamber

As you know, fast moving charged particles suchas those present in radioactive emissions can cause theformation of ions when they collide with molecules throughwhich they pass. If this process occurs in a container whichis saturated with water vapor, the water molecules cancondense on ions forming tiny spots of fog. This fog formsalong the paths of the radioactive emissions since that iswhere the ions form. These foggy paths are visible to theeye. They are called trails. Photographs of these particletrails enable scientists to study how certain decays occur.The device in which all this takes place is called a cloudchamber. In Figure 28.5, the curved vertical linerepresents the path of a subatomic particle passingthrough a thin sheet of lead. The path is curved due to thepresence of a strong magnetic field in the cloud chamber.

Scintillation Counter

When radiation strikes fluorescent substances (known as phosphors) it causes flashes of light tobe emitted. This is what happens in a fluorescent light bulb or on a television screen. There areinstruments which can count these small flashes of light, and in this way, measure radiation. The processof producing light flashes is called scintillation. The devices are called scintillation counters .

28-14 ©1997, A.J. Girondi

Section 28.5 More Practice With Nuclear Equations

Problem 3. Complete the equations below, and make sure that they are balanced.

+14

7Na. He

4

2----->

17

8O +

+ 9

4Beb. He

4

2----->

12

6C +

c. H3

1----->

3

2He +

+ 23

11Nad. He

3

2----->

1

1H +

13

7N+e. He

3

2----->

1

0n +

Now, complete the equation below. Does anything appear strange? An electron with a positive charge!

f. P30

15----->

0

+1e +

Yes, there is such a thing as an electron with a positive charge. It's call a positron. As you can imagine,there's a lot more to know about nuclear chemistry!


This is the end of Chapter 28. The subject of nuclear chemistry is continued in Chapter 26.Review the learning outcomes below. When you feel that you have mastered them, arrange to take theexam on Chapter 26, and then move on to Chapter 27.

_____1. Define and /or describe nuclear terms including: isotope, transmutation, alpha particle, beta particle, gamma rays, fission, fusion, radioactivity, Geiger counter, scintillation counter, and cloudchamber.

_____2. Write the nuclear notation of nuclear particles and of the nuclei of atoms given mass numbers, atomic numbers, or other relevant data.

_____3. Describe the historical contributions of Becquerel, Madame and Pierre Curie, and Rutherford.

_____4. Given sufficient information, complete and balance nuclear equations.

_____5. Be able to locate the radioactive elements on the periodic table.

28-15 ©1997, A.J. Girondi


Questions:

{1} number of neutrons; {2} yes; {3} it will also change by a value of 2; {4} U and Lr; {5} six; {6} two;{7} none; {8} 21D; {9} 31T; {10} hydrogen–3; {11} sum of protons and neutrons in nucleus; {12} yes;{13} no; {14} protons; {15} transmutation; {16} protons; {17} protons; {18} 236; {19} 236 (note that thereare three neutrons represented); {20} they are equal; {21} 226; {22} equal; {23} 88; {24} equal;{25} yes; {26} carbon; {27} C; {28} since some isotopes are decomposing, the average mass of theisotopes is changing; {29} 84; {30} 88; {31} two; {32} they contained particles with opposite charges;{33} it contains no particles; {34} almost none; {35} moderate; {36} high; {37} high; {38} moderate;{39} almost none; {40} alpha or beta; {41} alpha emission results in loss of 2 protons, while beta emissionresults in formation of one proton; {42} polonium, radium, uranium

Problems:

14

6C He

6

2U

233

92Sn

110

50a. b. c. d.1.

12

6Ca.2. B

29

5b. Na

24

11c. Fe

55

26d. Mg

24

12e. He

4

2f. n

1

0g. n

1

0h i.

0

-1e

1

+1Pa.3. b. c. B

11

5d. Mg

25

12e. Si

30

14f.n

1

0

0

-1e

28-16 ©1997, A.J. Girondi

NAME________________________________ PER ________ DATE DUE ___________________


CHAPTER 29

NUCLEARCHEMISTRY

(Part 2)

29-1 ©1997, A.J. Girondi

SECTION 29.1 The Zone of Stability

Transmutations occur naturally in radioactive elements, since the nuclei of the atoms are unstable.As these nuclei "transmute," they form nuclei of elements which are more stable. Let's examine why thisis so. Various factors are responsible for the stability of a nucleus. For now, let's consider just two of them:(1) the "zone of stability," and (2) the 1:1 neutron/proton ratio. These two factors are illustrated in Figure29.1 below.

0

20

40

60

80

100

120

140

10 20 30 40 50 60 70 80

zone of stability

nuclei with equal numberof protons and neutrons(1:1 ratio)

unstableregion

too manyneutrons

unstableregion

too manyprotons

Number of Protons

Numberof

Neutrons

Figure 29.1The Zone of Stability of Stable Nuclei

The neutron/proton ratios of the first 83 elements are shown as the zone of stability in figure29.1. Locate the shaded zone of stability now. The number of neutrons in a nucleus is plotted along thevertical y axis, and the number of protons in the nucleus is plotted on the horizontal x axis. What is anotherterm which refers to the number of protons in a nucleus?{1}___________________________________Find number 20 on the x axis and go straight up from there to the center of the zone of stability. From thisposition, move over to the y axis and determine the number of neutrons. Calculate the neutron/proton(N/P) ratio and enter it in Table 29.1. Repeat this procedure using 30, 40, 50, 60, and 70 protons on the xaxis. Record all N/P ratios in the table.

Data Table 29.1Neutron/Proton Ratio of Selected Elements

No. of Protons No. of Neutrons N/P Ratio

20 ____________ _______

30 ____________ _______

40 ____________ _______

50 ____________ _______

60 ____________ _______

70 ____________ _______

29-3 ©1997, A.J. Girondi

Are the N/P ratios which you calculated greater than 1 in value?{2}_______________ What can you conclude

about the relative number of protons and neutrons in these nuclei {3}___________________________

Although your calculations stopped at 70 protons, this conclusion holds true for elements 20 through 83.

Based on your calculations, how would you say the N/P ratio changes as the atomic mass (atomic weight)

of elements 20 through 83 increases? {4}________________________________________________

______________________________________________________________________________

Note in Figure 29.1 that the zone of stability lies very close to the 1:1 N/P ratio line for nuclei which

contain 1 to 20 protons. These nuclei are not radioactive. What conclusion can you draw about the

stability of those nuclei which have N/P ratios that are equal to or close to 1:1? {5}___________________

______________________________________________________________________________

(Hydrogen-1 has no neutrons and ,therefore, no N/P ratio. However, it is stable.)

Nuclei which do not lie in the zone of stability will undergo transmutations that will result in theformation of new nuclei (of different elements) that do lie in the zone of stability. The graph in Figure 29.1ends with atomic number 83, indicating that no stable nuclei exist in atoms with more than 83 protons.Find 50 protons on the x axis and go up from there to the center of the zone of stability. Next, look to theleft at the y axis and estimate the number of neutrons. About how many neutrons would a nucleus with 50protons have to lose to achieve a stable 1:1 N/P ratio?{6}_____________________________________

SECTION 29.2 Types of Radioactive Decay

Naturally-occurring radioactive elements have isotopes which may undergo radioactive decay.This decay can occur in one step or in a series of steps which end with the formation of a nucleus whichfalls within the zone of stability. Let's consider two kinds of decay: alpha and beta.

Alpha Decay

He4

2

As you may recall from Chapter 25, an alpha particle is identical to the nucleus ofthe helium-4 atom. So, an alpha particle is actually a bundle of protons and neutrons.How many of each compose an alpha particle? {7}________ protons, and {8}________neutrons.

Many isotopes with an atomic number of 84 or higher undergo transmutation by emitting an alphaparticle and form products which are closer to the zone of stability . The emission of an alpha particle froma nucleus is known as alpha decay. An atom which emits an alpha particle is called an alpha emitter.Radium-226 with an atomic number of 88 will serve as an example of an alpha emitter. During its decay, analpha particle is ejected which means that the nucleus loses {9}______ protons and {10}______ neutrons.After radium-226 undergoes an alpha decay, what will be the atomic number of the product? {11}________If this is true, what is the identity of the new element formed after alpha emission? {12}_______________(Check the periodic table.)

The equation for the alpha decay of radium-226 can be written:

Ra226

88Rn

222

86-----> + He

4

2

29-4 ©1997, A.J. Girondi

In an alpha decay, the atomic number is reduced by {13}________ and the mass number is reduced by{14}________. An alpha decay results in the formation of a more stable product with {15}______________mass and with {16}_______________ protons and neutrons. more/less

more/fewer

Beta Decay

neutron

proton

electron

neutrino

Natural radioactive decay in which betaparticles (electrons) are given off from the nucleus iscalled beta decay. An atom which emits a beta particleis called a beta emitter. Because a beta particle is anelectron, it is natural to wonder how a nucleus can emitan electron when the nucleus doesn't contain anyelectrons! It so happens that a neutron in the nucleuscan be transformed into a proton, an electron, and aneutrino.

The proton is a fundamental particle of the nucleus and remains in the nucleus after the neutrontransformation. So, a neutron transformation reduces the number of neutrons in the nucleus by 1, butincreases the number of protons by 1. Thus, the atomic number increases by 1, but the mass numberremains the same.

Why is it that when a neutron is lost and a proton is formed in a nucleus that the mass number does not

change? {17}_____________________________________________________________________

Since the electron is not a fundamental nuclear particle, it is emitted from the nucleus with a great amountof energy when a neutron is transformed.

The beta decay of thorium-234, which has an atomic number of 90, is shown in the followingequation:

Th234

90Pa

234

91-----> + e

0

-1thorium protactinium

beta particle(electron)

Since the atomic number is increased by 1, the element formed as the product of the transmutation is oneposition "higher" than thorium on the periodic table. In any transmutation by beta decay, the new elementformed as the product will be one position "higher" in the periodic table than the beta emitter. Note thatthe equation above is balanced. This means that the sum of the mass numbers of the products (0 + 234 =234) equals the mass number of the reactant (234). Note, too, that the sum of the subscripts of theproducts (-1 + 91 = 90) is equal to that of the reactant (90).

Calculate the number of neutrons present in a thorium-234 nucleus:{18}___________

Calculate the number of neutrons in a protactinium-234 nucleus: {19}____________

Calculate the N/P ratio for thorium-234: {20}___________

Calculate the N/P ratio for protactinium-234: {21}___________

Based on your calculations above, in beta decay does the N/P ratio move closer to or farther from a 1:1

ratio?{22}_____________________________________ If this is true, then does a beta decay form

products with more stability or less? {23}_____________

29-5 ©1997, A.J. Girondi

SECTION 29.3 Decay Series

A series of elements from the periodic table which are related by a series of alpha and beta decaysis called a decay series. Several such decay series are known among the naturally-occurring radioactiveelements. The uranium-238 decay series shown in Figure 29.2 is a good example.

Note that this series of transformations begins in the upper left corner with U-238 and continues down to an isotope of lead in the lower right corner. Lead-206(Pb-206) with atomic number 82 is stable – it lies within the zone of stability. You seethat a beta decay results in a vertical rise in Figure 29.2; this is because a beta decayresults in an increase in atomic number (y axis) and an unchanged mass number (xaxis). An alpha decay results in a diagonal lowering because of decreases in bothatomic number (y axis) and mass number (x axis). Note that on the x axis, the massnumbers are decreasing from left to right. If you carefully check Figure 29.2, you willsee that two elements in the series are missing! The first missing element is the onethat results from an alpha (α) emission from radium-226 (Ra-226). Write the nuclearnotation of this missing element in the slot at right. Now fill in the space in Figure29.2 with this nuclear notation.

{24}

{25}

The second missing element is found after a beta (ß) emission frombismuth-210 (Bi-210). Write the nuclear notation for this missing element in thespace at right. Now fill in the space in Figure 29.2 with this nuclear notation.

SECTION 29.4 The Half-Life of a Radioactive Element

At the present time, it is impossible to predict exactly when a particular radioactive nucleus willdecay. The uranium-238 nucleus that breaks down today is exactly like the one that will do so a billionyears from now. However, we can predict what fraction of a sample of nuclei will break down in a givenamount of time. For example, one-half of the nuclei in a given sample of carbon-14 will decay in 5570years. One-half of the remaining half (or one fourth of the original sample) will decay in another 5570years. One-half of what is left (or one eighth of the original sample) will decay in another 5570 years, etc.Thus, carbon-14 is said to have a half-life of 5570 years.

The half-life of a radioisotope is the time that is required forhalf of any given sample of its nuclei to decay.

As shown in Figure 29.3, the amount of radioactivity emitted is directly proportional to the numberof nuclei in any given sample of an isotope. In the case of the hypothetical radioisotope illustrated by thegraph in Figure 29.3, the amount of radioactivity emitted from any given sample decreases with time. Thisis because the number of nuclei in the sample decreases with time as a result of decay.

According to Figure 29.3, what percent of radioactivity is emitted at zero minutes? {26}_________How much time has passed when the amount of radioactivity emitted drops to 50%? {27}___________ Sohow much time was required for half of the nuclei to decay? {28}_________ What is the half-life (inminutes) of this radioisotope?{29}__________ If this is true, then how many total minutes should it take forthe number of nuclei to decrease by one-half again (1/4 of the original sample)? {30}__________ Whatshould the percent of radioactivity emitted be at this point?{31}__________ The graph in Figure 29.3should verify your prediction. Does it?{32}_________ What fraction of the original number of nuclei will beleft in this sample after twelve minutes have passed? {33}______________________________ Whatpercent of the original amount of radioactivity is still present after twelve minutes?{34}___________

29-6 ©1997, A.J. Girondi

Figure 29.2 The decay series for uranium-238 goes on this page. This page can be found on your ALICEdisk listed as a separate file (ALICE CHP 29 pg 29-7).

29-7 ©1997, A.J. Girondi

50%

25%

12.5%6.25%

100%

RadioactivityEmitted

0 2 4 6 8 10Minutes

Figure 29.3Emitted Radioactivity as a Measure of Half-Life

The rate at which unstable atoms decay is constant and is not influenced by external forces suchas temperature and pressure. At present, there is no method which can be used to alter the half-life of aradioisotope. The half-lives of different radioisotopes vary from a small fraction of a second to billions ofyears. Polonium-215 has a half-life of only 0.0018 second. Uranium-238, on the other hand, has one ofthe longest known half-lives, about 4.5 billion years (which is approximately the same as the age of theearth). Therefore, what fraction of uranium-238 remains on the earth today compared to when this planetwas formed? {35}____________________

Since the human lifespan is so short, you may wonder how we are able to determine a half-life of4.5 billion years. Various indirect methods are used, including chemical ones. The main problems whentrying to determine the length of a half-life are to identify the source of the radioactivity coming from asample of a radioisotope and to determine its rate of decay. It is, of course, possible that some isotopesthat we currently think are stable and do not decay actually have half-lives that are so long that we have notbeen observing them long enough to detect their radioactivity!

Let's take another look at the decay series for uranium-238 shown in Figure 29.2. Any rockcontaining U-238 will also contain all of the "daughter" elements of the series. (The daughter elementsare those that are formed as a result of decay process.) The relative amounts of each daughter elementpresent will depend on the age of the rock. Therefore, if we can measure the amounts of certain daughterelements in the rock, we can determine its age. The half-lives of the elements in the U-238 decay seriesare presented in Table 29.2.

Let's review now by answering the following questions and problems. Why is the atomic number

of the transmuted element from an alpha decay always two less than the atomic number of the alpha

emitter from which it came? {36}_______________________________________________________

Why is the atomic number of the transmuted element from a beta decay always one greater than the

atomic number of the beta emitter from which it came? {37}____________________________________

___________________________ Since the nucleus does not contain any electrons, how is it possible

for a nucleus to emit an electron in beta decay? {38}_________________________________________

29-8 ©1997, A.J. Girondi

Table 29.2Half-Lives of Isotopes in the U-238 Decay Series

Symbol Half-Life Symbol Half-Life

U-238 4.51 X 109 yrs Po-218 3.05 minutes Th-234 24.1 days Pb-214 26.8 minutes Pa-234 1.18 minutes Bi-214 19.7 minutes U-234 2.48 X 106 yrs Po-214 0.00016 sec Th-230 8.0 X 104 yrs Pb-210 19.4 yrs Ra-226 1.62 X 103 yrs Bi-210 5.0 days Rn-222 3.82 days Po-210 138.4 days

Pb-206 stable

What is the half-life of a radioisotope? {39}________________________________________________

How could a Geiger counter be used to measure the half-life of a given sample of a radioisotope? (Hint:

see Figure 27.3) {40}_______________________________________________________________

______________________________________________________________________________

Problem 1. Write a balanced nuclear equation for each of the following:

a) emission of a beta particle by protactinium -234 (Pa-234)

b) emission of an alpha particle by thorium-230 (Th-230)

c) emission of a beta particle by lead-214 (Pb-214)

Problem 2. If the half-life of a radioisotope is 120 days, how much of an 8.00 gram sample will be leftafter 360 days?

Problem 3. The bromine-35 nucleus has a half-life of 18 minutes. How many minutes are required forseven-eighths (7/8) of a sample to decay?

__________ min.

29-9 ©1997, A.J. Girondi

SECTION 29.5 The Mass Defect

The mass number of a proton or a neutron is taken to be 1, but their actual masses in atomic massunits (amu) are fractional values: proton = 1.00783 amu; neutron = 1.00867 amu. When writing nuclearequations we use mass numbers and the sum of the values of the mass numbers on both sides of theequation must be the same. However, if all calculations are performed using atomic mass units, thendiscrepancies occur. For example, when the mass of the protons and neutrons of a helium nucleus areexamined separately, they are found to have more mass than when they are combined together forming anucleus. The amu of the two protons and two neutrons in a helium nucleus, when not combined, can becalculated by simple multiplication as follows:

Particles mass in amu

2 protons 1.00783 X 2 = 2.01566 amu2 neutrons 1.00867 X 2 = 2.01734 amu

Sum of the masses = 4.03300 amu

So, the masses of 2 protons + 2 neutrons when not combined = 4.03300 amu. A helium-4 nucleuscontains 2 protons and 2 neutrons. If the total mass of a helium-4 nucleus is determined, it is found tohave less mass than the sum of the amu's of its protons and neutrons when they are not combined. (SeeTable 29.3 below.)

Table 29.3Mass of Particles in Helium Nucleus Separately and Combined

Helium nucleus Mass

Particles separate 4.03300 amu Particles combined 4.00260 amuDifference 0.0304 amu Mass defect 0.0304 amu

This difference between the sum of the separate masses of the nuclear particles and the mass resultingfrom their merging is called the mass defect. A mass defect exists for every atom; that is, the sum of themasses of the individual protons and neutrons of any atom is greater than the mass of these particleswhen they have merged in any given nucleus. As shown in Table 29.3, the helium nucleus has a massdefect of 0.0304 amu. What has happened to this mass? Where does the 0.0304 amu of mass go whenthe particles join together?

You may recall Albert Einstein's famous equation E = mc2 states that matter and energy are reallydifferent forms of the same thing. The missing matter of the helium nucleus has not been lost ordestroyed, but converted into a large amount of energy when the nuclear parts are joined to form thehelium nucleus. We know a relationship which allows us to convert units of mass (amu) into units ofenergy: 1 amu = 931 Mev. Mev stands for million electron volts. Since you are already familiar with joulesas units of energy, it may help to express the relationship as 1 Mev = 1.59 X 10-7 joules. Let's summarizeall this below:

1 amu = 931 Mev

1 Mev = 1.59 X 10-7 joules

29-10 ©1997, A.J. Girondi

So, to convert the mass defect in helium into the energy equivalent:

0. 0304 amu X

931 Mev

1 amu = 28.3 Mev of energy

OR

0. 0304 amu X

931 Mev

1 amu X

1.59 X 10-7 J

1 Mev = 4.50 X 10-6 Joule

Note that when a very tiny mass like 0.0304 amu is converted into energy, the energy equivalentexpressed in joules is very small. That's why a smaller unit of energy such as Mev is more suitable forthese kinds of calculations. Thus, when 2 protons and 2 neutrons combine to form a helium nucleus,28.8 Mev of energy (or 4.50 X 10-6 Joule) of energy are released. The protons and neutrons are foundinside the nucleus which occupies only a tiny fraction of the volume of the atom. The protons each have apositive charge and repel each other with a large amount of force, since they occupy the nucleus at closedistances. In order for the protons and neutrons to be held together in the nucleus, an even larger forcemust be present to overcome the repulsive forces. The amount of energy needed to hold the nuclearparticles together in the nucleus is called the nuclear binding energy.

(Binding energy can also be defined as the energy needed to separate a nucleus into its individual particles.)

We have seen that when the protons and neutrons of helium combine to form the heliumnucleus, there is a mass defect of 0.03040 amu. If you multiply this value by 931 Mev you obtain a bindingenergy of 28.3 Mev for the helium nucleus. The binding energy per individual proton or per individualneutron in the nucleus can be determined by dividing the bonding energy for the nucleus by the totalnumber of particles in the nucleus. Since the helium nucleus contains 2 protons and 2 neutrons, thebinding energy per particle is:

28.3 Mev

4 particles = 7.08 Mev/particle

The greater the binding energy per particle, the more stable the nucleus. Helium has a relatively highbinding energy per particle, as you might have guessed since it is a stable noble gas atom.

Problem 4. Suppose the nucleus of hypothetical element "X" has a mass defect of 0.0484 amu. Themass number of this element is 8. Determine the binding energy per particle in Mev for element X:

__________ Mev / particle

Is the nucleus of element X more or less stable than the helium nucleus? {41}______________________

Explain:{42}______________________________________________________________________

______________________________________________________________________________

29-11 ©1997, A.J. Girondi

SECTION 29.6 Man-Made Isotopes - Stable and Radioactive

In the remainder of this chapter you will be reading about artificial (man-made) transmutations,nuclear fission, and nuclear fusion. Before that, however, let's review and extend our knowledge of howto write nuclear equations. It is not difficult to determine what new element is formed in a transmutation ifthe "secondary" particle which is formed can be identified. Secondary particles include neutrons,protons, alpha particles, beta particles, and positrons which are positively charged electrons (they werementioned at the end of Chapter 25, remember?).

Problem 5. As a review, write the nuclear notation for each of the particles listed below.

a. neutron ________; b. proton ________; c. alpha particle ________; d. beta particle ________;

e. positron ________

We know that in nuclear reactions the total mass of the reactants only approximates the total massof the products because of the mass defect. However, because the mass defect is so small, it is does notshow up in the mass numbers (which are always whole numbers). Therefore, any nuclear equation,written as it is with mass numbers and atomic numbers both in whole numbers, must be in balance. That is,the total of the mass numbers (superscripts) must be equal on both sides of the equation, and the same istrue for the total of the subscripts on both sides.

Let's consider an example of "alpha capture." Some of this will be a review of what you havealready learned in Chapter 25. Examine the equation below. If a beryllium (Be) nucleus is bombardedwith an alpha particle, the beryllium nucleus will capture it, making the nucleus unstable. The unstablenucleus disintegrates, releasing a {43}_______________ and forming a new element. In the equationbelow, the symbol of the new element is shown as "X." "A" is the mass number, and "Z" is the atomicnumber.

+ 9

4Be He

4

2----->

A

ZX +

1

0n

What is the sum of the mass numbers on the left side of the equation?{44}______ So, what must the sumof the mass numbers be on the right side?{45}______ What mass number must the newly formed elementhave?{46}______ What is the sum of the atomic numbers on the left side of the equation?{47}______ So,what must the sum of the atomic numbers be on the right side?{48}______ Based on this atomic number,what is the symbol for the newly formed element?{49}______ Now rewrite this equation below,substituting the correct nuclear notation for the newly formed element:

{50}___________________________________________________

The nuclear equation you wrote above is an example of an artificial transmutation. Scientists produce thistype of nuclear reaction by bombarding a nucleus with particles (nuclear projectiles) so as to cause nucleardisintegration and the formation of a new element. Artificial isotopes, produced by this method, may bestable or they may be radioactive.

Artificial Stable Isotopes

In 1919, Rutherford bombarded nitrogen gas with alpha particles which he obtained from aradioactive source. The reaction can be pictured as the capture of an alpha particle by the nitrogennucleus to form an isotope of fluorine.

14

7N He

4

2+ -----> F

18

9Step 1:

29-12 ©1997, A.J. Girondi

This isotope of fluorine is unstable and emits a proton, leaving a stable isotope of oxygen-17:

----->F18

9Step 2: O

17

8+

1

1H

Steps 1 and 2 may be combined and written as an overall reaction:

14

7N He

4

2+ -----> O

17

8+

1

1H

Since this first artificial transmutation, many different isotopes have been produced, not only by usingradioactive sources, but also by using particles given very high velocities in special accelerators such as alinear accelerator and a cyclotron . When particles collide at very high velocities, transmutations can occur.

Artificial Radioactive Isotopes

In the 1930's the daughter of Madame Curie, Irene Joliot-Curie and her husband, Frederic Joliot,bombarded a stable aluminum nucleus (mass no. 27, atomic. no. 13) with an alpha particle emitted by aradioactive source. The products were a phosphorus nucleus (mass no. 30, atomic no. 15) and a neutron.Write this nuclear equation in the space below:

{51}________________________________________________

Note that this phosphorus-30 was produced by artificial transmutation. The Joliot-Curies expected thatupon removal of the alpha source, the radioactivity would soon stop. However, unlike the oxygen-17produced by Rutherford in the previous example above, the phosphorus produced by this reaction wasnot stable. Long after the removal of the radioactive source of bombarding particles, the sample ofphosphorus continued to exhibit radioactivity. The primary product of this reaction was an artificiallyproduced radioactive isotope (radioisotope). The ability to make a stable nucleus unstable (radioactive) bybombarding it with a high-energy projectile is known as induced radioactivity. A radioactive isotope may bethe product of natural radioactive decay (alpha or beta), or it may be an artificial isotope (man-made). By far,most of the known radioactive isotopes are man-made.

The value of radioisotopes can be appreciated when it is realized that these isotopes can bedetected wherever they are. Since a radioisotope reacts chemically in the same way as a stable isotope ofthe same element, a radioisotope can be substituted for the stable isotope in a chemical reaction. Sincethe radioisotope gives off radioactivity, its presence can always be detected no matter how it is chemicallycombined. Because of this, radioisotopes are called tracers or tagged atoms. Tracers are very valuable inhelping scientists to study the sequence of chemical reactions in living things. For example, fertilizercontaining many stable phosphorus atoms and some atoms of the radioisotope phosphorus-32 can be"fed" to plants. Radiation detection equipment can be used to determine where the greatestconcentrations of phosphorus occur in the plant. Experiments of this kind are simple to perform.

The radioisotope carbon-14 is useful in dating old objects containing carbon. It has a half-life of5570 years. An archeologist can use a Geiger counter to measure the radiation coming from an ancientwooden axe handle and thereby determine the amount of carbon-14 in a given sample (say one gram) ofthe wood. If the sample has only one-fourth as much carbon-14 as one gram of present-day wood, howmany half-lives has the ancient wood experienced?{52}___________ About how old is the ancient wood?{53}_______________ As the hunt for new isotopes continued, two German scientists, Otto Hahn andFritz Strassmann, made a startling discovery. When they bombarded uranium-235 (atomic number 92)with neutrons, two elements and a great deal of energy were produced. Their chemical analysis indicatedthat barium (atomic number 56) was and krypton (atomic number 36) were the elements that were

29-13 ©1997, A.J. Girondi

produced. It was difficult to believe these results. No transmutation had been previously observed inwhich the atomic numbers of the products differed from the atomic number of the reactant as much asthey did in this case:

n1

0n

1

0U +

235

92----> Ba +

138

56Kr +

95

363 + energy

big difference!

The German chemists Lise Meitner and Otto Frisch concluded that if a heavy element wereconverted into two middleweight elements, large amounts of binding energy would be given off. Theamount of energy liberated is roughly the difference in the total binding energy per nucleon (proton orneutron) of the elements in the reactants and the products. For example, suppose uranium-238 (with abinding energy of 7.6 Mev per nucleon) could be converted directly to iron (which has a high bindingenergy of about 8.8 Mev). This would result in 238 X (8.8 - 7.6)Mev = 286 Mev. It is not quite this simple inpractice because the neutron/proton ratio and other factors influence results. Nevertheless, a great dealof energy is released during fission, the process by which a heavy nucleus is split into two lighter nuclei.

Now let's return to the reaction discovered by Hahn and Strassman. The fission of U-235 beginswith a neutron capture. Write the nuclear equation below which shows what happens when a U-235nucleus absorbs a neutron. Hint: the only product is a uranium isotope.

{54}_______________________________________________

The product of the above reaction is unstable and exists for a very short period of time and thenundergoes spontaneous fission. The products of its fission are barium-141, krypton-92, 3 neutrons andenergy. Write this nuclear equation in the space below.

{55}_______________________________________________

Note that three neutrons are produced as a result of the fission of U-236. Perhaps some of theseneutrons could be used to continue the reaction. If the neutrons produced by this reaction are used tobegin other fissions, a chain reaction occurs. (See Figure 29.4.)

n1

0 U235

92+ U

236

92

Ba141

56

Kr92

36

n1

0

n1

0

n1

0

U235

92

U236

92

n1

03

Ba141

56

Kr92

36

U235

92U

236

92n

1

03

Kr92

36

Ba141

56

Figure 29.4Schematic Drawing of a Chain Reaction

29-14 ©1997, A.J. Girondi

A chain reaction is one in which a product of one reaction (for example, a neutron) is used tocause the reaction to occur again. To keep a chain reaction going, we must be sure that at least oneneutron from each fission causes another nucleus to undergo fission. A chain reaction has an efficiencyof 1 if each fission causes 1 additional fission. If the efficiency can be maintained at about 1, a continuoussupply of energy can be produced. When each reaction produces more than 1 reaction, the efficiency isgreater than 1 and the amount of energy constantly increases.

An atomic bomb is a device with an efficiency of greater than 1; an atomic power plant mustmaintain an efficiency of approximately 1. In either an atomic bomb or a nuclear power plant, a minimumamount of nuclear fuel (uranium or plutonium) is needed to maintain the chain reaction. This minimumamount of nuclear fuel is called the critical mass. It is not possible to continue the chain reaction unless atleast this amount of nuclear fuel is available to it.

A device capable of sustaining and regulating a chain reaction involving fission is an atomic pile, ormore correctly, a nuclear reactor. The efficiency of a chain reaction depends on the fate of the neutronsproduced by the fission reactions. The possible fates of these neutrons are:

1. They can escape from the sample of nuclear fuel.

2. They can be absorbed by nuclei of atoms other than those of the nuclear fuel.

3. They can be absorbed by the nuclear fuel and continue the chain reaction.

In nuclear reactors, moderators are used to assure a steady supply of neutrons, and control rodsare employed to control the rate of the chain reaction. (See Figure 29.5 below.)

pump

hot water flow

concrete containment building

steel reactor vessel

control rod

uranium fuel rod

hot water and moderator

heat exchanger

secondary watersystem

water flow to steam turbine

cool water return

Figure 29.5A Nuclear Reactor System

A moderator is a material which will not absorb neutrons but will slow their rate of speed. As theneutrons produced by the fission of uranium collide with the moderator, they are slowed down andprevented from making a rapid escape from the nuclear reactor. Thus, by slowing down the neutrons, amoderator makes neutron capture more likely. Commonly used moderators are: carbon in the form ofgraphite, and "heavy water." (The water molecules of heavy water contain two atoms of the hydrogenisotope deuterium, hydrogen-2, instead of two atoms of ordinary hydrogen.)

We have noted that moderators do not absorb neutrons. Control rods, however, do absorbneutrons. For example, cadmium and boron effectively absorb neutrons. Rods of these materials (controlrods) can be inserted into, maintained in, or removed from rows of uranium. When control rods areinserted, neutrons are absorbed by them and, consequently, the neutrons are no longer available to the

29-15 ©1997, A.J. Girondi

reaction. Accordingly, the insertion of control rods can be used to slow the reaction to an efficiency of 1.This efficiency can be maintained by simply keeping these control rods in position. The efficiency of areaction can be increased by the removal of control rods.

Problem 6. The amount of energy released from the conversion of matter into energy in a nuclearreactor can be appreciated when it is considered that the conversion of only 1 gram of matter into energyreleases the same amount of energy as the burning of over 2000 tons of coal! Calculate in the spacebelow how many grams of coal this would be! (Useful information: 1 ton = 2000 lbs; 454 grams = 1 1b)

Answer = ____________________ grams

SECTION 29.7 Fusion Reactions

Fusion is the combination of the nuclei of light elements to make a heavier nucleus. Because thetwo light nuclei are positively charged and, therefore repel each other, it is difficult to bring them closeenough together to make them fuse. An atomic bomb (fission bomb) can heat atoms so that they haveenough energy to overcome the repulsive forces and combine. This heat from fission reactions is used toachieve fusion in weapons such as the hydrogen bomb. In essence, an atomic bomb is the "fuse" for ahydrogen bomb. Because extremely high temperatures are required to achieve fusion, the fusionreaction is known as a thermonuclear reaction.

Scientists are trying to heat gases of light elements with electricity or lasers and keep them closetogether for a thermonuclear reaction by enclosing them in a magnetic field. Such a controlled fusionreaction, if achieved, would be of great value. It would have advantages over fission reactions, whichemploy uranium in nuclear reactors. Few radioactive waste products are produced by fusion reactions.The radiation hazard would not be as great, and the radioactive wastes are much less dangerous thanthose produced in fission reactors. Furthermore, since hydrogen-2 is much more abundant than uranium,the world would be assured of a continuous supply of fuel.

The sun and other stars show large amounts of hydrogen and helium in their spectra, indicatingthat they may produce their energy as a result of fusion reactions. There are two proposed mechanismsfor these fusion reactions - the carbon cycle and the proton-proton chain. Let's briefly study both of them.

In the carbon cycle, carbon-12 acts as a catalyst and is regenerated in the final step enabling it tocatalyze additional reactions. Parts of the six steps of the carbon cycle are shown in Problem 7 below.

Problem 7. Use your knowledge of nuclear reactions to predict the product of reaction "a" and put itsnuclear notation in the blank. Put this same nuclear notation in the blank on the left side of reaction "b,"and predict the notation of the product of reaction "b." Put this same notation in the blank on the left sideof reaction "c", and predict the notation of the product of that reaction. Continue this process until youhave completed all six reactions. If your work is correct, equation "f" will be properly balanced.

12

6C

1

1H+ -----> + energya.

0

+1e-----> +b.

29-16 ©1997, A.J. Girondi

----->c.1

1H+ + energy

----->d.1

1H+ + energy

0

+1e-----> +e.

He4

2

12

6C----->f.

1

1H+ +

Note in the carbon cycle shown in Problem 7, that carbon-12 is used in reaction "a" and reproduced in reaction "f" as itshould be since it is a catalyst.

Is equation "f" in Problem 7 balanced according to mass numbers and atomic numbers? _____________

The proton–proton chain pictures the fusion reaction as a continued combination of hydrogennuclei (protons) until helium is formed:

a. ----->+2

1H

1

1H

1

1H +

0

-1e + energy

b.2

1H

1

1H+ -----> He

3

2+ energy

+1

1Hc. He

3

2He

3

2+ -----> He

4

22 + energy

Note that both the carbon cycle and the proton-proton chain use hydrogen-1 as a fuel and producehelium-4 as a product. Note also that the hydrogen we started with in reaction "a" ultimately is turned intohelium in reaction "f." Fusion involves the conversion of hydrogen into helium.

The heaviest naturally-occurring nucleus is the isotope U-238, the most abundant isotope ofuranium. As previously noted, the artificial radioactive elements (those above uranium on the periodictable) are the transuranium elements. All of the isotopes of the transuranium elements are radioactive.Some of them are very unstable having half-lives of only a tiny fraction of a second! The transuraniumisotope, plutonium-239, is relatively stable having a half-life of 214,000 years. Uranium-238 can be usedas a target for projectiles or "atomic bullets." These projectiles may be positively-charged particles whichhave been speeded up in particle accelerators by passing them through charged fields toward theirtargets. In all of the equations presented in this section, a high speed positively-charged ion is the atomicbullet. Some transuranium elements with atomic numbers above 93 can be synthesized by using U-238as a target.

29-17 ©1997, A.J. Girondi

Problem 8. In the equations below, predict the products and enter the correct nuclear notations in theblanks. (Pay attention to the number of neutrons produced in each case.)

H2

1U +

238

92----> n

1

02a. +

Name the element formed in equation "a" above: ____________________

He4

2U +

238

92----> n

1

02b. +

Name the element formed in equation "b" above: ____________________

C12

6U +

238

92----> n

1

05c. +

Name the element formed in equation "c" above: ____________________

O16

8U +

238

92----> n

1

04d. +

Name the element formed in equation "d" above: ____________________

Some of these new elements are themselves used as targets for other bombardments. Thus, an isotopeof plutonium has been used as a target in the production of other elements. Complete the equationsbelow.

He4

2Pu +

238

94----> n

1

03e. +

Name the element formed in equation "e" above: ____________________

The element formed in equation "e" above can then be used to produce still another new element. Usethe element formed in equation 5 as the reactant in the equation below, and then complete the equation:

-----> ++f. He4

2

1

1H n

1

02+

Name the element formed in equation "f" above: ____________________

SECTION 29.8 Assorted Problems

Study the next two sample problems, and then solve those which remain.

29-18 ©1997, A.J. Girondi

Sample Problem: Polonium-210 undergoes alpha decay to form lead-206:

He4

2Po

210

84----> +Pb

206

82

The half-life of Po-210 is 138.4 days. How many moles of alpha particles will be emitted if a 105 gramsample of Po-210 decays over a period of 138.4 days? Hint: since this problem deals with a specific isotope(Po- 210) you should not use the atomic mass of polonium from the periodic table. In this case, it will be moreaccurate to use the mass number (or atomic mass, if known) of this specific isotope.

105 g Po -210 X 1 mole Po -210

210 g Po -210 X

0.50 mole Po -210 decayed

1 mole Po -210 in sample X

1 mole 24He produced

1 mole Po -210 decayed

= 0.25 mole 24He produced

Note in the problem above (as well as the one below) that 0.50 mole of the isotope has decomposed foreach 1 mole present in the original sample. This is because one half-life had elapsed.

Sample Problem: How many moles of Po-210 are needed in a sample so that its decomposition willproduce 4.2 moles of lead-206 in 138.4 days? Refer to the equation in the problem above.

4.2 moles Pb -206 formed X 1 mole Po -210 decayed

1 mole Pb -206 formed X

1 mole Po -210 in sample

0.50 mole Po -210 decayed

= 8.4 moles Po -210 in sample (needed)

Problem 9.

a) Write the nuclear equation which shows the beta decay of bismuth-210 (Bi-210) to form a new element.(A beta particle (electron) is emitted.)

b) Using the equation above, calculate how many moles of electrons will be produced as a 166 gramsample of Bi-210 decays over a period of 10 days. (Check Table 29.2 for the half-life of Bi-210.)

Problem 10.

a) C-14 undergoes beta decay to N-14. Express this as a nuclear equation.

29-19 ©1997, A.J. Girondi

b) How many moles of C-14 will produce 3.0 moles of N-14 in 11,000 years? (The half-life of C-14 is 5570years.)

Problem 11. Write the following nuclear equations:

a) an alpha capture in which C-12 and a neutron are the products.

b) a neutron capture in which U-236 is the only product.

SECTION 29.9 LEARNING OUTCOMES

This is the end of Chapter 29. Check the learning outcomes below when you are sure you havemastered them. Then arrange to take the exam on Chapter 29.

_____1. Define and/or describe nuclear terms including: zone of stability, half-life, mass defect, binding energy, chain reaction, decay series, induced radioactivity, critical mass, moderators, control rods, alpha decay, and beta decay.

_____2. Predict the products formed when a nucleus undergoes alpha or beta decay.

_____3. Solve problems involving mass defect and binding energy.

_____4. Solve problems involving the use of the half-life of an isotope.

_____5. Explain how alpha and beta decays affect the neutron/proton ratio, and explain the significance of a 1:1 N/P ratio.

_____6. Given sufficient information, predict the nuclear notation of a missing element in a decay series.

_____7. Describe the historical contributions of Hahn and Strassman, Meitner and Frisch, and Irene and Frederic Joliot.

29-20 ©1997, A.J. Girondi


Questions:

{1} atomic number; {2} yes (the first ratio = 1.0); {3} more neutrons than protons; {4} The N/P ratio getslarger; {5} They are stable; {6} about 25; {7} 2; {8} 2; {9} 2; {10} 2; {11} 86; {12} Radon (Rn); {13} 2;{14} 4; {15} less; {16} fewer; {17} the SUM of protons and neutrons does not change; {18} 144;{19} 143; {20} 1.6; {21} 1.57; {22} closer to; {23} more; {24} 22286Rn; {25} 21084Po; {26} 100%;{27} 2 min; {28} 2 min; {29} 2 min; {30} 4 min; {31} 25%; {32} yes; {33} 1/64; {34} 1.56%; {35} one-half;{36} Because of the loss of an alpha particle which contains 2 protons; {37} Because the loss of a betaparticle is accompanied by the formation of a proton (when a neutron decays); {38} Neutron decay formsan electron (beta particle) and a proton; {39} The time required for 1/2 of the atoms to decay; {40} Use it todetermine the time required for the measured radiation to drop by 50%; {41} less stable; {42} Lessenergy is required to break it apart; {43} neutron; {44} 13; {45} 13; {46} 12; {47} 6; {48} 6; {49} C;

+ 9

4Be{50.} He

4

2----->

12

6C + n

1

0

+ 27

13Al{51.} He

4

2----->

30

15P + n

1

0

{52} two; {53} about 1393 yrs;

{54.} n1

0U +

235

92---->

236

92U

{55.} n1

0U

235

92----> Ba +

141

56Kr +

92

363 + energy

Problems:

1. a. Pa 234

91---->

234

92U + e

0

-1

1. b. Th 230

90---->

226

88Ra + He

4

2

e0

-11. c. Pb

214

82---->

214

83Bi +

2. 1.00 gram3. 54 minutes4. 5.63 Mev/particle

5. n1

0p

1

1He

4

2e

0

-1e

0

+1

6. 18.16 grams

29-21 ©1997, A.J. Girondi

7. a.12

6C + H

1

1----->

13

7N + energy

13

7N +7. b. ----->

13

6C e

0

-1

H1

1+7. c. ----->

13

6C

14

7N + energy

H1

1+7. d. ----->

15

8O

14

7N + energy

+7. e. ----->15

8O

15

7N e

0

+1

+ H1

1+7. f. ----->

15

7N

12

6C He

4

2

H2

1U +

238

92----> n

1

028. a. +Np

238

93

He4

2U +

238

92----> n

1

028. b. +Pu

240

94

C12

6U +

238

92----> n

1

058. c. +Cf

245

98

O16

8U +

238

92----> n

1

048. d. +Fm

250

100

He4

2Pu +

238

94----> n

1

038. e. +Cm

239

96

-----> ++8. f. He4

2

1

1H n

1

02+Cm

239

96Bk

240

97

-----> +0

-1eBi

210

83Po

210

849. a.

9. b. 0.59 mole

----> +0

-1eC

14

6N

14

710. a.

10. b. 4 moles C-14

29-22 ©1997, A.J. Girondi

92

91

90

89

88

87

86

85

84

83

82

238 234 230 226 222 218 214 210 206

MASS NUMBER

ATOMIC NUMBER

U238

92

Th234

90

α

U234

92

ß

Pa23491

ß

Th23090

α

Ra22688

α

α

Po21884

α

Pb21482

α Bi214

83

ß

Po214

84

ß

Pb210

82

αBi

210

83

ß ßPb

206

82

α

what goes here?

what goes here?

Figure 29.2Decay Series of U-238

NAME_________________________________________________________ PERIOD_________


"ALICE"

APPENDICES

A-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS




[email protected]



APPENDIX A

SECTION A-1. Uncertainty in measurement

Measurement is the foundation of all experimental science. Accordingly, the degree of certaintyabout conclusions drawn from experiments depends upon the accuracy and precision of themeasurements made while conducting the experiments. The accuracy of a measurement indicates theamount by which it differs from a known, true value; the precision of a measurement is a measure of the"reproducibility" of the measurement. Let's use an analogy to make this more clear. Suppose weconsider two archers who are shooting arrows at the targets below. The small circles in the centers of thetargets are the bull's eyes, and the x's mark the points where the archers' arrows have struck the targets.Note that archer A shot arrows all over the place. Archer B shot arrows which landed very close together,but not near the bull's eye. Archer C shot arrows very close together too, and they also hit the desiredspot! Archer B is precise, but not accurate. Archer C is both precise and accurate. Archer A should sellhis bow before he hurts somebody!

xx

x

xx

xxxx x

x xxx x

Archer A Archer B Archer C

Figure A-1

Archer B is precise because he can reproduce his results. You can also think of precision as areflection of uncertainty. If precision is high, uncertainty is low. Since archer B is precise, the uncertaintyabout where his arrows will hit is small. Time after time, he gets the same result. (Even though he maynever hit the bull's eye.) Archer C always gets the desired result. Therefore, he is accurate. Since hisaccuracy is consistent, he is also precise. Archer A is not precise or accurate, poor guy.

The scientific investigator tries to be both accurate and precise. However, regardless of theexactness of the measurements, there is a degree of uncertainty. This uncertainty may be caused by theuse of inadequate instruments, by inexactness in the readings made by the investigator, or by acombination of these factors. It is customary in reporting scientific measurements to include all of thefigures known with certainty and one doubtful or estimated figure. For example, the metric ruler shownbelow in Figure A-2 can be read as 3.5 cm.

Since the 3 can be read directly from the ruler, it is known with certainty. The 0.5 cm reading is anestimate made by the observer and is only the estimate of the person reading the scale. Althoughestimated and therefore doubtful, the 0.5 cm is reported as part of the measurement . The digits knownwith certainty in any measurement and the first estimated digit are all part of the measurement and arecalled significant figures.

0 2 4 6 8 10

Figure A-2


It is important that a person does not report a figure such as 3.58 cm as having been read on theruler shown above. Since the 0.5 figure is only an estimate, the next figure (0.08) cannot even beestimated and, therefore, a measurement of 3.58 indicates more information than is really known.Similarly, a reported reading of 3 cm would convey too little information, because the reader would assumethat the 3 was an estimate, when in fact it is known with certainty.

A scientific measurement should include all of the information possible, but not more than cantruthfully be recorded. Generally speaking, most measurements can be read to one decimal place beyondthe smallest divisions on the scale being read. Thus, if an instrument is calibrated so that the closestmarked divisions are at intervals of 1 cm (the ones column), readings can usually be made to the tenthscolumn with the last digit being an estimate. Thus, we read 3.5 cm from the scale in Figure A-2.

There are two methods for expressing how much uncertainty there is in a measurement. In thefirst method, the uncertainty is specifically expressed. For example, the reading from the scale in FigureA-2 above was 3.5 cm. We are unsure of the 5. All we can say is that the real value is somewhere between3 and 4. Thus the "real" value could actually fall into a range which is 0.5 cm above or below the 3.5 cmreading. We would, therefore, say that the uncertainty in the measurement is ± 0.5 cm. Themeasurement would be expressed as 3.5 cm ± 0.5.

The second method of expressing uncertainty involves the use of the proper number ofsignificant figures when expressing the measurement. In this method, the amount of uncertainty is notexpressed. Instead, the column in which the uncertainty exists is implied. For example, when we write themeasurement as 3.5 cm, the reader will assume that the last significant figure (the 5) is an estimate. Thereader then understands that the uncertainty is in the tenths column. The reader understands that theamount of uncertainty is at least ± 0.1, but it could be more (such as ± 0.2 or ± 0.3, etc.). This method ofexpressing uncertainty is not as specific, but it is a lot more convenient.

In measurements that have been reported correctly, all nonzero numbers are always significantfigures. A zero may or may not be a significant figure. Why is that, and how can we tell whether a zero issignificant or not? First, we should try to establish if the zero in question is serving merely to determine orlocate a decimal point. When a zero serves as a determiner or indicator of a decimal point (that is, if itserves as a "place holder"), it is NOT considered to be a significant figure. For example, consider thefigures shown below. In Figure A-3, note that the arrow is located between 10,000 and 20,000. As weread the scale we are , therefore, sure that the first digit in the measurement should be a 1. This is not anestimate, and it is not, therefore, the final sig fig. Now, we estimate that the second digit might be a 2. Sowe read 12,000 as the location of the arrow. Since the 2 is an estimate, it is the final significant digit in themeasurement. We can read no more digits from the scale. However, in order to put the decimal in theproper location we need some place holders. Zeros are used for that purpose; but, these zeros will not beread or estimated from the scale. They are NOT significant. They serve merely as place holders. Look atFigure A-4. Note that we can be sure that the arrow is located between 10,000 and 10,001 on the scale.Thus, we are sure that the first digit should be a 1, and we are sure that the next four digits should bezeros. (We read them from the scale.) Since the arrow appears to be about half way between the twovalues, we estimate that the measurement should be read as 10,000.5. The 5, however, is an estimateand is, therefore, the final sig fig. The zeros in this case, are not merely place holders. They aresignificant!

10,000 20,000 30,000 40,000

Figure A-3


Figure A-4

10,000 10,001

The following 4 rules should be observed in determining whether a digit is a significant figure. You will beexpected to know these rules!

1. Nonzero digits are always significant.

2. "Leading zeros" (zeros which appear in the front portion of a number) are never significant. Forexample, the number 0.0039 has two "sig figs." The zeros are used to locate the decimal point. We oftenput a zero to the left of the decimal in numbers which have a value less than one. It is not significant.

3. "Trapped zeros" (zeros which appear between significant digits in a number) are always significant. Forexample, the numbers 0.0304 and 203 both have three sig figs., while the number 800006 has six sigfigs.

4. "Trailing zeros" (zeros at the end of a number) may or may not be significant. They are significant only ifthe decimal point is expressed. If the decimal is understood (not showing), the trailing zeros are notsignificant. Thus, the numbers 1900. and 16.00 both contain four sig figs since the decimal point isexpressed (showing) in both cases. However, in a number with an understood decimal point, the finalzeros are just used to locate the decimal point and are NOT significant. Thus, the number 16,000 has twosig figs (the 6 is uncertain). If we express the decimal, 16000., the final zero is the uncertain digit and thenumber now has five sig figs. (Sometimes when a number ends with several zeros, the last significant onehas a bar over it. For example, if the second zero in 16,000 had a bar over it, the number would thencontain 4 sig figs.)

When performing arithmetic calculations involving measured values, we must express our resultsso that they contain only the number of significant figures justified by the uncertainty of the originalmeasurements. Thus, it is frequently necessary to round off numbers so that a result does not appear tobe more certain than the original measurements.

The following rules should be carefully observed when rounding off a measurement:

a. When the digit dropped is less than 5, the preceding digit remains unchanged; for example, 8.3734when expressed to three sig figs becomes 8.37.

b. When the digit dropped is 5 or more, the preceding digit is increased by 1; for example, 3.6287expressed to three sig figs becomes 3.63.

When measurements are added or subtracted, the results of the calculations should be rounded off to thecolumn containing the leftmost uncertain digit. For example,

ADD SUBTRACT

28.6 cm 287.56 g 327.33 cm 76.4 g 5891.212 cm 211.16 g ---> (211.2 g)6247.142 cm ---> (6247.1 cm)

In the addition problem, since 28.6 contains the leftmost uncertain digit (the 6 in the tenths column),6247.142 is rounded off to 6247.1. In the subtraction problem, since the 76.4 contains the leftmost


uncertain digit (the 4 in the tenths column), 211.16 is rounded off to 211.2.

When measured values are multiplied or divided, count the number of sig figs in each measurement. Theone which has the least number of sig figs determines how many sig figs will appear in the answer.

Example: 28 cm X 4728 cm = 132,384 cm2 = 130,000 cm2 (rounded)

Since 28 contains two sig figs and 4728 contains 4 sig figs, the answer must be rounded to two sig figswhich would be 130,000. Since the zeros in 130,000 only indicate the position of the decimal point, theyare not sig figs. 130,000 cm2 contains only two sig figs.

Here are a couple of rules to remember:

1. When experimental quantities (measurements) are multiplied or divided, the result is rounded to thesame number of sig figs as the measurement which contains the least number of sig figs.

2. When experimental quantities (measurements) are added or subtracted, the uncertain digit in the resultmust be in the same column as the leftmost uncertain digit in the original measurements.

This sounds confusing so look at the addition problem illustrated on page A-5. The 6 is theuncertain digit in 28.6 and it is in the tenths column. The second 3 is the uncertain digit in 327.33 and it isin the hundredths column. The second 2 is uncertain in 5891.212 and it is in the thousandth column. Ofthe three uncertain digits, the leftmost one is the 6 in 28.6. So, the answer should be rounded to thetenths column and becomes 6247.1. Do you now see why the answer to the subtraction problem isrounded off to the tenths column?

When doing a problem which involves both multiplication (or division) AND addition (orsubtraction), you should round to the correct number of sig figs only when switching from multiplication (ordivision) to addition (or subtraction). Thus, you should not round off any intermediate results while you arejust multiplying and/or dividing or when you are just adding and/or subtracting.

For example, try to solve the problem below with your calculator.

1.113 cm X 4.3 cm X 8.11 cm

2.00 cm X 1.00 cm - 4.5 cm + 6.32 cm = ??? cm

The best way to handle this calculation is to do all the multiplication and division and then correctly roundthe result using the proper rule. Then, do all the addition and subtraction and round that result to thecorrect number of sig figs using the proper rule. Finally, the two calculations can be combined androunded to obtain the final result. The correct answer to the problem is 21 cm when rounded properly.

SECTION A-2. Problems

Problem 1. Underline any digits in the following measurements which are NOT significant.

a. 2.4421 cm h. 42.0040 m o. 7080.0940 mg

b. 200.41 m i. 3000 mm p. 0.8 mg

c. 3.00 L j. 5.300 g q. 674 L

d. 0.10004 cm3 k. 00.0050050 mL r. 4000200.080 m

e. 0.0020 m2 l. 240 kg s. 767003 cm

f. 00.0030030 m m. 23,000.010 cm t. 97600 g

g. 108,090 cm n. 0.0060 mL u. 8740. mg


Problem 2. Express answers to the following using the correct number of sig figs. Check your answers.You may be surprised!

a. 3.14 m X 3m =

b. 4.688 m / 2.0 m =

c. 3.4 m + 2.11 m + 0.8001 m =

d. (4.811 m)(3.1 m)(5 m) =

e. (3.4 m)(9.22 m) / 3.2 m =

f. (6.68 m2 / 2.2 m) - 3.4 m + 7.88 m =

g. (3.42 m2 / 2.1 m)(2.442 m / 2.10 m)(8.866 m / 2.14 m) + 4.532 m =

h. 34.5772 cm + 0.43 cm =

i. (345.2 m)(2.01m) =

j. 457.8865 mL / 3 mL =

k. 889 g - 2.886 g =

l. (45 g + 124 g) / 20. g =

m. (23,800 m)(2 m) =

3. Indicate how many sig figs are present in each of the following.

a. 671 m _____ j. 40003 m _____

b. 360 m _____ k. 0.00100 m _____

c. 059 m _____ l. 408.0 m _____

d. 609 m _____ m. 20000 m _____

e. 3040 m _____ n. 200130 m _____

f. 6009 m _____ o. 20.000 m _____

g. 3564.20 m _____ p. 0.050440 m _____

h. 0.042 m _____ q. 5744 m _____

i. 55600 m _____ r. 99100 m _____

4. When a number is converted into exponential form, only significant figures (digits) are used. Keepingthis in mind, express the following in scientific notation.

a. 381 cm ______________________ e. 6040 L ______________________

b. 80462 kg ______________________ f. 0.003400 g ______________________

c. 0.0055 g ______________________ g. 300 mL ______________________

d. 101000 mm ______________________ h. 0.52 kg ______________________

It is important to emphasize that significant figures only exist in measurements. Therefore, therules regarding significant figures only apply when you are working with measurements. At this point youneed to learn that there are three particular cases in which the rules regarding significant figures do notapply.

First, the rules do not apply when you are working with pure numbers. So, for example, if we were tomultiply 5 times 45, the answer would be: 5 X 45 = 225. These are pure numbers, they do not have units.Since they are not measurements, we do not round the answer using the rules for significant figures.


However, if we multiply 5 cm X 45 cm then the rules for sig figs do apply and the answer would be:

5 cm X 45 cm = 225 cm2 = 200 cm2 (rounded)

Second, the rules do not apply when you are dealing with definitions that relate units of measure withinthe same system - such as the metric system. For example, look at the problem below and calculate theanswer.

434.5mL X

1 L

1000 mL = ??? L

This problem involves multiplication and division. Therefore, we count the number of sig figs in each ofthe measurements. The only measurement in the problem is 434.5 mL. The ratio of 1L/1000mL is adefinition. Even though this ratio includes numbers with units, 1 liter is equal to 1000 mL by definition.Therefore, we look only at 434.5 mL. Seeing that it contains 4 sig figs, we round the answer to four sigfigs. If a ratio is used which changes units in one system of measurement to an equivalent measure inanother system of measurement, then the rules do apply. For example, calculate the answer to theproblem below.

6764 g X

1.00 pound

454 grams = ??? pounds

The ratio in the problem above converts pounds (English system) into grams (metric system).Measurements actually had to be made to determine this relationship, so the rules apply here. You cannotchange pounds into grams by definition since they belong to different systems. Following the rule formultiplication and division, the answer is rounded to 3 sig figs.

Third, the rules for significant figures do not apply to "counts," because "counts" are not measurements.For example, in the problem below, calculate the number of dozens of eggs we can obtain from 155 eggs:

8064 eggs X

1 dozen

12 eggs = ??? dozens

8064 eggs is a "count." There is no uncertainty. It means exactly 8064 eggs. Similarly, there are exactly12 eggs in a dozen. No measuring instrument is needed here, thus these cannot be measurements. Therules for sig figs do not apply. The answer is 672 dozens.

SECTION A-3. Answers to Problems

1. a. 2.4421 cm h. 42.0040 m o. 7080.0940 mg

b. 200.41 m i. 3000 mm p. 0.8 mg

c. 3.00 L j. 5.300 g q. 674 L

d. 0.10004 cm3 k. 00.0050050 mL r. 4000200.080 m

e. 0.0020 m2 l. 240 kg s. 767003 cm

f. 00.0030030 m m. 23,000.010 cm t. 97600 g

g. 108,090 cm n. 0.0060 mL u. 8740. mg

2. a. 9 m2; b. 2.3 m; c. 6.3 m; d. 70 m3; e. 9.8 m; f. 7.5 m; g. 12.3 m; h. 35.01 cm; i. 694 m2; j. 200;k. 886 g; l. 8.5 g; m. 50,000 m2

3. a. 3; b. 2; c. 2; d. 3; e. 3; f. 4; g. 6; h. 2; i. 3; j. 5; k. 3; l. 4; m. 1; n. 5; o. 5; p. 5; q. 4; r. 3

4. a. 3.81 X 102 cm; b. 8.0462 X 104 kg; c. 5.5 X 10-3 g; d. 1.01 X 105 mm; e. 6.04 X 103 Lf. 3.400 X 10-3 g; g. 3 X 102 mL; h. 5.2 X 10-1 kg


APPENDIX B

Actual Electron Arrangement Of The Elements

Sublevels ---> 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d

1 hydrogen 12 helium 23 lithium 2 14 beryllium 2 25 boron 2 2 16 carbon 2 2 27 nitrogen 2 2 38 oxygen 2 2 49 fluorine 2 2 510 neon 2 2 611 sodium 2 2 6 112 magnesium 2 2 6 213 aluminum 2 2 6 2 114 silicon 2 2 6 2 215 phosphorus 2 2 6 2 316 sulfur 2 2 6 2 417 chlorine 2 2 6 2 518 argon 2 2 6 2 619 potassium 2 2 6 2 6 120 calcium 2 2 6 2 6 221 scandium 2 2 6 2 6 2 122 titanium 2 2 6 2 6 2 223 vanadium 2 2 6 2 6 2 324 chromium 2 2 6 2 6 1 525 manganese 2 2 6 2 6 2 526 iron 2 2 6 2 6 2 627 cobalt 2 2 6 2 6 2 728 nickel 2 2 6 2 6 2 829 copper 2 2 6 2 6 1 1030 zinc 2 2 6 2 6 2 1031 gallium 2 2 6 2 6 2 10 132 germanium 2 2 6 2 6 2 10 233 arsenic 2 2 6 2 6 2 10 334 selenium 2 2 6 2 6 2 10 435 bromine 2 2 6 2 6 2 10 536 krypton 2 2 6 2 6 2 10 637 rubidium 2 2 6 2 6 2 10 6 138 strontium 2 2 6 2 6 2 10 6 239 yttrium 2 2 6 2 6 2 10 6 2 140 zirconium 2 2 6 2 6 2 10 6 2 241 niobium 2 2 6 2 6 2 10 6 1 442 molybdenum 2 2 6 2 6 2 10 6 1 543 technetium 2 2 6 2 6 2 10 6 2 544 ruthenium 2 2 6 2 6 2 10 6 1 745 rhodium 2 2 6 2 6 2 10 6 1 846 palladium 2 2 6 2 6 2 10 6 1047 silver 2 2 6 2 6 2 10 6 1 1048 cadmium 2 2 6 2 6 2 10 6 2 10 49 indium 2 2 6 2 6 2 10 6 2 10 150 tin 2 2 6 2 6 2 10 6 2 10 251 antimony 2 2 6 2 6 2 10 6 2 10 352 tellurium 2 2 6 2 6 2 10 6 2 10 4


B-1 ©1997, A.J. Girondi


53 iodine 2 2 6 2 6 2 10 6 2 10 554 xenon 2 2 6 2 6 2 10 6 2 10 655 cesium 2 2 6 2 6 2 10 6 2 10 6 156 barium 2 2 6 2 6 2 10 6 2 10 6 257 lanthanum 2 2 6 2 6 2 10 6 2 10 6 2 158 cerium 2 2 6 2 6 2 10 6 2 10 6 2 1 159 praseodymium 2 2 6 2 6 2 10 6 2 10 6 2 360 neodymium 2 2 6 2 6 2 10 6 2 10 6 2 461 promethium 2 2 6 2 6 2 10 6 2 10 6 2 562 samarium 2 2 6 2 6 2 10 6 2 10 6 2 663 europium 2 2 6 2 6 2 10 6 2 10 6 2 764 gadolinium 2 2 6 2 6 2 10 6 2 10 6 2 7 165 terbium 2 2 6 2 6 2 10 6 2 10 6 2 966 dysprosium 2 2 6 2 6 2 10 6 2 10 6 2 1067 holmium 2 2 6 2 6 2 10 6 2 10 6 2 1168 erbium 2 2 6 2 6 2 10 6 2 10 6 2 1269 thulium 2 2 6 2 6 2 10 6 2 10 6 2 1370 ytterbium 2 2 6 2 6 2 10 6 2 10 6 2 1471 lutetium 2 2 6 2 6 2 10 6 2 10 6 2 14 172 hafnium 2 2 6 2 6 2 10 6 2 10 6 2 14 273 tantalum 2 2 6 2 6 2 10 6 2 10 6 2 14 374 tungsten 2 2 6 2 6 2 10 6 2 10 6 2 14 475 rhenium 2 2 6 2 6 2 10 6 2 10 6 2 14 576 osmium 2 2 6 2 6 2 10 6 2 10 6 2 14 677 iridium 2 2 6 2 6 2 10 6 2 10 6 2 14 778 platinum 2 2 6 2 6 2 10 6 2 10 6 1 14 979 gold 2 2 6 2 6 2 10 6 2 10 6 1 14 1080 mercury 2 2 6 2 6 2 10 6 2 10 6 2 14 1081 thallium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 182 lead 2 2 6 2 6 2 10 6 2 10 6 2 14 10 283 bismuth 2 2 6 2 6 2 10 6 2 10 6 2 14 10 384 polonium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 485 astatine 2 2 6 2 6 2 10 6 2 10 6 2 14 10 586 radon 2 2 6 2 6 2 10 6 2 10 6 2 14 10 687 francium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 188 radium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 289 actinium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 190 thorium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 291 protactinium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 2 192 uranium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 3 193 neptunium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 4 194 plutonium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 695 americium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 796 curium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 7 197 berkelium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 8 198 californium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 1099 einsteinium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 11100 fermium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 12101 mendelevium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 13102 nobelium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14103 lawrencium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 1104 unnilquadium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 2105 unnilpentium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 3106 unnilhexium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 4107 unnilseptium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 5108 unniloctium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 6109 unnilennium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 7Sublevels ---> 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d


Appendix B. cont.-"Core Notation" of 100 Elements According to the Diagonal Rule

1 H 1s1 51 Sb [Kr]5s24d105p3

2 He 1s2 52 Te [Kr]5s24d105p4

3 Li [He]2s1 53 I [Kr]5s24d105p5

4 Be [He]2s2 54 Xe [Kr]5s24d105p6

5 B [He]2s22p1 55 Cs [Xe]6s1

6 C [He]2s22p2 56 Ba [Xe]6s2

7 N [He]2s22p3 57 La [Xe]6s24f18 O [He]2s22p4 58 Ce [Xe]6s24f29 F [He]2s22p5 59 Pr [Xe]6s24f310 Ne [He]2s22p6 60 Nd [Xe]6s24f411 Na [Ne]3s1 61 Pm [Xe]6s24f512 Mg [Ne]3s2 62 Sm [Xe]6s24f613 Al [Ne]3s23p1 63 Eu [Xe]6s24f714 Si [Ne]3s23p2 64 Gd [Xe]6s24f815 P [Ne]3s23p3 65 Tb [Xe]6s24f916 S [Ne]3s23p4 66 Dy [Xe]6s24f10

17 Cl [Ne]3s23p5 67 Ho [Xe]6s24f11

18 Ar [Ne]3s23p6 68 Er [Xe]6s24f12

19 K [Ar]4s1 69 Tm [Xe]6s24f13

20 Ca [Ar]4s2 70 Yb [Xe]6s24f14

21 Sc [Ar]4s23d1 71 Lu [Xe]6s24f145d1

22 Ti [Ar]4s23d2 72 Hf [Xe]6s24f145d2

23 V [Ar]4s23d3 73 Ta [Xe]6s24f145d3

24 Cr [Ar]4s23d4 74 W [Xe]6s24f145d4

25 Mn [Ar]4s23d5 75 Re [Xe]6s24f145d5

26 Fe [Ar]4s23d6 76 Os [Xe]6s24f145d6

27 Co [Ar]4s23d7 77 Ir [Xe]6s24f145d7

28 Ni [Ar]4s23d8 78 Pt [Xe]6s24f145d8

29 Cu [Ar]4s23d9 79 Au [Xe]6s24f145d9

30 Zn [Ar]4s23d10 80 Hg [Xe]6s24f145d10

31 Ga [Ar]4s23d104p1 81 Tl [Xe]6s24f145d106p1 32 Ge [Ar]4s23d104p2 82 Pb [Xe]6s24f145d106p2

33 As [Ar]4s23d104p3 83 Bi [Xe]6s24f145d106p3

34 Se [Ar]4s23d104p4 84 Po [Xe]6s24f145d106p4

35 Br [Ar]4s23d104p5 85 At [Xe]6s24f145d106p5

36 Kr [Ar]4s23d104p6 86 Rn [Xe]6s24f145d106p6

37 Rb [Kr]5s1 87 Fr [Rn]7s1

38 Sr [Kr]5s2 88 Ra [Rn]7s2

39 Y [Kr]5s24d1 89 Ac [Rn]7s25f140 Zr [Kr]5s24d2 91 Pa [Rn]7s25f342 Mo [Kr]5s24d4 92 U [Rn]7s25f443 Tc [Kr]5s24d5 93 Np [Rn]7s25f544 Ru [Kr]5s24d6 94 Pu [Rn]7s25f645 Rh [Kr]5s24d7 95 Am [Rn]7s25f746 Pd [Kr]5s24d8 96 Cm [Rn]7s25f847 Ag [Kr]5s24d9 97 Bk [Rn]7s25f948 Cd [Kr]5s24d10 98 Cf [Rn]7s25f10

49 In [Kr]5s24d105p1 99 Es [Rn]7s25f11

50 Sn [Kr]5s24d105p2 100 Fm [Rn]7s25f12


H Be B C N O F Ne

Group: 1A 2A 3A 4A 5A 6A 7A 8A

Dot Notations of Elements in Groups 1A - 8A


APPENDIX C

SECTION C.1 Freezing Point Depression And Boiling Point Elevation

As you should already know, the presence of a solute in a solution raises the boiling point. Themost scientific explanation for the increase in boiling point is that the solute tends to lower the vaporpressure of the solvent. That seems kind of vague, right? Maybe the graphs shown below will help. Thesolid lines represent the vapor pressure of pure water, while the dotted lines represent the vapor pressureof water in a solution. Note that the vapor pressure at any given temperature is lower when water iscombined with a solute in a solution. Also note that in a concentrated solution, the vapor pressure isdepressed more than it is in a dilute solution.

You will recall that water will boil only when its vapor pressure equals the atmospheric pressure (760mm Hg at sea level). Note that the vapor pressure of pure water equals atmospheric pressure at 100oC;thus, 100oC is the boiling point of water at sea level. (See T1 on the graphs.) However, note that thetemperature of the water in a solution has to go higher before its vapor pressure will equal atmosphericpressure. (See T2 on the graphs below.) Therefore, the boiling point of the water in the solution is above100oC. Also note that the boiling point of the water in the concentrated solution will be higher than that ofthe water in the dilute solution.

760 mm

T1 T2

Concentrated Solution

V.P.

Temperature

760 mm

T1 T2

Dilute Solution

V.P.

Temperature

A

B

A = vapor pressure of pure water at T1

B = vapor pressure of water in a solution at T1

vapor pressureof pure water

vapor pressure of water in a solution

A

B

A = vapor pressure of pure water at T1

B = vapor pressure of water in a solution at T1

C

C

A = boiling point of pure waterC = boiling point of water in a solution

A = boiling point of pure waterC = boiling point of water in a solution

C-1 ©1997, A.J. Girondi

Above 0oC water is a liquid rather than a solid because the molecules have enough kinetic energy(energy of motion) to overcome the attractive forces between them that could "lock" them into a nearlyfixed position. At 0oC the molecules have lost enough energy so that the attractive forces between themmove the molecules into a solid crystalline arrangement. The presence of a solute in the water interfereswith this process, and the result is a lower freezing point. You should have already completed an activity inwhich you determined the freezing point of a salt water solution. You should have noted that an ice bathof salt water is colder than an ice bath of pure water. Now, why does the addition of salt make ice watercolder? It seems kind of mysterious, doesn't it?

In ice water (at 0oC) the ice is melting and the water is freezing. This is because 0oC is both thefreezing point of water and the melting point of ice. As the ice melts, it absorbs heat from the water (anendothermic process). As the water freezes, the heat in it is released to the water (an exothermicprocess). It all makes sense if you keep in mind that an ice water bath represents an equilibrium condition.That's the secret to understanding this phenomenon. When we say that the system is at equilibrium, wemean that the heat is being released to and removed from the water at equal rates; thus, the temperatureremains constant at 0oC. However, when you add salt the equilibrium is upset. The salt will not affect therate at which the ice melts, but it will slow down the rate at which the water freezes. The particles from thesalt (solute) get in between the water (solvent) molecules and weaken the attraction between the watermolecules. This causes the freezing rate of the water to slow down. Thus, the melting rate of the ice isfaster than the freezing rate of the water. The salt has allowed the endothermic process (melting) to gofaster than the exothermic process (freezing). That means that heat is being removed from the waterfaster than it is being replaced. The result is that the temperature of the water goes down. When all of theice is consumed, the melting can no longer continue, and the temperature drop will cease.

water freezing

exothermicice

ice water melting

endothermic

water freezing (slower)

exothermicice

ice water melting (faster)

endothermic

Ice Water (equal rates) Salt Water (unequal rates)

Homemade ice cream requires that the cream be frozen in a mixture of rock salt and ice. Ice wateralone will not freeze the cream. If you know the concentration of a solution, you can actually calculate howmuch the boiling point will go up of how much the freezing point will go down. However, you cannot useconcentrations expressed as molarity (M). Remember that molarity (M) tells you nothing about the amountof solvent in a solution. The units for molarity are moles of solute per liter of solution. Solvent is not evenmentioned in the definition. Molarity is commonly used in chemistry because we usually don't care aboutthe amount of solvent in a solution. We are always concerned about the amount of solute. However,freezing point depression and boiling point elevation are properties that depend on the ratio of soluteparticles to solvent particles in a solution. Such properties are known as colligative properties. Osmoticpressure, which you may have read about in biology class, is another example of a colligative property ofsolutions. To calculate the change in the freezing point or boiling point of a solution, you will need to usea unit of concentration that contains information about both the solute and the solvent. Molality (m)contains this information. Molality (m) is defined as moles of solute per kilogram of solvent. A lowercase"m" represents molality.

1 mole solute

1 kg solvent1 m =

First, let's try some problems that involve only the concept of molality, and then we will look at problemswhich reveal how molality is used to calculate changes in freezing and boiling points.


Sample Problem: What is the molal concentration of a solution which contains 45 grams of NaOHdissolved in 400. mL of water?

First, note that the amount of water is given in the problem, whereas the amount of solution would havebeen given if we were working with molarity. We want to end up with units of moles NaOH/Kg H2O. Sincewe want to end with a ratio of information about solute divided by information about solvent, let's start withthe information about solute divided by the information about solvent: 45 g NaOH/400 mL H2O.

45 g NaOH

400. mL H 2O X

?

? X

?

? X

?

? =

?? moles NaOH

kg H2O

Remember that 1 g H2O = 1 mL H2O, and that 1 Kg H2O = 1000 g H2O. Complete the problem in thespace below. The correct answer is 2.8 moles NaOH /1 Kg H2O,or 2.8 m. Solve the problem after youcomplete the set-up and see if you can obtain this answer.

45 g NaOH

400. mL H 2O X

X

X

=

moles NaOH

kg H2O

Now try the problems below.

Problem 1. What is the molal concentration of a solution which contains 188 grams of BaCl2 and 250.mL of water?

__________ m

Problem 2. What is the molality of a solution if 45.0 grams of HCl are dissolved in 500. mL of water?

__________ m

Problem 3. If 0.360 kilogram of C12H22O11 is dissolved in 788 grams of water, what is the molalconcentration?

__________ m


Problem 4. How many grams of KOH must be added to 0.45 kg of water to make a 0.33 m (molal)solution of KOH?

__________ g KOH

Problem 5. How many grams of water are needed to make a 1.200 m solution of NaCl which contains75.00 grams of NaCl?

__________ g H2O

There are formulas which can be used to determine the amount that a solute lowers a freezing pointor raises a boiling point:

∆Tƒ = (kƒ ) (m) (i) and ∆Tb = (kb) (m) (i)

∆Tƒ and ∆Tb stand for the change in the freezing point and the change in the boiling point, respectively.The symbols kƒ and kb represent the molal freezing-point constant and the molal boiling-point constant,respectively. These quantities are experimentally determined, and are found in reference books. Like allscientific constants, they are present in the equations to "make them work." The lowercase "m"represents molality, while the "i" is called the activity coefficient. In simple terms, "i" represents thenumber of moles of particles which are formed when one mole of a particular solute dissolves.

Electrolytes are substances which conduct an electric current when dissolved in solvents, whilenonelectrolytes are substances which do not conduct an electric current when dissolved in solvents.Many ionic and polar covalent compounds are electrolytes, while nonpolar covalent substances arenonelectrolytes. For all nonelectrolytes, "i" always has a value of 1. For electrolytes, the value of "i"depends on the number of particles the substance breaks into (dissociates into) when dissolved insolution. (Remember that polyatomic ions do not break apart in solution.) For example, let's consider thefour examples below.

When one mole of table sugar molecules (a nonelectrolyte with formula C12H22O11) dissolves inwater, it forms only one mole of particles. This is true of all nonelectrolytes. One mole of anynonelectrolyte forms one mole of particles, since nonelectrolytes do not split up into ions when theydissolve.

nonelectrolyte: C12H22O11(s) -------> C12H22O11(aq) i = 1

When one mole of the electrolyte NaOH dissolves in water, it forms two moles of particles (1 mole of Na1+

ions and one mole of OH1- ions).

electrolyte: NaOH(s) -------> Na1+(aq) + OH1-(aq) i = 2


One mole of the electrolyte Sr(NO3)2 will form three moles of particles when it dissolves in water:

electrolyte: Sr(NO3)2(s) -------> Sr2+(aq) + 2 NO31-(aq) i = 3

The electrolyte AlCl3 will form four moles of particles per one mole of solute. Examine the equation below.

electrolyte: AlCl3(s) -------> Al3+(aq) + 3 Cl1-(aq) i = 4

This is important because it is the number of moles of solute particles in solution, rather than just the number of moles of solute, which are important when you are studying colligative properties. If you are working with a nonelectrolyte, "i" is always equal to one, because nonelectrolytes are composed of molecules which do not break up into smaller parts in solution. However, if you are working with an electrolyte, you must look at the formula to determine how many parts the substance breaks into and to determine the value of "i." Therefore, you must either know or be told whether or not the substance you are working with is an electrolyte. There are no units generally assigned to "i." In addition, "i" is not a measurement, so its value will not affect the number of significant digits in the answers to problems.

Problem 6. For the electrolytes listed below, show how they break into particles, and determine the value of "i" for each substance. Be sure to show the charges on ions which form.

a. Ca(OH)2(s) -----> ________________________________ i = ________

b. KMnO4(s) -----> ________________________________ i = ________

c. GaF3(s) -----> ________________________________ i = ________

d. MgBr2(s) -----> ________________________________ i = ________

e. LiCl(s) -----> ________________________________ i = ________

f. Na3PO4(s) -----> ________________________________ i = ________

g. (NH4)2SO4(s) -----> ________________________________ i = ________

As for the constants, for water Kƒ has a value of 1.86 and kb has a value of 0.52. Other solvents have different values for these constants. These values may be used only for water solutions. Constants in science can have strange units. The unit for these is (kg solute)(oC)/(moles solute). Why that, you ask? Well, we want ∆Tƒ and ∆Tb to have values expressed in oC. Look at the formulas for ∆Tƒ and ∆Tb given previously and figure it out for yourself.

For all water solutions: kƒ = 1.86 and kb = 0.52

Sample Problem: What are the freezing and boiling temperatures of a 0.10 m solution of NaOH (an electrolyte)?

We begin by calculating the change in the freezing point of water:

∆Tƒ = (kƒ ) (m) (i) so, ∆Tƒ = (1.86)(0.10)(2) = 0.37oC

Since the freezing point goes down, we subtract ∆Tƒ from the normal freezing point of water:

0oC - 0.37oC = -0.37oC = Tƒ = freezing point of the solution


Next, we calculate the change in the boiling point of water:

∆Tb = (kb)(m)(i) so, ∆Tb = (0.52)(0.10)(2) = 0.10

Since the boiling point goes up, we add ∆Tb to the normal boiling point of water:

100oC + 0.10oC = 100.10oC = Tb = boiling point of solution

Problem 7. Now, determine the freezing and boiling points of the solutions in problems 1, 2, and 3 which you solved earlier in this appendix, keeping in mind that BaCl2 and HCl in problems 1 and 2 are electrolytes, while C12H22O11 in problem 3 is a nonelectrolyte.

a.

Tƒ = __________oC; Tb = __________oCb.

Tƒ = __________oC; Tb = __________oCc.

Tƒ = __________oC; Tb = __________oC

Problem 8. The label fell off of a bottle of a 0.20 molal water solution in the laboratory, and an attempt isbeing made to identify the solute. It is known to be either KBr, CaCl2, or GaI3. Tests reveal that thesolution is an electrolyte. Its freezing point is -1.1oC. What is the identity of the unknown solid? Show anycalculations, and explain your conclusion.

Calculations:

Explanation: ____________________________________________________________________

______________________________________________________________________________

Problem 9. A car owner puts 301 grams of ethylene glycol, C2H6O2, in his car's radiator for every 750.grams of water in it. This is a commonly used antifreeze mixture. How cold will it have to get before thisguy will have to worry about a frozen cooling system? (Ethylene glycol is a nonelectrolyte.)

_________oC


SECTION C.2 Answers to Problems

1. 3.61 m

2. 2.47 m

3. 1.33 m

4. 8.3 g

5. 1069 g

6. a. Ca(OH)2(s) -----> Ca2+(aq) + 2 OH1-(aq); i = 3b. KMnO4(s) -----> K1+(aq) + MnO41-(aq); i = 2c. GaF3(s) -----> Ga3+(aq) + 3 F1-(aq); i = 4d. MgBr2(s) -----> Mg2+(aq) + 2 Br1-(aq); i = 3e. LiCl(s) -----> Li1+(aq) + Cl1-(aq); i = 2f. Na3PO4(s) -----> 3 Na1+(aq) + PO43-(aq); i = 4g. (NH4)2SO4(s) -----> 2 NH41+(aq) + SO42-(aq); i = 3

7. a. Tƒ = -20.1oC; Tb = 105.6oCb. Tƒ = -9.19oC; Tb = 102.6oCc. Tƒ = -2.47oC; Tb = 100.69oC

8. molality = 0.20m; i = 2.96 which rounds to 3 since "i" is a whole number; Since i = 3, the solute must be CaCl2 : CaCl2(s) ----> Ca2+(aq) + 2 Cl1-(aq)


SECTION C.3 Student Notes


APPENDIX D

SPECIFIC HEAT, HEAT CAPACITY, AND TEMPERATURETHE "SWIMMING POOL" ANALOGY

Let's begin with a few definitions. Temperature and heat are not the same. Temperature is ameasure of the average kinetic energy of the molecules of a system, while heat is a measure of the totalkinetic energy of a system. You can also think of temperature as a measure of the ability of a system totransfer heat to another system. Heat flows from high temperature systems to low temperature systems.

Specific heat is defined as the number of calories or joules of heat required to raise thetemperature of 1 gram of substance by 1oC. Therefore, specific heat has units of cal/g.oC or J/g.oC.Different substances have different specific heats. The larger the specific heat of a substance, the greaterthe number of calories or joules required to raise the temperature of 1 gram of the substance by 1oC. So,if a substance has a high specific heat, then a lot of heat is required to raise its temperature. Thetemperatures of substances with smaller specific heats can be increased a great deal with smaller amountsof heat.

As an analogy, let's consider two swimming pools - a large one and a small one. Think of the largepool as a substance with a large specific heat value, and of the small pool as a substance with a smallspecific heat value. Think of the water level in the pools as temperature and of the amount of water in thepools as heat. Now add an equal amount of water (heat) to each pool. The water level (temperature) of thesmall pool has gone up more than that of the large pool. Therefore, the temperatures of substances withsmall specific heat values go up more than those of substances with large specific heat values when equalamounts of heat are gained.

Large Pool(large specific heat)

Small Pool(small specific heat)

temp

temp

HEATHEAT

If a car is exposed to the sun on a frosty morning and begins to get warmer, the frost on the car willmelt off of the metal car body before it melts off of the glass windshield. Assume that both the metal andthe glass absorb equal amounts of heat, but that the temperature of the metal goes up more becausemetal has a smaller specific heat than glass does. Since temperature is a measure of the ability to transferheat, the metal is more able to transfer heat to the frost than the glass is. That's why the frost lingers onthe windshield and frequently has to be scraped off.

Molar heat capacity is very similar to specific heat. The only difference is that molar heat capacityrefers to 1 mole of a substance, while specific heat refers to 1 gram of a substance. The unit for molar heatcapacity is cal/mole oC or J/mole oC. Since the glass on the car has a higher specific heat than the metal, italso has a higher heat capacity than the metal does. Molar heat capacity can also be interpreted as theamount of heat that 1 mole of a substance can hold for every degree of temperature. For example, 1 moleof a substance with a molar heat capacity of 80 J/mole oC can hold 80 joules of heat for each 1 oC. If itstemperature is 10oC, then 1 mole of the substance can hold 800 joules of heat. If 1 moles of anothersubstance has a molar heat capacity of 30 J/mole oC, then at 10oC it can hold only 300 joules of heat. Inother words, the higher the heat capacity of a substance, the more heat it can hold at a given temperature.So, as the car sits out in the sun, the metal body will get hotter faster. But, if the temperatures of the metaland glass eventually equalize, the glass will have more heat in it than the metal will. Glass has a greatercapacity to hold heat (heat capacity) than the metal does.

D-1 ©1997, A.J. Girondi

Summary:

1. With a small heat capacity, a substance warms quickly when heated, but can't hold much heat. Addwater to the small pool and the water level (temperature) goes up fast, but it can't hold much water (heat).

2. With a large heat capacity, a substance warms slowly when heated, but can hold a lot of heat. Add waterto the large pool and the water level (temperature) goes up slowly, but it can hold lots of water (heat).

Note: Good cookware is made of materials with high heat capacities or specific heats. They take a littlelonger to get hot, but the large amount of heat that they hold tends to keep cooking temperatures moreeven. (The water level in the large pool tends to fluctuate less than that of the small pool.)

- End of Appendix D

D-2 ©1997, A.J. Girondi

APPENDIX E

SECTION E.1 The Solubility Product Constant

You have probably already studied equilibrium systems that involve chemical changes. However,

equilibrium systems can also involve physical changes. For example, consider the system illustrated bythe equation below:

NaCl(s) <===> Na1+(aq) + Cl1-(aq)

This system involves the physical processes of dissolving and crystallization. A solution in which thesetwo processes are in equilibrium is said to be {1}_________________. Notice, too, that there is a solidinvolved in this equilibrium. You will recall from your previous studies that solids and pure liquids are neverincluded in equilibrium expressions. With that in mind, examine the equilibrium expression for thissystem:

equilibrium expression = [Na1+] [Cl1-]

Notice that the solid NaCl has not been included in the expression. When the equilibrium system involvesthese physical processes of dissolving and crystallization, the constant is given a different name. Insteadof calling it the equilibrium constant, Keq, it is called the solubility product constant and is given the symbol,Ksp. Because solids are not included, Ksp, expressions have no denominator! So,

Ksp = [Na1+] [Cl1-]

When an ionic solid is dissolved in water in sufficient quantity to saturate the solution, anequilibrium is established between the ions in the saturated solution and the molecules of excess solid inthe container: NaCl(s) <===> Na1+(aq) + Cl1-(aq) Adding more solid to a solution which is alreadysaturated does not alter the amount of solid that is dissolved. Thus, adding more NaCl(s) to the systemabove will not change the concentrations of Na1+ or Cl1- ions in the solution. Since the amount of solidNaCl does not affect the equilibrium, it is omitted from the equilibrium expression. This is often the casewhen solids are involved in equilibrium systems.

Problem 1. Write the Ksp expressions for the systems defined by the four equations below. Rememberthat coefficients in the equations become exponents in the equilibrium expressions.

a. AgCl(s) <===> Ag1+(aq) + Cl1-(aq) Ksp =

b. CdS(s) <===> Cd2+(aq) + S2-(aq) Ksp =

c. PbCl2(s) <===> Pb2+(aq) + 2 Cl1-(aq) Ksp =

d. Ag3PO4(s) <===> 3 Ag1+(aq) + PO43-(aq) Ksp =

The size of the Ksp value indicates just how soluble a particular salt is in water. The larger the Ksp value, themore soluble the salt. However, it is wise to compare only salts with "analogous" Ksp expressions. Forexample, of the four salts listed in problem 1, the Ksp expressions of AgCl and CdS are analogous. In eachcase, two different ion concentrations are involved and they have comparable exponents. Theexpressions for PbCl2 and Ag3PO4 are not analogous to any of the other expressions listed.

The Ksp values of some selected salts are listed in Table E.1. The Ksp expressions in Table E.1are all analogous.

E-1 ©1997, A.J. Girondi

Table E–1Ksp Values of Selected Salts

Salt Equilibrium Ksp Expression Ksp Value

AgCl(s) <===> Ag1+(aq) + Cl1-(aq) Ksp = [Ag1+] [Cl1-] 1.7 X 10-10

CaCO3(s) <===> Ca2+(aq) + CO32-(aq) Ksp = [Ca2+] [CO32-] 8.8 X 10-9

PbCrO4(s) <===> Pb2+(aq) + CrO42-(aq) Ksp = [Pb2+] [CrO42-] 2.0 X 10-14

ZnS(s) <===> Zn2+(aq) + S2-(aq) Ksp = [Zn2+] [S2-] 1.0 X 10-23

BaSO4(s) <===> Ba2+(aq) + SO42-(aq) Ksp = [Ba2+] [SO42-] 1.4 X 10-9

Which salt listed in Table E.1 is most soluble in water? {2}______________________

Which salt listed in Table E.1 is least soluble in water? {3}______________________

The value of Ksp can be calculated from solubility data. In Chapter 16 you calculated thesolubilities of various substances. Now we will use those solubilities to calculate Ksp values. Study thesample problem below. Use the procedure shown to solve the problems that follow.

Sample Problem: The solubility of AgCl in water at 25oC is 1.26 X 10-5 mole / liter. Calculate Ksp.

Solution:

a. Write the equilibrium expression: AgCl(s) <===> Ag1+(aq) + Cl1-(aq)

b. Write the Ksp expression: Ksp = [Ag1+] [Cl1-]

c. Examine the mole ratio in the equation. In this case it is 1:1:1. So, if 1.26 X 10-5 mole of AgCl dissolvesper liter of solution, then it would form 1.26 X 10-5 moles per liter of Ag1+ ions and the same number of Cl1-

ions. Therefore, substituting into the Ksp expression:

Ksp = [1.26 X 10-5 ] [1.26 X 10-5 ] = 1.59 X 10-10

Problem 2. Calculate Ksp for PbS given that the solubility of PbS is water at 25oC is 7.07 X 10-5 M.

Problem 3. Calculate Ksp for PbCl2 given that the solubility of PbCl2 at 25oC is 1.59 X 10-3 M. (Becareful. The ratio here is not 1:1:1.)


The Ksp value of a substance can also be used to calculate the concentration of one ion if youknow the concentration of the other ion in a saturated solution. Study the next two sample problems.

Sample Problem: The Ksp value of CaCO3 is 5.00 X 10-9 at 25oC. Calculate [Ca2+] in a saturatedsolution in which [CO32-] = 0.400 M.

Solution:

a. Write the equilibrium expression: CaCO3(s) <===> Ca2+(aq) + CO32-(aq)

b. Write the Ksp expression: Ksp = [Ca2+] [CO32-]

c. Substitute the known information into the Ksp expression: 5.00 X 10-9 = [Ca2+] [CO32-]

[Ca2+] =

5.00 X 10-9 M2

0.400 M = 1.25 X 10 -8 Md. Solve for the unknown quantity:

Sample Problem: The K sp value of BaF2 is 2.4 X 10-5 at 25oC. Find [F1-] in a saturated solution ofBaF2 if [Ba2+] = 4.8 X 10-2 M.

Solution:

a. Write the equilibrium expression: BaF2(s) <===> Ba2+(aq) + 2 F1-(aq)

b. Write the Ksp expression: Ksp = [Ba2+] [F1-]2

c. Substitute the known information into the Ksp expression: 2.4 X 10-5 = [4.8 X 10-2] [F1-]2

d. Solve for the unknown quantity:

[F1− ]2 = 2.4 X 10-5 M3

4.8 X 10 -2 M = 5.0 X 10-4 M2

[F1− ] = 2.2 X 10 -2 M

Now try the problems below.

Problem 4. The Ksp of barium carbonate, BaCO3, at 16oC is 7.0 X 10-9. Calculate the concentration ofbarium ions, given that the equilibrium mixture contains 0.040 M of carbonate ions, CO32-.

Problem 5. The Ksp of silver sulfate, Ag2SO4 is 1.2 X 10-5 at 25oC. Find the [Ag1+] in a solution inwhich the [SO42-] = 6.2 X 10-4.


Ksp values are useful in much the same way as Keq values are. NaCl has a much larger Ksp valuethan AgCl. From this information we can conclude that NaCl is much {4}__________ soluble than AgCl.

ACTIVITY E.2 Qualitative Analysis of Solutions

When solutions containing two different soluble salts are mixed together, a chemical reaction mayoccur. As you saw in a earlier chapter, a chemical reaction may be evidenced by a change in color,evolution of a gas, a change in temperature, or by the formation of a precipitate. In cases where aprecipitate forms, the amount formed is directly related to the size of the Ksp value of the precipitatingsolid. In this activity, you will be seeing a number of chemical reactions as you mix various solutionstogether. Substances which have very small Ksp values are not very soluble. So, when you mix ionstogether which result in the formation of such a substance, most of the substance will take the form of anundissolved solid precipitate.

In this activity, you will be given six labeled solutions. You will mix all of the solutions (in pairs) witheach other, making careful observations of any chemical reactions that occur. You will then be given four"unknown" solutions. Three of the unknowns will be identical to three of the original solutions. The fourthunknown will contain a combination of two of the original solutions. Your task will be to establish theidentity of each unknown. When the purpose of an experiment is to identify the substance(s) present inunknowns, the procedure is called qualitative analysis. By mixing a little of each of the unknowns with thesix known solutions, you should be able to determine their identities. You will also be testing each of theknown and unknown solutions with litmus paper to identify each solution as an acid or base. (You will bestudying acids and bases shortly, if you haven't already.)

(Note: some of the solutions are naturally blue, so blue food coloring was added to the others solutionsso that they all look the same.)

Obtain a dropping plate and the six known solutions. Before you do any mixing, test eachsolution with litmus paper by adding a drop of each solution to a piece of red litmus paper and anotherdrop to a piece of blue litmus paper. If the red paper turns blue, the solution is a base. If the blue paperturns red, the solution is an acid. If no change occurs with either paper, then the solution is neutral. Enterthe results in Table E.1. Mix the solutions in all possible combinations by adding about 5 drops of onesolution to a well of the dropping plate, and then adding 5 drops of another solution to the same well.Observe the result and place your data in the proper place in Table E.1. Repeat this procedure until youhave completed all possible pair combinations of the six solutions.

Next, clean and dry your dropping plate. Obtain four small, clean test tubes and label them withnumbers 1 through 4. Place them in a beaker or test tube rack and give them to your instructor. He will filleach tube with an unknown solution for you to identify. Tubes 1 through 3 will each contain one of the sixsolutions which you studied previously. Tube four will contain a mixture of two of those six solutions. Bymixing and testing these four unknown solutions as you did with the known solutions previously, youmust identify the solution in each of the four tubes. Record your observations and results in Table E.2.

SECTION E.3 Answers To Questions and Problems

Questions: {1} saturated; {2} CaCO3; {3} ZnS; {4} more

Problems:

1. a. [Ag] [Cl1-]; b. [Cd2+] [S2-]; c. [Pb2+] [Cl1-]2; d. [Ag1+]3[PO43-]2. 5.00 X 10-9; 3. 1.6 X 10-8; 4. 1.75 X 10-7; 5. 1.4 X 10-1


Table E.1Qualitative Analysis of Six Known Solutions

NaHCO3 HCl BaCl2 CuSO4 KNO3 Cu(NO3)2 Acid or Base

NaHCO3

HCl

BaCl2

CuSO4

KNO3

Cu(NO3)2


Table E.2Qualitative Analysis of Four Unknown Solutions

NaHCO3 HCl BaCl2 CuSO4 KNO3 Cu(NO3)2 Acid or Base

Unknown #1

Unknown #2

Unknown #3

Unknown #4

Identities of Unknowns: #1: ________________

#2: ________________

#3: ________________

#4: ________________ and ________________

End of Appendix E


APPENDIX F

SECTION F.1 Use of Inverse Logarithms in Acid–Base Problems

Calculation of pH, given [H1+], is rather straightforward. However, doing the reverse (calculating[H1+], given pH) is a bit trickier, because it involves the use of inverse logarithms which are also known asantilogarithms. An in-depth discussion of "antilogs" is beyond the purpose of this discussion. Let's justrecognize that since an antilog is the inverse of a log, when you multiply a "log" by an "antilog," the resultin one: (antilog) X (log) = 1.

Antilogs are numbers, whereas logs are exponents. For example, consider the expressionbelow.

102 = 100

log

antilogbase

The 10 is called the base, the exponent 2 is the logarithm, and the number 100 is the antilogarithm or"antilog." An antilog is the number you get when you raise ten to a certain power.

Inverse logs (antilogs) can be found quickly using a scientific calculator. If your calculator has an[inverse] key, you may be able to get an inverse log by pressing the [inverse] key followed by the [log]key. You can also find antilogs by using the [10x] key or the [yx] key.

Let's try to find an antilog on your calculator. We will try to find the antilog of 2. Another way ofsaying this is, "what number do you get when 10 is raised to the 2nd power?" We already know that theanswer is 100, but let's try to find it using the calculator. If you have an [inverse] key, start by pressing thenumber 2. Then press the [inverse] key followed by the [log] key. If you got an answer of 100, you havesucceeded. If you got something else, check the instruction manual for your calculator to determine thecorrect procedure for finding antilogs.

To find the antilog of 2 using the [10x] key, first enter the number 2, then press the [10x] key. Thenumber 100 should appear in the display. (To use the [10x] key on some calculators, you may have topress the [second function] key first, followed by the [10x] key.

To find the antilog of 2 using the [yx] key, first enter the number 10, which is the base. Next, pressthe [yx] key followed by the number 2. Finally, press the [=] key. The display should read 100. You canraise any base to any power using the [yx] key.

Problem 1. Find the antilogs of the following logs.

a. 4 ___________ e. 67 ___________

b. 3.540 ___________ f. 0.24 ___________

c. 0.83 ___________ g. 1.09 ___________

d. –2.13 ___________ h. –0.88 ___________

F-1 ©1997, A.J. Girondi

According to part "g" of problem 1, if you want to get 12.3, you must raise 10 to the 1.09 power.Supposedly: 101.09 = 12.3 Let's see if that's correct. Enter the number 12.3 into yourcalculator. Now if you press the [log] key, you will find the power to which 10 must be raised to give you12.3. Try it. You should get 1.09 (rounded).

Now Let's try a problem involving pH. Suppose you know that the pH of a solution is 5.9, and youwant to find [H1+] for the solution. We know that: pH = - log [H1+], so let's substitute the 5.9 into theequation: 5.9 = - log [H1+]. To solve for [H1+], we need to isolate it on one side of the equation. First,multiply both sides by -1, which yields: -5.9 = log [H1+]. Next, multiply both sides by antilog:

antilog -5.9 = antilog (log [H1+])

All that is left to do is to find the antilog of -5.9. The answer you should get is 1.3 X 10-6 (rounded to 2 significant figures). Now try the problems below.

Problem 2. What is the [H1+] of a solution that has a pH of 8.22?

Problem 3. What is the [H1+] of a solution that has a pH of 1.23?

Problem 4. What is the [OH1-] of a solution which has a pH of 4.2? (In solving this problem, use the equation Kw = [H1+] [OH1-], and recall that Kw has a constant value of 1.0 X 10-14.)

Problem 5. What is the [OH1-] of a solution which has a pH of 9.11? (Again, use the Kw equation.)

SECTION F.2 Answers to Problems

1. a. 10,000; b. 3467; c. 6.8; d. 0.00741; e. 1 X 1067; f. 1.7; g. 12.3; h. 0.132. 6.03 X 10-9

3. 5.89 X 10-2

4. 1.6 X 10-10

5. 1.29 X 10-5

End of Appendix F

F-2 ©1997, A.J. Girondi

NAME________________________________________________________ PER_____________


"ALICE"

REFERENCENOTEBOOK

This document may be used (when directed) duringtests and quizzes, provided that you do not alter it inany way. Do not add any information to it.

R-1 ©A.J. Girondi

Reference Table 1 Common Oxidation Numbers of Selected Elements

(Note: Sometimes these elements can assume oxidation numbers other than those listed.)

Aluminum Al +3 Lithium Li +1Antimony Sb +3,+5 Magnesium Mg +2Arsenic As +3,+5 Manganese Mn +2,+4,+7Barium Ba +2 Mercury Hg +1,+2Bismuth Bi +3 Nickel Ni +2Boron B +3 Nitrogen N -3,+3,+5Bromine Br -1,+5 Oxygen O -2Calcium Ca +2 Phosphorus P +3,+5Carbon C +2,+4 Platinum Pt +2,+4Cesium Cs +1 Potassium K +1Chlorine Cl -1,+5,+7 Silicon Si +4Chromium Cr +2,+3,+6 Silver Ag +1Cobalt Co +2,+3 Sodium Na +1Copper Cu +1,+2 Strontium Sr +2Fluorine F -1 Sulfur S -2,+4,+6Gold Au +1,+3 Tin Sn +2,+4Hydrogen H +1 Titanium Ti +3,+4Iodine I -1,+5 Tungsten W +6Iron Fe +2,+3 Zinc Zn +2Lead Pb +2, +4

Reference Table 2Common Oxidation Numbers of Selected Polyatomic Ions

Name Formula Charge

Ammonium NH41+ +1Acetate C2H3O21- -1 Chlorate ClO31- -1Perchlorate ClO41- -1 Cyanide CN1- -1 Hydrogen carbonate HCO31- -1 (or bicarbonate) Hydrogen sulfate HSO41- -1 Hydroxide OH1- -1 Nitrate NO31- -1 Nitrite NO21- -1 Permanganate MnO41- -1 Thiocyanate SCN1- -1Carbonate CO32- -2 Chromate CrO4 2- -2 Dichromate Cr2O72- -2 Sulfate SO42- -2 Sulfite SO32- -2Phosphate PO43- -3

R-3 ©A.J. Girondi

Reference Table 3Latin Names of Four Selected Elements

Element Symbol Latin Name

copper (I) Cu1+ cuprouscopper (II) Cu2+ cupric

iron (II) Fe2+ ferrousiron (III) Fe3+ ferric

mercury (I) Hg1+ mercurousmercury (II) Hg2+ mercuric

tin (II) Sn2+ stannoustin (IV) Sn4+ stannic

Reference Table 4Four Kinds of Decomposition Reactions

1. Metallic carbonates decompose into metallic oxides and carbon dioxide.

CaCO3 ----> CaO + CO2

K2CO3 ----> K2O + CO2

H2CO3 ----> H2O + CO2

2. Many metallic hydroxides decompose into metallic oxides and water.

2 KOH ----> K2O + H2O

Ca(OH)2 ----> CaO + H2O

2 Al(OH)3 ----> Al2O3 + 3 H2O

3. Metallic chlorates decompose into metallic chlorides and oxygen gas.

2 KClO3 ----> 2 KCl + 3 O2

Ba(ClO3)2 ----> BaCl2 + 3 O2

2 Al(ClO3)3 ----> 2 AlCl3 + 9 O2

4. Many binary (two-element) compounds decompose into pure elements.

2 HgO ----> 2 Hg + O2

2 H2O ----> 2 H2 + O2

2 NaCl ----> 2 Na + Cl2

Note: There are many other kinds of decomposition reactions in addition to those

shown above.

R-4 ©A.J. Girondi

Reference Table 5The Activity Series

Metals Nonmetals

lithium fluorinepotassium chlorinecalcium brominesodium iodinemagnesiumaluminumzincchromiumironnickeltinleadhydrogen Although hydrogen iscopper not a metal, it ismercury compared to metals insilver the activity series.platinumgold

Reference Table 6SOLUBILITY RULES

Type of Compound Precipitate?

1. Compounds of nitrates (NO31-), acetates (C2H3O21-),and chlorates (ClO31-) ARE soluble.- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO

2. All common compounds of Na, K, and NH41+ ARE soluble. - - - - - - - - - - - - - - - - - - - - NO

3. Most chlorides, Cl1-, bromides, Br1-, and fluorides, F1-, ARE soluble -- - - - - - - - - - - - NO(But those of Ag, Hg, and Pb* are NOT soluble) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - YES

4. Most sulfate, SO42-, compounds ARE soluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO(But those of Pb, Ba, Sr, and Ca** are NOT soluble) - - - - - - - - - - - - - - - - - - - - - - - - - - - YES

5. Most carbonate (CO32-), phosphate (PO43-),and sulfide (S2-) compounds are NOT soluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - YES(But those of Na, K, and NH41+ ARE soluble) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - NO

6. Most hydroxides, OH1-, are NOTsoluble - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - YES(But those of all Group1A metals, and Ca, Ba and Sr ARE soluble) - - - - - - - - - - - - - - NO

*Lead chloride, PbCl2, is fairly soluble in hot water. Consider it a precipitate.**Calcium sulfate, CaSO4, is slightly soluble. Consider it a precipitate.

R-5 ©A.J. Girondi

ReferenceTable 7A SOLUBILITY TABLE

For our purposes, the combinations below designated with the letter P, are considered to be insoluble enough to form precipitates. If a P is not listed for a substance, it is soluble in water (it dissolves). If youcannot find the data you need in this table, check the solubility rules. P = PRECIPITATE

1 - acetate, C2H3O21- 6 - chromate, CrO42- 11 - phosphate, PO43-

2 - bromide, Br1- 7 - hydroxide, OH1- 12 - silicate, SiO42-

3 - carbonate, CO32- 8 - iodide, I1- 13 - sulfate, SO42-

4 - chlorate, ClO31- 9 - nitrate, NO31- 14 - sulfide, S2-

5 - chloride, Cl1- 10 - oxide, O2-

1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4

aluminum - - - - - - P - - P P P - Pammonium - - - - - - - - - - - - - -barium - - P - - P - - - P P P P Pbismuth - P P - - - P P - P P - P Pcadmium - - P - - - P - - P - - - Pcalcium - - P - - - P - - P P P - Pcopper(II) - - P - - - P - - P P P - Phydrogen - - - - - - - - - - - P - -iron(II) - - P - - - P - - P P P - Piron(III) - - - - - P P - - P P P - Plead(II) - P P - P P P P - P P P P Pmagnesium - - P - - - P - - P P P - -manganese(II) - - P - - - P - - P P P - Pmercury(I) P P P - P P - P - P P - P Pmercury(II) - P P - - P P P - P P - P Pnickel - - P - - - P - - P P - - Ppotassium - - - - - - - - - - - - - -silver P P P - P P - P - P P - P Psodium - - - - - - - - - - - - - -strontium - - P - - P - - - - P P P -tin(II) - P - - - P P - - P P - - Ptin(IV) - - - - - - P - - P - - - Pzinc - - P - - P P - - P P P - P

(Note that sodium, potassium, and ammonium never form precipitates.)

R-6 ©A.J. Girondi

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d

7s

Reference Table 9The Diagonal Rule

% Error = O - A

A X 100% O = observed value; A = accepted value

% Deviation = O - M

M X 100% O = observed value; M = mean (average) value

Reference Table 10Formulas for Percent Error and Percent Deviation

R-8 ©A.J. Girondi

Reference Table 11Standard Half-Cell Reduction Potentials

Reduction---> <---Oxidation E ored (Volts)

Li1+ + e- <====> Li -3.00Rb1+ + e- <====> Rb -2.92 K1+ + e- <====> K -2.92 Cs1+ + e- <====> Cs -2.92 Ba2+ + 2e- <====> Ba -2.90 Sr2+ + 2e- <====> Sr -2.89 Ca2+ + 2e- <====> Ca -2.87

Na1+ + e- <====> Na -2.71 Mg2+ + 2e- <====> Mg -2.37 Al3+ + 3e- <====> Al -1.66 Mn2+ + 2e- <====> Mn -1.18 Zn2+ + 2e- <====> Zn -0.76 Cr3+ + 3e- <====> Cr -0.74 Cd2+ + 2e- <====> Cd -0.40 Fe2+ + 2e- <====> Fe -0.44 Cr3+ + e- <====> Cr2+ -0.41 StrongerCo2+ + 2e- <====> Co -0.28 ReducingNi2+ + 2e- <====> Ni -0.25 AgentsSn2+ + 2e- <====> Sn -0.14 Pb2+ + 2e- <====> Pb -0.13

------------- 2 H1+ + 2e - <====> H2(g) 0.00 ------------- Sn4+ + 2e- <====> Sn2+ +0.15 Cu2+ + e- <====> Cu1+ +0.15

Stronger Cu2+ + 2e- <====> Cu +0.34 Oxidizing Cu1+ + e- <====> Cu +0.52 Agents I2 + 2e- <====> 2 I1- +0.53

Fe3+ + e- <====> Fe2+ +0.77 Hg2+ + 2e- <====> Hg(l) +0.78 Hg22+ + 2e- <====> 2 Hg(l) +0.79 Ag1+ + e- <====> Ag +0.80Br2(l) + 2e- <====> 2 Br1- +1.06Cl2(g) + 2e- <====> 2 Cl1- +1.36Au3+ + 3e- <====> Au +1.50F2(g) + 2e- <====> 2 F1- +2.87

Oxidizing Reducing Agents Agents

R-9 ©A.J. Girondi

Reference Table 12Vapor Pressure of Water

T (oC) P (mm) T (oC) P (mm)

0 4.6 26 25.25 6.5 27 26.710 9.2 28 28.4 12 10.5 29 30.014 12.0 30 31.8 16 13.6 35 42.217 14.5 40 55.318 15.5 45 71.9 19 16.5 50 92.5 20 17.5 55 118.0 21 18.6 60 149.422 19.8 65 187.523 21.2 70 233.7 24 22.4 80 355.125 23.8 90 525.8

100 760.0

Reference Table 13Selected Metric Equivalents

1 meter (m) = 100 centimeters (cm)1 centimeter (cm) = 10 millimeters (mm) 1 liter (l) = 1000 milliliters (mL) 1 milliliter (mL) = 1 cubic centimeter (cm3)1 kilogram (kg) = 1000 grams (g)1 gram (g) = 1000 milligrams (mg)1 g of H2O = 1 mL H2O

Reference Table 14Selected English-Metric Equivalents

1 meter (m) = 39.37 inches (in) 1 inch (in) = 2.54 centimeters (cm)1 kilogram (kg) = 2.2 pounds (lbs)1 pound (lb) = 453.6 grams (g) 1 ounce (oz) = 28.35 grams (g) 1 gram (g) = 15.43 grains (gr) 1 liter (l) = 1.06 quarts (qts) 1 cubic foot (cu. ft.) = 28.32 liters

R-10 ©A.J. Girondi

R-11

H

Li Be

Na Mg

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga

C B

Al Si

Ge

N

P

As

O

S

Se

He

F

Cl

Br

Ne

Ar

Kr

Xe

Rn

ITeSbSnIn

Tl Pb Bi Po At

Rb

Cs

Fr

Sr

Ba

Ra

Y

La

Ac

Zr

Hf

Unq

Nb

Ta

Mo

W

Tc

Re

Ru

Os

Rh

Ir

Pd

Pt

Ag

Au

Cd

Hg


Reference Table R-15The Periodic Table With Atomic Weights1

1.013

11

19

37

55

87

6.94

22.99

39.10

85.47

132.90

223

4

12

20

38

56

88

24.31

9.01

40.08

87.62

137.34

(226)

21 22 23 24 25 26 27 28 29 30 31 32 33

44.96 47.90 50.94 52.00 54.94 55.85 58.93 58.71 63.54 65.37 69.72 72.59 74.9239 40 41 42 43 44 45 46 47 48 49 50 51

88.91 91.22 92.91 95.94 (99) 101.07 102.91 106.40 107.87 112.40 114.82 118.69 121.7557 72 73 74 75 76 77 78 79 80 81 82 83

138.91 178.49 180.95 183.85 186.21 190.20 192.20 195.09 196.97 200.59 204.37 207.19 208.98

5 6 7

13 14 1510.81 12.01 14.01

26.98 28.09 30.97

89 104 105 106 107 108 109 110 111

227 261 262 263

8 9 10

2

16 17 18

4.00

16.00 19.00 20.18

32.06 35.45 39.95

78.96 79.91 83.80

127.60 126.90 131.30

34 35 36

52 53 54

84 85 86

(210) (210) (222)

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

58 59 60 61 62 63 64 65 66 67 68 69 70 71

90 91 92 93 94 95 96 97 98 99 100140.12 140.91 144.24 (147) 150.35 151.96 157.25 158.92 162.50 164.93 167.26 168.93 173.04 174.97

232.04 (231) 238.03 (237) (242) (243) (247) (249) (251) (254) (253) (256) (254) (260)

101 102 103

atomic weight (parentheses indicate atomic mass of most stable isotope)E

??

261

atomic numbersymbol

Reference Table R-16Common Oxidation Numbers of the Elements

1 H +1, (-1 in hydrides like NaH) 39. Y +3 2. He 0 40. Zr +4 3. Li +1 41. Nb +5, +4 4. Be +2 42. Mo +6, +4, +3 5. B +3 43. Tc +7, +6, +4 6. C +4, (less commonly +2, -4) 44. Ru +8, +6, +4, +3 7. N +5, +4, +3, +2, +1, -3 45. Rh +4, +3, +2 8. O -2 , (-1 in peroxides like H2O2) 46. Pd +4, +2

9. F -1 47. Ag +110. Ne 0 48. Cd +211. Na +1 49. In +312. Mg +2 50. Sn +4, +213. Al +3 51. Sb +5, +3, -314. Si +4, -4 52. Te +6, +4, -215. P +5, +3, -3 53. I -1, +7, +5, +3, +116. S +6, +4, +2, -2 54. Xe +6, +4, +217. Cl -1, +7, +5, +3, +1 55. Cs +118. Ar 0 56. Ba +219. K +1 57. La +320. Ca +2 58-71. Ce through Lu +321. Sc +3 72. Hf +422. Ti +4, +3, +2 73. Ta +523. V +5, +4, +3, +2 74. W +6, +424. Cr +6, +3, +2 75. Re +7, +6, +425. Mn +7, +6, +4, +3, +2 76. Os +8, +426. Fe +3, +2 77. Ir +4, +327. Co +3, +2 78. Pt +4, +228. Ni +2 80. Hg +2, +129. Cu +2, +1 81. Tl +3, +130. Zn +2 82. Pb +4, +231. Ga +3 83. Bi +5, +332. Ge +4, -4 84. Po +233. As +5, +3, -3 85. At -134. Se +6, +4, -2 86. Rn 035. Br -1, +7, +5, +3, +136. Kr +4, +237. Rb +138. Sr +2

R-12 ©A.J. Girondi

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