chemistry green lab manual
TRANSCRIPT
Lab Manual Introductory Chemistry: A Green Approach Version 3.1
© 2010, eScience Labs, Inc. All rights reserved esciencelabs.com • 888.375.5487
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Table of Contents
Chemistry and Experiments
Lab 1: Introduction and Safety
Lab 2: The Scientific Method
Measurements
Lab 3: Measurements
Lab 4: Properties of Matter ‐ Density
Matter and Energy
Lab 5: Mixtures and Solutions
Lab 6: Chemical and Physical Change
Lab 7: Heat and Calorimetry
Lab 8: Chemical Processes
Atomic Theory and Structure
Lab 9: Electron Configuration
Lab 10: Light Spectrum
Chemical Bonding
Lab 11: Molecular Models
Lab 12: Ionic and Covalent Bonds
Lab 13: Polar Bonding
Introducing Chemical Reactions
Lab 14: Chemical Reactions I
Lab 15: Chemical Reactions II
Lab 16: Metals and Oxidation
Classification of Elements
Lab 17: The Mole and Avogadro’s Number
Lab 18: The Periodic Table
Lab 19: Stoichiometry
Gas Laws
Lab 20: Ideal Gas Law
Reaction Rates
Lab 21: Reaction Rate
Lab 22: Catalysts
Acids and Bases
Lab 23: Acids and Bases
Lab 24: Titration
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Lab Safety
Always follow the instructions in your laboratory manual and these general rules:
Lab preparation
Please thoroughly read the lab exercise before starting!
If you have any doubt as to what you are supposed to be doing and how to do it safely, please STOP and then:
Double‐check the manual instructions.
Check www.esciencelabs.com for updates and tips.
Contact us for technical support by phone at 1‐888‐ESL‐Kits (1‐888‐375‐5487) or by email at [email protected].
Read and understand all labels on chemicals.
If you have any questions or concerns, refer to the Material Safely Data Sheets (MSDS) available at
www.esciencelabs.com. The MSDS lists the dangers, storage requirements, exposure treatment and disposal instruc‐
tions for each chemical.
Consult your physician if you are pregnant, allergic to chemicals, or have other medical conditions that may
require additional protective measures.
Proper lab attire
Remove all loose clothing (jackets, sweatshirts, etc.) and always wear closed‐toe shoes.
Long hair should be pulled back and secured and all jewelry (rings, watches, necklaces, earrings, bracelets,
etc.), should be removed.
Safety glasses or goggles should be worn at all times. In addition, wearing soft contact lenses while conducting
experiments is discouraged, as they can absorb potentially harmful chemicals.
When handling chemicals, always wear the protective goggles, gloves, and apron provided.
eScience Labs, Inc. designs every kit with safety as our top priority. Nonetheless, these are science kits and contain items which must be
handled with care. Safety in the laboratory always comes first!
Lab Safety
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Performing the experiment
Do not eat, drink, chew gum, apply cosmetics or smoke while conducting an experiment.
Work in a well ventilated area and monitor experiments at all times, unless instructed otherwise.
When working with chemicals:
Never return unused chemicals to their original container or place chemicals in an unmarked container.
Always put lids back onto chemicals immediately after use.
Never ingest chemicals. If this occurs, seek immediate help.
Call 911 or “Poison Control” 1‐800‐222‐1222
Never pipette anything by mouth.
Never leave a heat source unattended.
If there is a fire, evacuate the room immediately and dial 911.
Lab Clean‐up and Disposal
If a spill occurs, consult the MSDS to determine how to clean it up.
Never pick up broken glassware with your hands. Use a broom and a dustpan and discard in a safe area.
Do not use any part of the lab kit as a container for food.
Safely dispose of chemicals. If there are any special requirements for disposal, it will be noted in the lab man‐
ual.
When finished, wash hands and lab equipment thoroughly with soap and water.
Above all, USE COMMON SENSE!
Lab Safety
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Additional Materials Needed and Time Required to Complete Labs
** Note: If you are allergic to latex, please contact us and we will send you vinyl gloves**
Lab 1: Introduction and Safety
Time: 30 min.
Materials needed: none
Lab 2: The Scientific Method
Time: 30 min.
Materials needed: none
Lab 3: Measurements
Time: 60 min.
Materials needed: Water
Lab 4: Properties of Matter ‐ Density
Time: 60 min.
Materials needed: Distilled water
Lab 5: Mixtures and Solutions
Time: 60 min.
Materials needed: Distilled water, boiling water, ice, bowl
Lab 6: Chemical and Physical Change
Time: 60 min.
Materials needed: Boiling water
Lab 7: Heat and Calorimetry
Time: 60 min.
Materials needed: Snack food, mini marshmallow, aluminum can with tab
Lab 8: Chemical Processes
Time: 60 min.
Materials needed: none
Lab 9: Electron Configuration
Time: 60 min.
Materials needed: none
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Additional Materials Needed and Time Required to Complete Labs
Lab 10: Light Spectrum
Time: 60 min.
Materials needed: Book, white copy paper
Lab 11: Molecular Models
Time: 60 min.
Materials needed: none
Lab 12: Ionic and Covalent Bonds
Time: 60 min.
Materials needed: none
Lab 13: Polar Bonding
Time: 60 min.
Materials needed: Notebook paper, newspaper, variety of inks (Uni‐Ball, permanent marker, dry erase, and highlighter are suggested)
Lab 14: Chemical Reactions I
Time: 60 min.
Materials needed: none
Lab 15: Chemical Reactions I I
Time: 60 min
Materials needed: none
Lab 16: Metals and Oxidation
Time: 60 min.
Materials needed: none
Lab 17: The Mole and Avogadro’s Number
Time: 60 min.
Materials needed: none
Lab 18: The Periodic Table
Time: 60 min.
Materials needed: none
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Additional Materials Needed and Time Required to Complete Labs
Lab 19: Stoichiometry
Time: 60 min.
Materials needed: Distilled water, oven
Lab 20: The Ideal Gas Law
Time: 60 min.
Materials needed: Distilled water, warm water (ring stand and large ring are optional)
Lab 21: Reaction Rate
Time: 60 min.
Materials needed: Ice, boiling water
Lab 22: Catalysts
Time: 60 min.
Materials needed: Warm water, carrot, tomato, soil
Lab 23: Acids and Bases
Time: 60 min.
Materials needed: Tomato juice, distilled water, milk
Lab 24: Titration
Time: 60 min.
Materials needed: Distilled water
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Additional Online Content at www.esciencelabs.com
Student Portal Content
Introduction
Safety Video
The Scientific Method
Chemistry and Experiments
The Chemistry of Cooking
Measurements
Measuring Volume Using a Graduated Cylinder
Unit Conversions
Matter and Energy
Phase Diagrams
The Three States of Matter
Atomic Theory and Structure
The Structure of an Atom
Atomic Structure and Ionic Bonding
Atomic Symbols, Atomic Numbers, and Mass Numbers
Atomic Weight
Ion Charges
Ions
Chemical Bonding
Catching Crooks Green Handed
Covalent Bonds
Intermolecular Forces
Shapes of Simple Molecules—Parts I and II
Introducing Chemical Reactions
Balancing Oxidation Reduction Equations
Chemistry Math
Common Types of Oxidation‐Reduction Reactions
Oxidation‐Reduction Reactions
Classification of Elements
Conversion between Mass and Moles
Identifying Compounds and Ions
Moles
Stoichiometry
The Types of Elements in the Periodic Table and their Properties
Gas Laws
Ideal Gas Law
Acids and Bases
Acid base reactions
Log on to the Student Portal using these easy steps:
Visit our website, www.esciencelabs.com, and click on the green button (says “Register or Login”) on the top right side of the
page. From here, you will be taken to a login page. If you are registering your kit code for the first time, click the “create and account” hyperlink. Locate the kitcode, located on a label on the inside of the kit box lid. Enter this, along with other requested infor‐mation into the online form to create your user account. Be sure to keep
track of your username and password as this is how you will enter the Stu‐dent Portal for future visits. This es‐tablishes your account with the eScience Labs’ Student Portal.
Have fun!
Introductory Chemistry
Lab 1: Introduction and Safety
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Lab 1: Introduction and Safety
Introduction
Welcome to the Introductory Chemistry Lab Manual!
Chemistry can be fun! If you’ve ever seen the Frankenstein movies, then you may recall the dark, dungeon‐like laboratory with a mad scientist trying to come up with just the right chemicals to produce the Frankenstein character. Just as the mad scientist had to learn the safest chemicals and procedures to use to not destroy his labo‐ratory, you must also learn to correctly manage green, environmen‐tally friendly, chemicals and adhere to safety precautions. Before beginning these labs, you must first realize there are specific tech‐niques and precautions to learn. You will need to learn the location of certain items in your lab, certain rules, and what specific supplies are called. After you get acquainted with your surroundings, the chemistry lab will no longer seem confusing or even dangerous.
Playing a game is more enjoyable than just watching or reading about it. This is also true about learning chemistry. Getting to actually do experiments is a lot of fun. But chemistry can also be dangerous, and getting hurt is definitely not fun! For this reason there are safety rules to follow and protective equipment to identify and use.
This manual is written for a “Green Chemistry” approach—this means when compared to similar experiments, the activities in this manual are generally safer and produce less hazardous waste. The chemicals used are environmentally friendly and can be discarded down a household drain. Each of these experiments has been performed safely by other students. How‐ever, safer and less hazardous waste does not mean accidents, injuries, or damage cannot happen. Scrapes and burns can still happen. Because experiments will be performed in the home, be sure a working fire extinguisher is accessible. You are advised to always use the safety goggles and gloves provided in your kit.
Safety Equipment
Safety Goggles/Glasses – Safety goggles and glasses are used to protect your eyes, and should be worn at all times when in the chemistry laboratory, even if you are not currently working with chemicals.
Gloves – At times you may need to wear gloves to protect your hands from harmful chemicals or hot objects. The type of glove needed will depend on the application. For example, oven mitts are worn to remove hot objects from an oven while vinyl or latex gloves are used when working with acids and bases. It is very important not to touch your work area with gloves that have been contaminated with harmful chemicals.
Figure 1: Using the information in this section, you will be able to complete the chemistry labs in this manual safely and successfully.
Concepts to explore:
Learn how work to safely in the chemical laboratory
Learn when and how to use the safety equipment in the chemical laboratory
Learn the names of the equipment used in the experiments
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Safety Shower – A safety shower is used when a hazardous chemical is spilled on a person where they are unable to rinse it off thoroughly in the sink. It can also be used if a person’s clothes catch on fire. Most safety showers are operated by pulling a chain, though a standard home shower will also work. If a sink is insufficient to thoroughly rinse yourself, get to the nearest shower in your home as quickly as possible. This is not a time for modesty—remove the contaminated clothing while rinsing your skin with a copious amount of water
Eye Wash – An eye wash is used if a harmful chemical is splashed into your eyes or face. It is usually operated by pushing forward on a handle. In the home, find the nearest sink and flush water in your eyes for at least 20 min‐utes. If your sink has a sprayer, use it to rinse your eyes or face while making sure the water drains into the sink.
Fire Extinguisher – A fire extinguisher is used to put out small to medium fires.
Laboratory Fume Hood ‐ A laboratory fume hood removes harmful gases and fumes sometimes present when do‐ing an experiment. You should always work in a fume hood whenever you are working with corrosive, noxious, or flammable materials. Chemicals used in this kit will not require the use of a fume hood.
Besides knowing where the safety equipment is located and how and when to use it, there are general safety rules you will also need to follow in the laboratory. Some of the common safety rules are listed below. Your instructor may have a few others.
Laboratory Safety Rules
1. Always wear safety glasses or goggles. Never wear contact lenses.
2. Never attempt unauthorized experiments.
3. Always have someone available to help in the event of an accident.
4. Never have food, drink, or tobacco into the laboratory.
5. Always keep your work area free of clutter.
6. Always wear a protective apron and sensible clothing. This means no loose clothing, bare midriffs or open‐toe shoes.
7. Know the location of and how to use safety equipment in your home. This includes showers, fire extinguishers, and sinks.
8. Always read the experiment before doing it.
9. Always wash hands before leaving the lab.
10. Tie back long hair.
11. Never run or play practical jokes in the experiment area.
12. Place broken glass in a protective container, never loosely in a trash can.
In addition to following the general safety rules, chemicals need to be handled properly. Listed below are some guidelines on how to handle chemicals properly.
Lab 1: Introduction and Safety
17
Handling Chemicals
1. Always add acids to water, never water to acids.
2. Never return unused chemicals to the bottles from where they were first obtained.
3. Dispose of used chemicals in the proper waste containers and/or as instructed.
4. Always clean the work area, and put away extra equipment when laboratory work is completed.
5. Never leave anything unattended while it is being heated or is reacting rapidly.
6. Never carry out a reaction or heat a substance in a closed system.
7. Always be careful when working with previously heated objects.
8. Always replace stoppers or lids on bottles containing chemicals.
9. Weigh chemicals in weigh boats or on paper provided for that purpose. Never weigh chemicals by placing them directly on the scale.
10. Label all chemicals clearly and completely.
11. Read labels carefully before using chemicals.
12. Always lubricate glass tubing or thermometers before inserting them into rubber stoppers.
13. Material Safety Data Sheets (MSDS) for all chemicals provided can be found on our website at www.esciencelabs.com/educators/msds. These sheets contain all needed information regarding the dan‐ger, safety and disposal of every chemical.
Even though you follow all of the safety rules, accidents can still happen. This is why it is so important to know what to do for each type of accident.
How to Respond to Accidents
1. Chemical Spills on the bench or floor – Be sure to clean up the spill immediately. If the spill involves volatile or flammable materials, such as alcohol, make sure ALL flames in the lab area are extinguished and spark‐producing equipment is shutdown. In the case of an acid spill, pour baking soda on the acid before cleaning up. In the case of a base spill, pour vinegar on the base before cleaning it up. All other chemicals used in this manual can be cleaned up as you normally would. If you have any questions, check the MSDS.
2. Hazardous chemical spills on a person – If the spill covers a large area, the typical course of action is to re‐move all contaminated clothing while the person is under the safety shower. If it is a small area, flush the area immediately with a large amount of water and then wash it with soap. Check the MSDS for the spilled chemi‐cal and follow all instructions. Medical assistance may be necessary.
3. Chemicals spills in the eyes – If a harmful chemical is splashed on your face and/or in your eyes, immediate attention is critical. Call for help and get to the nearest sink. If the chemical splashes on your face, and you have goggles on, KEEP the goggles on. Remove the chemical from your face before you remove the goggles. If a chemical gets in your eyes, hold your eyes open in the eyewash for at least 20 minutes. Even though you should not be wearing contact lenses in the lab, if you are, rinse your eyes thoroughly, remove your contacts, and continue to rinse your eyes. A doctor should examine your eyes as soon as possible.
Lab 1: Introduction and Safety
18
4. Ingestion of chemicals – Check the MSDS immediately. Call 911 or “Poison Help” at 1‐800‐222‐1222.
5. Burns – Flush the area with cool running water for 20 minutes. Medical attention may be necessary.
6. Cuts and wounds – ‐ If a chemical gets into the cut or wound, rinse it off immediately with a large amount of water. Avoid contamination; check the MSDS.
7. Fire – Fires in a laboratory are often contained in pieces of glassware, such as a beaker. You should not move a beaker that has a chemical burning in it. Instead, extinguish the fire simply by covering the mouth of the beaker with a thin curved piece of glass called a watch glass and turning off the source of the flame. A plate or pie pan can also work in place of a watch glass. If the fire is not easily covered, you can use a fire extinguisher. If the fire is too large to extinguish quickly, clear the home and call the fire department immediately.
Clothing fires can be extinguished in a safety shower if it is close by. If it is not very close, you will need to STOP, DROP, and ROLL to quickly smother the fire.
Lab 1: Introduction and Safety
If you have ANY questions or concerns regarding a chemical, read the “Material Safety Data Sheet” (MSDS) for that chemical. The MSDS lists the dangers, storage requirements, exposure treatment and disposal instructions for every chemical.
The MSDS for any chemical supplied by eScience Labs Inc., can be found at www.esciencelabs.com/educators/msds
If you have any doubt as to what you are supposed to be doing and how to do it safely:
STOP!
Double‐check the manual
Check the website www.eScienceLabs.com
Email: [email protected]
Call for help 1‐888‐Esl‐Kits (1‐888‐375‐5487)
Contact your instructor
When you follow the safety rules the chemistry laboratory can be a
very interesting place to explore and learn.
Additional Resources
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Pre‐lab Questions
1. What should you always wear to protect your eyes when you are in the chemistry laboratory?
2. Should you add acid to water or water to acid?
3. Where should you dispose of broken glass?
4. What should you do if you spill a chemical on your hand?
Lab 1: Introduction and Safety
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Part 1: What is it?
A chemical laboratory contains special equipment to use while you are performing an experiment. Locate each of these items in your lab kit, and place a check mark in the appropriate place when you find it.
Beaker
50 mL
250 mL
600 mL
Erlenmeyer Flask
Funnel Wash Bottle Dropper
Lab 1: Introduction and Safety
Graduated Cylinder
10 mL
100 mL
Test Tube (11) Stir Rod
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Test Tube Rack
Test Tube Holder Petri Dish Watch Glass
2. Sketch a picture and name any other items that are located in your lab kit, classroom, or home that are likely to be useful for you in completing these labs.
Lab 1: Introduction and Safety
Well Plate Mortar and Pestle
Introductory Chemistry
Lab 2: The Scientific Method
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Introduction
Testable Observations: A testable observation is one that lends itself to further in‐
vestigation. The observation must be detailed enough to raise a question that can be
challenged.
For Example:
Plants grow well in soil.
Plants grow faster in soil with nutrients than in soil without nutrients.
The first observation makes a statement but provides no information to warrant further investigation. The second observa‐
tion is more detailed and provides a rational to investigate factors that make plants grow faster.
Hypothesis and Null Hypothesis
Hypothesis Generation: A hypothesis is an educated guess. By considering both the question asked and explanation posed,
a testable relationship can be designed—this is the hypothesis. A Null Hypothesis is a testable statement, that if proved
true means the hypothesis was incorrect. Both statements must be testable, but only one can be true.
From the second observation above we can develop both a Hypothesis and a Null Hypothesis:
Hypothesis:
The addition of nutrients to soil increases the speed of growth of plants.
Null Hypothesis:
The addition of nutrients to soil does not speed up the growth of plants.
Lab 2: The Scientific Method
Concepts to explore:
How to make testable observations
The roles of a hypothesis and null hypothesis How to conduct a successful experiment
The role of variables and control in an experiment
The importance of organized data collection
Recognize what makes a successful analysis
26
If the plants DO grow more quickly when nutrients are added:
Then, the hypothesis is supported and the null‐hypothesis is wrong.
Conducting an Experiment
Experimental Approach: There are often many ways to
test a hypothesis. When designing an experiment to test
a hypothesis there are 3 rules to follow:
1. The experiment must be replicable
2. Only test one variable at a time
3. Always include a control
Variables are defined and measurable components of an experiment. There are three types of variables:
Independent Variable: The variable that the scientist alters to test the hypothesis. For our example, the
independent variable would be the addition of nutrients.
Dependent Variable: This variable is measured in regards to conditions of the independent variable. In our
example, the dependent variable would be speed of growth.
Control: A group that is not subject to the independent variable being tested. This group becomes the
standard for comparison. For our example, the control would be soil without added nutrients.
Data Collection: In designing the experiment, establish a clear and concise procedure. Specify what data are needed and
when they should be collected in advance. As these data will be the basis for your conclusions, they must be accurately
recorded. Remember, replication is fundamental to scientific experiments.
Often, the best way to organize data for analysis is as a table or a graph. Remember, any table or graph should be able to
stand on its own. In other words, another scientist should be able to pick up the table or graph and have all of the informa‐
tion necessary to interpret it, with no other information.
Remember these 3 rules.
They are VERY Important!
Lab 2: The Scientific Method
27
Tables and Graphs
Table: A well organized summary of data collected. Only include information relevant to the hypothesis (e.g. don’t include
the color of the plant when it is not relevant to what is being tested). Always include a clearly stated title, label your col‐
umns and rows and include the measuring units. For our example:
Table 1: Plant Growth with and without Added Nutrients
Graph: A visual representation of the relationship between the independent and dependent variable. Graphs are useful in
identifying trends and illustrating findings. Rules to remember:
The independent variable is always graphed on the x axis (horizontal), with the dependent variable on the y
axis (vertical).
Use appropriate numerical spacing when plotting the graph, with the lower numbers starting on both the
lower and left hand corners.
Always use uniform (0, 5, 10, 15 . . . ) or logarithmic (1, 10, 100, 1000 . . .) intervals. For example, if you begin
by numbering, 0, 10, 20, do not jump to 25 and then to 32.
Title the graph and both the x and y axes such that they correspond to the table from which they come. For
example, if you titled your table “Heart Rate of those who eat vegetables and those who do not eat vegeta‐
bles”, be sure to title the graph the same.
Determine the most appropriate type of graph. Typically, line and bar graphs are the most common.
Line graph: A line graph shows the relationship between variables using plotted points that are connected with a line.
There must be a direct relationship and dependence between each point connected. More than one set of data can be
presented on a line graph. Figure 1 uses the data from our previous table:
Variable Height Day 1 (mm)
Height Day 2 (mm)
Height Day 3 (mm)
Height Day 4 (mm)
Control
(without nutrients) 3.4 3.6 3.7 3.8
Independent
(with nutrients) 3.5 3.8 4.3 4.9
Lab 2: The Scientific Method
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Figure 1: Rate of Plant Growth with and without Added Nutrients
Bar graph: A bar graph is used to compare results that are independ‐
ent from each other, as opposed to a continuous series. Since the re‐
sults from our previous example are continuous, they are not appropri‐
ate for a bar graph.
The following bar graph shows the top speeds of four cars. Since there
is no direct relationship between each car, each result is independent and a bar graph is appropriate.
Speed
Figure 2
3
3.2
3.4
3.6
3.8
4
4.2
4.4
4.6
4.8
5
1 2 3 4
Plant Height (m
m)
Plant Growth with and without Nutrients
Without Nutrients (Control)
Lab 2: The Scientific Method
29
Analysis
Interpretation: Based on the data you collected, is your hypothesis supported or refuted?
Based on the data, is the null hypothesis supported or refuted? If the hypothesis is sup‐
ported, are there other variables which should be examined? For instance, was the
amount of water and sunlight consistent between groups of plants; or, were all four cars
driven on the same road?
Exercise
Dissolved oxygen is oxygen that is trapped in a fluid, such as water. Since virtually every
living organism requires oxygen to survive, it is a necessary component of water systems
such as streams, lakes and rivers in order to support aquatic life. The dissolved oxygen is
measure in units of ppm—or parts per million. Examine the data in Table 2 showing the amount of dissolved oxygen pre‐
sent and the number of fish observed in the body of water the sample was taken from, then answer the questions below.
Table 2: Water quality vs. fish population.
Questions
1. Based on this information, what patterns do you observe?
2. Develop a hypothesis relating to the amount of dissolved oxygen measured in the water sample and the num‐
ber of fish observed in the body of water.
3. What would your experimental approach be to test this hypothesis?
Dissolved Oxygen (ppm) 0
Number of Fish Observed 0
2
1
4
3
6
10
8
12
10
13
12
15
14
10
16
12
18
13
Lab 2: The Scientific Method
30
5. What is your control?
6. What type of graph is appropriate for this data set? Why?
7. Graph the data from Table 2 (above) in the space below.
8. Interpret the data from the graph made in Question 7.
4. What are the independent and dependent variables?
Lab 2: The Scientific Method
31
9. Determine which of the following observations are testable.
For those that are testable:
Write a hypothesis and null hypothesis
What would be your experimental approach?
What are the dependent and independent variables?
What is your control?
How will you collect your data?
How will you present your data (charts, graphs, types)?
How will you analyze your data?
A) When a plant is placed on a window sill it grows faster than when it is placed on a coffee table in the middle of
the living room.
B) The teller at the bank with brown hair and brown eyes is taller than the other tellers.
Lab 2: The Scientific Method
32
C) John caught four fish at the seven o’clock in the morning but didn’t catch any at noon.
D) Salaries at Smith and Company are based on the number of sales, and Billy makes 3,000 dollars more than Joe.
E) When Sally eats healthy foods and exercises regularly, her blood pressure is lower than when she does not exer‐
cise and eats fatty foods.
F) The Italian restaurant across the street closes at 9 pm, but the one two blocks away closes at 10 pm.
Lab 2: The Scientific Method
33
G) Bob bought a new blue shirt with a golf club on the back for twenty dollars.
H) For the past two days the clouds have come out at 3 pm, and it has started raining at 3:15 pm.
I) George did not sleep at all last night because he was up finishing his paper.
J) Ice cream melts faster on a warm summer day than on a cold winter day.
Lab 2: The Scientific Method
Introductory Chemistry
Lab 3: Measurements
37
Introduction The loss of the Mars Climate Orbiter spacecraft in late September of 1999 was a tragic blow to NASA’s Mars exploration project. The $320 million dollar mission ultimately failed due to a miscalculation in the conversion of English measurements to metric measurements. NASA has reported the spacecraft’s engine appeared to have burned up while in orbit, most likely due to a miscalculation of altitude. The problem occurred when the supervisors of the project had measured in miles and feet, instead of kilometers and meters. During the project’s development all directions had specified that the metric system be used as the standard measurement. The mistake of using improper units cost NASA millions of dollars, much embarrass‐ment, and a wealth of potential information.
Throughout this lab, we will learn how to make measurements with both accuracy and precision while using significant fig‐ures and converting units. Precision is the agreement between repeated measurements under same conditions. Accuracy is how close a measurement is to an accepted measurement. For example, you throw three darts into a dartboard that all land in the same box, but they are nowhere near the bull’s eye. You are precise in your technique of hitting the dartboard in the same location, but your accuracy is bad (see Figure 1).
Along the same lines, a well‐designed chemistry experiment will use techniques that can be easily repeated and, under the same conditions, result in a high precision. As chemists, however, precision alone is not good enough; we want our data to be accurate or as close as possible to the accepted value. If, using a thermometer, you measure the temperature of boiling
High precision,
but low accuracy
Low precision,
but higher accuracy
High precision
and high accuracy
Figure 1: Diagrams comparing different combinations of accuracy and precision.
Lab 3: Measurements
Concepts to explore:
Learn how to use significant figures and understand their impor‐tance
Make effective and useful measurements in a chemistry lab
Convert measurements relevant to lab procedures and calcula‐tions
Become familiar with the relationship between volume and mass
38
The use of significant figures enables a chemist to decide whether certain digits in a measurement or calculation are im‐portant enough to include in data. The rule of thumb when measuring is to record all the decimal places of which you are certain, plus one estimated digit.
For example, say you are measuring the amount of liquid in millimeters in a graduated cylinder (see Figure 2). You can clearly see that there is at least 1 mL of liquid in the cylinder. To greater detail, you can see that the meniscus lies above the 1.7 mL mark, around half way between 1.7 and 1.8 mL. While there are no additional marks between these lines, you can make an accurate estimate for the precise value based on what you see. In this case, you might record the volume as 1.74 mL, since the meniscus lies slightly below the halfway point. As you can see, estimating the final digit gives a more ac‐curate measurement than simply recording 1.7 or 1.8 mL. In this case there are three significant figures: two are certain and one is estimated.
Figure 2
When measuring a liquid there is a certain place that one must measure ‐ the bottom of the me‐niscus. The meniscus is the curved line that a liquid makes when placed in a narrow con‐tainer. When looking for the bottom of the me‐niscus, one must look straight at it. When one’s line of sight is too high, then the reading that is received is too low. When one’s line of sight is too low, then the reading received is too high.
Is zero significant when using significant figures? Suppose a watermelon has a mass of 1050 grams. We know that the 1, the zero in between, and the 5 are significant figures, but what about the last zero? 1050 could be the exact measurement, or the zero on the end could simply act as a placeholder that indicates the value is somewhere close to but not precisely 1050 (and certainly not 105). To avoid confusion, scientists will often use exponential notation so that others can tell if the last zero is significant. If the zero is not significant in the above case, a scientist would write the number as 1.05 x 103 g. If the zero is in fact significant, the scientist would write the number as 1.050 x 103 g. Zeroes preceding decimal numbers (such as 0.000256) are not significant figures as they act as placeholders.
Now that we understand what to include as a significant figure, how do we keep track of significant figures in calculations? When it comes to multiplication and division, the rule is the result must have the same number of significant figures as the number with the lowest of number of significant figures in the calculations. When it comes to addition and subtraction, the number with the largest uncertainty or number with least decimal places is the deciding factor.
Lab 3: Measurements
1
2
39
Lab 3: Measurements
Example 2
Compute the sum of the following addition problems, then adjust your answer to the correct number of sig‐nificant digits.
Answers to Example 2:
Adding the numbers together in Example 2 item A, we get 607.95. However, this is not reported in the cor‐rect number of significant figures. The determining factor that decides the number of significant figures is the term with the fewest decimal places. In A this is the 155. It contains the lowest number of decimal places or the greatest amount of uncertainty. Therefore, the answer is reported to the ones place since that is the least place known for all three of the numbers being added.
Answers: A. 608 B. 632.3 C. 179
425.35
27.6
155 +
220.0
56.4565
5.28795
350.56 +
121
0.00365
57.8 +
Example 1
How many significant figures are in each of the measurements?
Answers to Example 1:
A. 6 significant figures (the digits before the number do not count)
B. 6 significant figures (the middle zeros count)
C. 1, 2, or 3 significant figures (We could represent this in exponential notation. For example, 3 x 102 would represent 1 significant figure, 3.0 x 102 would represent 2 significant figures, and 3.00 x 102 would repre‐sent 3 significant figures.)
D. 5 significant figures
E. 4 significant figures (the final digit in a decimal place counts).
A. 0.000256401 m
B. 17.0067 kg
C. 300 mm
D. 12,511 L
E. 2.020 cm
40
Lab 3: Measurements
It is important to understand that no measurement is 100% correct. There is a degree of uncertainty included, which is why significant figures are so crucial. For example, say you make a measurement of the width of a cube to be 2.36 cm. Your lab partner makes a measurement of the same cube to be 2.38 cm. Which measurement is correct? The answer is that both are “correct,” as any measurement contains a certain degree of uncertainty (remember that the last digit of the measure‐ment is an estimate). The best way to obtain an accurate answer is to take a large number of careful measurements, which then can be averaged to reduce errors.
Units are another crucial issue that affect what we measure. If we didn’t have a standard unit of measurement, what would we measure with? You could e‐mail your friend in Germany and tell them that you weigh 123. However, this would tell your friend nothing because she doesn’t understand what you’re comparing. For all your friend knows you could be com‐paring your weight to 123 horses. Realistically, your friend might not compare your weight to 123 horses, but she would probably assume you mean 123 kilograms, since Germany uses the metric system. If you weigh 123 pounds (lbs), this would be a substantial difference. Needless to say, units are important when making measurements. We must always be sure to use them and be very careful when converting.
Converting units can be tricky. When converting, make sure that the original and any intermediate units cancel out and that you are left with the desired unit.
Example 3: the answers to the following questions with the correct number of significant figures.
A. 6.3 x 56.3 = B. 62.3/68 = C. 1.65 x 1.54 x .57 =
Answer A: The answer is 354.69, however this is not reported in the correct number of significant figures. The deciding factor is 6.3 simply because it has the fewest number of significant figures. Therefore, we must round our answer off to 3.5 x 102. This answer has two significant figures just as the deciding factor.
Answers: A. 350 or 3.5 x 102 B. 0.92 C. 1.4
Example 4: You measure the length of your foot in order to buy shoes and find it to be 10 inches long. The shoe company you are ordering from wants to know your foot length in centimeters. Convert 10 inches to centimeters. (Hint: 1 cm = 0.39370 in. or 2.54cm = 1 in.)
OR
Notice how the units cancel to give the answer in cm.
10 inches x = 25 cm 1 cm
0.39370 in
10 inches x = 25 cm 2.54 cm
1.0 in
Always make sure that your units cancel out!
41
Lab 3: Measurements
The following tables contain information about unit conversion and prefixes commonly used in chemistry.
Table 1: Common Unit Conversions
Mass/Weight Length Volume
1 kg = 2.2 lb 1 km = 0.62 mi 3.744 L = 1 gal
907.185 kg = 1 ton * 1 m = 39.37 in 1 L = 1.06 qt
28.3 g = 1oz * 1m = 1.0963 yd 250 mL = 1 cup
453.59 g = 1 lb * 1 cm = 0.39370 in
101,325 Pa = 1 atm 4.184 J = 1 cal ⁰C = ( 5/9 )( ⁰F ‐ 32 )
Pressure Energy Temperature
Table 2: Prefixes with SI Units
Prefix Symbol Decimal Value Scientific Notation
giga G 1,000,000,000 109
mega M 1,000,000 106
kilo k 1,000 103
hecto h 100 102
deca da 10 101
— — 1 100
deci d 0.1 10‐1
centi c 0.01 10‐2
milli m 0.001 10‐3
micro μ 0.000 001 10‐6
nano n 0.000 000 001 10‐9
*Note: though we equate units of mass (kg) and units of weight (lb) here, this relationship is only true on Earth’s surface where mass has weight due to the pull of gravity. In outer space, an object has mass despite being “weightless.”
42
Pre‐lab Questions
1. What is the importance of significant figures in chemistry?
2. State the number of significant figures in each case.
3. Write the answer to the following problems in the correct number of significant figures.
A. 12.01
B. 56
A. 532.02 + 50.2 + 5601 =
B. 56.00 ‐ 5.3 =
C. 6.3 x 56.3 =
D. 58 / 63 =
Lab 3: Measurements
43
4. Convert the following values to the new units shown.
5. Calculate the density of an object that has a mass of 43 g and a volume of 56.0 mL.
A. 153 lb → kilograms
B. 62.0 in → centimeters
Lab 3: Measurements
44
Density is something that you will learn more in detail about in a later lab. However, we will briefly touch the subject here. Density can be expressed as the mass of a substance/object divided by its volume:
Density has the unit grams/milliliter (g/mL). Water is a special standard for density as well as several things in chemistry. In this experiment you will calculate the density of water and compare your value to the accepted value. Therefore, we need to know the accepted value for the density of water. It is 1.00 g/mL at 20oC.
In this kit you have been supplied a small scale to make mass measurements. In cases where more precision is needed, sci‐entists typically use a precision balance, which compares the weight of an object to a known mass.
Density = Mass
Volume
Procedure
Part 1: Density using l inear measurements
1. Use a scale to obtain the initial mass of the cell. Record it in the data table.
2. Measure the dimensions of the refraction cell using the ruler. Record these values in Table 3.
3. Fill the cell completely with water. Obtain the mass of the cell with the water. Record it in the data table. CAUTION: Water spills can cause a slick floor and work area. Do not have electrical plugs in close range due to possible electric shock.
4. Pour out the water and dry the cell. Repeat steps 1 through 3 for accuracy.
Part 2: Density using a graduated cylinder
5. Weigh a 10 mL graduated cylinder. Record the mass in the data table.
6. Fill the graduated cylinder with water to approximately 8 mL water. Measure the amount of water accurately. Remember to measure at the bottom of the meniscus, the circular bottom of the water. HINT: Refer to Figure 2 in the Introduction.
7. Weigh the graduated cylinder and water and record this mass in the data table.
8. Empty the water and dry the cylinder. Repeat steps 5 through 7 with the same cylinder.
9. Clean up and prepare for calculations.
Lab 3: Measurements
Experiment: Density of Water
Materials
Safety Equipment: Safety goggles, gloves 10 mL graduated cylinder
Refraction cell Scale
Metric ruler
Water* *You must provide
45
Data
Measurement Trial 1 Trial 2
Mass of cell (g)
Mass of water and cell (total, g)
Mass of water (total mass minus cell mass)
Cell dimensions
(cm)
Length (L) =
Width (W) =
Height (H) =
Length (L) =
Width (W) =
Height (H) =
Volume (mL) V = L x W x H = V =
Table 3: Procedure Part 1 Data
Trial 1 Trial 2
Exact volume of water (mL)
Graduated cylinder mass
(g)
Mass of graduated cylinder and water (total, g)
Mass of water (g) (total minus cylinder mass)
Table 4: Procedure Part 2 Data
Lab 3: Measurements
46
Calculations
Part 1: Density using l inear measurements
1. Calculate the density of water for both trials. Remember 1 cm3 = 1 mL. Find the average of the two densities. Be sure to use the correct number of significant figures. Show your calculations.
Trial 1
Volume of cell (L x W x H) =
Density = mass of water/volume of cube
Density =
Trial 2
Volume of cell =
Density = mass of water/volume of cube
Density =
Average density =
2. Use the equation below to calculate the percent error. Remember the accepted density for water is 1.00 g/mL. Your average density of water is the experimental density. Show your calculations.
Accepted value ‐ Experimental value Percent Error = x 100%
Accepted value
Lab 3: Measurements
47
Part 2: Density using a graduated cylinder
1. Calculate the density for both trials, and find the average of the two densities. Be sure to use the correct number of significant figures. Show your calculations.
Trial 1
Density = mass / volume =
Trial 2
Density = mass/volume =
Average Density =
2. Determine the percent error once again. The accepted density for water is 1.00 g/mL. Your average density of water is the experimental density.
Post‐lab Questions
1. What is the difference between precision and accuracy?
Lab 3: Measurements
48
2. Were your measurements precise in Part 1 and in Part 2? Explain your answer.
3. Were your measurements accurate in Part 1 and in Part 2? Explain your answer.
Lab 3: Measurements
Introductory Chemistry
Lab 4: Properties of Matter ‐ Density
51
Lab 4: Properties of Matter ‐ Density
Introduction
Have you ever wondered why a 12 ounce diet soda floats in water, but a 12 ounce regular soda sinks to the bottom?
Different substances come with a wide variety of physical and chemical prop‐erties. Physical properties can be observed without changing the chemical identity of the substance. Some physical properties include melting point, boiling point, and specific heat. For water, the melting point is 00 C, the boil‐ing point is 1000 C, and the specific heat is 1.00 cal/g·oC (see Lab 7). Chemical properties, on the other hand, have to do with how a substance’s chemical structure is altered in a chemical reaction. Some examples of chemical proper‐ties are reactivity and oxidation state. You will learn more about these proper‐ties in later labs.
The answer to the introductory question relates to the physical property of density. Both drinks are great to enjoy on a hot summer day. If you hold one in each hand, then it will feel as if they both weigh the same. So how does density come into play? Density is the ratio of mass to volume of a substance at a certain
MassDensity =
Volume
Water has a density of 1.00 g/mL at 4°C—this is because the gram was originally defined as the mass of one cubic centimeter of water (1 cm3 = 1 mL). For comparison, lead has a density of about 11.3 g/cm3, while the Earth’s atmosphere has an average density of 0.0012 g/cm3.
When objects and materials are mixed together, they separate according to density. For instance, an object will float in water if it is less dense than water, which has a density of one gram per milliliter of volume (1.00 g/mL). The regular soda, with 39 g of dissolved sugar, has a density of around 1.11 g/mL, while the diet soda, with 0 g sugar, has a density of nearly 1.00 g/mL. The regular soda is more dense than water on average and sinks to the bottom, while the diet soda’s density is about equal to water’s, allowing it to float.
Figure 2: How does this giant ship stay afloat? The answer has to do with density: though the ship is very large and heavy, it displaces more than its own weight in water, allowing it to easily stay above the surface.
Figure 1: Different ingredients mean different densities. What makes a diet soda chemically different from a regular soda?
Concepts to explore:
List the chemical properties of substances
Determine the densities of items using the density formula
Apply the density concept by making a density gradient column
Figure 1: Is an aluminum can more or less dense than water?
52
The density of a substance will change a small amount when the temperature changes. It will almost always increase when the temperature decreases. Water in the temperature range of 0oC to 4oC is an exception to this. Water’s density is actu‐ally a little higher at 4 oC than it is at 0oC. This explains why ice will float in a glass of water instead of sinking to the bottom.
The physical property of density is often used by chemists as an analytical tool. A chemist may check the density of a part to help verify it is made out of the correct material. For example, if an aluminum washer is set next to washers made out of galvanized or stainless steel of the same size, you might not be able to tell them apart. Also, dif‐ferent types of plastics such as polystyrene and polypro‐pylene can also look alike. Substances and materials are applied to different uses based on their physical proper‐ties. In many cases, substituting one material for an‐other, even if they look and feel similar, can have drastic and even disastrous results.
There are several methods frequently used to determine the density of an object or substance, and this lab you will take two different approaches. In the first part you are asked to determine what material a sample of metal washers are made from. To do this you will measure the mass of the washers and determine the volume by meas‐uring how much water they displace. You can then calcu‐late the density using the provided formula and compare this to a table that gives the density of several metals.
In the second part of this lab you will observe that different plastics have different densities through using a density gradi‐ent column. In a density gradient column the density of liquid in the column is gradually varied from one with a higher density at the bottom of the column to one with a lower density at the top of the column. After you make a density gradi‐ent column you will drop pieces of plastics of known compositions into the gradient and observe how they situate.
Pre‐lab Questions
1. What can cause the density of the same substance to change?
2. List two possible hazards in this laboratory experiment.
Lab 4: Properties of Matter ‐ Density
Figure 3: Water is unique in that it’s solid state (ice) is less dense than its liquid state. This property prevents bodies of water from fully freezing, as top layers of ice insulate warmer waters below. If this weren’t the case, the oceans would gradually freeze from the bottom up.
53
Experiment: Determination of Density
Procedure
Part 1: Density of Common Metals
1. Weigh out 10 washers and record their mass.
2. Determine the volume of the washers by their water displacement in a graduated cylinder. Fill a 100 mL graduated cylinder up to the 50.0 mL mark with distilled water. Tilt the graduated cylinder to provide a slope for the washers to come down. GENTLY drop the washers into the graduated cylinder so that no water is splashed out. Gently tap the graduated cylinder until there are no more air bubbles around the washers. Record the difference between the initial and final volume of water in the graduated cylinder to the 0.01 mL place.
3. Determine the density of the washers using the formula: D = m/V.
4. Compare the density of your washers to the densities of different metals found in Table1: Densities of Some Common Metals. Determine which metal your washers are made out of and record this in the Data section.
Lab 4: Properties of Matter ‐ Density
3. A gold‐colored ring is found to weigh 2.542 g at 20oC. The ring is dropped into a graduated cylinder filled with water, and is found to displace 0.33 mL of liquid. If the density of 14 carat gold is 12.9 to 14.6 g/mL at 20oC, is the ring made of gold? Explain your answer.
Materials
Safety Equipment: Safety goggles, gloves 100 mL Graduated cylinder
10 Metal washers (3/8 inch diameter) 50 mL Beaker
1M Calcium chloride solution Dropper (pipette)
Ethanol Scale
Mixture of plastics: coffee stirs, plastic silverware, Styrofoam cup
Distilled Water* *You must provide
54
Part 2: Density of Plastics
1. Obtain a piece of each of the different plastics.
2. Make a density column in the 50 mL beaker with the following steps:
a. First measure 10 mL of 1 M calcium chloride solution into the gradu‐ated cylinder, and pour into the beaker. This solution has a density of 1.1 g/mL.
b. Measure 20 mL of distilled water into a 100mL graduated cylinder. Slowly add the distilled water into the density column so that the wa‐ter layer rests on top of the calcium chloride layer. Water has a den‐sity of 1.00 g/mL.
c. Measure 10 mL of ethanol into a 100 mL graduated cylinder. Slowly add the ethanol into the density column so that the ethanol rests on top of the water layer. Ethanol has a density of 0.80 g/mL.
3. Gently drop pieces of each piece of the plastic in the plastic mixture one‐by‐one into the beaker.
4. Observe where each object situates in the beaker.
5. In the Data section, record the location of each type of plastic in the gradient.
HINT: Make observations quickly, but watch the objects for a couple minutes. They may make grad‐ual movement up and down the column.
6. Clean‐up the work area and dispose of column down the sink with plenty of water.
Lab 4: Properties of Matter ‐ Density
Type of Material Density at 20⁰C
Aluminum 2.70 g/mL
Copper 8.92 g/mL
Lead 11.34 g/mL
Nickel 8.90 g/mL
Silver 10.50 g/mL
Steel 7.80 g/mL
Zinc 7.14 g/mL
Table 1: Sample densities of common metals
Figure 4: Sample density column for Experiment 1.
1 M CaCl2
Water
Ethanol
55
Data and Calculations
Procedure 1:
Mass of sample washers:
Initial volume of water:
Final volume of water:
Change in volume water:
Density of washers:
Density =
Procedure 2:
Type of Plastic Location on Column (mL)
Table 2: Observations of plastics
Lab 4: Properties of Matter ‐ Density
56
Post‐lab Questions
1. What material did you determine your washers are made from? Why did you decide your sample washers are made out of this metal?
2. Why is the calcium chloride solution poured into the density column first?
3. Which type of plastic is most dense? Which type is the least dense? Explain.
4. How could you use a density gradient column to help identify whether a toy car is made with more metal or more plastic?
Lab 4: Properties of Matter ‐ Density
Introductory Chemistry
Lab 5: Mixtures and Solutions
59
Introduction
How does cereal containing a mixture of items re‐late to relief from a headache?
It is a lazy Saturday morning and you’ve just awoken to your favorite cereal “Morning Trails” and milk. As you pour your bowl of cereal you remember the part you don’t like: the almonds. “Morning Trails’ contains all of your favorite things like granola, raisins, oats, and crispy flakes; however, in your distaste for al‐monds, you pick them out. How does this breakfast relate to mixtures? You use different characteristics within the cereal to pick out what you don’t like: things like appearance, taste, smell, and texture. In the same way, we can use different char‐acteristics to separate mixtures in chemistry.
Lab 5: Mixtures and Solutions
Often a mixture contains two or more substances that can be separated from each other by their physical properties. There are two categories into which mixtures can be divided: heterogeneous and homogeneous. A heterogeneous mixture is a mixture that contains visibly different parts, such as a breakfast cereal in milk. A homogeneous mixture is one that does not contain visibly different parts, much like coffee after the creamer and sugar have been stirred in.
Whether heterogeneous or homogeneous, chemists often need to separate the components of a mixture. Many techniques have been developed for separating a mixture. These techniques are based on the different characteristics of the compounds or elements in the mix‐ture.
One technique used for the separation of a heterogeneous mixture is filtration. This is used when a heterogeneous mixture contains both a solid and a liquid component. In filtration, a mixture is poured through a piece of filter paper seated in a funnel. The solid components of the mixture are trapped on the paper while the liquid components drain into a receiving container.
Other common techniques used for separating mixtures by physical changes include distillation, chromatography, and crystallization. Dis‐tillation takes advantage of the characteristic of compounds that have different boiling points. When a mixture of two liquids with different boiling points is gradually heated, the liquid with the lower boiling
Figure 1: You encounter different types of mixture in your everyday life. What properties make this breakfast cereal a heterogeneous mixture and the cup of coffee a homogene‐ous mixture?
Figure 2: A common example of crystallization is the formation of ice crystals and snowflakes. How is this crystallization different from the formation of aspi‐rin crystals in this lab?
Concepts to explore:
Use physical properties to separate the components of a mix‐ture
Use laboratory techniques to separate the components of a mixture
60
Pre‐lab Questions
1. Name two homogeneous mixtures and two heterogeneous mixtures that you come across daily.
2. Explain how you would separate a solid mixture of sand and salt using any of the methods mentioned above.
Lab 5: Mixtures and Solutions
point will transition into a gas first. This gas can be collected in a separate container to cool, where it changes back into a liquid—separating the original mixture. Certain methods of distillation are so precise that they can separate liquids in a mixture that vary only by a few degrees Celsius.
Chromatography is a method that is used to separate a mixture whose components move along a strip of paper or up a column at different rates due to their different properties.
Crystallization is a technique that is often used to produce solids with very high purity. One way to do this is to make a solu‐tion with the substance, then allow the solvent to cool and evaporate so that pure crystals can form from the solute. In this lab we will separate the components of an aspirin pill using crystallization. Coated aspirin tablets are actually a mix‐ture—they consist pain‐relieving medicine along with binders and fillers—substances that hold the pill together to help it reach the appropriate part of your body before becoming active. The ingredients that make up binders and fillers may in‐clude cellulose, dyes, starch, as well as many other ingredients to accompany the active ingredient on its way through your body.
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Experiment: Separation of a Mixture
A large amount of aspirin will dissolve in hot water, but very little (or none at all) will dissolve in cold water. The other in‐gredients in a safety‐coated aspirin pill are soluble in both hot and cold water. We can use this solubility difference to separate the aspirin in a pill from the other ingredients. You will allow a hot solution containing a crushed pill to cool in an ice bath, where the aspirin crystals should begin to form. If the cooled solution is filtered through a new piece of filter pa‐per, almost all of the solid aspirin will be caught by the filter and the impurities that are soluble in cold water will remain in the water.
Procedure
1. Prepare a boiling water bath. HINT: This may be done over the stove. CAUTION: Always use a heat pad or gloves when handling hot materials.
2. Pre‐weigh a sheet of filter paper and a watch glass. Record each mass in the Data section.
3. Prepare an ice water bath by filling a large bowl with a mixture of ice and tap water. Fill the wash bottle with distilled water and place it in the bath.
4. Mark three test tubes as A and B and C. Place Test tubes A and B into the test tube rack.
Lab 5: Mixtures and Solutions
Materials
Safety Equipment: Safety goggles, gloves Test tube rack
Aspirin pills (safety‐coated, low dose) 3 Test tubes
Test tube holder Funnel
Filter paper Mortar and pestle
Plastic bag Stir rod
10 mL graduated cylinder Wash bottle
Scale Watch glass
Distilled water* Boiling water*
Bowl* Ice*
*You must provide
62
5. Prepare a filtering funnel as shown in Figure 3: fold a piece of filter paper in half twice to make quarters, and place the paper in the funnel so that three quarters are open on one side and one quarter is on the opposite side. Seat the filter paper into the funnel by moistening the paper with a small amount of water.
6. Insert the funnel with filter paper into test tube A supported in the rack. Discard any water that filters into this test tube.
7. Place four safety‐coated, low‐dose (81 mg) aspirin pills into the mortar. Gently crush the pills with the pestle into a fine powder.
8. Pour all of the ground aspirin powder into test tube C. HINT: Pour the powder onto a piece of paper, then shape the paper to more easily funnel the powder into the tube.
9. Add 10 mL water to the powder in the test tube and stir with a stirring rod. CAUTION: The test tube bottom is easily broken if too much force is applied with a stirring rod.
10. Use a test tube holder to place test tube C (with the aspirin powder) into the boiling water bath. Keep the test tube in the bath for 5 minutes. Stir the solution in the test tube frequently while it is heated. Record your observations in Table 1. Leave the test tube in the bath until the next step. CAUTION: This solution will be extremely hot. Always use a test tube holder to handle the heated test tube. Do not touch it directly with your hands.
Lab 5: Mixtures and Solutions
Figure 3: Filtering funnel preparation diagram
a b c
d e f
63
Data and Calculations
1. Pre‐weighed filter paper mass:
2. Pre‐weighed watch glass mass:
3. Filter paper, watch glass, and aspirin mass:
4. Mass of aspirin obtained:
5. Calculate the percent of recovery using the formula below:
Actual amount (g aspirin obtained)100%
Theoretical yield (0.081 g/pill # pills used)
× %
Lab 5: Mixtures and Solutions
11. Use a test tube holder to remove test tube C from the boiling water. Pour half the hot contents in the filter‐ing funnel above test tube A, then immediately place test tube B back into the boiling water. Allow all the liquid to completely filter into the test tube.
12. Pour the rest of the hot contents into the funnel, and allow the solution to completely filter into test tube A. HINT: If filtering the first half of the solution went very slowly, it will help if you replace the filter paper for this new solution.
13. Allow the hot, filtered solution in test tube A to cool until it is no longer warm. Remove the filter paper and clean the funnel for later use.
14. Place test tube A with the cooled, filtered liquid into the ice water bath. Continue to cool the solution until generous amounts of crystals form. This should take about 5 minutes. If crystals do not form, try scratching the sides of the test tube with a glass stirring rod until the crystals begin to form. Record your observations in Table 1.
15. Prepare another folded filter paper (pre‐weighed) inside the cleaned funnel as you did in Step 5. Insert the cleaned filtering funnel into test tube C in the rack. Discard any water that filters into this test tube.
16. Transfer the crystals from cooled test tube A into the filtering funnel above test tube C. Use small amounts of ice‐cold distilled water from the wash bottle to rinse the crystals. DO NOT USE MORE THAN A SMALL AMOUNT OF ICE COLD WATER TO TRANSFER AND RINSE THE CRYSTALS OR TOO MANY WILL DISSOLVE.
64
Procedure Number Observations
10. Crushed aspirin and water mixture in hot water bath
14. Aspirin mixture in ice water bath
19. Aspirin crystals
Table 1: Observations for aspirin experiment
Post‐lab Questions
1. What important characteristics of aspirin did you use to separate it from the other pill components?
Lab 5: Mixtures and Solutions
65
Lab 5: Mixtures and Solutions
2. What is the purpose of rinsing the final aspirin crystals with a small amount of ice‐cold distilled water (Procedure 15)?
3. Evaluate your percentage recovery. Provide an explanation if your recovery was more than 100% or very low. How could you improve your results?
Introductory Chemistry
Lab 6: Chemical and Physical Change
69
Lab 6: Chemical and Physical Change
Introduction
What do chemistry and superheroes have in com‐mon?
Movie producers often make an action movie where an every‐day person will dramatically turn into a superhero. In the mov‐ies when a crisis hits, a normal guy or girl will disappear in a flash. Suddenly he or she will radically reappear disguised as a superhero who then destroys the horrifying villain and saves the day! In a similar way, molecules can transform their identity. Under different conditions, molecules in a substance can reor‐ganize into something that appears physically different or acts very different chemically. Sometimes these changes occur in appearance, while other times, these changes create a different substance altogether.
Within the science of chemistry there are many types of changes that can take place. These changes can be categorized into two primary types, physical changes and chemical changes.
A physical change alters the state of a substance without altering the identity of the substance. This includes dissolving, crushing, tearing, heating, melting, evaporating, sublimation or cooling. Sublimation is a type of physical change that oc‐curs when a solid goes directly into a vapor state without going through a liquid state. Only a small number of substances sublime; an example is carbon dioxide. You can observe this by setting out a piece of frozen carbon dioxide (dry ice). The carbon dioxide does not melt, but moves directly to a gas.
A chemical change alters the identity of the molecular identity of the substance. One example is rusting iron. The iron metal changes to iron oxide when oxygen is added to its chemical formula.
This experiment involves changing the pain medicine aspirin both physically and chemically. The aspirin is altered physi‐cally through the process of dissolving it in hot water. It is also changed chemically by adding sodium bicarbonate (NaHCO3). When sodium bicarbonate is added to aspirin, it changes into a new substance, and the new chemical has a dif‐ferent kind of bond known as an ionic bond. This makes it much more water soluble. You will learn more about this in a later laboratory experiment. We will call the new substance Aspirin*Na+. The Aspirin*Na+ is very soluble in water and will dissolve in a lot less water at room temperature. You will then change the Aspirin*Na+ back to the original aspirin by add‐ing citric acid to it. After it is changed back, it is again not very soluble. Many medicines undergo a similar type of reaction to allow the human body to absorb and use them. Usually the medicine is made more water soluble by adding chemicals such as hydrogen chloride (HCl) or hydrogen bromide (HBr).*
*Note: dissolved in water, these become hydrochloric acid and hydrobromic acid, respectively.
Figure 1: Unlike most substances you are probably familiar with, solid carbon dioxide (dry ice) does not melt at typical atmospheric pressures. Instead, dry ice sublimates directly
Concepts to explore:
Recognize the differences between chemical and physical changes
Determine if a change is physical or chemical
70
Pre‐lab Questions
1. Define physical change in your own words, and list two examples of types of physical changes.
2. Define chemical change in your own words, and list two examples of chemical changes.
3. What is sublimation, and what makes it either a physical or chemical change?
Lab 6: Chemical and Physical Change
71
Experiment: Chemical and Physical Changes of Caffeine
After the aspirin crystals obtained from the “Separation of Mixtures” lab have been weighed and recorded in the previous Lab, divide them in half. You will now use half of the mixture to demonstrate the physical change of dissolving and the other half to demonstrate a chemical change.
Procedure
1. Obtain the aspirin crystals from the Mixtures and Solutions Lab (about 0.2 to 0.3 g), and divide them in half. You will now use half of the mixture to show the physical change of dissolving and the other half to show a chemical change.
2. Place half of the aspirin (about 0.1 g) into a test tube labeled P for physical change and the other half into a test tube labeled C for chemical change.
3. Add 1.5 mL of distilled water to the test tube labeled P and swirl to mix. Record your observations in Table 1.
4. Very slowly add 1.5 mL of saturated sodium bicarbonate solution (NaHCO3) to the test tube labeled C and swirl to mix. If the crystals are not almost completely dissolved, add additional saturated sodium bicarbonate solution with a pipette until almost all of the crystals have dissolved. You have now witnessed a chemical change to the aspirin*Na+. Record your observations in the data table.
5. Compare the mixtures in the two test tubes. Record your observations in Table 1.
6. Add 1.5 mL saturated citric acid solution to the test tube labeled C and swirl to mix. If crystals do not re‐form, add additional saturated citric acid solution with a pipette until almost all of the crystals have dis‐solved. You have now witnessed a chemical change back to aspirin. Record your observations in the data table.
7. Add an additional 1.5 mL of distilled water to the test tube labeled P and swirl to mix. Record your observa‐tions in Table 1.
Lab 6: Chemical and Physical Change
Materials
Safety Equipment: Safety goggles, gloves 2 Test tubes
0.2—0.3 g Aspirin crystals (from Mixtures and Solutions Lab) Stir rod
10 mL graduated cylinder Test tube holder
2mL Saturated citric acid solution 250 mL beaker
2 mL Saturated bicarbonate solution (NaHCO3) Distilled water*
Pipette Boiling water bath*
*You must provide
72
Data and Observations
Table 1: Physical and chemical changes observations
Lab 6: Chemical and Physical Change
8. Place the test tube labeled P into the hot water bath and stir.
9. Allow the aspirin to dissolve until no more particles can be seen in the test tube. You have now witnessed a physical change to the aspirin. Record your observations in the data table.
10. Cautiously remove the test tube from the hot water using the test tube holder and place it in the test tube rack to cool. CAUTION: The test tube is extremely hot and can cause severe injury.
Procedure Observations
Adding distilled water to the aspirin crystals
Adding NaHCO3 to the aspirin crystals
Comparison of the two mixtures
Adding citric acid to the aspirin*Na+ solution
Adding 1.5 mL more wa‐ter to the aspirin crystals
Dissolving the aspirin
Cooling the aspirin
solution
73
Post‐lab Questions
1. The following are common changes that often take place. Define each as a chemical or physical change.
a. Igniting a Bunsen burner:
b. Boiling water:
c. Making popsicles:
d. Melting ice cream:
e. Rusting iron:
f. Subliming moth balls:
2. Does the chemical structure of aspirin change when it is dissolved in hot water?
3. What are the crystals that formed when the mixture cooled in Part 1?
Lab 6: Chemical and Physical Change
74
4. Why is adding NaHCO3 to the aspirin crystals a chemical change?
5. Why does adding a citric acid solution cause the aspirin crystals to reform?
Lab 6: Chemical and Physical Change
Introductory Chemistry
Lab 7: Heat and Calorimetry
77
Lab 7: Heat and Calorimetry
Introduction
Have you ever roasted marshmallows over a campfire?
You have probably heard that foods have calories, but do you know what a calo‐rie is? A calorie is defined as the amount of energy it takes to raise the tempera‐ture of one gram of water by one degree Celsius. Heat is a transfer of energy, and measuring the amount of heat given off by burning food is one way to determine how much energy (calories) that food contains.
Specific heat is a physical property of a substance. It is the amount of en‐ergy needed to raise the temperature of 1.00 g of a substance by 1oC. The specific heats of many common substances are known; for example, glass has a specific heat of 0.22 cal/g·oC and copper has a value of 0.092 cal/g·oC. From the definition of the calorie, you already know the specific heat of wa‐ter, which is defined as 1.00 cal/g·oC. As you can see in Table 1, water has a very high specific heat compared to most common substances. This means that it takes more energy to increase the temperature of water than an equal mass of copper or glass. The high specific heat of water makes it use‐ful as a coolant in everything from engine radiators to simple cold com‐presses.
In this lab you will determine the caloric content of different “junk” foods by measuring the amount of heat a sample gives off when burned. The follow‐ing equation relates the amount of heat energy added to a substance to the change in temperature of that substance:
In this equation E is energy measured in calories, m is the mass of the heated substance in grams, ΔT is its change in temperature (⁰C or K), and c is its specific heat in cal/g·oC.
Since it is difficult to measure the change in temperature of a food directly, we can use a calorimeter to obtain the heat indirectly. A simple form of calorimeter is a metal container filled with water and suspended above a combusting material. The energy given off by the burning material transfers as heat to the water sample, increasing its temperature by ΔT. Knowing the mass of the water sample and it’s specific heat (1.00 cal/g·oC), we can calculate the energy contained by the original food sample using the above equation. The assumption is that energy is always conserved, so that all the heat given off by the food sample is gained by the water. In other words:
where Q is the quantity of heat energy lost by substance A and gained by substance B.
E = mc T
Figure 1: You have probably looked over the nutrition facts label on many of your favorite foods—but what does the number of Calories actually mean, and how does it relate to chemistry?
lost, A gained, BQ = Q
Concepts to explore:
Gain applicable knowledge about calories Compare the calorie content of food samples
78
An important thing to remember is that food calories are often expressed as “Calories” with a capital C. One Calorie is actu‐ally equivalent to one kilocalorie, or 1000 calories with a lowercase c. This means that the number calories found using the formula above must be divided by 1000 to convert to the food Calories listed in the nutrition facts of your favorite food. The conversion can also be written out:
You now have all the information needed to determine the caloric content of a substance. For example, if 0.500g of a mini marshmallow is completely burned underneath a can containing 100.0 mL of water and the water temperature changes from 24oC to 28.5oC, the number of Calories per gram contained in the marshmallow can be calculated by the following steps:
Example
1. Use the density of water to determine the mass of water heated:
Mass = Density x Volume
m = (1.00 g/mL)*(100.0 mL) = 100 g
2. Use the constant for the specific heat of water:
c = 1.00 cal/g˚C
3. Determine the change in temperature of the water:
ΔT = (28.5˚C – 24.0˚C) = 4.5˚C
4. Calculate the energy gained by the water, which equals the energy contai‐ned (lost) by the marshmallow:
E = mc∆T
E = (100.0 g)*(1.00 cal/g˚C)*(4.5˚C) = 450 cal
5. Calculate the number of calories per gram of marshmallow:
6. Convert to Calories (kilocalories) per gram of marshmallow:
450 calcalories/g marshmallow = = 900 cal/g
0.500 g
900 cal 1 Calcalories/g marshmallow = x = 0.90 Calories/g
g 1000 cal
Lab 7: Heat and Calorimetry
Substance In cal/g∙K
Aluminum 0.215
Copper 0.0923
Gold 0.0301
Lead 0.0305
Ethyl Alcohol 0.580
Water 1.000
Iron/Steel 0.110
Glass 0.200
Wood 0.400
Table 1: Specific heats of some common substances
1 Cal = 1 kcal = 1000 cal
79
Pre‐lab Questions
1. In terms of food, what is a calorie?
2. A piece of chocolate chunk cookie was tested the same way as is used in this procedure. The cookie sample had a mass of 0.851 g and the temperature of a 100.0 mL of water increased by 7.87 ⁰C. How many Calories per gram did the cookie contain? Show all calculations.
Lab 7: Heat and Calorimetry
80
Procedure
1. Measure between 75 mL of distilled water into a 100 mL graduated cylinder. Record the exact amount of wa‐ter in Table 2.
2. Carefully pour the measured water into an empty aluminum can that still has a tab attached.
3. Set the aluminum can on the tripod as shown in Figure 2.
4. Insert a thermometer into a split rubber stopper. Place the thermometer into the can as shown in Figure 2, with the stopper resting on the top of the can. This allows you to adjust the height of the thermometer by slid‐ing the stopper up or down. The thermometer should touch the water but not the bottom of the can.
5. Gently straighten the outside fold of another large paper clip. This will be used as a holder for the food sam‐ple.
6. Insert the end of the paper clip that is still folded into a hunk of clay. The hunk of clay should be approxi‐mately the size of a ping pong ball.
7. Tear a junk food sample into a piece that is about 1 cm2. Determine the mass of the piece of junk food and record it in the data table. Remember NO eating in the lab!
Lab 7: Heat and Calorimetry
Experiment: The Calorimetry of Junk Food
In this experiment, you will measure the caloric content of “junk” food using a calorimeter made from an aluminum can. The experimental number of calories can then be compared to the actual number of calories reported by the manufacturer on the package. Make sure to treat the samples the same in order to get a more accurate comparison.
CAUTION: This experiment can produce excessive smoke. If possible, perform the following procedure under a stove’s ventilation hood or near a window that can be opened.
Do not use foods that contain ingredients to which you have a known allergy, such as peanut products.
Materials
Safety Equipment: Safety goggles, gloves Split rubber stopper
2 Large metal paper clips Tripod
50 mL beaker Scale
100 mL graduated cylinder Distilled water*
Butane lighter Cheetos® or similar snack food*
Thermometer Mini marshmallow*
Clay Aluminum can with tab (12oz)*
*You must provide
81
Figure 1: Apparatus for calorimetry experiment
Lab 7: Heat and Calorimetry
8. Insert the straightened end of the food holder into the sample. HINT: If this does not hold the sample you can make a loop at the end of the paper clip to rest the sample in.
9. Place a 50 mL beaker approximately half full of water near the apparatus to extinguish smoke after the sample has finished burning.
10. Record the initial temperature of the water inside the aluminum can in Table 2.
11. Light the butane lighter. CAUTION: Burns can occur with the use of flames.
12. Hold the clay end of the sample holder and carefully bring the sample into the flame until it ignites. HINT: The sample should be held in the flame for a few seconds to assure the sample will burn strongly.
13. Immediately and carefully bring the burning food approximately 1 cm below the bottom of the aluminum can in order to minimize the amount of heat lost. CAUTION: Excessive smoke can result from the ignited sample and can be a respiratory irritant. If there is excessive smoke the sample should be relit immediately.
14. Watch the thermometer as the food sample completely burns to ash. If the food sample goes out before it is completely burned or is producing only a little flame and excessive smoke, quickly relight it in the lighter flame and place it back under the aluminum can. Record the maximum temperature that is reached.
15. Immediately after the sample has completely burned, dip it into the beaker of water and wait for it to cool.
16. Place the remains of the sample in the trash. Wash end of the paper clip, and then dry it with a paper towel.
17. Repeat the procedure from steps 7‐16 using other foods from your pantry. (Hint: Use a sample size that has a similar mass to the previous sample.)
Thermometer
Aluminum can with tab
Burning food sample
Food holder (clay and paperclip) Tripod
82
Data
Junk Food Volume of
Water (mL)
Mass of Junk Food (g)
Initial water
Temperature (oC)
Maximum Temperature
(oC)
Water ΔT (oC)
Sample A:
Cheetos
Sample B:
Marshmallow
Mass of
water (g)
Sample C:
Sample D:
Table 2: Data for Calorimetry Experiment
Calculations
1. Assuming that all the heat from each food went into heating the water in the aluminum can, calculate how many Calories per gram each sample contains. HINT: Food calories are measured in kilocalories!
A. Food Sample A
Lab 7: Heat and Calorimetry
83
Lab 7: Heat and Calorimetry
B. Food Sample B
C. Food Sample C
D. Food Sample D
84
Lab 7: Heat and Calorimetry
2. Using the nutrition facts and serving size reported by the manufacturer, calculate how many Calories per gram each food contains as reported by the company.
A. Food Sample A
B. Food Sample B
C. Food Sample C
85
Post‐lab Questions
1. Which food sample had the most calories per gram? Was this what you expected? Why or why not?
2. Were your measured Calories/g values the same or close to the ones you calculated from the information pro‐vided on the manufacturers’ labels?
Lab 7: Heat and Calorimetry
D. Food Sample D
86
Lab 7: Heat and Calorimetry
3. What are the primary sources of error in this experiment that could explain any differences between your data and the label information? (HINT: How do these sources of error relate to our assumption that Qlost = Qgained ?)
4. Why is it important for the food to burn completely?
5. How might you alter the experiment to allow for a more accurate transfer of heat between the burning sam‐ple and the calorimeter water?
Introductory Chemistry
Lab 8: Chemical Processes
89
Lab 8: Chemical Processes
Introduction
Have you ever needed to place a cold pack on a sprained muscle?
It’s the final seconds of the community league champion‐ship basketball game, and your team is behind by one point. One of your team’s players takes a shot and scores. The game is over, and your team won! But something is wrong: the player is sitting on the floor, and appears to be in a lot of pain. The coach quickly brings a cold pack to the player, squeezes it, and places it on the swelling ankle. The bag immediately becomes cold—but how?
Though we often use them interchangeably, heat and tempera‐ture have different definitions—though they are closely related in the study of thermodynamics. Heat is the transfer of energy from one object to another due to a difference in tempera‐ture. Temperature, on the other hand, describes how much energy the atoms and molecules in a substance have. This en‐ergy, often called internal energy, describes how quickly the atoms or molecules in a substance move or vibrate around. When an object gains heat its molecules vibrate with more energy, which we can sense or measure as an increase in tem‐perature. When you touch a hot object, it feels hot because a heat moves from the hot object (higher energy) to your skin (lower energy). Similarly, an object feels cold when heat is lost by your hand and gained by the cold object. Heat always transfers in the direction of high temperature to low temperature—high energy to low energy.
Both physical processes and chemical reactions can release or absorb energy in the form of heat. When a reaction or physical change gives off energy it is called an exothermic process. To remember exothermic, think of ‘exiting’ as in leaving or going out. An endothermic process does just the opposite—it takes in energy from its surroundings. The generalized chemical equations for exo‐thermic and endothermic reactions are:
The direction energy moves determines whether the process is considered endothermic or exothermic, and tells you how the tem‐perature of a system changes. In an endothermic reaction or physical change, energy is absorbed and the overall temperature of the system decreases. Some examples of endothermic processes include the melting of water in a soft drink or the evaporation of a liquid. Similarly, an endothermic reaction takes in energy for chemical changes to occur. One example is what occurs in an instant cold pack like the ones used to decrease the swelling caused from a sports injury. These types of cold packs utilize the chemical
exothermic:
endothermic:
reactants → products + energy
reactants + energy → products
Figure 1: The combustion of fuel, such as wood or coal, is a com‐mon example of an exothermic reaction. Under the right condi‐tions (usually the application of enough heat), a chemical reac‐tion occurs between wood and the oxygen in air. Fire is the re‐
Concepts to explore:
Understand the difference between endothermic and exothermic processes
Understand the concept of enthalpy
90
Pre‐lab Questions
1. Define enthalpy:
2. What is the relationship between the enthalpy of a reaction and its classification as endothermic or exother‐mic?
3. With instant hot compresses, calcium chloride dissolves in water and the temperature of the mixture in‐creases. Is this an endothermic or exothermic process?
Lab 8: Chemical Processes
process of ammonium nitrate (NH4NO3 ) dissolving in water. The ammonium nitrate needs to absorb heat from the surrounding water to dissolve, so the overall temperature of the mixture decreases as the reaction occurs.
In contrast, energy is released in an exothermic process. An example of an exothermic reaction is what occurs in common hand warmers. The increase in temperature is the result of the chemical reaction of rusting iron:
4 Fe(s) + 3 O2(g) 2 Fe2O3(s) + energy
Iron usually rusts fairly slowly so that any heat transfer is not easily noticed. In the case of hand warmers, common table salt is added to iron filings as a catalyst to speed up the rate of the reaction. Hand warmers also have a permeable plastic bag that regulates the flow of air into the bag, which allows just the right amount of oxygen in so that the desired tempera‐ture is maintained for a long period of time. Other ingredients that are found in hand warmers include a cellulose filler, carbon to disperse the heat, and vermiculite to insulate and retain the heat.
Enthalpy is a quantity of energy contained in a chemical process. In the cases we will be dealing with, the energy released or ab‐sorbed in a reaction is in the form of heat. Enthalpy by itself does not have an absolute quantity, but changes in enthalpy can be observed and recorded. For example, if you stick your finger into a glass of cold tap water, it probably feels pretty cold. However, after being outside on a freezing winter day for a long period of time, the same glass of water might actually feel warm to touch. It would be difficult to measure the absolute quantity of energy in the water in either case, but it is relatively easy to notice the move‐ment of energy from one object to another. In exothermic reactions, heat energy is released and the change in enthalpy is negative, while in endothermic reactions, energy is absorbed and the change in enthalpy is positive.
Note: the energy term on the right side shows that the reaction is exothermic, but is not required.
91
Lab 8: Chemical Processes
Experiment: Cold Packs vs. Hand Warmers
In this lab you will observe the temperature changes for cold packs and hand warmers. Since temperature is defined as the average kinetic energy of the molecules, changes in temperature indicate changes in energy. You will use simply a Styro‐foam cup as a calorimeter to capture the energy. The customary lid will not be placed on the cup since ample oxygen from the air is needed for the hand warmer ingredients to react within a reasonable amount of time.
Procedure
Part 1: Cold Pack
1. Measure 10 mL of distilled water into a 10 mL graduated cylinder.
2. Place about 1/4 (or approximately 10.0 g) of the ammonium nitrate crystals found in the solid inner contents of a cold pack into a Styrofoam cup. The Styrofoam cup is used as a simple calorimeter.
3. Place a thermometer and a stirring rod into the calorimeter (Styrofoam cup). CAUTION: Hold or secure the calorimeter AND the thermometer to prevent breakage.
4. Pour the 10 mL of water into the calorimeter containing the ammonium nitrate, (NH4NO3) taken from the cold pack.
5. Immediately record the temperature and the time.
6. Quickly begin stirring the contents in the calorimeter.
7. Continue stirring and record the temperature at thirty second intervals in Table 1. You will need to stir the reaction the entire time you are recording data.
8. Collect data for at least five minutes and until after the temperature reaches its minimum and then begins to rise. This should take approximately 5 to 7 minutes.
9. Record the overall minimum temperature in the appropriate place on the data table.
Part 2: Hand Warmer
1. Wash and dry the thermometer. HINT: Remember to rinse it with distilled water before drying.
Materials
Safety Equipment: Safety goggles, gloves Scale
Entire contents of a hand warmer Stir rod
1/4 contents of a cold pack Spatula
Calorimeters (2 Styrofoam cups) Stopwatch
Thermometer (digital) Distilled water*
10mL Graduated cylinder *You must provide
92
Lab 8: Chemical Processes
2. Carefully place and hold the thermometer in another Styrofoam cup.
3. Cut open the inner package of a hand warmer and quickly transfer all of its contents into the calorimeter. Immedi‐ately record the initial temperature of the contents and being timing the reaction. HINT: Data collection should start quickly after the package is opened because the reaction will be activated as soon as it is exposed to air.
4. Quickly insert the stirring rod into the cup and begin stirring the contents in the calorimeter.
5. Continue stirring and record the temperature at thirty second intervals in Table 2. You will need to stir the reac‐tion the entire time you are recording data.
6. Let the reaction continue for at least five minutes and until the temperature has reached its maximum and then fallen a few degrees. This should take approximately 5 to 7 minutes.
7. Record the overall maximum temperature in the appropriate place in the data table.
Data
Time (sec) Temp. (0C) Time (sec) Temp. in (0C)
Initial 240
30 270
60 * 300
90 330
120 360
150 390
180 420
210 450
Minimum Temperature (0C) : __________
Table 1: Cold pack data
93
Lab 8: Chemical Processes
Time (sec) Temp. (0C) Time (sec) Temp. in (0C)
Initial 240
30 270
60 * 300
90 330
120 360
150 390
180 420
210 450
Maximum Temperature (0C) : __________
Table 2: Hand warmer data
94
Graph
Graph the data from Tables 1 and 2 as two sepa‐rate lines on the same chart. Be sure to title your graph and label each axis. Include the units in the axis labels. An example is shown here.
Cold Packs vs. Hand Warmers
-10
0
10
20
30
40
50
60
70
0 30 60 90 120 150 180 210 240 270 300 330
Time in seconds
Tem
pera
ture
in o
C
Cold Packs Hand Warmers
Use the following space to graph your data:
Lab 8: Chemical Processes
95
Calculations
Calculate the overall temperature change for the cold and hot pack substance. HINT: This is the difference in the maximum temperature and minimum temperature of each.
a. Cold pack ΔT:
b. Hand warmer ΔT:
Post‐lab Questions
1. Which pack works by an exothermic process? Use experimental data to support your answer.
2. Which pack works by an endothermic process? Use experimental data to support your answer.
Lab 8: Chemical Processes
96
3. Which pack had the greatest change in enthalpy? How do you know?
Lab 8: Chemical Processes
Introductory Chemistry
Lab 9: Electron Configuration
99
Lab 9: Electron Configuration
Introduction
Have you ever wondered how fireworks burst into different col‐ors?
Atoms radiate light when a change in energy state of the electrons occurs. Various salt compounds can be added to produce a specific color. Fireworks and their awe‐some colors and brilliance are based around this concept. The main ingredients of fireworks are potassium chlorate or perchlorate, charcoal, and sulfur. With each different ingredient or compound, a different color can be produced to give a beautiful multi‐colored light show.
The energy of atoms is quantized, meaning there are specific energy levels. This is similar to climbing up a ladder: you can only step on the rungs, not in between. In a similar fashion, the changes in energy of electrons must be whole energy level changes, not between two levels.
Picture your foot on the rung of a ladder closest to the ground. This is like an elec‐tron in the first principle energy level (n = 1). Scientists invented a naming system for describing the location of a specific electron in an atom. Each principle energy level has a sublevel or sublevels (s, p, d, or f). The system for naming is called the electron configuration. For example, the electron configuration for helium is 1s2 because it is in the first principal energy level (1), the first sublevel (s), and has two electrons.
The lowest energy configuration for an atom is called its ground state; as energy is added from the ground state, electrons tend to fill energy levels and sublevels in a specific order (see Figure 2). The first sublevel in any orbital is the “s” sub‐level, which can hold up to two electrons. The next sublevel filled is the “p” sub‐level, which can hold up to six electrons. Continuing, the “d” sublevel can hold up to 10 electrons and “f” sublevel can hold up to 14 electrons. Electrons will always occupy the lowest available energy level first, meaning that an electron will not exist in a higher energy level or sublevel when a lower one is vacant.
He 1s2
Principle energy level Sublevel
Number of electrons Color Shades Wavelength Range (nm)
Red 650 – 750
Orange 595 – 650
Yellow 580 – 595
Yellow ‐ Green 560 – 580
Green 500 ‐560
Green ‐ Blue 490 – 500
Blue ‐ Green 480 – 490
Blue 435 – 480
Violet 400 – 435
Table 1: Wavelength ranges for the
visible spectrum
Figure 1: The flame of a burning lithium salt demonstrates the bright colors that can be achieved by burning different compounds.
Concepts to explore:
Observe energy emitted from different energy levels when salt compounds are ignited
Obtain a general knowledge of what produces different colors in fireworks
100
When an atom receives the right amount of energy from an outside source, an electron will absorb that energy and move to a higher energy level corresponding to that energy. This electron will not usually stay at the higher energy for long; instead it will emit or re‐lease that extra energy in the form of light, and in the process re‐turn to its original energy level. This outside energy source might be in the form of heat or electromagnetic waves (light). As it turns out, the wavelength of light emitted depends directly on the quantity of energy absorbed and released. Higher energies correspond to smaller wavelengths, and lower energies correspond to longer wavelengths of light. Table 1 summarizes several color shades and their wavelength ranges.
Electrons of different elements absorb and emit different amounts of energy. This means that by observing the color (wavelength) of light emitted by a substance when its atoms absorb and release en‐ergy, scientists can determine its chemical composition. This phe‐nomena helps explain how fireworks produce different colors of light. When the atoms of different materials in a firework are ex‐cited by heat, they will absorb and release energy in particular wavelengths, producing a particular variety of colors. n = 1
n = 2
n = 5
n = 3
n = 4
1s
2s
2p
3s
3p
4s
3d
4p
5s
4d
5p
4f
Incr
easi
ng E
nerg
y overlap
{
overlap
{
overlap
{
Principal energy level Sublevel
Pre‐lab Questions
1. What is an electron configuration?
2. How is the light emitted by an atom related to its electron configuration?
Figure 2: Energy level diagram for electron configuration. Energy levels are filled starting from the bottom and moving upwards. Note the positions that overlap.
Lab 9: Electron Configuration
101
Experiment: Chemistry of Fireworks
In this experiment the flame from a tea light candle is the outside energy source. The flame emits a broad range of energy, but the electrons of the atom being heated will only absorb specific amounts of energy.
Procedure
1. Straighten out five large paper clips and make a small (approximately 2‐5 mm in diameter) loop in the end of each one by gently bending it.
2. Place a round piece of clay on the straight end of the paperclip as you did in the Heat and calorimetry Lab. This will act as a holder.
3. Place in order the LiCl, NaCl, KCl, and CaCl2 saturated solutions along with the activated charcoal. Set one of the paper clip wire wands next to each sample.
4. Light the candle using the butane lighter. CAUTION: Both lighter and candle can cause fire or burns to skin and/or clothing if the flame comes into contact with skin or clothes. Be sure you have your safety goggles on!
5. Hold the paper clip wire wand for the LiCl at the very end of the non‐looped end in order to avoid burns. Heat the looped end in the candle flame until its loop is faintly orange and any coating is burned off.
6. Dip the loop into the LiCl solution. CAUTION: The loop will remain extremely hot for several minutes fol‐lowing being in the flame. Do not touch the loop!
7. Bring the loop of the paper clip into the flame. Make observations about what is happening, especially any color changes. HINT: The color change will be most apparent around the edges of the flame. You may have to try this a few times to see the color change.
8. Extinguish the candle and record your observations in Table 1.
9. Repeat the steps 4‐7 for each of the other solutions and the activated charcoal. Use a different paper clip for each one.
10. To clean‐up, you may throw away the paper clips after they have cooled to room temperature.
Lab 9: Electron Configuration
Materials
Safety Equipment: Safety goggles, gloves
Lithium chloride solution (LiCl) Butane lighter
Sodium chloride solution (NaCl) Tea light candle
Potassium chloride solution (KCl) 5 Paperclips
Calcium chloride solution (CaCl2) Modeling clay
Activated charcoal
102
Data
Substance Observations
Lithium chloride
Sodium chloride
Potassium chloride
Calcium chloride
Activated charcoal
Table 1: Results of firework material ignition
Post‐lab Questions
1. Write out the electron configurations of each of the metals of the salt compounds used and of carbon. Potassium is already done as an example for you. HINT: The periodic table is very helpful and can be used as guide.
Element Electron Configuration
K 1s2 2s2 2p6 3s2 3p6 4s1
Li
Na
Ca
C
Lab 9: Electron Configuration
103
2. What is the approximate wavelength of light emitted by each of the salts?
Salt Color Wavelength
LiCl
NaCl
KCl
CaCl2
C
3. Why does a salt compound give off light or a colored flame when burned?
4. Did sodium chloride and charcoal give off similar colors? Why or why not?
Lab 9: Electron Configuration
Introductory Chemistry
Lab 10: Light Spectrum
107
Lab 10: Light Spectrum
Introduction
Why do we see a beautiful rainbow paint the sky just after it rains?
The gentle rain that has been falling gradually comes to an end. The gray sky is replaced with a light blue color and a few wispy white clouds. Off in the distance a beautiful rainbow can be seen. People have wondered throughout history what causes the colors of the rainbow and why are the colors always in a certain order?
Sir Isaac Newton deeply investigated the nature of light in the 17th century, making many important discoveries. Newton dem‐onstrated that sunlight, or “white light” could be separated into a range of colors when passed through a prism. From this, Newton theorized that white light is actually made up of a range of many different colors of light. Although we see sunlight (or white light) as a single color, it is composed of a broad spectrum of colors, as you can see in a rainbow (Figure 1).
Light is a form of electromagnetic radiation, which travels through space in the form of electromagnetic waves. The energy of electromagnetic waves varies over time just like with other types of waves, such as waves on a string, waves on water, or sound waves. Figure 2 depicts a wave traveling in two dimensions. The wavelength of light (or any other wave) is a use‐ful characteristic that is defined as the distance between adjacent peaks (or troughs) as shown in Figure 2. It might be measured in meters, centimeters or even nanometers (1 nm = 10‐9 meters) depending on the type of wave.
Figure 2: Sample graph of a simple wave moving along the x‐axis (horizontal arrow). In the case of light waves, the peaks represent the places where the energy is greatest, and the points that cross the x‐axis are where the energy is zero. Visible light waves oscillate very quickly and oscillate too rapidly for the human eye to notice.
1 wavelength
Figure 1: A rainbow is an example of a natural continu‐ous spectrum formed by the reflection and refraction of sunlight through water droplets in the atmosphere. What do you notice about the order of the colors in the image above?
Concepts to explore:
Learn the difference between continuous and line spectra Determine the wavelength measurement of a red laser light.
108
An observable difference between different materials is the color they naturally ap‐pear under a light source. The human eye detects the wavelengths of light reflected from the surface of a solid or passing through a liquid, and interprets different wave‐lengths as different colors. Visible light (the light our eyes are capable of detecting) has wavelengths between approximately 400 nm and 800 nm. The longest visible wavelength is red and the shortest is violet. Table 1. lists approximate wavelength ranges of different colors of light.
The visible spectrum is just a small part of the electromagnetic spectrum. The sun not only emits the radiation we can see as visible light, but also forms of electromag‐netic radiation our eyes do not detect, such as infrared and ultraviolet light. The electromagnetic spectrum ranges from very short wavelengths (including gamma and x‐rays) to very long wavelengths (including microwaves and broadcast radio waves). Figure 3 displays many of the important regions of this spectrum. It can be seen that the longer the wavelength, the lower the energy. Many instruments are designed to detect electromagnetic waves of a particular range.
The light shone through a prism is an example of a continuous spectrum, where the bands of colors blend into one another. Another way to separate and view different wavelengths of light is by using a diffraction grating. With a diffraction grating it is possible to create a bright line spectrum, where
light disperses into a series of bright lines that can be shown on a screen. This can be achieved by shining light of a single wavelength through the grating, such as the light from a laser. Typically there is one distinct center bright spot or maximum, followed by a series of gradually dimmer lines moving outward from center.
In both these cases the observed spectrum is a result of wave interference. You can think of the diffraction grating as splitting the light source into many small sources as light travels through each groove. These waves disperse as they transmit through the grating and interact with each other on their way to the screen. When two waves intersect at their respective maximums, the energy at that point will be larger (constructive interfer‐ence); at other points, one wave’s energy cancels out that of another, resulting in lower energy (destructive interference). The spectrum lines seen on the screen are points where constructive interference occurs, forming a bright spot at that particular location on a screen (see Figure 4).
We can utilize this phenomenon to calculate the wavelength of a certain light. To do this light is shone directly through a diffraction grating with a known groove separation. The distance between the center maximum and the first bright spot (X) and the distance be‐tween the diffraction grating and the first bright spot (L) are measured. The modified “grating equation” shown below can then be used to calculate the light’s wavelength(s):
Color Shades Wavelength Range (nm)
Red 650 – 750
Orange 595 – 650
Yellow 580 – 595
Yellow ‐ Green
560 – 580
Green 500 ‐560
Green ‐ Blue 490 – 500
Blue ‐ Green 480 – 490
Blue 435 – 480
Violet 400 – 435
Table 1: Wavelength ranges for the
visible spectrum
Xnλ =d
L
Figure 3: The full spectrum of electromagnetic radiation ranges from gamma radiation to radio waves. While the figure has divisions, the spectrum is continuous and infinite. Note that this figure is not drawn to scale.
Lab 10: Light Spectrum
109
When you use this equation it is easier to keep all calculations in millimeters and convert the answer to nanometers (10‐9 meters) in the end. The quantities in the above equation are:
λ = wavelength
d = the distance between the lines on the diffraction grating
n = the order of the image (most of the time n = 1 since the measuring is made to the first side spot)
X = the distance between the center spot and first side spot
L = the distance from the diffraction grating to the first side spot
Example 1: Calculating the Wavelength of Laser Pointer Light
You shine a laser pointer light through a diffraction grating that has 500 lines/mm and find the distance between the center spot and first side spot is 40 mm and the distance between the dif‐fraction grating and the first side spot is 126 mm.
Calculations
To find the laser pointer light’s wavelength you need to first calculate d.
d = space between lines = (1mm / 500 lines) = 0.002 mm
Next solve the modified grating equation for λ. For the first bright spot we can use n = 1.
nλ = d X/L
(1)λ = [(0.00200 mm)*(40.0 mm) / (126.0 mm)]
λ = 0.000635 mm
Converting from millimeters to nanometers:
l = 0.000635 mm (1 nm/10‐6 mm)
l = 635 nm
The red light emitted by laser pointers consists of waves that are shorter than a millionth of a meter. The precise wave‐length of laser light often varies from one laser pointer to another, even if each one appears red. Red laser pointer light typically ranges between 633 to 690 nm.
Lab 10: Light Spectrum
Laser light
Diffraction Grating
X
L
Figure 4: A diffraction grating creates spots of interfer‐ence, or a bright line spectrum.
110
Pre‐lab Questions
1. What is a diffraction grating?
2. You see a girl with a beautiful jeweled ring reflecting light brightly off her finger. If the reflected light appears purple and green on a piece of paper near her hand, what wavelengths of light do you suspect the jewel re‐flects?
3. A laser light was shone through a diffraction grating whose lines were 1/1000 mm apart. The distance was measured between the center spot and the first side spot and found to be 99 mm. The distance from the dif‐fraction grating to the first side spot was found to be 154 mm. Calculate the wavelength of light in nm that the laser pointer was emitting.
Lab 10: Light Spectrum
111
Experiment: Measuring the Wavelength of Laser Light
A diffraction grating has thousands of parallel grooves etched into its surface. The closer the grooves are etched, the further the spectrum will spread out. A compact disc (CD) behaves like a diffraction grating in that it has many parallel lines etched into it, which create a spectrum when light reflects from its surface.
In this experiment you will first compare the spectrum that results from shining light onto the surface of a CD and from shining light through a diffraction grating. Second, you will analyze the spectra formed by shining a laser pointer through diffraction gratings with different groove separation (d). The modified grating equation can be used to calculate the wave‐length of light emitted by the laser. See the example on the previous page for help.
Procedure
Part 1: Observation of spectra formed by CD
1. Take a CD and shine a flashlight on the back of it. This will show the colors of the visible spectrum. Describe these colors in the Data section. Also, use colored pencils or makers to draw the spectrum that you see.
2. Take the same CD and shine a laser pointer on the back of it. Describe and draw what you observe in Data
section.
Part 2: Measuring the wavelength of laser l ight
1. Set up for this experiment as shown in Figure 5. Elevate the laser pointer using a book, and place the 1000 lines/mm diffraction grating in the grating stand.
2. Position the kit box on the opposite side of the grating as the laser, approximately 20 cm from the grating. You will use this as a screen to mark the position of diffraction maxima. NOTE: You may also affix a piece of paper to the box surface for easy removal and measurement.
3. Begin with the 500 line/mm diffraction grating. Turn on the laser pointer and direct it through the grating onto the screen. CAUTION: DO NOT point the laser pointer into anyone’s eyes.
Lab 10: Light Spectrum
Materials
Safety Equipment: Safety goggles, gloves
Laser pointer Grating stand (mirror support)
String Kit box/lid
Ruler Compact Disc*
Flashlight Book*
Colored pencils White paper*
Diffraction grating (1000 lines/mm and 500 lines/mm)
*You must provide
112
Lab 10: Light Spectrum
Figure 5: Setup for diffraction experiment
4. You should see a central bright red dot and two somewhat dimmed red dots to both sides of this dot (some distance away). These dots are the first‐order diffraction image of the laser beam. You may also see some dimmer dots spaced further away—these are the second‐order diffraction image of the laser beam.
5. If necessary, adjust the apparatus so that the distances between each first‐order dot and the center dot are the same (In other words, make sure your beam is perpendicular to the screen). Record your observations.
6. When shining the laser pointer through the 500 lines/mm diffraction grating, mark the locations of the cen‐tral dot and the first order dot with a small “x” on the screen.
7. Measure L by stretching a string from the first‐order dot to where the laser light passes through the diffrac‐tion grating. Use a ruler to determine the amount of string used, making sure it is stretched the same amount as before. HINT: Tie a knot in one end of the string to help hold it down with your finger. Record this measurement in Table 2.
8. Measure the distance between the central dot and the first order dot using the marks you made on the screen. This is distance X. Record this measurement in millimeters in Table 2.
9. Replace the 500 lines/mm diffraction grating slide with the 1000 lines/mm slide and repeat Steps 3‐6. Turn on the laser light and make observations on how the projected image has changed. Record your observa‐tions in the Data section.
10. Turn off the laser light and remove it.
11. Replace the laser pointer with a flashlight, and dim the room lights. Turn on the flashlight and record all ob‐servations. Repeat with the second grating. In the Data section, draw the resulting spectra using colored pencils or markers.
Laser pointer
Diffraction grating
Diffraction maxima (red spots)
Ruler
Book
Kit box
Screen
113
Data
Part 1
Observations and drawing for the spectrum produced by the CD and flashlight:
Observations and drawing for the spectrum produced by the CD and laser pointer:
Lab 10: Light Spectrum
114
Lab 10: Light Spectrum
Part 2
Observations for the laser pointer shone through 1000 and 500 lines/mm diffraction gratings:
Observations and drawings for the flashlight shone through the 500 and 1000 lines/mm diffraction gratings:
Grating Measurement of X Measurement of L
500 lines/mm
1000 lines/mm
Table 2: Distance Measurements
115
Calculations
Calculate the wavelength of the laser light for a) d = 1/500 mm and b) d = 1/1000 mm. Use the modified “grating equation” given below:
nλ = d (X/L)
Remember to keep all calculations in millimeters (mm) until the end and then convert your answer to nanometers (1 nm = 10‐6 mm). Show all your work below:
Lab 10: Light Spectrum
116
Post‐lab Questions
1. Why did the spectra for the laser light and flashlight differ when reflected from the CD surface?
2. What was the result of shining the flashlight through the diffraction grating compared to the laser? Explain this difference.
3. Is the value you obtained for the wavelength of the laser light consistent with the range of wavelengths for the red light? Explain any sources of error that might have caused a large deviation.
Lab 10: Light Spectrum
Introductory Chemistry
Lab 11: Molecular Models
119
Lab 11: Molecular Models
Introduction
Why can’t you play basketball with a football?
The obvious answer is because a football isn’t the right shape. A football can’t be dribbled and would be very diffi‐cult to shoot. On the other hand, a fairly large spherical ball can be easily dribbled and is needed to successfully play basketball. The shape of a molecule is also extremely im‐portant in how it can be used. Just like sports need objects of a particular shape, certain reactions or applications need molecules of a particular shape.
The shapes of the different objects used in sports can be described in a few words. For example, an arrow can be described as straight or linear, a hockey puck as a flat disc, a football as oblong, and a basketball as spherical. Mole‐cules also have definite shapes that can be described in a few words. A few molecular shape examples are linear, bent, trigonal planar, pyramidal, and tetrahedral. Some of these words might already give you an idea of how such molecules are shaped.
But how do you know the shape of a particular molecule since you can’t see it? Fortunately a model has been devel‐oped that helps us predict the shapes of molecules. It is called the VSEPR model. VSEPR stands for valence‐shell electron‐pair repulsion.
When you play a team sport one of the first things you need to decide is who is go‐ing to be the captain. When using the VSEPR model a central atom has to first be decided. This is usually the atom in the molecule with the most bonds. A single bond counts as one bond, a double bond counts as two bonds, etc. For example, the car‐bon atom in methane, CH4, is the central atom since it has 4 bonds and each hydro‐gen atom has just one bond. In water, H2O, the oxygen has two bonds so it is chosen as the central atom. In carbon dioxide, CO2, the carbon has 2 double bonds or 4 total bonds so it is chosen as the central atom. In the case where two atoms have the same number of bonds, one can be chosen. For example, the oxygen atoms in hy‐drogen peroxide, HOOH, each have 2 bonds to them whereas the hydrogen atoms each have only one bond. In this case either one of the oxygen atoms can be de‐picted as the central atom.
Figure 1: Molecular structure of a caffeine molecule. Can you identify instances where atoms act as the central atom of their grouping?
Figure 2: Molecular model for meth‐ane, CH4. Note the four regions of electron density—one for each single bond to the central carbon atom
Concepts to explore:
Understand why molecules have a particular shape
Determine the shapes of molecules using the VSEPR model
120
Example Atom (central atom is underlined)
Number of Bonds on Cen‐tral Atom
(multiple bonds count as 1)
Unshared elec‐tron pairs on central atom
Regions of Electron Den‐
sity
Molecular Ge‐ometry
Structure
CO2 2 0 2
Linear
C OO
AlCl3 3 0 3
Trigonal Planar
Al
C l
C l
C l
H2O 2 2 4
Bent
O
HH :
..
NH3 3 1 4
Trigonal pyrami‐dal
. .
NH
HH
CCl4 4 0 4
Tetrahedral
C
Cl
ClCl
Cl
PCl5 5 0 5
Trigonal bipyra‐midal
P
Cl
Cl
ClCl
Cl
SF6 6 0 6
Octahedral
S
F
F
FF
FF
Table 1: Molecular Models
Lab 11: Molecular Models
121
After the central atom is decided, the number of regions of electron density around it is determined. This is a region that generally describes the location of negative charge due to surrounding electrons. A single, double, or triple bond or a pair of unshared electrons counts as one region of electron den‐sity. The carbon in methane, CH4, has four single bonds so it has four regions of electron density. However, the number of regions can be different than the number of bonds: in H2O, the oxygen molecule has two single bonds and two pairs of unshared electrons—resulting in four regions of electron den‐sity.
If you have played with magnets, you probably know that when the same poles of magnets are brought together they repel each other. The VSEPR model is based on a similar concept. Valence electrons, shared or unshared, create a negative region that repels the other negative regions. For this rea‐son, the regions of electron density are placed the same distance from the central atom, but as far away from each other as possible.
The last thing to be done is to describe the shape the atoms make. The un‐shared electrons influence the shape, but only the resulting shape of the atoms is described. Certain terminology is used for this. The shape is called the molecular geometry. Listed in Table 1 are some common molecular shapes and the words used to describe them.
All of this can be summarized in 4 steps.
1. Determine the central atom.
2. Determine the regions of electron density around the central atom by adding the number of bonded atoms and the number of unshared electrons.
3. Place the regions the same distance away from the central atom and as far as possible away from each other.
4. Describe the shape only the atoms make.
Pre‐lab Questions
1. Identify the central atom and determine how many regions of electron density there are around the central atom in each of the following molecules:
Figure 3: Regions of electron density for a water molecule (H2O).
Lab 11: Molecular Models
Molecule Central Atom # Regions of Electron Density
BeCl2
BH3
CBr4
Table 2: Pre‐lab exercise
122
2. Explain why the shape of H2O is bent and not linear.
Experiment: Molecular Structures
Procedure
Part 1: Magnets
1. Bring together the same poles of two bar magnets.
2. Observe and describe what happens.
3. Explore the linear shape of molecules by bringing the same poles as close together as you can in a straight line. Record your observations in the Data section.
4. Explore why a molecule makes a trigonal planar shape by placing the same pole of 3 bar magnets as the points of an equilateral triangle with the rest of the magnet behind them.
5. Slowly bring the same pole of the 3 magnets as close as it is possible while maintaining the equilateral triangular appearance. Now try to find another shape that will be as stable with all of the magnets at least as close as they were in the equilateral triangle position. Record your observations in Table 3.
Figure 4: Magnet setup for experiment Procedure 1.
Lab 11: Molecular Models
Materials
3 Bar magnets
Toothpicks
Modeling clay
Protractor
123
Part 2: Modeling Clay
1. Make a model of a molecule made up of two atoms such as H2 by attaching balls of modeling clay to both ends of a toothpick. Describe the shape in Table 3.
2. Make a model of a linear molecule that is made up of three atoms (such as CaCl2) using the following steps:
a. First attach two toothpicks to a larger ball of modeling clay representing the central atom. Locate them so that they are as far away from each other as possible.
b. Next attach two smaller balls of modeling clay to the ends of the two toothpicks.
c. Describe the shape in Table 3.
d. Determine the bond angle from one of the smaller balls of modeling clay to the large ball to
another small ball using a protractor. Record the bond angle in Table 3.
3. Make models of the other molecular shapes listed in Table 4 by the following:
a. First attach a toothpick to a large ball of modeling clay (central atom) for each region of electron density from Table 1. Place them so that they are as far away from each other as possible.
b. Next attach a small ball of modeling clay for each bond in the molecule. You should have one ball for each bond listed in Table 1.
c. Remove the toothpicks that do not have small balls attached.
d. Describe and record the shape in Table 4.
e. Use a protractor to determine the bond angle from one of the small balls to the large ball to another small ball. Record the bond angle in the Data section.
Lab 11: Molecular Models
Data
Procedure Step Observations
Step 1
Step 2
Step 3
Table 3: Observations for Procedure 1
124
Molecular Shape Observations Bond Angle
Linear (2 atoms)
Linear (3 atoms)
Trigonal Planar
Bent
Trigonal Pyramidal
Tetrahedral
Trigonal Bipyrami‐dal
Octahedral
Table 4: Observations and data for Procedure 2
Lab 11: Molecular Models
125
Post‐lab Questions
1. In Part 2 of the procedure, did your models fit the molecular shape description or match the geometry in the introduction? Explain why or why not.
2. Predict the shapes and bond angles of the following molecules:
a. BeCl2
b. BH3
c. CBr4
Lab 11: Molecular Models
Introductory Chemistry
Lab 12: Ionic and Covalent Bonds
129
Lab 12: Ionic and Covalent Bonds
Introduction
Have you ever accidentally used salt instead of sugar?
Drinking tea that has been sweetened with salt or eating vegeta‐bles that have been salted with sugar tastes awful! Salt and sugar may look the same, but they obviously taste very different. They are also very different chemically. Salt is made up of sodium and chloride and is ionically bonded. Sugar, on the other hand, is com‐posed of carbon, oxygen, and hydrogen and has covalent bonds.
A salt (sodium‐chloride) molecule is made up of one sodium atom and one chlorine atom. In order for the atoms to combine, the sodium atom must lose an electron, while the chlorine atom must gain an electron; the resulting ions have opposite charges and at‐tract one another.
When sodium loses an electron it becomes a positively charged ion (Na+), called a cation.
Na Na+ + e-
The chlorine atom adds this free electron, becoming a negatively charged anion.
Cl + e- Cl-
A bond can now form between the negatively‐charged Cl– and the positively‐charged Na+. This type of bond is called an ionic bond. Ionic bonds typically form between one metal and one non‐metal ion. The above reaction can be written as:
Na+ + Cl- Na+Cl-
Table sugar (sucrose) differs from salt in the bonding between its atoms. The atoms in sugar do not form ions; instead, they are held together because of shared electrons. This is an example of covalent bonding. Table sugar has a
Figure 1: In order to undergo ionic bonding, an electron must transfer between the Na and Cl atoms. This gives each atom an opposite charge, resulting in attraction.
Figure 2: Covalent bonding diagram for meth‐ane (CH4). Note that both the carbon and hydrogen atoms have full outer shells.
Concepts to explore:
Understand the differences between ionic and covalent bonding Link ionic and covalent bonding with the physical properties of
matter
130
much more complex chemical structure than salt (see Figure 3). A covalent bond between one carbon atom and one hydrogen atom forms when one of the valence electrons of the carbon atom groups with one of the valence electrons of the hydrogen atom, forming an electron pair.
Note: This is normally written C‐H.
Ionically bonded compounds behave very differently from covalently bonded compounds. In the first part of this lab you will investigate how ionically bonded and covalently bonded substances behave differently in their conduction of electricity. You will do this by using a simple anodizing apparatus that uses a stainless steel screw and an iron nail as electrodes. In an anodiz‐ing apparatus the water between the electrodes must contain enough ions to conduct electricity. As this happens, the water will react to form hydrogen and oxygen gases.
2H2O 2H2 + O2
In the second part of this lab you will explore the differences in melting points between ionically bonded and covalently bonded compounds.
Figure 3: Chemical structure for table sugar (sucrose). Here, different atoms combine via covalent bonding, as opposed to ionic bonding. Notice the complexity of this molecule compared to NaCl.
Lab 12: Ionic and Covalent Bonds
131
Pre‐lab Questions
1. What is an ionic bond?
2. What is a covalent bond?
3. Do you think sugar or salt will melt at a higher temperature? Explain your answer.
Lab 12: Ionic and Covalent Bonds
132
Experiment: Sugar or Salt?
Stainless steel is not very reactive, while iron will react with oxygen to form iron oxide, commonly called rust. You will use this fact to help determine how well dissolved sugar and salt conduct electricity.
The melting points of sugar and salt can be tested by placing a small amount of substance in a test tube and heating it at differ‐ent heights over a burner or lighter flame. These experiments will help you draw comparisons between ionically and covalently bonded materials.
Procedure
Part 1: Nail Test for Ionic Bonding
1. Rinse a clean 250 mL beaker several times with distilled water to prevent contamination from ions that may be on the beaker. Fill the beaker about ¾ full with distilled water.
2. Pour a packet of sugar (about 3 g) into the 250 mL beaker. Stir the solution with a clean stirring rod until the sugar is dissolved and the solution is well mixed.
3. Stretch two rubber bands around the 250 mL beaker. Be careful not to spill any of the solution. The rubber bands should loop from the top to the bottom of the beaker. Position the 2 rubber bands next to each other (Figure 4). HINT: Do not position the bands around the circumference of the beaker.
4. Attach the first wire lead to just underneath the flat head of an iron nail (using the alligator clip). Place the iron nail between the 2 rubber bands on one side of the 250 mL beaker so that it is suspended in the water. The end of the nail should be in the solution while the head with clip is resting on the rubber bands. (See Figure 4).
Lab 12: Ionic and Covalent Bonds
Materials
Safety Equipment: Safety goggles, gloves
2 Sugar packets Test tube holder
2 Salt packets Test tube rack
9‐Volt battery Stir rod
2 Rubber bands Butane lighter
Iron nail (uncoated) Tea light candle
Stainless steel screw Spatula
Wire leads with alligator clips on each end Permanent marker
250 mL beaker Ruler
2 Test tubes Distilled water*
*You must provide
133
Part 2: Melting Points
1. Place a spatula tip full of sugar into a test tube. The sugar should just coat the bottom of the test tube. CAUTION: Be sure the test tube does not have any small cracks or chips in it.
2. Light the candle using the butane lighter. CAUTION: Long hair should be tied up and loose clothing re‐strained when around an open flame to prevent fire and burns. Be sure you are wearing your safety gog‐gles.
3. Place the test tube containing the sugar in a test tube holder. Hold the test tube at a slight angle over the candle flame. Position the test tube so that it contacts the blue inner core of the flame.
4. Continue to hold the test tube in the flame until the sugar just begins to melt. HINT: If you keep the sugar in the flame until it turns dark brown or black, you will not be able to clean the test tube. Stop heating test tube as soon as it begins to melt. Extinguish the candle.
Lab 12: Ionic and Covalent Bonds
Figure 4: Apparatus for Procedure Part 1
+ _
Iron Nail
Steel Screw
5. Attach the second wire to just below the head of the stainless steel screw. Place the screw between the 2 rubber bands on the opposite side of the 250 mL beaker, next to the nail. Make sure the end of the screw is in the solution and the head with the clip is resting on the rubber bands.
6. Connect the wire coming from the iron nail to the positive (+) terminal of the 9‐volt battery (usually the cir‐cular terminal). CAUTION: Be careful when using energy sources such as batteries around water.
7. Connect the wire coming from the steel screw to the negative (‐) terminal of the battery (usually the hexago‐nal terminal). CAUTION: Be careful when using energy such as batteries around water.
8. Allow the apparatus to stand for two minutes and make observations. Record your observations in Part 1 of the Data section.
9. Thoroughly clean the glassware, nail and screw with distilled water.
10. Repeat the procedure using a salt packet (approximately 0.65 g) instead of a sugar packet.
134
Lab 12: Ionic and Covalent Bonds
Data
Part 1
Observations for the sugar solution:
Observations for the salt solution:
Part 2
Observations for the melting of sugar:
Observations for the melting of salt:
5. Allow the test tube to cool to room temperature before touching it. CAUTION: The test tube will be very hot and can burn your skin if touched before it cools. Hint: After the test tube has cooled for a few sec‐onds, place it in the test tube rack to finish cooling and continue with the procedure.
6. Record your observations in the Data section.
7. Repeat the procedure using salt instead of the sugar.
8. Make sure the test tubes have cooled to room temperature before touching them. CAUTION: The test tube will be very hot and can burn your skin if touched before it cools.
9. Record your observations in the Data section.
10. Clean‐up: The sugar and salt solutions can be poured down the drain. Rinse the beaker, screw, nail, and stir‐ring rod several times with distilled water. Clean the test tubes with water first and then rinse them with distilled water. They may need to soak for a few minutes in hot water in order to remove the melted sub‐
135
Lab 12: Ionic and Covalent Bonds
Post‐lab Questions
1. Why is distilled water instead of tap water used in Part 1?
2. In Part 1, why did you not observe a stream of bubbles coming off the stainless steel screw in the sugar solu‐tion?
3. Did any bubbles form off the screw in the sugar solution at all? Why might this happen, despite your answer to Question 2?
136
Lab 12: Ionic and Covalent Bonds
4. In Part 1, why did you observe a stream of bubbles coming off the steel screw in the salt solution?
5. Explain any changes that took place on the nail.
6. In Part 2, which of the substances has the lower melting point? Was this what you expected? Explain your results.
Introductory Chemistry
Lab 13: Polar Bonding
139
Lab 13: Polar Bonding
Introduction
Have you ever transferred newspaper print to silly putty?
This oddity is possible because of the chemical characteristic called polarity. Polarity is based on two primary factors, electronegativity and the shape of the molecule. Polarity is an important aspect of chemistry and it is everywhere. Loads of household substances are examples of both polar and nonpolar molecules. To explore polarity, let’s experiment with two favorite toys—slime and silly putty!
The root word for polarity is “pole”. There are strong attractions in a molecule that induce poles, similar to the North and South poles of the earth or of a magnet. A molecule can be polar too. Whether a molecule is polar or nonpolar depends primarily on its electronegativ‐ity and its shape.
Electronegativity is a term that describes the attraction an atom has for electrons. Fluorine has the strongest attraction for electrons, and therefore has the greatest electronegativity. Elements on the right side of the Periodic Table close to where fluorine is located have larger electronegativities than elements located on the left side of the Periodic Table. If all of the atoms in a molecule have similar electronegativities, the molecule is non‐polar. Hexane is an example of a non‐polar mole‐cule—it has only carbon and hydrogen atoms. The electronegativities of carbon and hydrogen atoms are very similar. They have only a 0.4 difference in their electronegativity values.
Figure 1: Transferring ink from a newspaper to silly putty. What other unique characteristics does this material have?
Figure 2: The oxygen atoms in carbon dioxide are positioned symmetrically about the central carbon atom. This is an example of a non‐polar molecule.
Concepts to explore:
Compare and contrast the chemical bonding properties of slime and silly putty
Gain knowledge about polar and non‐polar bonding by support‐ing or rejecting his/her experimental hypothesis
Learn about the technique of chromatography and how it is used to separate the components of a mixture
140
Lab 13: Polar Bonding
A polar bond is one in which there is a large difference in electronegativity between covalently bonded atoms. In order for a covalent bond to be considered polar, the atoms bonded together must have an electronegativity difference between 0.5 and 1.9. Simply because atom’s bond exhibits this difference in electronegativity does not automatically mean the mole‐cule is polar. The shape of a molecule is also important, and must be considered.
To understand how the shape of a molecule affects its polarity, imagine the atoms within a molecule playing tug‐of‐war. If there is an equal pull on each side of the rope, neither team will move. A similar phenomenon occurs in CO2, which has two bonds that pull on the central carbon with equal force from each side. The differences in electronegativity on either side balance each other out.
Now consider the water molecule. The electronegativity of oxygen is much greater than that of hydrogen, and the un‐shared electrons in the molecule are pulled closer to the oxygen atom than the electrons in the covalent bonds. This cre‐ates a large region of electron density that forces the bonds (along with the hydrogen atoms) apart, resulting in the “bent” shape shown above (see Figure 3 of the Molecular Models Lab to help you visualize this phenomenon). The result is an overall polarity that gives one side of the molecule a slightly negative charge, and the other side a slightly positive charge.
Polar molecules have a number of interesting properties. One of the consequences of polarity is that polar molecules dis‐solve other polar molecules. Some examples of polar substances include water, vinegar, and ethanol At the same time, non‐polar molecules such as those in oil or gasoline dissolve other non‐polar molecules. This trend is known as “like dis‐solves like,” which can be generalized for most polar and nonpolar substances. Since sugar dissolves in water, is sugar polar or non‐polar? If you said polar, you are correct. Sugar is a polar molecule since it can be dissolved in water, which we know is a polar molecule.
Paper chromatography is a laboratory technique that uses “like dissolves like” to separate different chemicals. A sample is first spotted on an absorbent paper such as filter paper and allowed to dry. A solvent is then allowed to travel up the pa‐per by capillary action. The components of the sample most like the solvent tend to dissolve and travel the farther up the paper, while the components least like the solvent will travel the least.
Figure 3: The hydrogen atoms in water are pulled together more strongly on one side than the other, giving the mole‐cule polarity. In water’s case, this bending is due to the negative charge of unpaired electrons around the oxygen: these unpaired negative charges repel the negative charge in the covalent OH bonds.
Negative Region
Positive Region
141
Lab 13: Polar Bonding
Pre‐lab Questions
1. What two conditions are considered when determining whether a molecule is polar or non‐polar?
2. What determines if a bond is polar?
3. List several examples of polar molecules.
4. List several examples of non‐polar molecules.
5. What is the rule when using polar and non‐polar solvents?
142
Lab 13: Polar Bonding
Experiment: Slime Time
Some inks are polar while others are non‐polar. A polar solvent will pick up polar inks, while a non‐polar solvent will pick up non‐polar inks. In this lab you will use inks to identify slime and silly putty as polar or non‐polar. You will also use paper chromatography to verify the inks are correctly identified as polar or non‐polar.
Procedure
Part 1: Making Slime
1. Weigh out 0.5 g of guar gum into a 250 mL beaker.
2. Measure 50.0 mL of distilled water into a 100 mL graduated cylinder and pour it into the 250 mL beaker that contains the guar gum.
3. Rapidly stir the mixture with a stirring rod for at least 3 minutes and until the guar gum is dissolved.
4. Measure 4.00 mL of a 4% Borax solution into a 10 mL graduated cylinder and add it to the guar gum and wa‐ter.
5. Stir the solution until it becomes slime. This will take a few minutes. If the slime remains too runny, add an additional 1.0 mL of the 4.0% Borax solution and continue to stir until the slime is the right consistency.
6. Once you are satisfied with the slime, pour it into your hands. Be sure not to drop any of it on to the floor.
7. Manipulate the slime in your hands. Write down observations made about how slime pours, stretches, breaks, etc. CAUTION: Slime is slippery and if dropped it can make the work area slick.
8. Place the slime back into the beaker and WASH YOUR HANDS.
Materials
Safety Equipment: Safety goggles, gloves Scale
Borax solution (4%) Water
Guar gum Uni‐ball® roller pen
Silly putty Permanent marker
10 and 100 mL graduated cylinders Highlighter
250 mL beaker Dry‐erase marker
Filter paper Distilled water*
Stirring rod Notebook paper*
Spatula Newspaper*
*You must provide
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Lab 13: Polar Bonding
Part 2: Slime and Putty Ink Tests
1. On a piece of notebook paper make one 20‐25 mm long mark of each of the inks you are testing. Space the marks at least one inch apart. Use a pencil to label each mark with its description.
a. Water soluble inks include those in highlighters and certain pens.
b. Water insoluble inks include those in a permanent pen/markers, newsprint, and a dry‐erase markers.
2. While the inks are drying, select a passage or a picture in the newspaper to test with the slime.
3. Break off a small piece that is 3‐5 cm in diameter of slime. Gently place this piece on top of the newspaper print, then carefully pick it up again.
4. Observe and record in Table 1 whether or not the ink was picked up onto the slime.
5. Break off another small piece of slime. Once the inks from Step 1 have dried gently place the slime on top of the first spot on the notebook paper, then carefully pick it up. Repeat this for each of the inks. Observe and record which inks were picked up (dissolved) by the slime in Table 1.
6. Repeat this ink testing two more times for accuracy.
7. Before performing ink tests on silly putty, in the Data section hypothesize which inks the silly putty will pick up.
8. Perform ink tests on silly putty in the same manner as above. Record your results in Table 2
Part 3: Chromatography of Ink Samples
1. Use a pencil or scissors to poke a small hole in the center of a piece of filter paper (see Figure 4).
2. Spot the filter paper evenly spaced approximately 2 cm from the small hole with the two insoluble inks and the two soluble inks that were used in Part 2.
3. Obtain a ½ piece of filter paper. Fold the paper in half sev‐eral times so that it makes a narrow wick.
4. Insert the wick into the hole of the spotted paper so that it is above the top of the filter paper by approximately 2 cm.
5. Fill a 250 mL beaker 3/4 full with water.
6. Set the filter paper on top of the beaker so that the bottom of the wick is in the water. The paper should hang over the edge of the beaker with the spotted side up.
7. Allow water to travel until it is approximately 1 cm from the edge of the filter paper. Remove the filter paper from the beaker.
8. Observe which inks moved from where they were originally spotted. Record your observations in Part 3 of the Data section.
Figure 4: Chromatography apparatus for Proce‐dure Part 3.
Folded wick
Ink spots
144
Lab 13: Polar Bonding
Data
Part 1
Slime Observations:
Part 2
Name of Ink Picked up (dissolved) Did not pick up
Test 1 Test 2 Test 3 Test 1 Test 2 Test 3
Newsprint
Highlighter
Roller ball pen
Sharpie marker
Dry‐erase marker
Table 1: Results of Ink Testing for Slime
Hypothesis for Silly Putty (Procedure Part 2, Step 7):
145
Lab 13: Polar Bonding
Part 3
Observations of inks following chromatography:
Name of Ink Picked up (dissolved) Did not pick up
Test 1 Test 2 Test 3 Test 1 Test 2 Test 3
Newsprint
Highlighter
Roller ball pen
Sharpie marker
Dry‐erase marker
Table 2: Results of Ink Testing for Silly Putty
Post‐lab Questions
1. Did the slime pick up water soluble or water insoluble inks? From these results, what can you conclude about the polarity of slime molecules?
146
2. Explain how you determined your hypothesis about whether or not silly putty would pick up water. Was your hypothesis correct?
3. Were the inks you used properly classified as soluble and insoluble? Explain your answer.
Lab 13: Polar Bonding
Introductory Chemistry
Lab 14: Chemical Reactions I
149
Introduction
Have you ever wondered how toothpaste helps prevent cavities?
Remember your last visit to the dentist to have your teeth cleaned and a check‐up? You most likely waited anxiously as the dentist finished checking your teeth and breathed a big sigh of relief if he pronounced, “No cavities!” But what causes cavities anyway, and how does brushing your teeth help to prevent them?
Your teeth, like many other bones in the body, are mostly made of a substance called hydroxyapatite. The empirical formula of hydroxyapatite is Ca10(PO4)6(OH)2. Plaque organ‐isms that live in your mouth produce acids that dissolve min‐erals, such as the calcium and phosphorous found in hy‐droxyapatite. The active ingredient in toothpaste, fluoride, help replace those lost minerals. It has been found that the hydroxyapatite in your teeth can easily replace lost minerals
with fluoride. Because of this, a tooth’s crystal structure that has incorporated fluoride is more resistant to decay than the original structure, which is why we need to brush our teeth with toothpaste that contains fluoride every day!
Ions have very different chemical and physical properties than atoms. For in‐stance, fluoride (F‐) as an ion is safe enough to add to drinking water and toothpaste to help prevent cavities; while fluorine (F2) is a very poisonous gas. Throughout history, scientists have developed many ways to test for the pres‐ence of different types of ions. These are called qualitative tests. In many of these tests, one ion replaces another to produce a solid precipitate. There are also some useful generalizations that can help a chemist know whether or not a particular chemical is soluble in water. One of the useful solubility rules for ionic compounds states that compounds containing Li+, Na+, K+, or NH4
+ are soluble in water. The water solubility of many other compounds can also be found in handbooks.
Figure 1: Brushing with a fluoride toothpaste can help prevent tooth decay. Fluoride mouth rinses are also popular dental hygiene products that contain fluoride. Other products in‐clude fluoride gels and foams that are applied to teeth for a period of time.
Lab 14: Chemical Reactions I
Figure 2: Diagram showing the molecular structure of a nitrate ion (NO3
‐)
Concepts to explore:
Understand qualitative tests for ions that are based on solubility Learn that Na+, K+, NH4
+, NO3–, and CH3COO
– are water soluble
Recognize balanced chemical equations
Qualitatively test for fluoride ions in mouth rinses
150
You can see that nitrates (NO3‐), acetates (CH3COO
– or C2H3O2‐), and ammonium ions (NH4
+) all have a charge like the fluo‐ride ion, but they are made up of more than one atom. These are called polyatomic ions, which are molecules with a net charge.
Calcium acetate can be used to qualitatively test for the existence of fluoride in a solution. This relates to another impor‐tant solubility rule, which states that nitrates and acetates are water soluble—therefore calcium acetate, Ca(C2H3O2)2 , is soluble in water. A solution of calcium acetate will react with fluoride to make calcium fluoride, which is not very soluble in water and will form a solid precipitate that is easily observed.
Chemists have a short way to write out what is happening in a reaction. It is called a chemical equation. The chemicals that are reacting are called the reactants and placed on the left side of an arrow. The chemicals that result from the reac‐tion are called products and are placed on the right side of the arrow. The reaction that you will be observing in this lab can be written as the following chemical equation:
2 NaF(aq) + Ca(C2H3O2)2(aq) CaF2(s) + 2 Na (C2H3O2) (aq)
The abbrevation “aq” by the chemical formula stands for “aqueous,” meaning that reactant or product is dissolved in wa‐ter, while the “s” indicates that the substance is a solid.
This chemical equation has also been balanced. This means there is the same number of each of the different atoms on either side of the equation. This may not be very clear when you first look at it. Notice that there is a 2 in front of the NaF and no number in front of the Ca(C2H3O2)2. A number in front of a molecule, such as the 2 in front of the NaF, is called a coefficient. A coefficient of 1 is usually not written, but can be assumed. As you can see, one molecule of a sub‐stance does not always react with just one molecule of another sub‐stance. The coefficients tell us the ratio of how the molecules react. In this reaction 2 NaF react with 1 Ca(C2H3O2)2.
A subscript after an element’s symbol in a chemical formula indicates the quantity of atoms of that element in the compound. For example, the CaF2 molecule has one calcium (Ca) and two fluorine molecules (F). This is also true for polyatomic ions. In Ca(C2H3O2)2, the ion (C2H3O2)
‐ is in parentheses and has a subscript 2. This means there are two (C2H3O2)
‐ ions in calcium acetate. You can think of it as a shorter way of writing Ca(C2H3O2)( C2H3O2). In total, Ca(C2H3O2)2 has one Ca atom, four C at‐oms, six H atoms, and four O atoms. If a molecule with a polyatomic ion in it does not have it in parenthesis it means that there is only one of them for each molecule. The NaC2H3O2 has only one (C2H3O2)
‐ .
Now that you know how a chemical equation is written, you can count the number of each elements on the reactants side and on the products side for the reaction above. Did you get 2 Na, 2 F, 1 Ca, 4 C, 6 H, and 4 O on each side? If so, you are correct! Since there is the same number of each atom on both sides of the arrow, the chemical equation is balanced.
Lab 14: Chemical Reactions I
Figure 3: A three‐dimensional model of an acetate (CH3COO
‐) anion—can you identify the structures pictured above?
151
Pre‐lab Questions
1. Name the chemical that makes up teeth.
2. How does plaque harm teeth?
3. How does fluoride promote dental health?
4. Write two solubility rules that are used in this lab.
Lab 14: Chemical Reactions I
152
Experiment: Battle of the Mouth Rinses
Some mouth rinses also contain fluoride, usually in the form of sodium fluoride which is soluble in water. In this lab, you will determine which one of two mouth rinses would be better at preventing cavities by replacing lost minerals with fluo‐ride. You will do this by determining which mouth rinse contains fluoride.
Procedure
1. Label the two test tubes with a Sharpie: A, and B. HINT: Make sure to write down which rinse is A and which is B.
2. Pour 10 mL of Rinse A into the test tube marked A. HINT: If using the same graduated cylinder, rinse WELL to prevent cross contamination.
3. Pour 10 mL of Rinse B into the test tube marked B.
4. Pour 3 mL of 1 M Ca(C2H3O2)2 solution into each of the test tubes. Gently stir each test tube with a stirring rod to mix. Be sure to clean your stirring rod each time before placing it in a solution. CAUTION: Mixing should be done gently to prevent glass breakage and injury.
5. Observe the reactions for at least 10 minutes to insure it is finished. HINT: A positive test is indicated by a cloudy appearance of the solution. The precipitate formed can be more easily seen if the test tube is held up to the light. The precipitate will eventually settle to the bottom of the test tube.
6. Record all observations in the Data section.
7. To clean up, you can rinse the small amount of precipitate down the drain.
Lab 14: Chemical Reactions I
Materials
Safety Equipment: Safety goggles, gloves
Mouth rinse A (10 mL)
Mouth rinse B (10 mL)
1 M Calcium acetate Ca(C2H3O2)2
Test tube rack
2 Test tubes
10 mL Graduated cylinder
Stirring rod
Permanent marker
153
Data
Observations of NaF and Ca(C2H3O2)2 :
Observations of Rinse A and Ca(C2H3O2)2 :
Observations of Rinse B and Ca(C2H3O2)2 :
Lab 14: Chemical Reactions I
154
Post‐lab Questions
1. Did either of the mouth rinses contain fluoride? How did you know?
2. Which mouth rinse would be better at fighting cavities? Why?
3. Based on the solubility rules learned in this lab, could you use potassium nitrate to test for fluoride in mouth rinses? Explain your answer.
Lab 14: Chemical Reactions I
Introductory Chemistry
Lab 15: Chemical Reactions II
157
Lab 15: Chemical Reactions II
Introduction
Chemical reactions and laundry: what’s the connec‐tion?
Laundry sometimes involves different colors and fabrics that re‐quire special treatment for washing and drying. You might not want to wash your white dress shirt with a bright and new red sweater—not unless you want your white shirt to end up not‐so‐white. To avoid this, it helps to sort your laundry into colors, darks, and whites. In a similar fashion, chemical reactions can be sorted into categories based on the characteristics of the reac‐tions.
There are many types of reaction and many ways to compare and contrast them. A common way of classifying chemical reactions is to use the following five categories: combustion, synthesis, single replacement, double replacement, and decomposition. Most re‐actions can be placed into these categories, including the reactions we will observe in this lab.
The first reaction you will perform is a combustion reaction. During combustion, a hydrocarbon and oxygen break into two simple compounds: water vapor and carbon dioxide gas. A hydrocarbon is a molecule that contains only carbon and hydro‐gen atoms. To illustrate, methane (CH4), sometimes called “natural gas,” will combust with the oxygen in the air. The equa‐tion for this reaction is:
CH4(g) + O2(g) → CO2(g) + H2O(g)
In the chemical reaction given above, all of the reactants and products are given, but is the equation balanced? From the law of conservation of mass, we know that atoms are never destroyed or created. In the chemical equation above there is 1 C on both sides, but there is 4 H and 2 O on the left side (reactant side) of the arrow and 2 H and 3 O on the right side (product side) of the arrow. This means that the equation is not balanced. If you answered no, you are correct!
There are four steps used to balance a chemical equation:
1. Count the number of atoms of each element on both the reactant and the product sides.
2. Determine which atoms are not the same for both sides.
3. Balance one element at a time by changing the coefficients for the molecules in the reaction and not their chemical formulas.
4. After you think the chemical equation is balanced, check it as in step 1.
Figure 1: Methane, the main component of natural gas, is a common fuel used for home heating and cooking. Other gases that combust in a similar way (forming CO2 and H2O) are propane and butane.
Concepts to explore:
Observe chemical reactions and identify the reactants and products
Classify types of chemical reactions
Practice balancing chemical equations
158
Now let’s use these four steps to balance the chemical equation.
Step 1:
Step 2: The hydrogen and oxygen are not balanced.
Step 3: Insert a 2 before the O2 on the reactants side and a 2 before the H2O on the products side.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Step 4:
Now the chemical equation is balanced! Some of the chemical equations for the reactions that you will observe are given in this introduction as unbalanced. Try to balance each one as they are given.
The combustion reaction you will observe is the reaction of butane (C4H10) from a lighter with the oxygen in air. The unbal‐anced reaction equation is
C4H10 + O2 → CO2 + H2O
The next type of reaction you will perform is a synthesis reaction. A synthesis reaction takes two or more substances and combines them to create a more complex substance. A general reaction equation for this type of reaction is:
A + B → C
The chemical equation for the synthesis reaction that you will peform is:
Hb(s) + O2 HbO2 (s)
Reactants Products
1 C 1 C
4 H 2 H
2 O 3 O
Reactants Products
1 C 1 C
4 H 4 H
4 O 4 O
Hemoglobin Oxygen Oxyhemoglobin
Lab 15: Chemical Reactions II
159
In this reaction, oxygen is taken up through the lungs and enters into blood‐stream. When oxygen levels are high in the lung, oxygen binds to the hemes in hemoglobin molecules (abbreviated Hb), resulting in the formation of the prod‐uct, oxyhemoglobin. In the tissues, where the oxygen level is slightly lower, the reverse reaction occurs, releasing oxygen from the hemoglobin. This reaction allows the blood to transport and exchange oxygen throughout bodily tissue.
Following the synthesis reaction, you will perform a single replacement reac‐tion. This type of reaction takes place when a more reactive element replaces one component of a compound. Two or more reactants will produce two or more products. A common example of a single replacement reaction is when one metal replaces another in a compound. A general reaction equation for this type of reaction is:
A + BC → AC + B
The unbalanced chemical equation for the single replacement reaction that you will observe is:
Zn(s) + H3C6H5O7 (aq) → Zn3(C6H5O7)2 (aq) + H2 (g)
This reaction uses a zinc coated (galvanized) washer and citric acid. Zinc is more reactive than the hydrogen it replaces. The zinc citrate stays in the solution and hydrogen gas is given off. When this reaction is allowed to continue to completion, the zinc coating almost disappears.
The fourth type of reaction you will perform is a double replacement reaction. It involves two different ionic compounds that exchange components in the reaction. Most single or double replacement reactions take place in an aqueous solution where the free ions can float around and react. The general format for the reaction is:
AB + CD → AD + CB
The unbalanced chemical equation for the double replacement reaction that you will observe in this lab is:
Zn(C2H3O2)2 (aq) + Na3PO4 (aq) → NaC2H3O2 (aq) + Zn3(PO4)2 (s)
In this reaction the zinc and sodium change places. The zinc bonds to phosphate ions and sodium bonds to acetate ions. Zinc phosphate is not soluble in water like the two reactants, and after the two reactants are mixed a precipitate of zinc phosphate is produced.
A decomposition reaction is the last type of reaction you will do. It is much like a synthesis reaction running in reverse. In this type of reaction, a more complex compound breaks down into a less complicated compound or elements.
C → A + B
Figure 2: Oxyhemoglobin molecule. Hemo‐globin molecules transport oxygen in the bloodstream, and are vital to the circula‐tory system.
Lab 15: Chemical Reactions II
160
Pre‐lab Questions
1. Michelangelo used fresco painting when he painted the Sistine Chapel. Fresco painting involves most of the types of chemical reactions you just studied. Listed below are some of the reactions used in creating a Fresco painting. Identify the type of chemical reaction used for each step and balance the chemical equation if needed.
Initially, some sort of heat must be generated. Propane is an example of a common fuel source used for heat‐ing.
C3H8 + O2 → CO2 + H2O
This is a reaction.
Next, quicklime (calcium oxide) is made by roasting calcium carbonate (limestone).
CaCO3 → CaO + CO2
This is a reaction.
The quicklime is slaked to form lime plaster.
CaO + H2O → Ca(OH)2
This is a reaction.
The lime plaster is cured or dried.
Ca(OH)2 + CO2 → Ca(OH)(HCO3)
This is a reaction.
Lab 15: Chemical Reactions II
The unbalanced chemical equation for the decomposition reaction that you will observe in this lab is:
(NH4)2CO3 (s) → NH3 (g) + H2O(g) + CO2(g)
As shown, the ammonium carbonate decomposes when heated to form three gases: ammonia, water vapor, and carbon dioxide. These five categories of reactions will give you a good foundation to understand reaction processes throughout chemistry.
161
This quickly continues to react to form calcium carbonate and water.
Ca(OH)(HCO3) → CaCO3 + H2O
This is a reaction.
Frescos will deteriorate over time when exposed to the damp, acidic environments typical of modern urban city atmospheres.
CaCO3 + H2SO4 → CaSO4 + H2O + CO2
This is a reaction.
2. Balance each of the chemical equations you will be doing in this laboratory exercise.
Combustion:
C4H10 (g) + O2 (g) → CO2 (g) + H2O(g)
Synthesis:
Hb (s) + O2 → HbO2 (s)
Single Replacement:
Zn (s) + H3C6H5O7 (aq) → Zn3(H3C6H5O7)2 (aq) + H2 (g)
Double Replacement:
Zn(C2H3O2)2 (aq) + Na3PO4 (aq) → NaC2H3O2 (aq) + Zn3(PO4)2 (s)
Decomposition:
(NH4)2CO3 (s) → NH3 + H2O + CO2 (g)
Lab 15: Chemical Reactions II
162
Experiment: Getting to Know Your Reactions
Procedure
Part 1: Combustion
C4H10 (g) + O2(g) → CO2 (g) + H2O (g)
1. Light a butane lighter and observe the flame. (The ignition of the flame is a reaction between butane and the oxygen in the air you breathe.)
2. Record your observations in the data table from when the lighter is turned on until it is turned off.
Part 2: Synthesis (A + B → C)
Hb(s) + O2 → HbO2 (s)
1. Take a deep breath, hold it as long as possible and then exhale. Visualize the reaction occurring.
2. Record your observations before inhaling and after you exhale.
3. Construct an oxyhemoglobin molecule with modeling clay and toothpicks. HINT: See figure of molecule in intro‐duction as a guide.
Part 3: Single Replacement (A + BC → AC + B)
Zn (s) + H3C6H5O7 (aq) → Zn3(C6H5O7)2 (aq) + H2 (g)
1. Place a test tube in a test tube rack or small beaker.
Lab 15: Chemical Reactions II
(unbalanced)
(unbalanced)
(unbalanced)
Materials
Safety Equipment: Safety goggles, gloves Spatula
2 mL 0.1M Zinc acetate (Zn(C2H3O2)2) Tea light candle
2 mL 0.1M Sodium phosphate tribasic (Na3PO4) 3 test tubes
Ammonium carbonate powder, (NH4)2CO3 Test tube holder
2 mL Saturated citric acid solution Test tube rack
Zinc‐coated (galvanized) washer Modeling clay
Baking soda Toothpicks
Butane lighter
163
2. Slightly tilt a test tube and slide a small zinc‐coated (galvanized) washer down the side.
3. Use a 10 mL graduated cylinder to measure out approximately 2 mL of saturated citric acid and carefully pour it into the test tube containing the zinc washer. CAUTION: Citric acid is irritating to the eyes and skin.
4. Observe the reaction for several minutes, and record your observations in the data table.
5. To clean up, separate the acid solution from the washer by pouring it into a small beaker while leaving the washer in the test tube. This is called decanting. Rinse the test tube containing the washer several times with water and add each rinse to the beaker—CAUTION: Do not pour the acid directly down the drain. To neu‐tralize the acid, add small amounts of baking soda to the solution in the beaker and stir with a stirring rod.
6. Continue stirring and adding small amounts of baking soda until gas no longer forms. Pour the liquid down the drain and throw the washer in the trash.
Part 4: Double Replacement (AB + CD → CB + AD)
Zn(C2H3O2)2 (aq) + Na3PO4 (aq) → NaC2H3O2 (aq) + Zn3(PO4)2 (s)
1. Pour approximately 2 mL of 0.1 M zinc acetate (Zn(C2H3O2)2) into a clean test tube.
2. Add approximately 2 mL of 0.1 M sodium phosphate tribasic (Na3PO4) into the test tube.
3. Record your observations before and after the addition of Na3PO4 in the data table.
4. To clean up, pour the contents of the test tube down the drain.
Part 5: Decomposition (AB → A + B)
(NH4)2CO3 (s) → NH3 + H2O + CO2 (g)
1. Place a spatula tip full (approximately 0.02 g) of ammonium carbonate (NH4)2CO3, powder into a test tube. CAUTION: Do not inhale the strong ammonia odor. Try to work in well ventilated area.
2. Light the candle using the butane lighter. CAUTION: Long hair should be tied up and loose clothing re‐strained when around an open flame to prevent fire and burns. Be sure you are wearing your safety gog‐gles.
3. Use a test tube holder to hold the test tube containing the ammonium carbonate at a slight angle in the can‐dle flame. Keep the open end of the tube pointed away from you and other students. Continue to heat the sample until the reaction is finished. Hint: Remember the products of this reaction are all gases.
4. Record your observations in the Data section.
5. Allow the test tube to cool to room temperature before touching it. CAUTION: The test tube will be very hot and can burn your skin if touched before it cools. Hint: After the test tube has cooled for a few sec‐onds, place it in a small beaker or test tube rack to finish cooling.
6. Extinguish the candle. Wash out the test tube with soap and water.
Lab 15: Chemical Reactions II
(unbalanced)
(unbalanced)
164
Data
Reaction Before the Reaction After the Reaction
Combustion
Synthesis
Single
Replacement
Double
Replacement
Decomposition
Table 1: Reaction Observations, Procedures 1‐5
Lab 15: Chemical Reactions II
165
Post‐lab Questions
1. Write the combustion reaction that occurs when you cook out on a propane gas grill. Propane has the chemi‐cal formula C3H8. Make sure to balance the reaction equation.
Balance the following equations and identify the type of reaction.
BaCl2 (s) + K2SO4 (aq) → BaSO4 (s) + KCl (aq)
KClO3 (s) → KCl (s) + O2(g)
H2 (g) + O2 (g) → H2O (l)
F2 (g) + LiCl (aq) → LiF (aq) + Cl2 (l)
a.
b.
c.
d.
Lab 15: Chemical Reactions II
Introductory Chemistry
Lab 16: Metals and Oxidation
169
Lab 16: Metals and Oxidation
Introduction
Have you ever wondered why some rings will turn a fin‐ger green, while others won’t?
The ring you just bought a couple of weeks ago is already turning your finger green. Another ring that you have had for years still looks al‐most new. Why is this? Knowing how reactive different metals are is extremely important. This property helps us to decide which metal to use for a particular application. Some metals are so nonreactive that they will not react with even the strongest acids. Many will only react with certain acids, while others are so reactive that they will react with water. This is why jewelry is often made out of gold and not ordinary iron. Iron is more reactive, and will gradually rust when exposed to the oxygen and moisture in the air. Gold, on the other hand, will not react easily with anything.
You know that gold is less reactive than iron, but what about other metals? How do you know which ones are more reactive? The activity series of metals places elements in order by their reactivity. Metals higher on the list give up their valence electrons more easily than the metals below them, meaning that any metal on this list will displace a metal below it in a reaction. This can easily be observed if the less re‐active metal is in an aqueous solution during the reaction, as it will precipitate out of the solution when replaced. A partial activity series list is shown in Figure 2.
As a part of this lab, you will observe the reaction of an acid and zinc metal and compare it to the reaction of the same acid with iron metal. The reaction between the zinc and the acid can be written as follows:
2 H+( a q ) + Zn(s) → Zn2+
(aq) + H2(g)
In this reaction, the zinc transfers its valence electrons to the hydrogen in an acid solution. (Remember a superscript plus sign indicates that the compound is missing a valence electron a minus sign indicates that it has an extra valence electron.) This causes the hydrogen in the acid solution to separate as hydrogen gas while the zinc metal forms zinc ions dissolved in the solution. The zinc is said to be more electropositive than the hydrogen it displaces.
Figure 1: The Statue of Liberty in New York Harbor is made of copper, and was originally the color of a penny. The statue has gradually acquired its green color due to the natural oxidation of copper when exposed to air and water.
Concepts to explore:
Observe an oxidation‐reduction reaction
Use the properties of a reaction product to verify its identity
Rank the reactivity of certain metals in a weak acid, and compare it to their order in the Activity Series of Metals
170
Reactions like this that involve the transfer of electrons are called oxidation‐reduction reactions, or “redox” for short. Oxi‐dation generally describes the loss of electrons by a molecule, while reduction describes the gain of electrons. In this reac‐tion, since zinc is losing electrons, it is being oxidized.
Zn (s) → Zn2+ (aq) + 2e-
Hydrogen is gaining electrons in this reaction, so it is being reduced:
2 H+(aq) + 2e- → H2 (g)
This is often remembered by the phrase “LEO the Lion goes GER.” LEO stands for Losing Electrons Oxidation, and GER stands for Gain Electrons Reduction.
In Lab 15 we called this same reaction a single replacement reaction. How can this be? It has to do with how the reactions are sorted. Just like laundry, there are several different ways to categorize and sort reactions depending on the applica‐tion. For example, when doing laundry, you may need to separate red clothes from white ones. Other times you separate the light colored clothes from the dark colored clothes. Sometimes reactions are categorized as they were in Lab 15, and sometimes they are categorized as redox and non‐redox reactions.
But how do we know just by looking at this reaction equation that it is an redox reaction and that electrons are being trans‐ferred? The oxidation number must first be assigned to each of the atoms involved on both sides of the reaction equation. Let’s look at a reaction between aluminum and hydrochloric acid.
6 HCl ( a q ) + 2 Al(s) → 2 AlCl3 (aq) + 3 H2(g)
There are several rules to help you determine oxidation numbers in a reaction:
The oxidation number of an element by itself is zero. This means that the aluminum metal (Al) on the right side of the above equation has an oxidation number of 0. The hydrogen gas (H2) on the right side of the equation also has an oxidation number of 0.
When an atom exists as a simple ion in a substance, the oxidation number is the same as its charge in the com‐pound. The chloride (Cl) ion has a negative charge, so its oxidation number in is ‐1 in each instance it appears.
For a neutral compound, the sum of the oxidation numbers is always zero. A polyatomic ion’s oxidation number equals the charge on the ion. Since HCl is neutral, the sum of the oxidation numbers should equal zero. We know from above that the Cl ion has an oxidation number of ‐1; For the sum of the oxidation numbers in HCl to equal 0, the H must have an oxidation number of +1. Similarly , for the oxidation of AlCl3 to be zero, the Al must have an oxidation number of +3 to balance out the three Cl atoms (each with oxidation numbers of ‐1).
Lab 16: Metals and Oxidation
171
Lithium Potassium Barium Calcium Sodium
Magnesium Aluminum Zinc Chromium Iron
Cobalt Nickel Tin Lead
Hydrogen Copper Mercury Silver Platinum Gold
Release hydrogen from cold water, steam, and acids
Release hydrogen from steam and acids
Release hydrogen from acids
Do not release hydro‐gen from acids
Loses electrons easily (more easily oxidized)
Do not lose electrons easily (not easily oxidized)
Figure 2: Activity Series of Metals (Partial List)
+1 ‐1 0 +3 3(‐1) 0
Lab 16: Metals and Oxidation
Sometimes it is helpful to write the oxidation numbers over the atoms as is shown below:
6 HCl ( a q ) + 2 Al (s) → 2 AlCl3 (aq) + 3 H2(g)
The activity series (Figure 2) of metals also indicates how easily a metal will cause a release of hydrogen in redox reactions. Metals at the top portion of the list release hydrogen merely by being placed in cold water. As you go down the list more harsh conditions are required to release hydrogen. Metals towards the bottom do not release hydrogen even when placed in acids.
Sometimes a metal forms a thin oxide coating that protects it from reacting any further with its surroundings. Aluminum is fairly reactive and does this quickly. On the other hand, when iron is exposed to oxygen it corrodes entirely into iron oxide (rust) over time. For this reason, iron is commonly coated in zinc to prevent oxidation.
In this laboratory exercise, you will compare the reactivity of the redox reactions of zinc and iron with a weak acid solution: saturated citric acid. Citric acid is a different type of acid than HCl, but works in a similar way.
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Pre‐lab Questions
1. What is the oxidation state for each atom in the following reaction:
4Fe + 3O2 → 2Fe2O3
a. Elemental iron (Fe)
b. Elemental oxygen (O2)
c. One iron atom in Fe2O3
d. One oxygen atom in Fe2O3
2. Which element was oxidized and which element was reduced in the above reaction equation?
a. Element oxidized
b. Element reduced
3. From the Activity Series of Metals, determine the order of reactivity of the following metals: Ni, Au, Fe, Ca, Zn, and Al.
Most Reactive → Least Reactive
Lab 16: Metals and Oxidation
173
Procedure
1. Label two test tubes Zn and Fe with the permanent marker, and place them in a test tube rack.
2. Tip the test tube labeled Zn slightly and let a galvanized nail gently slide into it with the top of the nail going down first.
3. If necessary, lightly sand the iron nail that is not galvanized to remove any rust and wipe it clean. Gently place it in the test tube labeled Fe the same way you did previously.
4. Add approximately 3 to 5 mL of saturated citric acid solution to each test tube.
5. Make initial observations and continue recording observations after one minute, three minutes, and five minutes. HINT: The most notable observations are how quickly bubbling occurs and how violently the bub‐bling of each continues.
6. After observing the reactions for five minutes, rank the two metals in order of their reactivity. Compare your results with the actual reactivity series.
7. To clean up, separate the acid solution from the metals by pouring them into a 250 mL beaker while leaving the metals in their test tubes. This is called decanting. Rinse the test tube containing the metals several times with water and add the rinses to the beaker. To neutralize the acid, add small amounts of baking soda to the acid solution in the beaker and stir. Continue this until no more gas forms. Pour the liquid down the drain, and throw the metals in the trash.
Experiment: Metal Reactivity
In this lab you will use what you know about chemical reactions and oxidation to examine how two metals (zinc and iron) react differently with a citric acid solution. You will then draw conclusions about the reactivity of the metals based on what you observe.
Lab 16: Metals and Oxidation
Materials
Safety Equipment: Safety goggles, gloves
10 mL graduated cylinder Iron nail (uncoated)
250 mL beaker Saturated citric acid solution
2 test tubes Sandpaper
Test tube rack Stopwatch
Galvanized nail Baking soda
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Time (minutes)
Observations
Initial
1
3
5
Table 1: Observations of Reactions
Summarize your observations and list the order of reactivity of the metals that you observed:
Data
Lab 16: Metals and Oxidation
175
Post‐lab Questions
1. Based on what you observed, what is one of the products formed in Part 1? How do you know?
2. Did the order of reactivity you determined in Part 2 match the order given in the Activity Series of Metals? Explain.
2. Do you think the following reaction would occur? Explain your answer.
FeCl2 ( a q ) + Cu(s) → Fe(s) + CuCl2 (aq)
4. How do you think acid rain might affect the rate of rusting of metal?
Lab 16: Metals and Oxidation
Introductory Chemistry
Lab 17: The Mole and Avogadro’s Number
179
Lab 17: The Mole and Avogadro’s Number
Introduction
“Avocados” number: How many avocados and artichokes are there in a mole?
A recipe calls for two avocados and two artichokes. Would you say that equal amounts of avocados and artichokes are used? The answer to that question depends on how you define “the same amount”. If you consider the quantity, yes, there are two of each. What if instead the recipe told you to use 250 grams of artichokes and avocados? There might be one and a half avocados used for every arti‐choke—would you call this the same amount as well?
One way to solve this problem would be to ask for more specific instructions. The phrase “the same amount” is be‐ing used to describe quantity (number) in one instance and mass in another. Similar situations come up in chemistry. If you are supposed to put the same mass of two different substances into a beaker, all you would have to do is weigh equal amounts using a scale. Things become more complicated, however, when the ratio of the total number of molecules or atoms is important in an experiment—you might know the number of molecules in the reactants and products, but not the actual masses. How would you measure an exact number of NaCl molecules?
The mole is an important unit in chemistry that you will use often. The modern definition of the mole is based around the carbon‐12 (12C) atom: 12 grams of 12C is exactly one mole of substance. It turns out that 1 gram of 12C has approximately 6.02 x 1023 individual atoms—a value called Avogadro’s Number after the chemist Amedeo Avogadro. The mole is a similar concept to a dozen: one mole of a substance will always have 6.02 x 1023 atoms or molecules.
Atomic weight is a measurement of the mass of each element, and can be easily found on most periodic tables. Atoms that have a large number of protons and neutrons have more total mass than atoms with only a few protons and neutrons, and this difference is reflected by the atomic weight. Conveniently, the mole and atomic weight are defined so that one mole of a substance will have a mass equal to the atomic weight of that substance in grams. This number is called molar mass. For instance, the atomic weight of potassium (K) is 39.098. This means that one mole of potassium will have a mass equal to 39.098 grams. Calcium (Ca) is a larger atom than potassium, and has an atomic weight of 40.078. One mole of calcium will have the same number of atoms as a mole of potassium, but the Calcium will weigh more due to its larger atomic weight. See the next lab for more detail on atomic weigh and atomic mass.
So how do you measure 1 mole of NaCl? We know that this molecule is made up of one sodium ion and one chlorine ion (chloride). Looking up sodium (Na) on the Periodic Table tells you its atomic weight is 22.99, meaning it has a molar mass of 22.99 g/mol. Chlorine, meanwhile, has a molar mass of 35.45 g/mol. Since one mole of sodium chloride consists of one mole sodium ions and one mole chlorine ions, we can add these together to find the molecular mass of NaCl.
Figure 1: You can think of avocados and artichokes in this exam‐ple as two different types of molecule—each with a different molar mass.
Concepts to explore:
Understand the importance of Avogadro’s Number
Approximate the value of Avogadro’s Number
180
What if we weighed out 1.00 grams of NaCl—how many molecules is this? We use the molar mass of NaCl, which we already know from above, to convert from mass to a number of moles. We can then use the fact that there are 6.02 x 1023 molecules in a mole to find the number of molecules NaCl in one gram:
Notice how the molar mass is inverted in the second term. The value is the same, but we “flip” the fraction so that the gram units cancel. You can go through and cross out the units that cancel to verify that the resulting units are molecules. You can use similar calculations to convert between mass, moles, and the number of molecules fairly easily.
Through this lab procedure, we will determine the experimental value for Avogadro’s number. You will float cinnamon, evenly distributed, on the surface of water in a Petri dish. The dishwashing liquid you will use in this Lab is about 1% sodium stearate, and a solution with a known concentration of the liquid will be dropped onto the water. The sodium stearate molecules will form a single layer and spread out, pushing the cinnamon toward the edges of the Petri dish, allowing the surface area to be
2322
1 mol NaCl 6.02 10 molecules
1.00 g NaCl 1.03 10 molecules NaCl58.44 g NaCl 1 mol NaCl
1 Na = 22.99 g/mol
+ 1 Cl = 35.45 g/mol
58.44 g/mol of NaCl
This is how to determine molar mass of a compound.
54 88 .58.44 g NaCl
1 mol NaCl grams NaCl1 mol NaCl
Pre‐lab Questions
1. How many grams of H2O do you need to weight out to have 1 mole of H2O?
Lab 17: The Mole and Avogadro’s Number
181
2. How many molecules of water are there in one mole of H2O?
3. How many moles of H2O are there in 1.0 g of H2O?
4. How many molecules of H2O are there in 1.0 g of H2O?
Lab 17: The Mole and Avogadro’s Number
182
Experiment: Avogadro’s Number
Procedure
Part 1: Preparing the Sodium Stearate Solution
1. Measure exactly 1.50 mL of dishwashing liquid into a 10 mL graduated cylinder.
2. Fill a wash bottle with distilled water. Gently rinse the 1.50 mL of dishwashing liquid with distilled water and pour it into a 100 mL graduated cylinder. Rinse the 10 mL graduated cylinder several times to make sure all the dishwashing liquid has been transferred to the 100 mL graduated cylinder. HINT: Try not to create suds.
3. Add enough additional distilled water to get to the 100.0 mL.
4. Gently stir the solution with a stirring rod until it is mixed well.
Part 2: Calibrating a Dropper
1. Fill a 50 mL beaker half full with distilled water. Use a pipette to fill a 10 mL graduated cylinder to 1.00 mL with water. HINT: Make sure the 10 mL graduated cylinder is clean of dishwashing liquid.
2. Next, draw up water from the 50 mL beaker into the pipette. Add water dropwise into the graduated cylin‐der. Hold the pipette consistently at a 45o angle and drop at a rate of about one drop per second. Count the drops it takes to reach the 2.00 mL mark. HINT: It should take about 25 drops. If you feel that your measure‐ment is incorrect, repeat until you achieve consistent readings.
3. Record in the Data section the number of the drops it takes to add 1 mL water to the graduated cylinder.
4. Repeat calibration for a second trial, and record the number of drops in the Data section. Average the two results.
Part 3: Calculating the Number of Molecules
1. Rinse and then fill a petri dish with 20 mL distilled water. Allow the water to settle and remain motionless.
Lab 17: The Mole and Avogadro’s Number
Materials
Safety Equipment: Safety goggles, gloves
Ground cinnamon 10 mL graduated cylinder
Dishwashing liquid Stirring rod
Dropper (pipette) 50 mL beaker
Petri dish (bottom) Wash bottle
Ruler Distilled water*
100 mL graduated cylinder *You must provide
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2. Lightly sprinkle cinnamon onto the surface of the water in the Petri dish. HINT: Add just enough to barely cover the water.
3. Draw up the dishwashing liquid solution with the calibrated pipette. Hold the pipette at a 45o angle about 1 inch above the center of the Petri dish. Slowly deliver one drop of the solution. HINT: A clear circle should form, spreading the cinnamon outward.
4. Quickly use a ruler to measure the diameter of the cleared circle in cm.
5. Record the diameter in the Data section. Wash out the Petri dish.
Data
Part 2: Calibrating a Dropper
1. The number of drops in 1 mL water (drops used to move from the 1.00 mL to 2.00 mL mark):
Trial 1: Trial 2:
2. The number of drops on average per one milliliter:
Part 3: Calculating the Number of Molecules
1. The diameter of the circle formed (cm):
Calculations
1. Calculate the surface area of the circle formed ( πd2 /4 ) :
2. Calculate the number of molecules on the top layer. We must convert the surface area in centimeters squared to nanometers squared and then multiply that by the surface area of a sodium stearate molecule.
Convert the surface area of the circle formed (#1) to molecules per layer:
Surface area =
cm2 1 m2 1 x 1018 nm2 1 molecule
Top layer SA (Question 1)
10,000 cm2 1 m2 0.210 nm2
= molecules
top layer
Lab 17: The Mole and Avogadro’s Number
184
3. Calculate the concentration of grams of sodium stearate per milliliter of diluted solution. To do this, multiply the concentration of sodium stearate in the dishwashing liquid by the dilution of the solution (1.50 mL dishwashing liquid per 100 mL solution).
4. Calculate the number of moles of sodium stearate in a single layer. To do this, first take the number of drops used to achieve the monolayer (1 drop) and convert it to mL using the calibrated number of drops per mL. Then multi‐ply the number of grams of sodium stearate per milliliter of solution. Finally, convert to moles through the molar mass of sodium stearate. HINT: The molar mass of sodium stearate is 296.4 g/mol.
1 g sodium stearate 1.50 mL dish liquid
100 mL dish liquid 100 mL diluted solution
= g /ml
= mol / top layer
5. Finally, we can calculate the Avogadro’s number through the comparison of molecules of sodium stearate in the top single layer to the moles of sodium stearate in the monolayer.
Avogadro’s number (experimental) =
# molecules / top layer (#2)
# moles / top layer (#4)
molecules
mole =
Lab 17: The Mole and Avogadro’s Number
1 drop
(added to dish)
1 mL dish liquid solution
g
sodium stearate
(from #3 calculation) 1 mol
top layer drops
(avg # calibrated per mL from Data
Part 2)
296.4 g
(molar mass of sodium stearate)
1 mL dish liquid solution
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Post‐lab Questions
1. Why do you think that Avogadro’s number, 6.02 x 1023, was probably not the exact number you obtained? Was your experimental value close to the actual value (i.e., was your experimental value on the order of 1023 molecules)?
2. How many moles are in 0.289 g of methane (CH4)?
3. How many moles are in 1,000,000,000 molecules of H2 02?
4. What is the mass in grams of 1,000,000,000 (109) molecules of H2O2?
Lab 17: The Mole and Avogadro’s Number
Introductory Chemistry
Lab 18: The Periodic Table
189
Lab 18: The Periodic Table
Introduction
What is the importance of an element’s location on the periodic ta‐ble?
The periodic table contains almost endless useful information for performing chemis‐try. If you know how to interpret it, its organization alone can provide a wealth of knowledge about the elements and their relationships. Without the periodic table, chemists would have to memorize a lot of details and facts. Now memorizing is very difficult and not a lot of fun. But chemistry is fun! So, let’s blow up some balloons to show how the periodic table can be a party.
The periodic table is organized similar to a typical data table. It contains individual cells that are arranged logically in rows and columns according to specific trends. The rows on the periodic table are called periods, while the columns are referred to as families or groups. Each cell contains information about a particular element. In the middle of the cell is a one or two letter abbreviation for a particular element called a chemical symbol. For example Fe is the symbol for iron and Na for sodium. Above an element’s symbol is the atomic number, which is the number of protons (and thus electrons) that exist in an atom of that element. Each element has its own unique number of protons.
The elements are arranged in such a way that both chemical and physical trends run across the rows and down the columns. Trends found in the periodic table can demonstrate uniform increases or decreases or represent similarity. As you go across a period moving left to right, you will find that the atomic number of the elements increases by one. This means the number of protons in the elements increases by one. On the other hand, when moving down a group of the periodic table you will find that the number of electrons in the outer shell remains the same. This is why the atoms within a group have similar chemical behavior.
Below an element’s symbol is the atomic weight. This is the average mass of a given element after taking into account the existence of natural isotopes. The molar mass is obtained by multiplying the atomic weight(s) by the molar mass constant (Mu = 1 g/mol). As you have seen in the Mole and Avogadro’s Number Lab, knowing the mo‐lar mass of a substance allows you to convert between the mass of a sample and the total number of moles/molecules of a sample.
Atoms are often bonded together to form compounds. The molar masses of each element in a compound can be added together to
Figure 1: A typical periodic table cell with atomic number, the ele‐
Atomic Weight vs. Atomic Mass
The names are similar, but atomic mass and atomic weight refer to different quantities. Atomic weight is sometimes called “relative atomic mass” to avoid confusion. While atomic weight is technically a di‐mensionless value (no units), it is common to hear atomic weight given in grams per mole. In this case it is more accurate to use the term molar mass, but the concept is the same. Atomic mass gives the mass of a specific isotope of an element in unified atomic mass units (u).
Both atomic mass and atomic weight are measured relative to 12C, which is defined as having an atomic mass of 12 u. Since a typical sample of carbon con‐tains small amounts of 13C (7 neutrons) and 14C (8 neutrons), the atomic weight is slightly larger at 12.0107 (see Figure 1). Atomic mass is more accu‐rate when a sample is known to be purely one iso‐tope, but the percentage difference is very small.
Concepts to explore:
Understand the periodic table and its uses
Relate the characteristics within a group on the periodic table
Understand the use of molecular weight and molar ratios in chemical reactions
190
Lab 18: The Periodic Table
find the molar mass of the compound as a whole. For example, to find the molar mass of Li2CO3 you would add the individual atomic weights of two lithium atoms, one carbon atom, and 3 oxygen atoms:
2 Li = 2 (6.94) (1 g/mol) = 13.88 g/mol
1 C = 1 (12.01) (1 g/mol) = 12.01 g/mol
3 O = 3 (16.00) (1 g/mol) = 48.00 g/mol
13.88 g/mol + 12.01 g/mol + 48.00 g/mol = 73.89 g/mol
This means if you weigh out 73.89 gram of Li2CO3, you would have one mole (or 6.02 x 1023
molecules) of Li2CO3. In the following experiment you only want 0.025 moles of Li2CO3. To find out how many grams you need to weigh out to have 0.025 moles of Li2CO3, you simply multiply the molar mass of lithium carbonate by 0.025 moles.
73.89 g/mol x 0.0250 mol = 1.85 g Li2CO3
Notice that the moles cancel out leaving grams as the final unit.
In this experiment, you will investigate Group I elements by reacting their carbonates with cit‐ric acid to produce water, a citrate, and carbon dioxide gas (CO2). The CO2 gas will be captured by a balloon which will cause the balloon to inflate. The reaction for the lithium carbonate is shown below:
3 Li2CO3 + 2 H3C6H5O7 → 2 Li3C6H5O7 + 3 H2O + 3 CO2
Since you are using only Group I carbonates, all the compounds are similar and will react in a similar manner. For instance, the carbonates all have the same ratio of two Group I elements to one CO3. Also, they will all form a salt, water and carbon dioxide gas when citric acid is added to them.
You will first compare the reactions when the same fraction of a mole of Li2CO3, Na2CO3, and K2CO3 react with the same amount of citric acid solution. You will then compare the reactions when the same mass of Li2CO3 and K2CO3 react with the same amount of citric acid solution.
Figure 2: The first two groups on the periodic ta‐ble (far left section). Group I is commonly called the alkali metals (minus hydro‐gen), and Group II is com‐monly called the alkaline earth metals.
HHydrogen
1.0079
1
LiLithium
6.941
3
NaSodium
22.9898
11
KPotassium
39.0983
19
RbRubidium
85.4678
37
CsCesium
132.905
55
FrFrancium
(223)
87
BeBeryllium
9.0122
4
MgMagnesium
24.3050
12
CaCalcium
40.078
20
SrStrontium
87.621
38
BaCarbon
137.327
56
RaRadium
(226)
88
Group I
Group II
191
Lab 18: The Periodic Table
Pre‐lab Questions
1. Complete the three reactions shown below that are taking place throughout this experiment.
___ Li2CO3 + ___ H3C6H5O7 →
___ Na2CO3 + ___ H3C6H5O7 →
___ K2CO3 + ___ H3C6H5O7 →
2. Calculate the molecular weights of Na2CO3 and K2CO3. (See the Introduction for the Li2CO3 example.)
a. Na2CO3 :
b. K2CO3 :
3. Calculate how many grams of Na2CO3 and K2CO3 you will need to weigh out to have 0.0250 mol of each of the substances. (See the Introduction for the Li2CO3 example.)
a. Na2CO3 :
b. K2CO3 :
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Experiment: Periodic Table Fun
Procedure
1. Determine the masses of each substance in the Data and Calculations section. If you are unsure of your cal‐culations, contact your instructor before proceeding.
2. Inflate four balloons to approximately the same size and then let the air out. Label each balloon with a per‐manent marker. Label the first one Li2CO3, the second Na2CO3, the third K2CO3, and the fourth 2
nd K2CO3.
3. Place the calculated number of grams of 0.025 moles of Li2CO3, Na2CO3, and K2CO3 in the appropriate bal‐loons. Place 1.8 g of K2CO3 into the balloon marked 2nd K2CO3. HINT: Place a 50 ml beaker on the scale, zero the scale, then weigh out the correct amount of grams. Pour the substance from the beaker into each bal‐loon using the funnel. HINT: Be sure to clean the spatula before placing it into a different substance.
4. Measure out 20 mL of saturated citric acid solution into each of the four Erlenmeyer flasks using the gradu‐ated cylinder. CAUTION: Remember safety goggles! Citric acid is an irritant and should be washed off skin or eyes immediately with plenty of water.
5. Attach the four balloons to the Erlenmeyer flasks without turning the balloons upright or allowing any of the powder to fall into the flasks. CAUTION: Remember safety goggles! Make sure the balloons are well se‐cured to the flasks. If not, a balloon could fly off and cause injury.
6. Turn only one balloon upright at a time. Gently shake the balloon until all of the carbonate falls into the flask. Swirl the flask until the reaction is complete Record your observations of the reaction in Data Table 1.
7. After each reaction is complete, quickly measure the circumference of the largest part of the balloon. Do this by using a string to determine the distance around the largest part of the balloon and then measure the corresponding length of the string with a ruler. Record the length in Data Table 1.
8. Repeat steps 6 and 7 with each flask.
Lab 18: The Periodic Table
Materials
Safety Equipment: Safety goggles, gloves
4 Balloons 100 mL graduated cylinder
Saturated citric acid solution Spatula
Lithium carbonate, Li2CO3 String
Sodium carbonate, Na2CO3 Ruler
Potassium carbonate, K2CO3 Funnel
4 Erlenmeyer flasks 50 mL beaker
Permanent marker Baking soda
Scale pH paper
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Substance Moles of Substance
(mol) Equivalent Mass
(g) Reaction Observations
Inflated Balloon Circumference
(cm)
Li2CO3 0.0250
Na2CO3 0.0250
K2CO3 0.0250
2nd K2CO3
(1.8 g) 1.8
Table 1: Data and observations for carbonate‐citric‐acid reactions
1. Fill in the corresponding mass of 0.025 moles of each carbonate in Table 1 (from Pre‐lab Question 3).
Mass of Li2CO3:
Mass of Na2CO3:
Mass of K2CO3:
2. Calculate the number of moles there are in the 1.8 g sample of K2CO3 and record this value in the Table 1.
Data and Calculations
Lab 18: The Periodic Table
9. Remove the balloons from the flasks and throw them in the trash. CAUTION: When removing the balloons avoid direct inhalation of gases. Hold the balloons away from your or another student’s face and body. Remove the lip of the balloon that is opposite from your body first.
10. To clean up, place a waste beaker in the sink. Rinse each flask several times with water and add each rinse to the waste beaker—CAUTION: Do not pour the acid directly down the drain. To neutralize the acid, add small amounts of baking soda to the solution in the beaker and stir with a stirring rod until no more gas forms. It is then safe to pour down the drain.
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Post‐lab Questions
1. What was the source of the gas that causes the balloons to inflate?
2. Does a trend of similarity appear in the data and/or observations among the three reactions with 0.0250 moles of each substance? Explain.
3. Considering the organization of the periodic table, explain why there were similarities among these reactions?
4. In the experiment, you used the same number of moles of each of the first three salt carbonates. Explain what happened when you put the same amount of grams of the potassium carbonate in the fourth balloon as you used of the lithium carbonate.
Lab 18: The Periodic Table
Introductory Chemistry
Lab 19: Stoichiometry
197
Lab 19: Stoichiometry
Introduction
Have you ever wondered why hot dogs are sold in packages of 10, but hot dog buns are sold in packages of 8?
This is an example of a ratio. Stoichiometry is the scientific word for ratios of moles. It is the study of the quantitative relationships between reactants and products in a chemical reaction. These relationships are often referred to as mole‐to‐mole ratios. Mole‐to‐mole ratios help scientists know how much of each re‐actant is needed to produce a sufficient amount of product. In this lab, you will observe the importance of mole‐to‐mole ratios when making the main compo‐nent in lime that is used in gardens!
The manufacturers of hot dogs and hot dog buns would not make very good chemists. Hot dogs are sold in packages of 10 where hot dog buns are sold in packages of only eight. If one package of each is purchased, how many hot dog‐on‐buns can you make? The answer of course is eight, meaning there are two hot dogs left over. Now examine the equation below and try to fix their mistake. How many packages of hot dogs and hot dog buns must you purchase in order to not waste any hot dogs? How many hot dog‐on‐buns does this give you?
___ 10 Hot dogs + ___ 8 Hot dog buns → ___ Hot dog-on-buns
If you buy four packages of hot dogs and five packages of buns you would get 40 hot dog‐on‐buns.
Balancing a chemical equation is done the same way: you place numbers, called coefficients, in front of each type of mole‐cule. Just like you cannot change how many hot dogs there are in a package of hot dogs, you cannot change how many of each type of atom there is in a chemical substance. You can, however, change the ratio of the chemical substances so that the same number of each type of atom is on both sides of the reaction equation. In the hot dog example there are 40 hot dogs and 40 hot dog buns on both sides of the arrow.
But what if instead you buy only two packages of hot dogs, and three packages of hot dog buns, which would you run out of first? You would have a total of 20 hot dogs and 24 hot dog buns, so you would run out of the hot dogs first. This means you could only make 20 of the hot dog‐on‐buns final product. The number of hot dogs limits the number of hot dog‐on‐buns you can make.
This is similar to what in chemistry is called the limiting reagent. By using stoichiometry it is possible to calculate the limit‐ing reagent of a reaction. Knowing which reactant will be consumed first in a reaction allows chemists to calculate how much product can be theoretically made. This can be compared to the actual amount produced by calculating the percent yield. The percent yield is found by using the following equation:
Actual YieldPercent Yield = 100
Theoretical Yield
Figure 1: Buying hot dogs and buns can be complicated!
Concepts to explore:
Demonstrate the use of stoichiometry to synthesize calcium carbonate
Practice using a scale and proper lab techniques
Find the limiting reagent, the theoretical yield, and the percent yield
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Using the hot dog example, the percent yield would be like only making 19 hot dogs when there were enough hot dogs and buns to make 20 hot dog‐on‐buns. In this example the percent yield would be calculated as follows:
Chemists will often try making a desired product many different ways. They will then compare the percent yields to deter‐mine which of the methods tried gives the greatest yield.
In this laboratory exercise you will explore the concepts of limiting reagents and percent yields through making calcium carbonate, the main component in lime. Lime is used in many different industries such as farming, where it is used to ad‐just soil pH, making crops more productive.
Calcium carbonate, CaCO3, can be made by mixing CaCl2 and K2CO3 in water. The trick is to be sure to use the correct amount of each. Look at the following reaction and balance it to determine the mole‐to‐mole ratio that would be necessary to have the same number of each kind of atom on both sides of the equation.
___ CaCl2(aq) + ____ K2CO3(aq) → ____ KCl(aq) + ____ CaCO3(s)
When this chemical equation is balanced there is one mole of calcium chloride for every one mole of potassium carbonate and together these quantities make two moles of potassium chloride and one mole of calcium carbonate. The potassium chloride is soluble in water so it remains in solution, while the calcium carbonate will precipitate out as a white solid.
CaCl2(aq) + K2CO3(aq) → 2KCl(aq) + CaCO3(s)
What if instead you added 2 moles of CaCl2 and 1 mole of K2CO3. How much CaCO3 could you make? The answer is still only one mole of CaCO3. The K2CO3 is the limiting reagent since it limits how much of the product can be made.
The following calculation illustrates how the limiting reagent and percent yield are determined for the above reaction. De‐termining the limiting reagent can be complex, so breaking the calculation into several steps can be useful. The following example demonstrates how to determine the limiting regent for a typical reaction.
19 Actual Hotdog‐on‐bunsPercent Yield = 100
20 Theoretical Hotdog‐on‐buns
Percent Yield = 95%
Lab 19: Stoichiometry
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Example:
A procedure calls for 8.25 g of CaCl2 to be dissolved in water and reacted with 5.00 g of K2CO3. How many grams of CaCO3 can theoretically be made? If 3.25 g of CaCO3 were actually made, what is the percent yield?
a. First write the balanced reaction equation:
CaCl2 + K2CO3 → 2KCl + CaCO3
b. Calculate the formula weight for each reactant:
c. Calculate the number of moles used of each reactant:
CaCl2 : (8.25 g) (1 mol/110.98 g) = 0.0743 mol
K2CO3: (5.00 g) (1 mol/138.21 g) = 0.0362 mol
d. Determine how many moles of the product could be made by each of the reactants.
CaCl2 :
K2CO3:
e. Determine the limiting reagent. It is the reactant that will make the least number of moles of CaCO3.
K2CO3 is the limiting reagent!
This means that only 0.0362 moles of CaCO3 can be made when 8.25 g of CaCl2 and 5.00 g of K2CO3 are mixed in water. In this example, the reactant that had the lowest mass in the procedure turned out to be limiting. However, this is not always true! The only way to know for sure which reactant is limiting is to calculate it.
Ca = 40.08 g/mol
+ 2 Cl = 70.90 g/mol
= 110.98 g/mol CaCl2
2 K = 78.20 g/mol
3 O = 48.00 g/mol
+ C = 12.01 g/mol
= 138.20 g/mol K2CO3
32 3
2
1 mol CaCO(0.0743 mol CaCl ) = 0.0743 mol CaCO
1 mol CaCl
32 3 3
2 3
1 mol CaCO(0.0362 mol K CO ) = 0.0362 mol CaCO
1 mol K CO
Lab 19: Stoichiometry
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f. Determine how many grams of CaCO3 can theoretically be produced:
First calculate the formula weight for the product, CaCO3:
Next calculate the theoretical yield by calculating how many grams there are in 0.0362 moles of CaCO3:
g. Calculate the percent yield:
This means that in this example only 89.8% of what theoretically could have been made was ac‐tually made.
Ca = 40.08 g/mol
3 O = 48.00 g/mol
+ C = 12.01 g/mol
= 100.09 g/mol CaCO3
33 3
3
100.09 g CaCOTheoretical Yield = (0.0362 mol CaCO ) = 3.62 g CaCO
1 mol CaCO
3.25 gPercent Yield = 100 = 89.8%
3.62 g
Lab 19: Stoichiometry
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Pre‐lab Questions
1. What is a limiting reagent?
2. A student used 7.15 g of CaCl2 and 9.25 g of K2CO3 to make CaCO3. The actual yield was 6.15 g of CaCO3. Cal‐culate the limiting reagent and the percent yield.
Lab 19: Stoichiometry
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Experiment: Synthesis of Garden Lime
Procedure
1. An example set of data has been provided for 1.0 g CaCl2.
2. For Trial 1, weigh into a 250 mL beaker the amount of cal‐cium chloride (CaCl2) shown in Table 1. Record the exact mass you weigh out in the Trial 1 column of the Data sec‐tion.
3. Measure 50.0 mL of distilled water into a 100 mL graduated cylinder. Pour the water into the 250 mL beaker with the calcium chloride.
4. Stir the solution with a stirring rod until all of the calcium chloride is dissolved.
5. Weigh out 2.5 g of potassium carbonate (K2CO3) in a 50 mL beaker. Record the exact mass in the Data sec‐tion.
6. Measure 25.0 mL of distilled water into a 100 mL graduated cylinder. Add the water into the 50 mL beaker containing the potassium carbonate.
7. Stir the potassium carbonate in the distilled water with a stirring rod until it is all dissolved.
8. Pour the K2CO3 solution into the 250 mL beaker that has the CaCl2 solution. Rinse the beaker that contained the K2CO3 with a few mL of water and add this to the CaCl2 solution. Stir the mixture.
9. As soon as the reaction begins, record your observations in the Data section. Continue stirring until you see no more precipitate forming.
10. Set up the funnel in the Erlenmeyer flask as shown in Figure 2. HINT: Do NOT begin filtering yet!
11. Zero the scale and weigh a piece of filter paper and a watch glass. Record the masses of both items in the Data section.
Example About 1.0 g CaCl2
Trial 1 About 2.0 g CaCl2
Trial 2 About 3.0 g CaCl2
Table 1: Approximate CaCl2 amounts
Lab 19: Stoichiometry
Materials
Safety Equipment: Safety goggles, gloves Wash bottle
Calcium Chloride (CaCl2) Scale
Potassium Carbonate (K2CO3) Funnel
Ethanol 3 Filter papers
100 mL graduated cylinder Spatula
Erlenmeyer flask Stirring rod
3 250 mL Beakers Distilled water*
3 50 mL Beakers Oven*
Watch glass *You must provide
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Lab 19: Stoichiometry
12. Prepare a filtering funnel as shown in Figure 2: fold a piece of filter paper in half twice to make quarters, and open the paper to make a small cone (three quarters are open on one side and one quarter is on the opposite side). Place the paper cone into the funnel and hold it in place with your fingers. Pour a small amount of distilled water through the paper to secure it inside the funnel.
13. Filter the mixture by pouring it into the filter paper in the fun‐nel. Use the stirring rod and distilled water in a wash bottle to transfer the entire solid into the filter paper. HINT: For best results, be sure to transfer all of the precipitate into the filter paper. Use a rubber policeman if it is available to help with the transfer.
14. Rinse the remaining solid in the filter paper twice with distilled water from a wash bottle to rinse off excess sodium chloride (NaCl). After all the liquid has filtered through, rinse the prod‐uct with approximately 5 mL of ethanol to aid in its drying. Al‐low the ethanol to completely finish filtering through the pa‐per.
15. Remove the filter paper carefully so as to not lose any product. Gently unfold the filter paper and lay it flat on the pre‐weighed watch glass to dry.
16. Allow the product to air dry completely. This may take 24 hours or more. Once dry, weigh the dry product on the filter paper and watch glass. Record the total mass in the Data section. Calculate the mass of the product.
17. Repeat the above procedure for Trial 2 using the amount of CaCl2 mass indicated in Table 1.
18. To clean up, wash any dirty glassware, pour liquids down the drain, and throw the product on the filter paper in the trash.
Filter Paper
Funnel
Flask
Figure 3: Filter Setup
Figure 2: Filtering funnel preparation diagram
a b c d e f
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Mass (g) Example Trial 1 Trial 2
Mass of CaCl2 1.0 g
Mass of K2CO3 2.5 g
Mass of filter paper 0.8 g
Mass of watch glass 38.5 g
Combined mass of prod‐uct, filter paper, and
watch glass 40.2 g
Mass of dry product 0.9 g
Table 2: Reaction product data
Lab 19: Stoichiometry
Data
1. Record your data for each of the trials in Table 2.
2. Record your reaction observations (Step 8) below:
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Calculations
1. Determine the limiting reagent for each trial. Show your calculations. (Hint: See the example in the Introduc‐tion.)
Example:
Trial 1:
Note: These should be about the same and either CaCl2 or K2CO3 can be the limiting reagent depending on their initial masses.
Trial 2:
Lab 19: Stoichiometry
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2. Calculate the theoretical yield of CaCO3 that could be produced by each trial and then fill in Table 2.
3. Find the percent yield each trial obtained for the CaCO3.
Table 4: Comparison of theoretical and actual yields for CaCO3
Trial # Limiting Reagent Theoretical Yield
of CaCO3 Actual Yield of
CaCO3
Trial 1
Trial 2
Trial 3
% Yield
Lab 19: Stoichiometry
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Lab 19: Stoichiometry
Post‐lab Questions
1. Compare the results of the different trials. How does the amount of grams of CaCO3 compare?
2. Were the results of the trials as you expected? Why or why not?
3. Predict what would happen if 6.0 grams of CaCl2 were used for the reaction and the amount of K2CO3 re‐mained the same.
Introductory Chemistry
Lab 20: Ideal Gas Law
211
Introduction
Have you ever opened a container of milk after the expiration date and found that it had gone bad?
It is very easy to tell if milk has gone bad: it looks, smells, and tastes awful. The putrid odor lets you know a gas is being formed as it decomposes. Many items you purchase are not nearly as easy to tell if they have degraded. Some form a gas that is impossible to detect just by smelling it. Hydrogen peroxide, a common household item that is used to clean minor cuts, is like this. The bottle you buy in the store says it contains 3% hydrogen peroxide, but it will very slowly decompose over time to form water and oxygen gas. Since we breathe oxygen every second, we can’t easily detect this. But we can use the ideal gas law and yeast to find this out!
Lab 20: Ideal Gas Law
The ideal gas law is very valuable when dealing with gases since it establishes a relationship between temperature, pressure, volume, and amount of a gas.
In this equation:
P is the gas pressure in atmospheres
V is the volume of the gas in liters
n is the number of moles of the gas
R is the constant value of 0.0821 L·atm/mol·K
T is for the temperature of the gas in Kelvin.
Since hydrogen peroxide forms oxygen gas when it decomposes, we can use the ideal gas law to check the percent hydrogen peroxide in a bottle of it purchased at the store. To find this out we need to take a small sample out of the bottle and accelerate its decomposition through using a catalyst.
2 H2O2 catalyst 2 H2O + O2
PV = nRT
Figure 1: Pressure gauges are found wherever moni‐toring the pressure of a gas is important—such as this fire extinguisher above. Simple gauges are commonly used to measure the pressure of air in automobile and bicycle tires.
0
0.1
0.2
0.3
0.4
0.5
0.6
0 2 4 6 8 10
Pressure
Volume
Pressure vs. Volume of a Typical Gas
(Arbitrary Units)
Figure 2: The relationship between pressure and vol‐ume of an ideal gas. With constant T and n, pressure decreases as volume increases. Can you verify this using the equation to the left?
Concepts to explore:
Use the ideal gas law to determine the percentage of hydrogen peroxide in a commercially available hydrogen peroxide solution
Observe how a catalyst affects a reaction
Determine the decomposition rate of the hydrogen peroxide solution
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Lab 20: Ideal Gas Law
PVn =
RT
Once the number of moles of O2 gas is calculated, the percent of H2O2 present in the solution can be determined. To do this, you first need to calculate the theoretical number of moles of O2 there would be if the solution was 100% hydrogen peroxide. This can be found by using the following equation:
For this experiment:
mL H2O2 used is the volume of H2O2 you actually use (approximately 5 mL).
H2O2 density is 1.02 g/mL
1 mol H2O2 / 34.0 g H2O2 is the reciprocal (inverted fraction). of the molar mass of H2O2 . The molar mass of H2O2 is 34.0 g /mol, so this is equal to 1 mol H2O2 / 34.0 g H2O2.
1 mol O2 / 2 mol H2O2 is used since the decomposition produces 1 mole O2 from 2 moles of H2O2 .
The units in the entire equation cancel to give moles of O2.
The percent hydrogen peroxide can now be found. To do this, divide (n), the actual number of moles you calculated, by the theoretical moles of O2 there would be if the hydrogen peroxide were 100%. This number is then multiplied by 100%.
This value can now be compared to the 3% hydrogen peroxide shown on the label to see if any decomposition has occurred.
2 2 22 2 2 2 2
2 2 2 2
1 mol H O 1 mol OTheoretical moles O = H O used × H O density × ×
34.0 g H O 2 mol H O
2
22 2
Actual moles O (n)% H O = 100
Theoretical moles O
In this experiment, we will use yeast to accelerate the decomposition of the hydrogen peroxide into water and O2 gas. Yeast contains the enzyme catalase, which is a catalyst for this reaction. You will add yeast activated in warm water to a known amount of hydrogen peroxide and quickly seal off the system so that the O2 gas formed is collected in a graduated cylinder. After measuring the total volume of gas produced, its temperature, and the atmospheric pressure, the ideal gas law can then be used to calculate how many moles of O2 gas is formed. We can do this by solving the ideal gas law equation for n.
Figure 3: Carbonated beverages contain dissolved CO2 at high pressure. When the container is opened, this pressure can create a powerful burst, such as with this sparkling wine bottle, or when your soda “explodes.”
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Lab 20: Ideal Gas Law
Pre‐Lab Questions
1. What is it in yeast that aids in the decomposition of hydrogen peroxide?
2. List the ideal gas law and define each term with units.
3. How many moles of O2 were produced in a decomposition reaction of H2O2 if the barometric pressure was 0.980 atm, the temperature was 298 K and the volume of O2 gas collected was 0.0500 L?
4. If you decomposed 10.00 mL of 100% H2O2, how many moles of O2 could you theoretically obtain?
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Lab 20: Ideal Gas Law
Experiment: Finding Percent H2O2 with Yeast
Procedure
1. Prepare the materials for the apparatus as shown in Figure 1. Insert the smaller rigid tubing into one end of the larger, flexible tubing. Insert the free end of the rigid tubing securely into the rubber stopper hole.
2. Bend the free end of the flexible tubing into a U shape, and use a rubber band to hold this shape in place. This will allow you to more easily insert this end of the flexible tubing into the inverted graduated cylinder. Make sure the tubing is not pinched and that gas can flow freely through it.
2. Fill the 600 mL beaker with 400 mL distilled water.
3. Fill the 100 mL graduated cylinder with distilled water slightly over the 100 mL mark.
Figure 3: Gas Collection Apparatus (not to exact scale)
Rigid tubing
Stopper
Erlenmeyer flask
Flexible tubing
Graduated cylinder
Ring Stand
600 mL Beaker
Collected gas
Rubber Band
Materials
Safety Equipment: Safety goggles, gloves
Yeast Rubber band Flexible tubing (18 in.)
10 mL Hydrogen peroxide 2 Droppers (pipettes) 250 mL Beaker
10 and 100 mL Graduated cylinders Stir rod 600 mL Beaker
Erlenmeyer flask Thermometer Stopwatch
Stopper with hole Warm water* Ring stand*
Rigid plastic tubing (3 in.) Large ring* Distilled water*
*You must provide *Optional Materials (not provided)
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Lab 20: Ideal Gas Law
4. Take the temperature of the water in the 600 mL beaker, and record it in the Data section. Also, determine the barometric pressure in the room, and record it in the Data section. HINT: The pressure in your region may be found online—if necessary, convert this value to mm Hg.
5. Mix 100 mL of warm water (45°C) and 1 packet of baker’s yeast in a 250 mL beaker. This will activate the yeast from the dormant (dry) state. Be sure to mix well with a stir rod until the yeast is completely dissolved.
6. Use a 10 mL graduated cylinder and pipette to measure out 5.00 mL of hydrogen peroxide. Pour this hydrogen peroxide into the Erlenmeyer flask, and place the stopper with stopper tube over the top.
7. Clean the 10 mL graduated cylinder by rinsing it at least three times with distilled water. Dispose of the rinse down the drain.
8. Cover the opening of the graduated cylinder with two or three fingers and quickly turn it upside down into the 600 mL beaker already containing 400 mL of water. DO NOT remove your fingers from the opening until the graduated cylinder is fully submerged under the water. If the amount of trapped air exceeds 10 mL, refill the cylinder and try again.
9. Insert the U shaped side of the flexible tubing into the beaker, and carefully snake it into the submerged opening of the graduated cylinder. You want as little air as possible to be in the graduated cylinder.
10. Secure the graduated cylinder to the ring stand by sliding a ring under the submerged cylinder, then attach‐ing the ring to the stand.
OPTIONAL PROCEDURE: If your kit does not include a ring stand, you will hold the graduated cylinder in place while gas is collected. Make sure to keep the open end of the cylinder completely submerged to pre‐vent additional gas from entering. Rest the graduated cylinder against the side of the beaker during experi‐mental setup.
11. With the cylinder vertical, record the volume of air inside (the line at which the water reaches in the cylinder) in the Data section in Table 1.
12. Using the pipette, measure out 5.00 mL of yeast solution into the rinsed 10 mL graduated cylinder. NOTE: Do not immediately pour the yeast solution into the Erlenmeyer flask.
13. Prepare to place the stopper (still connected to the hose) on the Erlenmeyer flask. Reset the stopwatch.
14. Quickly pour the 5.0 mL of yeast solution into the Erlenmeyer flask. Immediately place the stopper securely in the opening of the Erlenmeyer flask by twisting it down into the flask gently.
15. Start timing the reaction with the stopwatch.
16. Swirl the Erlenmeyer flask to mix the two solutions together.
17. You will begin to see bubbles coming up into the 100 mL graduated cylinder. HINT: If gas bubbles are not immediately visible, make sure the stopper is on tight enough and the tubing is not leaking. You will need to start over after correcting any problems.
18. Continue to swirl the Erlenmeyer flask and let the reaction run until no more bubbles form to assure the reaction has gone to completion. This should take approximately 6‐10 minutes. HINT: Catalase works best around the temperature of the human body. You can speed the reaction up by warming the Erlenmeyer flask with your hands.
19. Record the time when the reaction is finished in Table 2 of the Data section, along with the final volume of air in Table 1. Remember to read it at eye‐level and measure from the bottom of the meniscus.
20. Pour all other liquids down the drain and clean the labware.
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Lab 20: Ideal Gas Law
Data
Water temperature: ⁰C
Barometric Pressure: mm Hg
Initial volume of air (mL) Final volume of air after reac‐tion (mL)
Volume of O2 collected
(Final volume ‐ initial volume)
Table 1: Volume data
Time reaction started Time reaction ended Reaction time (s)
Table 2: Reaction time data
Calculations
The goal is to find the percentage of hydrogen peroxide in the solution! This can be found by working through the following steps.
1. Convert the temperature of the water from ⁰C to Kelvin (K). Use the equation K = ⁰C + 273. This will be your value for absolute T or the temperature in Kelvin.
2. If necessary, convert the barometric pressure in the room from mm Hg to atmospheres (atm).
Divide the measured pressure from the Data section by 760 mm Hg. This will give you pressure (P) in atmospheres.
T = ⁰C + 273 = K
P = mm Hg * = atm 1 atm
760 mm Hg
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Lab 20: Ideal Gas Law
3. Convert the volume of oxygen from mL to liters (L).
4. Rearrange the ideal gas law to solve for n.
5. You are now ready to solve for the number of moles of O2. Be sure the units cancel so that you end up with only the moles of O2 left. Use the value for the constant R given:
Actual number of moles of O2 (n) = moles
V = mL * = L 1 L
1000 mL
R = 0.0821 L∙atm/mol∙K
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6. Calculate the theoretical number of moles of O2 there would be if the hydrogen peroxide were 100%, and not an aqueous solution.
To use the above equation, calculate the following:
— H2O2 volume is the volume (mL) of hydrogen peroxide used: Volume = mL H2O2
— H2O2 density is known: Density = 1.02 g/mL
Molar mass of H2 O2 = g H2O2/1 mol H2O2
Molar mass of H2 O2 reciprocal =
Now you have all of the information needed to solve the equation for the theoretical moles of O2. All you need to do is fill in the blanks and do the calculations.
Theoretical moles of O2 =
Theoretical moles of O2 = mol
mol H2O2
g H2O2
is the reciprocal of the molar mass of H2O2. First write the molar mass of H2O2 then find the reciprocal.
* * *
mol H2O2
g H2O2
1 mol O2
2 mol H2O2 Theoretical moles of O2 = H2O2 volume * H2O2 density * *
—
Lab 20: Ideal Gas Law
219
7. Find the percent hydrogen peroxide.
% H2O2 = * 100% = %
8. You can also easily determine the reaction rate. To do this, divide the total volume of oxygen collected by the total time of the reaction.
Reaction rate = = mL/sec
Actual moles O2
Theoretical moles O2
Volume O2 (mL)
Reaction time (s)
Post‐Lab Questions
1. Was the calculated percentage of hydrogen peroxide close to the same as the percentage on the label?
2. Considering that catalysts are not consumed in a reaction, how do you think increasing the amount of catalyst would affect the reaction rate for the decomposition of hydrogen peroxide?
Lab 20: Ideal Gas Law
Introductory Chemistry
Lab 21: Reaction Rate
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Lab 21: Reaction Rate
Introduction
Why is it easier to sweeten hot tea with granular sugar than cold tea with sugar cubes?
As you begin to stir sugar cubes into cold tea, the sugar particles sluggishly separate from the cube while following the flow of your spinning stirrer. Eventually, after much stressful stirring, the little sugar crystals make their way into solution. Then you notice that your friend has almost finished his sweet tea and has already returned for more. You complain that you’ve been stirring forever just to dissolve the sugar in your cold tea. Your friend responds, “I only had to stir mine for a few seconds, and it was good to drink!” Why do you think your friend’s sweet tea was finished so much faster? Well, his sweet tea was made from hot water and granular sugar.
A reaction rate is the time that it takes for the reactants to be changed into products. The rate is given as the change of the concentration of a reactant or product in a certain amount of time, and can be described using various units. Reaction rates are affected by several factors which include the fol‐lowing: the nature of the reactants, surface area, concentration, tempera‐ture, pressure, and the presence of a catalyst. Whether a reaction rate will increase or decrease depends on the rate that the molecules involved effec‐tively collide to result in a reaction.
Throughout this laboratory exercise, you will use calcium carbonate and cit‐ric acid to discover how temperature, surface area, and concentration affect reaction rates. Calcium carbonate is the main compound found in marble. Marble is often used to make statutes or as decorative rock chips in flower beds. Citric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide gas, and water. This is similar to how acid rain degrades marble statutes.
In this laboratory exercise you will compare how two different surface areas of calcium carbonate, a powder and a solid rock piece, react with different concentrations of citric acid at various temperatures. The powder has a large overall surface area due to its many individual parts. In contrast, the solid, a crystallized rock, has a much smaller surface area. You will record how long it takes for each reaction to complete, then calculate and compare the reaction rates.
Figure 1: Calcium carbonate is abundant in nature, and is the primary component of lime‐stone and marble rock (top). Calcium carbon‐ate is commonly found in antacids and cal‐cium supplements.
Concepts to explore:
Understand how temperature, surface area, and concentration influ‐ence the rate of a reaction
Relate reaction rates on a molecular level
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Lab 21: Reaction Rate
Pre‐lab Questions
1. Name five factors that can affect the rate of a reaction.
3. In the opening paragraph example, it took more time to make sweet tea with cold water and sugar cubes than to sweeten hot tea with granular sugar. Why?
4. What is the primary factor that determines whether a reaction rate increases or decreases?
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Lab 21: Reaction Rate
Experiment: Comparing Reaction Rates
Procedure
1. Label four test tubes 1, 2, 4, and 5 (Reaction 3 takes place in the 50mL beaker, which is why you number the test tubes this way).
2. Break off a piece of CaCO3 rock with a mass of approximately 0.2 grams. Record the actual mass in the Data section. Place the rock in test tube 1.
3. Weigh out three more pieces of approximately 0.2 grams of CaCO3 rock. These should be as close as possi‐ble to the mass of the first rock sample. Place the pieces of rock into test tubes 2, 4 and 5. Record each of their masses Table 1.
4. Into a 50 mL beaker, weigh out approximately 0.2 grams of CaCO3 powder. This should be as close as possi‐ble to the amount of the previously weighed pieces of marble rock. Record the mass in Table 1.
Reaction # 1
6. Measure 10 mL of saturated citric acid solution into a 10 mL graduated cylinder. Transfer the acid to test tube 1 (with the CaCO3 rock), and place this test tube immediately in an ice bath. Record the start time. Check on this reaction frequently and record when the reaction no longer produces bubbles (gas). Record all values in Table 1, along with your observations.
Reaction # 2
6. Measure 5 mL of saturated citric acid solution (60%) into a 10 mL graduated cylinder and dilute to 30% by adding 5 mL distilled water. Transfer the diluted acid to test tube 2, and place this test tube in the rack. Re‐cord the start time. Check on this reaction frequently and record the time when the reaction no longer pro‐duces bubbles (gas). Record all values in Table 1, along with your observations.
Materials
Safety Equipment: Safety goggles, gloves Baking soda
40 mL Saturated citric acid solution (60%) pH Paper
Calcium Carbonate rock (CaCO3) Permanent marker
Calcium Carbonate powder (CaCO3) Scale
10 mL graduated cylinder Stopwatch
4 test tubes Stir rod
50 mL beaker Distilled water*
250 mL beaker Ice*
Test tube holder Boiling water bath*
Test tube rack *You must provide
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Lab 21: Reaction Rate
Reaction # 3
6. Measure 5 mL of saturated citric acid solution (60%) into a 10 mL graduated cylinder and dilute it to 30% by adding 5 mL distilled water. Transfer this diluted acid to the 50 mL beaker that contains CaCO3 powder. Use a stopwatch to time the reaction from when the acid is poured onto the powder and until the reaction no longer produces bubbles (gas). Record all values in Table 1, along with your observations.
Reaction # 4
6. Measure 10 mL of saturated citric acid solution into a 10 mL graduated cylinder. Transfer the acid to test tube 4 and place this test tube in the rack. Record the starting time. Check on the reaction frequently and record the time when the reaction is no longer fizzing. Record all values in Table 1, along with your observa‐tions.
Reaction # 5
6. Measure 10 mL of saturated citric acid solution into a 10 mL graduated cylinder. Transfer the acid to test tube 5, and place this test tube in the previously started hot water bath. Record the starting time. Check on this reaction frequently and record the time when the reaction stops (no longer fizzing). Record all values in
Data
Substance, Reaction #
Variable
Mass of the CaCO3 (g)
(these should be close)
Time of the reaction
(sec) (start/stop)
% Citric Acid
CaCO3 Rock
#1
Saturated acid solution, iced
CaCO3 Rock
#2
Diluted acid solution, room
temp.
CaCO3
Powder
#3
Diluted acid solution, room
temp.
CaCO3 Rock
#4
Saturated acid solution, room
temp.
CaCO3 Rock
#5
Saturated acid solution, heated
Observations
Table 1: Reaction rate data and observations
227
Lab 21: Reaction Rate
Calculations
Calculate the rate (g/sec) of each of the reactions you observed.
Reaction 1 (rock, iced, saturated solution):
Reaction 2 (rock, room temperature, diluted solution):
Reaction 3 (powder, room temperature, diluted solution):
Reaction 4 (rock, room temperature, saturated solution):
Reaction 5 (rock, heated, saturated solution):
228
Post‐lab Questions
1. All the reactions that you performed were chemically the same. You just varied several factors. What were the factors that were varied?
2. Which factor do you think made the biggest influence on the reactions? Why?
3. Out of the five different reactions, which reaction was the slowest? Was this what you expected? Why?
4. Why do you think marble statues require long periods of time to degrade in regions that are affected by acid rain?
Lab 21: Reaction Rate
Introductory Chemistry
Lab 22: Catalysts
231
Lab 22: Catalysts
Introduction
Why do bubbles form when you put hydrogen peroxide on a wound?
Ouch! You just scraped your knee. After you hobble to a sink to wash it off, you apply some hydrogen peroxide. Small bub‐bles start forming almost immediately. Why? The reason is because blood and tissue contain a certain enzyme that accel‐erates the decomposition reaction of hydrogen peroxide forming oxygen gas and water. The enzyme is a biological catalyst. When there is a need for speed in a reaction, using a catalyst is often the best method.
Many chemicals the human body needs are made within the cells. This means the human body has a need for super‐fast chemical reactions. A high reaction temperature or a large concentration of the reactants will often sufficiently speed up a reaction in the laboratory. But our cells cannot rapidly increase their temperature or suddenly increase the availability of certain chemicals. Instead, the human body uses catalysts. A catalyst is a substance that speeds up a reaction, but is not consumed during the reaction. Biological catalysts are called enzymes. There are many different types of enzymes, and each type speeds up a certain reaction your body needs to have happen right then! Without catalysts your body could not do even the simplest task.
Reactions have a minimum amount of energy required to occur. This is called the activation energy. A catalyst will lower the activation energy by requiring less energy for the reaction to occur. Chemists cannot always speed up a reaction by changing the usual variables, and reactions that take a long time are seldom very useful. For this reason, chemists often add catalysts to speed up reactions.
After a chemist decides to use a catalyst, there are several things that have to be studied. One of the biggest challenges is finding the ideal catalyst for a particular reaction. Frequently several catalysts are found that will work, and they are com‐pared to determine which is best. Some of the factors that a chemist will take into consideration include the desired speed of the reaction, the cost of the catalyst, how long the catalyst will work, and if it is toxic or harmful to the environment. Another characteristic to consider is a catalyst’s phase. A heterogeneous catalyst is in a different state of matter (phase) than the reactants when it is applied, while a homogeneous catalyst is applied in the same phase. Generally, homogeneous catalysts will react faster, but heterogeneous catalysts are easier to separate from the products.
In addition to the enzymes in blood and tissue, there are several other catalysts that can be used to accelerate the reaction to decompose hydrogen peroxide into water and oxygen gas. Manganese dioxide, many fruits and vegetables, household bleach, and even soil can all be used to catalyze this reaction. A piece of a carrot put into a solution of hydrogen peroxide is an example of a heterogeneous catalyst, as the carrot is in a solid phase and the hydrogen peroxide is in a liquid phase. If
Figure 1: Adding a catalyst can drastically speed up the rate of a chemical reaction. Catalysts play a direct role in the envi‐ronment and in biology, and are often used in industrial ap‐plications for food processing and chemical refinement.
Concepts to explore:
Evaluate different catalysts to determine which one is the best choice
Illustrate the differences between using heterogeneous and homoge‐neous catalysts
Demonstrate how varying quantities of a catalyst affect the reaction
232
instead the carrot is made into a juice and added to the hydrogen peroxide solution, it is a homogeneous catalyst since it is then in the liquid phase.
Chemists also have to determine the optimal amount of a catalyst present in the reaction. If a catalyst is very expensive, toxic, or hard to remove from the product, they may use the least amount of catalyst that will work. If instead there is a need to have the reaction happen more rapidly, a chemist may choose to add more catalyst. There is a point however, where adding more catalyst will not increase the reaction rate. This is because there is as much or more catalyst than the limiting reactant.
Pre‐lab Questions
1. What is a catalyst?
2. If you continue to add more catalyst will the speed of a reaction always continue to increase? Explain your answer.
Lab 22: Catalysts
233
3. In this lab you will produce oxygen and water from hydrogen peroxide (H2O2). Write a balanced reaction equation for this reaction.
4. What causes the bubbles to form in this reaction?
5. The exhaust gas from car engines pass through catalytic converters that contain very small amounts of solid platinum, palladium, and rhodium catalysts. Are these metals homogeneous or heterogeneous catalysts?
Lab 22: Catalysts
234
Experiment: Reactions with Catalysts
In this laboratory exercise, you will evaluate carrots, tomatoes, yeast, and soil as catalysts to decompose hydrogen peroxide. You will also observe the differences between using a piece of carrot or carrot juice as a catalyst for this reaction. Finally, you will observe the effects of adding different amounts of carrot juice on the reaction rate.
Procedure
Part 1: Comparison of Different Catalysts
1. Place 11 test tubes in the test tube racks.
2. Use the permanent marker to label 5 of the test tubes C, T, D, and Y, symbolizing carrot, tomato, soil, and yeast. HINT: It’s best to clearly label glassware to prevent cross contamination.
3. Mix 100 mL warm water (45⁰C) with one packet of yeast in a 250 mL beaker. Stir with the stir rod until dis‐solved.
4. Use a 10 mL graduated cylinder and a pipette to add 3 mL of hydrogen peroxide to EACH of the 5 labeled test tubes.
5. Carefully add a small piece (approximately 1 cm2) of carrot to the test tube labeled C, and a small piece of tomato to the test tube labeled T. Add a similar small amount of soil to the test tube labeled D.
6. Measure 3 mL of yeast solution into a clean 10 mL graduated cylinder. Slowly add the yeast solution to the test tube labeled Y. Swirl until all bubbling and foaming stops. This will indicate the completion of the reac‐tion.
7. Record observations for each of the reactions in the Initial Observations column in Table 1 in the Data sec‐tion. Let the reactions continue until the end of the next part of the procedure.
Part 2: Catalyst Quantity Comparison
7. Use a permanent marker to label the 6 remaining test tubes, 1, 5, 10, 1A, 5A, and 10A.
Lab 22: Catalysts
Materials
Safety Equipment: Safety goggles, gloves
11 test tubes 250 mL beaker
10 mL graduated cylinder Stir rod
Hydrogen peroxide (H2O2) Soil
Yeast Warm water (45⁰C)*
3 Droppers (pipettes) Small piece of a carrot*
2 Test tube racks Small piece of a tomato*
*You must provide
235
8. Fill test tube 1A, 5A, and 10A with 1 mL, 5 mL, and 10 mL of yeast solution respectively.
9. Use a 10 mL graduated cylinder and pipette to add 1 mL of hydrogen peroxide to the test tubes labeled 1, 5, and 10.
10. Simultaneously (or as close to the same time as possible) pour the yeast solution from test tubes 1A, 5A, and 10A into the corresponding test tubes 1, 5, and 10 containing the hydrogen peroxide.
11. Record your observations of the three reactions in Table 2 of the Data section. HINT: Be sure to note obser‐vations both of similarities and differences in the bubbling and foaming among the reactions.
12. Return to the test tubes in Part 1 and make final observations. Record your observations in the Final Obser‐vations Column in Table 1 of the Data section.
Data
Type of Catalyst
Initial Observations Final Observations
Carrot
Tomato
Soil
Yeast
Table 1: Catalyst comparison observations
Lab 22: Catalysts
236
Amount of yeast
Observations
1 mL
5 mL
10 mL
Table 2: Catalyst quantity observations
Post‐lab Questions
1. Classify each catalyst you used as homogeneous or heterogeneous.
Homogeneous catalysts:
Heterogeneous catalysts:
2. Which catalyst made the reaction go the fastest? Is it a homogeneous or heterogeneous catalyst?
Lab 22: Catalysts
237
3. Which catalyst would be the easiest to remove from the water that was formed? Is it a homogeneous or het‐erogeneous catalyst?
4. Which amount of yeast solution you tested would be the best to use? Explain your answer.
Lab 22: Catalysts
Introductory Chemistry
Lab 23: Acids and Bases
241
Lab 23: Acids and Bases
Introduction
Have you ever had a drink of orange juice after brushing your teeth?
What do you taste when you brush your teeth and drink orange juice afterwards? Yuck! It leaves a really bad taste in your mouth. But why? Orange juice and toothpaste by them‐selves taste good. The terrible taste is the result of an acid/base reaction that occurs in your mouth. Orange juice is a weak acid and the toothpaste is a weak base. When they are placed together they neutralize each other and produce a product that is unpleasant to taste. In this lab we will discover how to distinguish between acids and bases.
Two very important classes of compounds are acids and bases. But what exactly makes them different? Acids and bases have physical and chemical differences that you can ob‐serve and test. According to the Arrhenius definition, acids ionize in water to produce a hydronium ion (H3O
+), and bases dissociate in water to produce hydroxide ion (OH‐).
Physical differences between acids and bases can be detected by the senses, including taste and touch. Acids have a sour or tart taste and can produce a stinging sensation to broken skin. For example, if you have ever tasted a lemon, it can often result in a sour face. Bases have a bitter taste and a slippery feel. Soap and many cleaning products are bases. Have you accidentally tasted soap or had it slip out of your hands?
Reactions with acids and bases vary depending on the particular reactants, and acids and bases each react differently with other substances. For example, bases do not react with most metals, but acids will react readily with certain metals to pro‐duce hydrogen gas and an ionic compound—which is referred to as a salt. An example of this type of reaction occurs when magnesium metal reacts with hydrochloric acid. In this reaction, magnesium chloride (a salt) and hydrogen gas are formed.
Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2(g)
metal + acid → a salt + hydrogen gas
Acids may also react with a carbonate or bicarbonate to form carbon dioxide gas and water. The general reaction equation for a reaction between an acid and a carbonate can be represented in this manner:
CO32-
(aq) + 2 H3O+
(aq) → CO2 (g) + 3 H2O (l)
carbonate + acid → carbon dioxide + water
The general equation for a reaction between an acid and a bicarbonate is similar and can be represented in this manner:
Figure 1: Orange juice has a pH of around 3.5. Dairy milk, by comparison, is much less acidic, with a pH of around 6.5.
Concepts to explore:
Understand the properties and reactions of acids and bases
Relate these properties to common household products
242
Lab 23: Acids and Bases
HCO3-
(aq) + H3O+
(aq) → CO2 (g) + 2 H2O (l)
Acids and bases can also react with each other. In this case, the two opposites cancel each other out so that the product formed has neither acidic nor basic (also called alkaline) properties. This type of reaction is called a neutralization reaction. The general equation for the reaction between an acid and a base is represented in this manner:
H3O+ + OH - → 2 H2O
An example of a neutralization reaction is when an aqueous solution of HCl, a strong acid, is mixed with an aqueous solution of NaOH, a strong base. HCl, when dissolved in water, forms H3O
+ and Cl‐. NaOH in water forms Na+ and OH‐. When the two solu‐tions are mixed together the products are water and common table salt (NaCl). Nei‐ther water nor table salt has acid or base properties. Generally this reaction is writ‐ten without the water solvent shown as a reactant:
HCl + NaOH → H2O + NaCl
There is another group of acids called organic acids. Acetic acid found in vinegar and citric acid found in citrus fruit are ex‐amples of organic acids. These acids are all much weaker than HCl. Organic acids have at least one –CO2H group in their molecular formula. When a base is added, the –H of the –CO2H group is replaced just like the –H in HCl. In this lab you will use citric acid as the acid and sodium bicarbonate as the base. Citric acid has three –CO2H groups and only each of the H’s on these groups react with a sodium bicarbonate. The other H’s in the formula do not react. This reaction can be repre‐sented in this manner:
HOC(CO2H)(CH2CO2H)2 + 3 NaHCO3 → HOC(CO2-Na+)(CH2CO2
-Na+ )2 + 3 CO2 + 3 H2O
Acids and bases are measured on a scale called pH. The pH of a substance is defined as the negative log of its hydronium ion concentration. An aqueous (water) solution that has a lot of hydronium ions but very few hydroxide ions is considered to be very acidic, while a solution that contains many hydroxide ions but very few hydronium ions is considered to be very basic.
pH = ‐ log [H3O+]
pH values range from less than 1 to 14, and are measured on a logarithmic scale (equation above). This means that a sub‐stance with a pH of 2 is 10‐times (101) more acidic than a substance with a pH of 3. Similarly, a pH of 7 is 100‐times (102)more basic than a pH of 5. This scale lets us quickly tell if something is very acidic, a little acidic, neutral (neither acidic nor basic), a little basic, or very basic. A pH of 1 is highly acidic, a pH of 14 is highly basic, and a pH of 7 is neutral.
Table 1: Approximate pH of various common foods.
Food pH Range
Lime 1.8 ‐ 2.0
Soft Drinks 2.0 ‐ 4.0
Apple 3.3 ‐ 3.9
Tomato 4.3 ‐ 4.9
Cheese 4.8 ‐ 6.4
Potato 5.6 ‐ 6.0
Drinking Water 6.5 ‐ 8.0
Tea 7.2
Eggs 7.6 ‐ 8.0
bicarbonate + acid → carbon dioxide + water
Acid + Base → Water
243
Lab 23: Acids and Bases
Pre‐lab Questions
1. What is a neutralization reaction?
2. Hydrochloric acid (HCl) is a strong acid. About what pH would you expect it to be?
3. Sodium hydroxide (NaOH) is a strong base. About what pH would you expect it to be?
pH indicators, which change color under a certain pH level, can be used to determine whether a solution is acidic or basic. Litmus paper is made by coating a piece of paper with litmus, which changes color at around a pH of 7. Either red or blue litmus paper can be purchased depending on the experimental needs. Blue litmus paper remains blue when dipped in a base, but turns red when dipped in an acid, while red litmus paper stays red when dipped in an acid, but turns blue when in contact with a base.
A more precise way to determine acidity or basicity is with pH paper. When a substance is placed on pH paper a color ap‐pears, and this color can be matched to a color chart that shows a wide range of pH values. In this way, pH paper allows us to determine to what degree a substance is acidic or basic and can provide an approximate pH value.
244
Lab 23: Acids and Bases
Experiment: Acidity of Common Household Products
In this experiment, we will observe the neutralization of acids and bases using grape juice as an indicator. We will also test common household products for their acidity or alkalinity.
Procedure
Part 1: Acid‐Base Neutralization
1. Label three test tubes 1, 2, and Standard.
2. Prepare 50 mL of a 10% grape juice solution by first pouring 5 mL of grape Juice into a 100 mL graduated cylinder. Add distilled water until the total volume of liquid is 50 mL. Mix well by stirring the solution with a stirring rod.
3. Pour 10 mL of the dilute grape juice solution into each test tube.
4. Note the color of the juice in the test tube labeled Standard in Table 2.
5. Using a pipette, add 15 drops of saturated citric acid solution into test tube 1. Record your observations con‐cerning the color change in Table 2 of the Data section. Use the juice in the test tube labeled standard for com‐parison.
6. Using a pipette, add 15 drops of saturated sodium bicarbonate solution into test tube 2. Record your observa‐tions concerning the color change in Table 2 of the Data section. Use the juice in the test tube labeled standard for comparison.
7. Use pH paper to determine the pH of the solution in each of the 3 test tubes. Record the pH values in Table 2.
8. Using a pipette, add drops of saturated sodium bicarbonate solution to test tube 1 until it returns to its original color. Record your observations in Table 3.
9. Using a pipette, add drops of saturated citric acid solution to test tube 2 until it returns to its original color. Re‐cord your observations in Table 3.
HINT: If the grape juice
is not dilute enough or
the base is not as
strong as needed, you
may continue adding
drops of base.
Materials
Safety Equipment: Safety goggles, gloves Vinegar Household ammonia
Grape Juice** 3 test tubes pH paper
Saturated citric acid solution (60%) Test tube rack Neutral litmus paper
Saturated sodium bicarbonate solution (15%) 2 50 mL beakers Tomato juice
Sodium bicarbonate 12‐well plate Powdered milk
Lemon juice 10 Droppers (pipettes) Baking soda
Dishwashing liquid Stirring rod Distilled water*
*You must provide ** Used in the next lab—refrigerate after opening
245
Lab 23: Acids and Bases
10. Use pH paper to test the pH of the 3 solutions. Record the pH values in Table 3.
Part 2: Testing acidity and basicity of common household products
1. Use the pipettes to place into different wells of your 12‐well plate a couple of drops of each of the following items: tomato juice, household ammonia, milk (mix powdered milk with 50mL water until dissolved), vine‐gar, lemon juice, and diluted dishwashing liquid (mix 1mL dishwashing liquid with 5mL water). Be sure to label or write down where each item is located in the 12‐well plate. CAUTION: Do not contaminate the items being tested. Be sure to use only a clean pipette for each item.
2. Guess the pH of each of the items before you find the experimental value and record your guess in Table 4.
3. Test each item with litmus paper and pH paper. Record your results in Table 4.
4. To clean up rinse all chemicals into a waste beaker. Neutralize the waste to a pH between 4 and 8 using either baking soda or vinegar. Wash the waste solution down the drain.
Data
Test tube 1 Test tube 2 Standard
Step 1 Add acid Add base Neutral
Color
pH value
Test tube 1 Test tube 2 Standard
Step 1 Add base Add acid Neutral
Color
pH value
Product Hypothesized pH Color of Litmus
Paper Color of pH Paper Actual pH
Table 2: Acid‐Base Neutralization for Part 1, Steps 5 & 6 Table 3: Acid‐Base Neutralization for Part 1, Steps 8 & 9
Table 4: Acidity and basicity testing for household products data
pH Paper Key
246
Lab 23: Acids and Bases
Post‐lab Questions
1. Why did the grape juice change color when an acid or base was added?
2. You added a base, sodium bicarbonate, to test tube 1 that contained citric acid and an acid to test tube 2 that con‐tained base. Why did the grape juice return to its original color?
3. Name 2 acids and 2 bases you often use.
Introductory Chemistry
Lab 24: Titration
249
Lab 24: Titration
Introduction
When do you use something known to find something unknown?
Every day we face things that are unknown to us, and every day we use what we do know to determine the unknown. When you wake up in the morning and see that it is cloudy, you speculate that you might need an umbrella because there is a possibility of rain. You do not know that it is going to rain, but you do know that it is cloudy and cloudy days often lead to rainy days. You use what you know to deter‐mine what you do not know. In the same way, in an acid/base titration, we determine the unknown concentration of an acid or base with a known concentration of the opposite acid or base.
You will use candy in this lab exercise—not to eat, but to determine its acid concentration. In general, to find an un‐known one must know how it relates to a known. Remem‐ber when acids and bases are combined, a neutralization reaction occurs. To determine the concentration of acid in an acidic candy solution, a base of known concentration, called a standard solution, is added in small quantities until complete neutralization occurs. This procedure is called an acid‐base titration. When neutralization occurs, the equivalence point of the titration has been reached.
When the standard solution is reacted with an acid or base, how do you know when the neutralization reaction is com‐plete? An additional substance is required for a titration ‐ an acid‐base indicator. An indicator provides a sudden color change when a certain pH is reached. The pH at which the indicator changes color, known as the end point of the titration, depends on the chemical nature of the indicator. An indicator for a titration should be chosen that changes color at a pH near the equivalence point. At the equivalence point, one can assume that the total number of H+ ions donated by the acid equals the total number of H+ ions accepted by the base.
But even more information can be learned from the titrations. You can even calculate how many moles of acid there is per gram of candy if the type of acid in the candy is known. To do this, the following equation is used:
In the above equation:
b b
acid
a x M x Vmoles =
b
Figure 1: Swimming pool test kits are like a small titration ex‐periment. Kits like this first use a phenol red indicator to meas‐ure the acidity of the pool water. Next, a titrant is gradually added by drop; the number of drops tells you how much acid or base mixture needs to be added to the pool water to achieve the ideal pH.
Concepts to explore:
Understand the process of titration.
Use titration with a standard Na2CO3 solution to determine whether Swee‐Tarts® or Smarties® candies requires more standard base per gram of candy to reach the endpoint.
Use titration with a standard Na2CO3 solution to determine whether Swee‐Tarts® or Smarties® candies has more moles of acid per gram of candy.
250
Lab 24: Titration
a = Reaction coefficient of the acid
Mb = Molarity of the basic solution
Vb = Volume of base used in titration converted to Liters
b = Reaction coefficient of the base
To determine the reaction coefficients of the acid and base, you must first balance the reaction equation. The reaction co‐efficients are just the number in front of the acid or the base in the balanced equation. In this experiment, the tart candies you will analyze are SweeTarts® which contain malic acid and Smarties© which contain citric acid. The reactions are shown below.
After the reaction coefficients are determined for a titration, you then calculate how many moles of acid there are in each gram of candy. An example is on the next page.
SweeTarts®: Malic Acid
(acid ↓) (base ↓) (salt↓)
HO2CCH2CH(OH)CO2H + Na2CO3 → Na+O2- CCH2CH(OH)CO2
- Na+ + CO2 + H2O
Notice that for the SweeTarts, one molecule of Na2CO3 is required for every one molecule of malic acid. This is because malic acid has 2 acid groups (‐CO2H). For this reaction shown above, a is 1 and b is 1.
Smarties®: Citric Acid
(acid ↓) (base ↓) (salt↓)
2 HOC(CO2H)(CH2CO2H)2 + 3 Na2CO3 → HOC(CO2- Na+)(CH2CO2
- Na+)2 + 3 CO2 + 3 H2O
For Smarties, 3 NaOH molecules are required for every one citric acid molecule. This is because citric acid has 3 acid groups (‐CO2H). Here, a is 2 and b is 3.
251
Lab 24: Titration
1 L
1000 mL
a x Mb x Vb
b
(1) (0.250 mol/L) (0.00288 L)
1
0.00115 mol
0.455 g
Less concentrated solutions are often prepared by diluting a more concentrated solution. Dilutions of molar concentra‐tions are made by using the equation:
where
M1 = Molarity of the starting, more concentrated solution
V1 = Volume needed of the starting more concentrated solution
M2 = Molarity of the final less concentrated solution
V2 = Final volume wanted of the final less concentrated solution
1 1 2 2M V = M V
Example 1: Calculating the Moles of Acid in SweeTarts®
If 0.455 g of SweeTarts requires 2.88 mL of 0.25 M Na2CO3 to reach the endpoint, how many moles of acid is there in each gram of the candy?
Answer: Since 1 molecule of malic acid requires 1 molecule of Na2CO3 , a = 1 and b = 1.
Next, 2.88 mL of Na2CO3 must be converted to liters of Na2CO3 .
Vb = 2.88 mL x = 0.00288 L
Mb is given to be 0.250 mol/L Na2CO3 .
Solving for moles acid, we get:
Molesacid = = = 0.00115 mol
Molesacid/g candy = = 0.00253 mol acid/g candy
252
In this experiment you will start with 0.500 M Na2CO3 and you will need to make 40.0 ml of 0.250 M Na2CO3. This equation can be used to find out how much of the 0.500 M Na2CO3 is needed. M1 would be 0.500 M, V1 is what needs to be calcu‐lated, M2 is 0.250 M, and V2 is 40.0 mL. The equation then becomes:
One way to make this solution is to first carefully measure out 20.0 mL of 0.50 M Na2CO3 into a 100.0 mL graduated cylin‐der. The volume is brought up to 40.0 mL using distilled water and the solution is mixed well. Notice that the volume is always brought up to the total volume. This is because if you put two different liquids together, the total amount of liquid you will end up with is not always the sum of the two volumes you started with. The molecules may move closer or farther apart.
Graduated cylinders are often accurate enough for many dilutions. However, when very accurate measurements are needed, chemists will use more accurate equipment such as volumetric flasks and volumetric pipettes.
2 3
2 21 1 2 2 1
1
1
2 3
M VM V = M V V =
M
(0.250 M Na CO ) (40 mL)V = = 20.0 mL
0.500 M Na CO
Lab 24: Titration
253
Pre‐lab Questions
1. What is the difference between equivalence points and end points?
2. What would happen if you forgot to put the indicator in?
3. If 2.078 g of Smarties® requires 2.11 mL of 0.250 M Na2CO3 to reach the endpoint, how many moles of acid is there in each gram of the candy?
Lab 24: Titration
254
Lab 24: Titration
Experiment: Titration of Sour Candy
In this experiment, you will dissolve SweeTarts® and Smarties® in water and titrate the solutions with 0.25 M Na2CO3. The end point will be determined with grape juice as the indicator. It changes color at approximately pH 8. You will then deter‐mine which candy requires more Na2CO3 per gram of candy to reach the endpoint. Finally, you will calculate the moles of acid per gram for each type of candy.
Procedure
Part 1: Titration of Grape Juice Indicator
1. Begin heating about 300 mL of distilled water to around 50oC.
2. While the water is heating, prepare the titrant: 40.0 mL of 0.250 M Na2CO3. Do this by first carefully measur‐ing out 20.0 mL of 0.500 M Na2CO3 into a 100 mL graduated cylinder. Use distilled water to bring the total volume up to 40.0 mL. Stir the solution to mix well.
3. Pour the solution into a small beaker and record 0.250 M as the concentration of Na2CO3, Mb, in Table 1.
4. Place a waste beaker below the syringe. Carefully rinse 0.5 mL of the 0.250 M Na2CO3 through the syringe into a waste beaker. Rinse the syringe twice.
5. Remove the syringe piston and place the stopcock on the end. Set the stopcock to close (no liquid drains out). Fill the syringe slightly over the 5.00 mL mark.
6. Open the stopcock and drain the syringe into a waste beaker until the bottom of the meniscus or curve touches the 5.00 mL mark. Close the stopcock. HINT: Make sure there are no air bubbles trapped in the sy‐ringe or the stopcock. If there are, drain the syringe into the waste beaker until the bubbles are washed out, then refill the syringe and repeat Step 6.
7. Record 5.00 mL as the initial syringe volume in Table 1.
8. Add 50 mL of distilled water and 5.00 mL of grape juice into an Erlenmeyer flask. Place the Erlenmeyer flask directly under the syringe/stopcock.
Materials
Safety Equipment: Safety goggles, gloves
0.5 M Na2CO3 2 Erlenmeyer flasks
Smarties® candies (1 pack) Scale
SweeTarts® candies (8 candies) Stirring rod
Grape juice Mortar and pestle
Syringe Funnel
Stopcock Dropper (pipette)
2 250 mL beakers Distilled water (heated, 50°C)*
*You must provide
255
Lab 24: Titration
9. Hold the syringe over the flask, and open the stopcock to begin to add 0.250 M Na2CO3 solution from the syringe drop by drop using a pipette.
10. The moment that the solution starts to turn green, close the stopcock and swirl the flask. Set the syringe in an empty beaker if needed, being careful not to lose any of the titrant.
11. Continue the titration by adding one drop at a time, then swirling the flask. When it looks like the color change remains after swirling, stop the titration.
HINT: You want to add just barely enough of the 0.250 M Na2CO3 for the color change to stay. This can be hard to see. When you think you are there, look at the syringe and see exactly how much you have added (to the 0.00 mL decimal place). Remember to take the syringe reading where the bottom of the meniscus is seen. Add 1 more drop while looking into the flask where it falls. Be sure you have your goggles on! If there is no color change where the drop falls, you are finished and record the reading you took before adding the last drop. If there is a color change, swirl the flask to mix and then repeat adding a drop at a time and swirling the flask until there is no color change.
12. Record the final syringe volume in Table 1 under Grape Juice Trial 1. Record your measurement to the 0.00 mL decimal place.
13. Calculate the volume of Na2CO3 required to titrate the grape juice, and record this value in Table 1. This is the difference between the initial and final syringe volumes. Maintain a precision to the 0.00 mL decimal place.
14. Fill the syringe to 5.00 mL and repeat the procedure. Record this data under Grape Juice Trial 2.
15. Average the results of the total volume of Na2CO3 required for the titration, and record this value in Table 1.
Part 2: Titration of Candy
1. Crush 8 SweeTarts candies using the mortar and pestle. Weigh between 2.5 and 3.0 g of crushed candy and place into an Erlenmeyer flask. Record the exact mass in Table 2. HINT: Candies that are of a lighter color work better.
2. Add about 50 mL of hot distilled water to the flask with the candies. Gently crush and stir the candy with a stirring rod until the candy is completely dissolved. Allow the solution to cool.
3. Use your 10 mL graduated cylinder to add 5.00 mL of grape juice into the Erlenmeyer flask.
4. Titrate using the 0.250 M Na2CO3 in the same way as in Part 1, except look for a definite color change. HINT: The color may vary depending on the original color of the solution.
5. Once a definite color change has been achieved, record the final amount of 0.250 M Na2CO3 (to the 0.00 mL place) in Table 1 under Trial 1 for the SweeTarts®.
6. Refill the syringe to exactly 5.00 mL and repeat the procedure for the SweeTarts®. Record the data in Table 2 under Trial 2 for the SweeTarts®.
7. Repeat this procedure for the two Smarties® trials. Using the mortar and pestle, crush approximately 15 Smarties©, or close to 1 pack. Weigh between 2.5 and 3.0 g of powder into a flask, and titrate as before.
8. To clean up, pour solutions down the drain.
9. Calculate the volume of Na2CO3 required for each of the trials in Part 2. This is the difference between the initial and final syringe volumes.
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Data
Measurement SweeTarts®
Trial 1
SweeTarts®
Trial 2
Smarties®
Trial 1
Smarties®
Trial 2
Mass of candy (g)
Concentration of Na2CO3, Mb
(on the bottle)
Initial syringe
volume (mL)
Final syringe
volume (mL)
Volume Na2CO3 solu‐tion required, Vb
(mL)
Table 2: Titration of candy
Lab 24: Titration
Measurement Grape Juice Trial 1 Grape Juice Trial 2
Molarity of Na2CO3
Initial syringe volume (mL)
Final syringe volume (mL)
Volume Na2CO3
required (mL)
Average volume Na2CO3
required (mL)
Table 1: Titration of grape juice indicator
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Calculations
1. Determine the volume of Na2CO3 used to titrate only the candy. HINT: Take the Volume of Na2CO3 used in Part 2 (candy and juice mixture) and subtract the volume required to titrate the grape juice alone (Part 1 average).
SweeTarts Trial 1:
SweeTarts Trial 2:
Smarties Trial 1:
Smarties Trial 2:
2. Determine the amount of Na2CO3 used per gram of candy (mL NaOH/g). HINT: Divide the volumes obtained above by the mass of each candy sample.
SweeTarts Trial 1:
SweeTarts Trial 2:
Smarties Trial 1:
Smarties Trial 2:
Lab 24: Titration
258
3. Determine the average amount of base used per gram of candy for each set of trials (mL Na2CO3/g).
Sweet Tarts Average:
Smarties Average:
4. Determine the molesacid per gram of candy. HINT: This is found by dividing the moles of acid found for each type of candy by the mass of the candy initially weighed out and recorded in the data table. Complete in the calculation in a stepwise process.
Moles of acid SweeTarts® Trial 1:
a. Write the balanced reaction equation.
b. Determine the mol to mol ratio of the acid to base (What are the values a and b?).
Lab 24: Titration
259
c. Apply the equation using your data in Table 1. (HINT: molesacid = a x Mb x Vb / b . Be sure to convert mL to L.)
d. Divide the number of moles of acid by the mass in grams of the candy used.
Moles of acid in SweeTarts® Trial 2:
Repeat Steps a through d for Trial 2.
Find the average of the two trials: Average Moles acid =
Lab 24: Titration
260
Moles of acid in Smarties® Trial 1:
Repeat Steps a through c for Smarties Trial 1.
Moles of acid in Smarties® Trial 2:
Repeat Steps a through c for Smarties Trial 2.
Find the average of the two trials: Average Moles acid =
Lab 24: Titration
261
Lab 24: Titration
Post‐lab Questions
1. Which candy required more base per gram of candy? Is this what you expected? Explain your answer.
2. Which candy contained more moles of acid per gram of candy? Is this what you expected? Explain your an‐swer.
3. How would your Na2CO3/g of candy reported change if there was a large air bubble in the stopcock tip when you started your titration?
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