Chapter 1

Chemistry: Methods and Measurement

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

1.1 The Discovery Process

• Chemistry - The study of matter…– Matter - Anything that has mass and

occupies space• A table• A piece paper

– What about air?• Yes, it is matter

1.1

The

Dis

cove

ry P

roce

ssChemistry:

• the study of matter• its chemical and physical properties• the chemical and physical changes it

undergoes• the energy changes that accompany

those processes

• Energy - the ability to do work to accomplish some change

1.1

The

Dis

cove

ry P

roce

ss MAJOR AREAS OF CHEMISTRY

• Biochemistry - the study of life at the molecular level

• Organic chemistry - the study of matter containing carbon and hydrogen

• Inorganic chemistry - the study of matter containing elements, not organic

• Analytic chemistry - analyze matter to determine identity and composition

1.1

The

Dis

cove

ry P

roce

ss• Physical chemistry - attempts to

explain the way matter behaves

CHEMISTRY

medical practitioners

pharmaceutical industry

forensic sciences

food science

public health

1.1

The

Dis

cove

ry P

roce

ss THE SCIENTIFIC METHOD• The scientific method - a systematic

approach to the discovery of new information

Characteristics of the scientific process Observation Formulation of a question Pattern recognition Developing theories Experimentation Summarizing information

1.1

The

Dis

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roce

ss

Models in Chemistry• To aid in understanding of

a chemical unit or system– a model is often used– good models are based on

everyday experience• Ball and stick methane

model– color code balls– sticks show attractive forces

holding atoms together

1.1

The

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ss

1.2 Matter and Properties

• Properties - characteristics of matter– chemical vs. physical

• Three states of matter1. gas - particles widely separated, no definite

shape or volume solid2. liquid - particles closer together, definite

volume but no definite shape3. solid - particles are very close together, define

shape and definite volume

Three States of Water

(a) Solid (b) Liquid (c) Gas

Comparison of the Three Physical States

1.2

Mat

ter a

nd P

rope

rties

1.2

Mat

ter a

nd P

rope

rties • Physical property - is observed

without changing the composition or identity of a substance

• Physical change - produces a recognizable difference in the appearance of a substance without causing any change in its composition or identity- conversion from one physical state to

another- melting an ice cube

Separation by Physical Properties

Magnetic iron is separated from other nonmagnetic substances, such as sand. This property is used as a large-scale process in the recycling industry.

1.2

Mat

ter a

nd P

rope

rties • Chemical property - result in a

change in composition and can be observed only through a chemical reaction

• Chemical reaction (chemical change) - a process of rearranging, removing, replacing, or adding atoms to produce new substances

hydrogen + oxygen water

reactants products

1.2

Mat

ter a

nd P

rope

rties Classify the following as either a

chemical or physical property:

a. Color

b. Flammability

c. Hardness

d. Odor

e. Taste

1.2

Mat

ter a

nd P

rope

rties Classify the following as either a

chemical or physical change:

a. Boiling water becomes steam

b. Butter turns rancid

c. Burning of wood

d. Mountain snow pack melting in spring

e. Decay of leaves in winter

• Intensive properties - a property of matter that is independent of the quantity of the substance- Density- Specific gravity

• Extensive properties - a property of matter that depends on the quantity of the substance- Mass- Volume1.

2 M

atte

r and

Pro

perti

es

1.2

Mat

ter a

nd P

rope

rties

• Pure substance - a substance that has only one component

• Mixture - a combination of two or more pure substances in which each substance retains its own identity, not undergoing a chemical reaction

Classification of Matter

1.2

Mat

ter a

nd P

rope

rties

• Element - a pure substance that cannot be changed into a simpler form of matter by any chemical reaction

• Compound - a substance resulting from the combination of two or more elements in a definite, reproducible way, in a fixed ratio

Classification of Matter

1.2

Mat

ter a

nd P

rope

rties

• Mixture - a combination of two or more pure substances in which each substance retains its own identity

• Homogeneous - uniform composition, particles well mixed, thoroughly intermingled

• Heterogeneous – nonuniform composition, random placement

Classification of Matter

Classes of Matter1.

2 M

atte

r and

Pro

perti

es

1.3 Significant Figures and Scientific Notation

• Information-bearing digits or figures in a number are significant figures

• The measuring devise used determines the number of significant figures a measurement has

• The amount of uncertainty associated with a measurement is indicated by the number of digits or figures used to represent the information

1.3

Sign

ifica

nt F

igur

es a

nd

Scie

ntifi

c N

otat

ion

Significant figures - all digits in a number representing data or results that are known with certainty plus one uncertain digit

1.3

Sign

ifica

nt F

igur

es a

nd

Scie

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c N

otat

ion

Recognition of Significant Figures

• All nonzero digits are significant• 7.314 has four significant digits

• The number of significant digits is independent of the position of the decimal point• 73.14 also has four significant digits

• Zeros located between nonzero digits are significant• 60.052 has five significant digits

1.3

Sign

ifica

nt F

igur

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nd

Scie

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ion

Use of Zeros in Significant Figures

• Zeros at the end of a number (trailing zeros) are significant if the number contains a decimal point.• 4.70 has three significant digits

• Trailing zeros are insignificant if the number does not contain a decimal point.• 100 has one significant digit; 100. has three

• Zeros to the left of the first nonzero integer are not significant.• 0.0032 has two significant digits

1.3

Sign

ifica

nt F

igur

es a

nd

Scie

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c N

otat

ion

How many significant figures are in the following?

1. 3.400

2. 3004

3. 300.

4. 0.003040

1.3

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nt F

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nd

Scie

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ion

Scientific Notation• Used to express very large or very small

numbers easily and with the correct number of significant figures

• Represents a number as a power of ten

• Example:

4,300 = 4.3 x 1,000 = 4.3 x 103

• To convert a number greater than 1 to scientific notation, the original decimal point is moved x places to the left, and the resulting number is multiplied by 10x

• The exponent x is a positive number equal to the number of places the decimal point moved

5340 = 5.34 x 104

• What if you want to show the above number has four significant figures?

= 5.340 x 104

1.3

Sign

ifica

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igur

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nd

Scie

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ion

• To convert a number less than 1 to scientific notation, the original decimal point is moved xplaces to the right, and the resulting number is multiplied by 10-x

• The exponent x is a negative number equal to the number of places the decimal point moved

0.0534 = 5.34 x 10-2

1.3

Sign

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nd

Scie

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ion

Types of Uncertainty• Error - the difference

between the true value and our estimation– Random– Systematic

• Accuracy - the degree of agreement between the true value and the measured value

• Precision - a measure of the agreement of replicate measurements1.

3 Si

gnifi

cant

Fig

ures

and

Sc

ient

ific

Not

atio

n

1.3

Sign

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Scie

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ion

correct answer 152.83 liters

Significant Figures in Calculation of Results

Rules for Addition and Subtraction• The result in a calculation cannot have greater

significance than any of the quantities that produced the result

• Consider:

37.68 liters6.71862 liters

108.428 liters152.82662 liters

1.3

Sign

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nd

Scie

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)calculator(on 109688692.210255.2

)94.15(102.4 84

3−

− ×=×

×

Which number has the fewest significant figures? 4.2 x 103 has only 2

The answer is therefore, 3.0 x 10-8

Rules for Multiplication and Division

• The answer can be no more precise than the least precise number from which the answer is derived

• The least precise number is the one with the fewest significant figures

1.3

Sign

ifica

nt F

igur

es a

nd

Scie

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ion

Exact and Inexact Numbers• Inexact numbers have uncertainty by

definition

• Exact numbers are a consequence of counting

• A set of counted items (beakers on a shelf) has no uncertainty

• Exact numbers by definition have an infinite number of significant figures

1.3

Sign

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Scie

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ion

=3.35 x 104

Rules for Rounding Off Numbers

• When the number to be dropped is less than 5 the preceding number is not changed

• When the number to be dropped is 5 or larger, the preceding number is increased by one unit

• Round the following number to 3 significant figures: 3.34966 x 104

How Many Significant Figures?

Round off each number to 3 significant figures:

1. 61.40

2. 6.171

3. 0.066494

1.3

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1.4 Units and Unit ConversionData, Results, and Units• Data - each piece is an individual result of a single

measurement or observation– mass of a sample– temperature of a solution

• Results - the outcome of the experiment• Data and results may be identical, however usually

related data are combined to generate a result• Units - the basic quantity of mass, volume or

whatever quantity is being measured– A measurement is useless without its units

1.4

Uni

ts a

nd U

nit

Con

vers

ion

English and Metric Units• English system - a collection of

functionally unrelated units – Difficult to convert from one unit to another – 1 foot = 12 inches = 0.33 yard = 1/5280 miles

• Metric System - composed of a set of units that are related to each other decimally, systematic – Units relate by powers of tens– 1 meter = 10 decimeters = 100 centimeters =

1000 millimeters

1.4

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Con

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ion

Basic Units of the Metric System

Mass gram gLength meter mVolume liter L

• Basic units are the units of a quantity without any metric prefix

1.4

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Con

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1.4

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Con

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UNIT CONVERSION

• You must be able to convert between units

- within the metric system - between the English system and metric system

• The method used for conversion is called the Factor-Label Method or Dimensional Analysis

!!!!!!!!!!! VERY IMPORTANT !!!!!!!!!!!

1.4

Uni

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nd U

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Con

vers

ion

• Let your units do the work for you by simply memorizing connections between units.– For example: How many donuts are in

one dozen? – We say: “Twelve donuts are in a dozen.”– Or: 12 donuts = 1 dozen donuts

• What does any number divided by itself equal?

• ONE! 1 dozen 1donuts 12 =

1.4

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nd U

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Con

vers

ion

1 dozen 1donuts 12 =

• This fraction is called a unit factor

• What does any number times one equal?

• That number• Multiplication by a unit factor does

not change the amount – only the unit

• We use these two mathematical facts to use the factor label method– a number divided by itself = 1– any number times one gives that number

back• Example: How many donuts are in 3.5

dozen?• You can probably do this in your head

but try it using the Factor-Label Method.1.4

Uni

ts a

nd U

nit

Con

vers

ion

1.4

Uni

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Con

vers

ion

Start with the given information...

3.5 dozen

Then set up your unit factor...

dozen 1donuts 12×

See that the units cancel...

Then multiply and divide all numbers...

= 42 donuts

1.4

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ts a

nd U

nit

Con

vers

ion

Common English System Units

• Convert 12 gallons to units of quarts

1.4

Uni

ts a

nd U

nit

Con

vers

ion

Intersystem Conversion Units

• Convert 4.00 ounces to kilograms

1.4

Uni

ts a

nd U

nit

Con

vers

ion 1. Convert 5.5 inches to millimeters

2. Convert 50.0 milliliters to pints

3. Convert 1.8 in2 to cm2

1.5 Experimental Quantities

• Mass - the quantity of matter in an object– not synonymous with weight– standard unit is the gram

• Weight = mass x acceleration due to gravity

• Mass must be measured on a balance (not a scale)

1.5

Expe

rimen

tal Q

uant

ities

• Units should be chosen to suit the quantity described – A dump truck is measured in tons– A person is measured in kg or pounds– A paperclip is measured in g or ounces– An atom?

• For atoms, we use the atomic mass unit (amu)– 1 amu = 1.661 x 10-24 g

1.5

Expe

rimen

tal Q

uant

ities

• Length - the distance between two points– standard unit is the meter– long distances are measured in km– distances between atoms are measured in nm,

1 nm = 10-9 m• Volume - the space occupied by an object

– standard unit is the liter– the liter is the volume occupied by 1000

grams of water at 4 oC– 1 mL = 1/1000 L = 1 cm3

1.5

Expe

rimen

tal Q

uant

ities

The milliliter (mL) and the cubic centimeter (cm3) are equivalent

1.5

Expe

rimen

tal Q

uant

ities • Time

- metric unit is the second

• Temperature - the degree of “hotness” of an object

1.5

Expe

rimen

tal Q

uant

ities

1.832-FC

oo =

32C)(1.8F oo +×=

1. Convert 75oC to oF

2. Convert -10oF to oC

1. Ans. 167 oF 2. Ans. -23oC

Conversions Between Fahrenheit and Celsius

1.5

Expe

rimen

tal Q

uant

ities

K = oC + 273

Kelvin Temperature Scale

• The Kelvin scale is another temperature scale.

• It is of particular importance because it is directly related to molecular motion.

• As molecular speed increases, the Kelvin temperature proportionately increases.

1.5

Expe

rimen

tal Q

uant

ities

• Energy - the ability to do work• kinetic energy - the energy of motion

• potential energy - the energy of position (stored energy)

• Energy is also categorized by form:• light• heat• electrical• mechanical• chemical

Energy

1.5

Expe

rimen

tal Q

uant

ities Characteristics of Energy

• Energy cannot be created or destroyed

• Energy may be converted from one form to another

• Energy conversion always occurs with less than 100% efficiency

• All chemical reactions involve either a “gain” or “loss” of energy

1.5

Expe

rimen

tal Q

uant

ities Units of Energy

• Basic Units:• calorie or joule• 1 calorie (cal) = 4.184 joules (J)

• A kilocalorie (kcal) also known as the large Calorie. This is the same Calorie as food Calories.• 1 kcal = 1 Calorie = 1000 calories

• 1 calorie = the amount of heat energy required to increase the temperature of 1 gram of water 1oC.

1.5

Expe

rimen

tal Q

uant

ities Concentration

Concentration:– the number of particles of a substance

– the mass of those particles

– that are contained in a specified volume

Often used to represent the mixtures of different substances– Concentration of oxygen in the air

– Pollen counts

– Proper dose of an antibiotic

1.5

Expe

rimen

tal Q

uant

ities

Vmd ==

volumemass

Density and Specific Gravity• Density

– the ratio of mass to volume– an extensive property– use to characterize a substance as

each substance has a unique density

– Units for density include:• g/mL• g/cm3

• g/cc

1.5

Expe

rimen

tal Q

uant

ities

liquid mercury

brass nutwater

cork

Calculating the Density of a Solid

• 2.00 cm3 of aluminum are found to weigh 5.40g. Calculate the density of aluminum in units of g/cm3.– Use the formula– Substitute our values

5.40 g2.00 cm3

= 2.70 g / cm3

1.5

Expe

rimen

tal Q

uant

ities

Vmd ==

volumemass

1.5

Expe

rimen

tal Q

uant

ities

Air has a density of 0.0013 g/mL. What is the mass of 6.0-L sample of air?

Calculate the mass in grams of 10.0 mL if mercury (Hg) if the density of Hg is 13.6 g/mL.

Calculate the volume in milliliters, of a liquid that has a density of 1.20 g/mL and a mass of 5.00 grams.

1.5

Expe

rimen

tal Q

uant

ities

(g/mL) water ofdensity (g/mL)object ofdensity gravity specific =

Specific Gravity• Values of density are often related to a standard

• Specific gravity - the ratio of the density of the object in question to the density of pure water at 4oC

• Specific gravity is a unitless term because the 2 units cancel

• Often the health industry uses specific gravity to test urine and blood samples

Chapter 2

The Structure of the Atom and the Periodic Table

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2.1 Composition of the Atom

• Atom - the basic structural unit of an element

• The smallest unit of an element that retains the chemical properties of that element

2.1

Com

posi

tion

of th

e Ato

m

• Nucleus - small, dense, positively charged region in the center of the atom

- protons - positively charged particles

- neutrons - uncharged particles

Electrons, Protons and Neutrons• Atoms consist of three primary particles

• electrons• protons• neutrons

2.1

Com

posi

tion

of th

e Ato

m Characteristics of Atomic Particles

• Electrons are negatively charged particles located outside of the nucleus of an atom

• Protons and electrons have charges that are equal in magnitude but opposite in sign

• A neutral atom that has no electrical charge has the same number of protons and electrons

• Electrons move very rapidly in a relatively large volume of space while the nucleus is small a dense

MassNumber

Atomic Number

Charge of particle

Symbol of the atom

2.1

Com

posi

tion

of th

e Ato

m Symbolic Representation of an Element

CAZ X

• Atomic number (Z) - the number of protons in the atom

• Mass number (A) - sum of the number of protons and neutrons

Atomic Calculations

number of protons + number of neutrons = mass number

2.1

Com

posi

tion

of th

e Ato

m

number of neutrons = mass number - number of protons

number of protons = number of electrons IF positive and negative charges cancel, the atom Charge = 0

Name Charge Mass(amu) Mass (grams)Electrons (e) -1 5.4 x 10-4 9.1095 x 10-28

Protons (p) +1 1.00 1.6725 X 10-24

Neutrons (n) 0 1.00 1.6750 x 10-24

Selected Properties of the Three Basic Subatomic Particles

2.1

Com

posi

tion

of th

e Ato

m

Calculate the number of protons, neutrons and electrons in each of the following:

B115

Fe5526

2.1

Com

posi

tion

of th

e Ato

m Determining the Composition of an Atom

4

Hydrogen(Hydrogen - 1)

Deuterium(Hydrogen - 2)

Tritium(Hydrogen - 3)

2.1

Com

posi

tion

of th

e Ato

m

Isotopes of Hydrogen

• Isotopes - atoms of the same element having different masses– contain same number of protons– contain different numbers of neutrons

Isotopes

2.1

Com

posi

tion

of th

e Ato

m Isotopic Calculations• Isotopes of the same element have identical

chemical properties

• Some isotopes are radioactive

• Find chlorine on the periodic table

• What is the atomic number of chlorine?17

• What is the mass given35.45

• This is not the mass number of an isotope

2.1

Com

posi

tion

of th

e Ato

m Atomic Mass• What is this number, 35.34?

• The atomic mass - the weighted average of the masses of all the isotopes that make up chlorine

• Chlorine consists of chlorine-35 and chlorine-37 in a 3:1 ratio

• Weighted average is an average corrected by the relative amounts of each isotope present in nature

2.1

Com

posi

tion

of th

e Ato

m Determining Atomic Mass

Calculate the atomic mass of naturally occurring chlorine if 75.77% of chlorine atoms are chlorine-35 and 24.23% of chlorine atoms are chlorine-37

Step 1: convert the percentage to a decimal fraction

0.7577 chlorine-35

0.2423 chlorine-37

Step 2: Multiply the decimal fraction by the mass of that isotope to obtain the isotope contribution to the atomic mass.

For chlorine-35:0.7577 x 35.00 amu = 26.52 amu

For chlorine-370.2423 x 37.00 amu = 8.965 amu

Step 3: sum these partial weights to get the weighted average atomic mass of chlorine:

26.52 amu + 8.965 amu = 35.49 amu

2.1

Com

posi

tion

of th

e Ato

m

2.1

Com

posi

tion

of th

e Ato

m Atomic Mass Determination• Nitrogen consists of two naturally occurring

isotopes– 99.63% nitrogen-14 with a mass of 14.003 amu– 0.37% nitrogen-15 with a mass of 15.000 amu

• What is the atomic mass of nitrogen?

2.1

Com

posi

tion

of th

e Ato

m Ions

• Ions - electrically charged particles that result from a gain or loss of one or more electrons by the parent atom

• Cation - positively charged– result from the loss of electrons– 23Na 23Na+ + 1e-

• Anion - negatively charged– results from the gain of electrons– 19F + 1 e- 19F-

2.2 Development of Atomic Theory

• Dalton’s Atomic Theory - the first experimentally based theory of atomic structure of the atom.

2.2

Dev

elop

men

t of A

tom

ic

Theo

ryPostulates of Dalton’s Atomic Theory

1. All matter consists of tiny particles called atoms

2. An atom cannot be created, divided, destroyed, or converted to any other type of atom

3. Atoms of a particular element have identical properties

4. Atoms of different elements have different properties

5. Atoms of different elements combine in simple whole-number ratios to produce compounds (stable aggregates of atoms)

6. Chemical change involves joining, separating, or rearranging atomsPostulates 1, 4, 5 and 6 are still regarded as true.2.

2 D

evel

opm

ent o

f Ato

mic

Th

eory

• Electrons were the first subatomic particles to be discovered using the cathode ray tube

Indicated that the particles were negatively charged.2.

2 D

evel

opm

ent o

f Ato

mic

Th

eory

Evidence for Subatomic Particles: Electrons, Protons and Neutrons

2.2

Dev

elop

men

t of A

tom

ic

Theo

ryEvidence for Protons and Neutrons

• Protons were the next particle to be discovered, by Goldstein– Protons have the same size charge but opposite in sign– Proton is 1837 times as heavy as electron

• Neutrons – Postulated to exist in 1920’s but not demonstrated to

exist until 1932– Almost the same mass as the proton

2.2

Dev

elop

men

t of A

tom

ic

Theo

ryEvidence for the Nucleus

• Initial assumed protons and neutrons were uniformly distributed throughout the atom

• Earnest Rutherford’s “Gold Foil Experiment” lead to the understanding of the nucleus– Most alpha particles pass through the foil

without being deflected

– Some particles were deflected, a few even directly back to the source

Rutherford’s Gold Foil Experiment

• Most of the atom is empty space• The majority of the mass is located in a

small, dense region

2.2

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Theo

ry

Models of the Atom

(a) Thomson (b) Rutherford2.3

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2.3 Light, Atomic Structure, and the Bohr Atom

• Rutherford’s atom – tiny, dense, positively charged nucleus of protons surrounded by electrons

• How do we describe the relationship of the electrons to each other and the nucleus?

• Use the measurement of particle energy rather than position

Light and Atomic Structure

• Spectroscopy - absorption or emission of light by atoms. – Used to understand the electronic structure.

• To understand the electronic structure, we must first understand light, Electromagnetic Radiation– travels in waves from a source– speed of 3.0 x 108 m/s

2.3

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Wavelengths2.

3 Li

ght,

Ato

mic

Stru

ctur

e,

and

the

Boh

r Ato

m • Light is propagated (moves) as a collection of sine waves

• Wavelength is the distance between identical points on successive waves

• Each wavelength travels at the same velocity, but has its own characteristic energy

high energyshort wavelength

low energylong wavelength

Electromagnetic Spectrum

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Bohr Theory• Atoms can absorb and emit energy via

promotion of electrons to higher energy levels and relaxation to lower levels

• Energy that is emitted upon relaxation is observed as a single wavelength of light

• Spectral lines are a result of electron transitions between allowed levels in the atoms

• emission spectrum - light emitted when a substance is excited by an energy source.

The emission spectrum of hydrogen lead to the modern understanding of the electronic structure of the atom2.

3 Li

ght,

Ato

mic

Stru

ctur

e,

and

the

Boh

r Ato

m

Electrons exist in fixed energy levels surrounding the nucleus

Quantization of energy

Promotion of electron occurs as it absorbs energy

Excited State

Energy is released as the electron travels back to lower levels

Relaxation

The Bohr Atom

Electronic Transitions• Amount of energy absorbed in jumping

from one energy level to a higher energy level is a precise quantity

• Energy of that jump is the energy difference between the orbits involved

• Orbit - what Bohr called the fixed energy levels

• Ground state - the lowest possible energy state

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2.3

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Bohr Theory• Allowed levels are quantized energy levels,

orbits• Electrons are found only in these energy levels• Highest-energy orbits are farthest from the

nucleus• Atoms

– absorb energy by excitation of electrons to higher energy levels

– release energy by relaxation of electrons to lower energy levels

• Energy differences may be calculated from the wavelength of light emitted

Modern Atomic Theory• Bohr’s model of the atom when applied to

atoms with more than one electron failed to explain their line spectra

• One major change from Bohr’s model is that electrons do not move in orbits

• Atomic orbitals - regions in space with a high probability of finding an electron

• Electrons move rapidly within the orbital giving a high electron density

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2.4 The Periodic Law and the Periodic Table

• Dmitri Mendeleev and Lothar Meyer - two scientists working independently developed the precursor to our modern Periodic Table.

• They noticed that as you list elements in order of atomic mass, there is a distinct regular variation of their properties.

• Periodic Law - the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Classification of the Elements2.

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Parts of the Periodic Table• Period – a horizontal row of elements in

the periodic table. They contain 2, 8, 8, 18, 18, and 32 elements,

• Group – also called families are columns of elements in the periodic table.

• Elements in a particular group or family share many similarities, as in a human family.2.

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Category Classification of Elements

• Metals - elements that tend to lose electrons during chemical change, forming positive ions.

• Nonmetals - a substance whose atoms tend to gain electrons during chemical change, forming negative ions.

• Metalloids - have properties intermediate between metals and nonmetals.

Classification of Elements Metals

• Metals: – A substance whose atoms tend to lose

electrons during chemical change– Elements found primarily in the left 2/3 of

the periodic table• Properties:

– High thermal and electrical conductivities– High malleability and ductility– Metallic luster– Solid at room temperature

2.4

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Classification of Elements Nonmetals

• Nonmetals: – A substance whose atoms may gain

electrons, forming negative ions– Elements found in the right 1/3 of the

periodic table• Properties:

– Brittle– Powdery solids or gases– Opposite of metal properties2.

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2.4

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Law

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Atomic Number and Atomic Mass

• Atomic Number: – The number of protons in the nucleus of

an atom of an element

– Nuclear charge or positive charge from the nucleus

• Most periodic tables give the element symbol, atomic number and atomic mass

2.4

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Element Information in the Periodic Table20 atomic number

Ca symbolCalcium name

40.08 atomic mass

Using the Periodic Table• Identify the group and period to

which each of the following belongs:a. P b. Cr c. Element 30

• How many elements are found in period 6?

• How many elements are in group VA?2.

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2.5 Electron Arrangement and the Periodic Table

• The electron arrangement is the primary factor in understanding how atoms join together to form compounds

• Electron configuration - describes the arrangement of electrons in atoms

• Valence electrons - outermost electrons– The electrons involved in chemical bonding

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Valence Electrons• The number of valence electrons is

the group number for the representative elements

• The period number gives the energy level (n) of the valence shell for all elements

Valence Electrons and Energy Level

• How many valence electrons does Fluorine have?– 7 valence electrons

• What is the energy level of these electrons?– Energy level is n = 2

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Valence Electrons - Detail• What is the total number of electrons in

fluorine? – Atomic number = 9– 9 protons and 9 electrons

• 7 electrons in the valence shell, (n = 2 energy level), so where are the other two electrons?– In n = 1 energy level– Level n=1 holds only two electrons

Determining Electron ArrangementList the total number of electrons, total number of

valence electrons, and energy level of the valence electrons for silicon.

1. Find silicon in the periodic table• Group IVA • Period 3• Atomic number = 14

2. Atomic number = number of electrons in an atom

• Silicon has 14 electrons2.5

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Determining Electron Arrangement #2List the total number of electrons, total number of valence

electrons, and energy level of the valence electrons for silicon.

3. As silicon is in Group IV, only 4 of its 14 electrons are valence electrons

• Group IVA = number of valence electrons4. Energy levels:

• n = 1 holds 2 electrons• n = 2 holds 8 electrons (total of 10)

• n = 3 holds remaining 4 electrons (total = 14)

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Determining Electron ArrangementPractice

List the total number of electrons, total number of valence electrons, and energy level of the valence electrons for:

• Na• Mg• S• Cl• Ar

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The Quantum Mechanical Atom

• Bohr’s model of the hydrogen atom didn’t clearly explain the electron structure of other atoms– Electrons in very specific locations,

principal energy levels

– Wave properties of electrons conflict with specific location

• Schröedinger developed equations that took into account the particle nature and the wave nature of the electrons

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• Equations that determine the probabilityof finding an electron in specific regionin space, quantum mechanics

– Principle energy levels (n = 1,2,3…)

– Each energy level has one or more sublevels or subshells (s, p, d, f)

– Each sublevel contains one or more atomic orbitals

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PRINCIPAL ENERGY LEVELS

• n = 1, 2, 3, …

• The larger the value of n, the higher the energy level and the farther away from the nucleus the electrons are

• The number of sublevels in a principal energy level is equal to n– in n=1, there is one sublevel

– in n = 2, there are two sublevels

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Principal Energy Levels• The electron capacity of a principal

energy level (or total electrons it can hold) is 2(n)2

– n = 1 can hold 2(1)2 = 2 electrons

– n = 2 can hold 2(2)2 = 8 electrons

• How many electrons can be in the n = 3 level?– 2(3)2 = 18

• Compare the formula with periodic table…..

n=1, 2(1)2=2

n=2, 2(2)2=8n=3, 2(3)2=18

n=4, 2(4)2=32

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Sublevels• Sublevel: a set of energy-equal orbitals

within a principal energy level

• Subshells increase in energy:

s < p < d < f• Electrons in 3d subshell have more energy

than electrons in the 3p subshell

• Specify both the principal energy level and a subshell when describing the location of an electron

Principle energylevel (n)

Possiblesubshells

1 1s

2 2s, 2p

3 3s, 3p, 3d

4 4s, 4p, 4d, 4f

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• Orbital - a specific region of a sublevel containing a maximum of two electrons

• Orbitals are named by their sublevel and principal energy level– 1s, 2s, 3s, 2p, etc.

• Each type of orbital has a characteristic shape– s is spherically symmetrical

– p has a shape much like a dumbbell

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• s is spherically symmetrical

• Each p has a shape much like a dumbbell, differing in the direction extending into space

Subshell Number oforbitals

s 1

p 3

d 5

f 7

• How many electrons can be in the 4d subshell?

•10

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Quantum Mechanical Model• Each orbital within a

sublevel contains a maximum of 2 electrons

• Energy increases as n, shell number increases, but ALSO increases as move from s to p to d to f sublevels

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Incr

easi

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nerg

y4s

4p

4d

4f

••

•• •• ••

•• •• •• •• ••

••••••••••••••

Electron

Orbital

Sublevel

Shell 4

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Electron Spin• Electron Configuration - the

arrangement of electrons in atomic orbitals• Aufbau Principle - or building up

principle helps determine the electron configuration– Electrons fill the lowest-energy orbital that is

available first– Remember s<p<d<f in energy– When the orbital contains two electrons, the

electrons are said to be paired

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Rules for Writing Electron Configurations

• Obtain the total number of electrons in the atom from the atomic number

• Electrons in atoms occupy the lowest energy orbitals that are available – 1s first

• Each principal energy level, n contains only nsublevels

• Each sublevel is composed of orbitals• No more than 2 electrons in any orbital• Maximum number of electrons in any principal

energy level is 2(n)2

Electron Distribution• This table lists the number of electrons in each

shell for the first 20 elements• Note that 3rd shell stops filling at 8 electrons even though if

could hold more

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• H– Hydrogen has

only 1 electron– It is in the

lowest energy level & lowest orbital

– Indicate number of electrons with a superscript

– 1s1

• Li– Lithium has 3

electrons– First two have

configuration of Helium – 1s2

– 3rd is in the orbital of lowest energy in n=2

– 1s2 2s1

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• Give the complete electron configuration of each element

– Be

– N

– Na

– Cl

– Ag

What noble gas configuration is this?•Neon•Configuration is written: [Ne]3s23p1

Shorthand Electron Configurations

• Uses noble gas symbols to represent the inner shell and the outer shell or valance shell is written after

• Aluminum- full electron configuration is: 1s22s22p63s23p1

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• Remember:– How many subshells are in each

principle energy level?

– There are n subshells in the n principle energy level.

– How many orbitals are in each subshell?

– s has 1, p has 3, d has 5, and f has 7

– How many electrons fit in each orbital?

– 22.5

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Shorthand Electron Configuration Examples

• N

• S

• Ti

• Sn

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Classification of Elements According to the Type of

Subshells Being Filled

2.6 The Octet Rule

• The noble gases are extremely stable– Called inert as they don’t readily bond to other

elements• The stability is due to a full complement of

valence electrons in the outermost s and psublevels:– 2 electrons in the 1s of Helium – the s and p subshells full in the outermost shell of

the other noble gases (eight electrons)

Octet of Electrons• Elements in families other than the noble

gases are more reactive– Strive to achieve a more stable electron

configuration– Change the number of electrons in the atom to

result in full s and p sublevels • Stable electron configuration is called the

“noble gas” configuration2.6

The

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2.6

The

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The Octet Rule

• Octet Rule - elements usually react in such a way as to attain the electron configuration of the noble gas closest to them in the periodic table– Elements on the right side of the table move right to the

next noble gas– Elements on the left side move “backwards” to the

noble gas of the previous row• Atoms will gain, lose or share electrons in

chemical reactions to attain this more stable energy state

2.6

The

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NaSodium atom

11e-, 1 valence e-

[Ne]3s1

Na+ + e-

Sodium ion10e-

[Ne]

Ion Formation and the Octet Rule

• Metallic elements tend to form positively charged ions called cations

• Metals tend to lose all their valence electrons to obtain a configuration of the noble gas

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AlAluminum atom13e-, 3 valence e-

[Ne]3s23p1

Al3+ + 3e-

Aluminum ion10e-

[Ne]

• All atoms of a group lose the same number of electrons

• Resulting ion has the same number of electrons as the nearest (previous) noble gas atom

Ion Formation and the Octet Rule

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The

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Using the Octet Rule• The octet rule is very helpful in predicting

the charges of ions in the representative elements

• Transition metals still tend to lose electrons to become cations but predicting the charge is not as easy

• Transition metals often form more than one stable ion– Iron forming Fe2+ and Fe3+ is a common example

Examples Using the Octet Rule

• Give the charge of the most probable ion resulting from these elements– Ca– Sr– S– P

• Which of the following pairs of atoms and ions are isoelectronic?– Cl-, Ar– Na+, Ne– Mg2+, Na+

– O2-, F-2.6

The

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+K3919

-23216S

+22412 Mg2.

1 C

ompo

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the A

tom Calculating Subatomic Particles

in Ions• How many protons, neutrons and electrons

are in the following ions?

2.7 Trends in the Periodic Table

• Many atomic properties correlate with electronic structure and so also with their position in the periodic table– atomic size– ion size– ionization energy– electron affinity

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eAtomic Size

• The size of an element increases moving down from top to bottom of a group• The valence shell is higher in energy and

farther from the nucleus traveling down the group

• The size of an element decreases from left to right across a period• The increase in magnitude of positive charge in

nucleus pulls the electrons closer to the nucleus

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eVariation in Size of Atoms

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eCation Size

Cations are smaller than their parent atom• More protons than electrons creates an increased

nuclear charge• Extra protons pulls the remaining electrons

closer to the nucleus• Ions with multiple positive charges are even

smaller than the corresponding monopositive ions– Which would be smaller, Fe2+ or Fe3+? Fe3+

• When a cation is formed isoelectronic with a noble gas the valence shell is lost decreasing the diameter of the ion relative to the parent atom

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eAnion Size

Anions are larger than their parent atom.

• Anions have more electrons than protons• Excess negative charge reduces the pull

of the nucleus on each individual electron• Ions with multiple negative charges are

even larger than the corresponding monopositive ions

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eRelative Size of Select Ions and

Their Parent Atoms

ionization energy + Na Na+ + e-2.7

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eIonization Energy

• Ionization energy - The energy required to remove an electron from an isolated atom

• The magnitude of ionization energy correlates with the strength of the attractive force between the nucleus and the outermost electron

• The lower the ionization energy, the easier it is to form a cation

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eIonization Energy of Select Elements

• Ionization decreases down a family as the outermost electrons are farther from the nucleus

• Ionization increases across a period because the outermost electrons are more tightly held

• Why would the noble gases be so unreactive?

Br + e- Br- + energy2.7

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eElectron Affinity

• Electron Affinity - The energy released when a single electron is added to an isolated atom

• Electron affinity gives information about the ease of anion formation– Large electron affinity indicates an atom

becomes more stable as it forms an anion

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ePeriodic Trends in Electron

Affinity

• Electron affinity generally decreases down a group

• Electron affinity generally increases across a period

Chapter 3

Structure and Properties of Ionic and Covalent Compounds

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3.1 Chemical Bonding

• Chemical bond - the force of attraction between any two atoms in a compound

• This attractive force overcomes the repulsion of the positively charged nuclei of the two atoms participating in the bond

• Interactions involving valence electrons are responsible for the chemical bond

3.1

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represent atoms using the element symbol and valence electrons as dots

• As only valence electrons participate in bonding, this makes it much easier to work with the octet rule

• The number of dots used corresponds directly to the number of valence electrons located in the outermost shell of the atoms of the element

3.1

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• Each “side” of the symbol represents an atomic orbital, which may hold up to two electrons

• Using Lewis symbols– Place one dot on each side until there are four dots

around the symbol– Now add a second dot to each side in turn– The number of valence electrons limits the number of

dots placed– Each unpaired dot (unpaired electron of the valence

shell) is available to form a chemical bond

Lewis Dot Symbols for Representative Elements

3.1

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3.1

Che

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Ionic and Covalent• Ionic bond - a transfer of one or more

electrons from one atom to another• Forms attractions due to the opposite charges of

the atoms

• Covalent bond - attractive force due to the sharing of electrons between atoms

• Some bonds have characteristics of both types and not easily identified as one or the other

Ionic Bonding• Representative elements form ions that

obey the octet rule• Ions of opposite charge attract each other

creating the ionic bond• When electrons are lost by a metal and

electrons are gained by a nonmetal– Each atom achieves a “Noble Gas”

configuration– 2 ions are formed; a cation and anion, which

are attracted to each other3.1

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3.1

Che

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ng Consider the formation of NaCl

Na + Cl NaCl

Sodium has a low ionization energy itreadily loses this electron

Na Na+ + e-

When sodium loses the electron, it gains the Ne configuration

Chlorine has a high electron affinity

When chlorine gains an electron, it gains the Ar configuration

−−

→+⋅ :

..

..Cl: e ....Cl:

Ionic Bonding

3.1

Che

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• Atoms with low I.E. and low E.A. tend to form positive ions

• Atoms with high I.E. and high E.A. tend to form negative ions

• Ion formation takes place by electron transfer

• The ions are held together by the electrostatic force of the opposite charges

• Reactions between metals and nonmetals (representative elements) tend to be ionic

Ion Arrangement in a Crystal• As a sodium atom loses one electron, it becomes a

smaller sodium ion• When a chlorine atom gains that electron, it

becomes a larger chloride ion• Attraction of the Na cation with the Cl anion

forms NaCl ion pairs that aggregate into a crystal

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3.1

Che

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Let’s look at the formation of H2:H + H H2

• Each hydrogen has one electron in its valance shell

• If it were an ionic bond it would look like this:

• However, both hydrogen atoms have an equal tendency to gain or lose electrons

• Electron transfer from one H to another usually will not occur under normal conditions

[ ]−+ +→⋅+⋅ :H H H H

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The shared electron

pair is called a Covalent Bond

Each hydrogen atom now has two electrons around it and attained a He configuration

• Instead, each atom attains a noble gas configuration by sharing electrons

H:H H H →⋅+⋅

Covalent Bonding in Hydrogen3.

1 C

hem

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Bon

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:....F:

..

..F: :....F

..

..F: →⋅+⋅Each fluorine is surrounded by 8 electrons – Ne configuration

Features of Covalent Bonds• Covalent bonds form between atoms with

similar tendencies to gain or lose electrons• Compounds containing covalent bonds are

called covalent compounds or molecules• The diatomic elements have completely

covalent bonds (totally equal sharing)– H2, N2, O2, F2, Cl2, Br2, I2

H:....O:H

..

..O 2H →⋅⋅+⋅

H H:

..C:H

.C 4H

H

⋅⋅→⋅⋅⋅+⋅

3.1

Che

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2e– from 2H 2e– for H6e– from O 8e– for O

4e– from 4H 2e– for H4e– from C 8e– for C

3.1

Che

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Electronegativity• The Polar Covalent Bond

– Ionic bonding involves the transfer of electrons

– Covalent bonding involves the sharing of electrons

– Polar covalent bonding - bonds made up of unequally shared electron pairs

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These two electrons are not shared equally

somewhat negatively chargedsomewhat positively charged

:..F:H :

..F H ⋅⋅→⋅⋅⋅+⋅

• The electrons spend more time with fluorine

• This sets up a polar covalent bond

• A truly covalent bond can only occur when both atoms are identical

Polar Covalent Bonding in Water• Oxygen is electron rich = -

• Hydrogen is electron deficient = +

• This results in unequal sharing of electrons in the pairs = polar covalent bonds

• Water has 2 covalent bonds

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Electronegativity• Electronegativity - a measure of the

ability of an atom to attract electrons in a chemical bond

• Elements with high electronegativity have a greater ability to attract electrons than do elements with low electronegativity

• Consider the covalent bond as competition for electrons between 2 positive centers– The difference in electronegativity determines

the extent of bond polarity3.1

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3.1

Che

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electronegativity increases elec

trone

gativ

ity in

crea

ses

Electronegativities of Selected Elements

• The most electronegative elements are found in the upper right corner of the periodic table

• The least electronegative elements are found in the lower left corner of the periodic table

3.1

Che

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• The greater the difference in electronegativity between two atoms, the greater the polarity of their bond

• Which would be more polar, a H-F bond or H-Cl bond?

• H-F … 4.0 - 2.1 = 1.9

• H-Cl … 3.0 - 2.1 = 0.9

• The HF bond is more polar than the HCl bond

3.2 Naming Compounds and Writing Formulas of Compounds

• Nomenclature - the assignment of a correct and unambiguous name to each and every chemical compound

• Two naming systems:– ionic compounds– covalent compounds

3.2

Nam

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Com

poun

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f Com

poun

ds Formulas of Compounds• A formula is the representation of the

fundamental compound using chemical symbols and numerical subscripts– The formula identifies the number and type

of the various atoms that make up the compound unit

– The number of like atoms in the unit is shown by the use of a subscript

– Presence of only one atom is understood when no subscript is present

3.2

Nam

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ds Ionic Compounds• Metals and nonmetals usually react to form

ionic compounds• The metals are cations and the nonmetals

are anions• The cations and anions arrange themselves

in a regular three-dimensional repeating array called a crystal lattice

• Formula of an ionic compound is the smallest whole-number ratio of ions in the substance

3.2

Nam

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ds Writing Formulas of Ionic Compounds from the Identities of the Component Ions

• Determine the charge of each ion– Metals have a charge equal to group number– Nonmetals have a charge equal to the group

number minus eight

• Cations and anions must combine to give a formula with a net charge of zero• It must have the same number of positive

charges as negative charges

3.2

Nam

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ds Predict FormulasPredict the formula of the ionic compounds

formed from combining ions of the following pairs of elements:

1. sodium and oxygen

2. lithium and bromine

3. aluminum and oxygen

4. barium and fluorine

3.2

Nam

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Com

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f Com

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dsWriting Names of Ionic Compounds from the Formula of the Compound• Name the cation followed by the name

of the anion

• A positive ion retains the name of the element; change the anion suffix to -ide

3.2

Nam

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Com

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f Com

poun

dsWriting Names of Ionic Compounds from the Formula of the Compound

• If the cation of an element has several ions of different charges (as with transition metals) use a Roman numeral following the metal name

• Roman numerals give the charge of the metal

• Examples:• FeCl3 is iron(III) chloride

• FeCl2 is iron(II) chloride

• CuO is copper(II) oxide

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ds Common Nomenclature System• Use -ic to indicate the higher of the

charges that ion might have

• Use -ous to indicate the lower of the charges that ion might have

• Examples:• FeCl2 is ferrous chloride

• FeCl3 is ferric chloride

Stock and Common Names for Iron and Copper Ions

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ds Common Monatomic Cations and Anions

• Monatomic ions - ions consisting of a single charged atom

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ds Polyatomic Ions• Polyatomic ions - ions composed of 2 or

more atoms bonded together with an overall positive or negative charge– Within the ion itself, the atoms are bonded

using covalent bonds

– The positive and negative ions will be bonded to each other with ionic bonds

• Examples:• NH4

+ ammonium ion• SO4

2- sulfate ion

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ds Common Polyatomic Cations and Anions

3.2

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ds Name These Compounds1. NH4Cl

2. BaSO4

3. Fe(NO3)3

4. CuHCO3

5. Ca(OH)2

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dsWriting Formulas of Ionic Compounds from the Name of the Compound

• Determine the charge of each ion

• Write the formula so that the resulting compound is neutral

• Example:Barium chloride:Barium is +2, Chloride is -1Formula is BaCl2

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ds Determine the Formulas from Names

Write the formula for the following ionic compounds:

1. sodium sulfate

2. ammonium sulfide

3. magnesium phosphate

4. chromium(II) sulfate

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ds Covalent Compounds

• Covalent compounds are typically formed from nonmetals

• Molecules - compounds characterized by covalent bonding• Not a part of a massive three-dimensional

crystal structure

• Exist as discrete molecules in the solid, liquid, and gas states

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ds Naming Covalent Compounds1. The names of the elements are written

in the order in which they appear in the formula

2. A prefix indicates the number of each kind of atom

3. If only one atom of a particular element is present in the molecule, the prefix mono- is usually omitted from the first element

Example: CO is carbon monoxide

4. The stem of the name of the last element is used with the suffix –ide

5. The final vowel in a prefix is often dropped before a vowel in the stem name

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ds Naming Covalent Compounds

3.2

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ds Name These Covalent Compounds

1. SiO2

2. N2O5

3. CCl4

4. IF7

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ds Writing Formulas of Covalent Compounds

• Use the prefixes in the names to determine the subscripts for the elements

• Examples:• nitrogen trichloride NCl3• diphosphorus pentoxide P2O5

• Some common names that are used: – H2O water– NH3 ammonia– C2H5OH ethanol– C6H12O6 glucose

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Covalent Compounds

1. nitrogen monoxide

2. dinitrogen tetroxide

3. diphosphorus pentoxide

4. nitrogen trifluoride

3.3 Properties of Ionic and Covalent Compounds

• Physical State– Ionic compounds are usually solids at room

temperature– Covalent compounds can be solids, liquids, and

gases• Melting and Boiling Points

– Melting point - the temperature at which a solid is converted to a liquid

– Boiling point - the temperature at which a liquid is converted to a gas

Physical Properties• Melting and Boiling Points

– Ionic compounds have much higher melting points and boiling points than covalent compounds

– A large amount of energy is required to break the electrostatic attractions between ions

– Ionic compounds typically melt at several hundred degrees Celsius

• Structure of Compounds in the Solid State– Ionic compounds are crystalline– Covalent compounds are crystalline or amorphous –

having no regular structure

3.3

Prop

ertie

s of I

onic

and

C

oval

ent C

ompo

unds

• Solutions of Ionic and Covalent Compounds– Ionic compounds often dissolve in water,

where they dissociate - form positive and negative ions in solution

– Electrolytes - ions present in solution allowing the solution to conduct electricity

– Covalent solids usually do not dissociate and do not conduct electricity - nonelectrolytes

Physical Properties3.

3 Pr

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Comparison of Ionic vs. Covalent Compounds

Ionic Covalent

Composed of Metal + nonmetal 2 nonmetals

Electrons Transferred Shared

Physical state Solid / crystal Any / crystal OR amorphous

Dissociation Yes, electrolytes No, nonelectrolytes

Boiling/Melting High Low

3.3

Prop

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onic

and

C

oval

ent C

ompo

unds

3.4 Drawing Lewis Structures of Molecules and Polyatomic Ions

Lewis Structure Guidelines1. Use chemical symbols for the various

elements to write the skeletal structure of the compound

– The least electronegative atom will be placed in the central position

– Hydrogen and halogens occupy terminal positions

– Carbon often forms chains of carbon-carbon covalent bonds

Lewis Structure Guidelines2. Determine the number of valence

electrons associated with each atom in the compound

– Combine these valence electrons to determine the total number of valence electrons in the compound

– Polyatomic cations, subtract one electron for every positive charge

– Polyatomic anions, add one electron for every negative charge

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Lewis Structure Guidelines3. Connect the central atom to each of the

surrounding atoms using electron pairs • Next, complete octets of all the atoms

bonded to the central atom• Hydrogen needs only two electrons• Electrons not involved in bonding are

represented as lone pairs• Total number of electrons in the structure

must equal the number of valence electrons in step 2

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4. Count the number of electrons you have and compare to the number you used

• If they are the same, you are finished• If you used more electrons than you have,

add a bond for every two too many you used• Then, give every atom an octet• If you used less electrons than you have….see

later exceptions to the octet rule

5. Recheck that all atoms have the octet rule satisfied and that the total number of valance electrons are used

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Lewis Structure Guidelines

Drawing Lewis Structures of Covalent Compounds

Draw the Lewis structure of carbon dioxide, CO2

Draw a skeletal structure of the molecule1. Arrange the atoms in their most probable order

C-O-O and/or O-C-O2. Find the electronegativity of O=3.5 & C=2.53. Place the least electronegative atom as the central

atom, here carbon is the central atom4. Result is the O-C-O structure from above3.

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5. Find the number of valence electrons for each atom and the total for the compound

1 C atom x 4 valence electrons = 4 e-

2 O atoms x 6 valence electrons = 12 e-

16 e- total

6. Use electron pairs to connect the C to each O with a single bond

O : C : O 7. Place electron pairs around the atoms

: O : C : O :This satisfies the rule for the O atoms, but not for C

Drawing Lewis Structures3.

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8. Redistribute the electrons moving 2 e- from each O, placing them between C:O

C::O::C9. In this structure, the octet rule is satisfied

• This is the most probable structure• Four electrons are between C and O

• These electrons are share in covalent bonds• Four electrons in this arrangement signify a double

bond10. Recheck the electron distribution

• 8 electron pairs = 16 valence electrons, number counted at start

• 8 electrons around each atom, octet rule satisfied

Drawing Lewis Structures of Covalent Compounds

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Lewis Structures PracticeUsing the guidelines presented, write Lewis structures for the following:

1. H2O

2. NH3

3. CO2

4. NH4+

5. CO32-

6. N2

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Lewis Structures of Polyatomic Ions

• Prepare Lewis structures of polyatomic ions as for neutral compounds, except:

• The charge on the ion must be accounted for when computing the total number of valence electrons

Lewis Structure of Polyatomic Cations

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Draw the Lewis structure of ammonium ion, NH4+

Draw a skeletal structure of the molecule1. Ammonium has this structure and charge:2. The total number of valence electrons is determined by

subtracting one electron for each unit of positive charge1 N atom x 5 valence electrons = 5 e-

4 H atoms x 1 valence electron = 4 e-

- 1 electron for +1 charge = -1 e-

8 e- total3. Distribute these 8 e- around the skeletal structure

Draw the Lewis structure of carbonate ion, CO32-

Draw a skeletal structure of the molecule1. Carbon is less electronegative than oxygen

• This makes carbon the central atom• Skeletal structure and charge:

2. The total number of valence electrons is determined by adding one electron for each unit of negative charge1 C atom x 4 valence electrons = 4 e-

3 O atoms x 6 valence electron = 18 e-

+ 2 negative charges = 2 e-

24 e- total3. Distribute these e- around the skeletal structure

Lewis Structure of Polyatomic Anions

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Draw the Lewis structure of carbonate ion, CO32-

4. Distributing the electrons around the central carbon atom (4 bonds) and around the surrounding O atoms attempting to satisfy the octet rule results in:

5. This satisfies the octet rule for the 3 oxygen, but not for the carbon

6. Move a lone pair from one of the O atoms to form another bond with C

Lewis Structure of Polyatomic Anions

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• Single bond - one pair of electrons are shared between two atoms

• Double bond - two pairs of electrons are shared between two atoms

• Triple bond - three pairs of electrons are shared between two atoms

• Very stable

Lewis Structure, Stability, Multiple Bonds, and Bond Energies

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NNor ..N

..N

OOor ..O::

..O H - Hor H:H

≡

=⋅⋅⋅⋅

Bond energy - the amount of energy required to break a bond holding two atoms together

triple bond > double bond > single bond

Bond length - the distance separating the nuclei of two adjacent atoms

single bond > double bond > triple bond

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3.4

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Lewis Structures and Resonance

• Write the Lewis structure of CO32-

• If you look around you, you will probably see the double bond put in different places

• Who is right? All of you!

• In some cases it is possible to write more than one Lewis structure that satisfies the octet rule for a particular compound

• Experimental evidence shows all bonds are the same length, meaning there is not really any double bond in this ion

• None of theses three Lewis structures exist, but the actual structure is an average or hybrid of these three Lewis structures

• Resonance - two or more Lewis structures that contribute to the real structure

:..O:C::

..O: :

..O:C:

..O: O::C:

..O:

: :: : :

..O: :O: :

..O:

⋅⋅↔⋅⋅⋅⋅↔⋅⋅

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Lewis Structures and Exceptions to the Octet Rule

1. Incomplete octet - less then eight electrons around an atom other than H– Let’s look at BeH2

1 Be atom x 2 valence electrons = 2 e-

2 H atoms x 1 valence electrons = 2 e-

total 4 e-

– Resulting Lewis structure:H : Be : H or H – Be – H

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Odd Electron2. Odd electron - if there is an odd number of

valence electrons, it is not possible to give every atom eight electrons• Let’s look at NO, nitric oxide• It is impossible to pair all electrons as the

compound contains an ODD number of valence electrons

N - O

3. Expanded octet - an element in the 3rd period or below may have 10 and 12 electrons around it

• Expanded octet is the most common exception• Consider the Lewis structure of PF5

• Phosphorus is a third period element

1 P atom x 5 valence electrons = 5 e-

5 F atoms x 7 valence electrons = 35 e-

40 e- total

• Distributing the electrons results in this Lewis structure

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Lewis Structures and Molecular Geometry: VSEPR Theory

• Molecular shape plays a large part in determining properties and shape

• VSEPR theory - Valance Shell Electron Pair Repulsion theory

• Used to predict the shape of the molecules

• All electrons around the central atom arrange themselves so they can be as far away from each other as possible – to minimize electronic repulsion

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VSEPR Theory• In the covalent bond, bonding electrons

are localized around the nucleus

• The covalent bond is directional, having a specific orientation in space between the bonded atoms

• Ionic bonds have electrostatic forces which have no specific orientation in space

Molecular Bonding

• Bonding pair = two electrons shared by 2 atoms– H:O

• Nonbonding pair = two electrons belonging to 1 atom, pair not shared– N:

• Maximal separation of bonding pairs = 4 corners of a TETRAHEDRON3.

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A Stable Exception to the Octet Rule

• Consider BeH2

– Only 4 electrons surround the beryllium atom

– These 2 electron pairs have minimal repulsion when located on opposite sides of the structure

– Linear structure having bond angles of 180°

3.4

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Another Stable Exception to the Octet Rule

• Consider BF3 – There are 3 shared electron pairs around the central

atom– These electron pairs have minimal repulsion when

placed in a plane, forming a triangle– Trigonal planar structure with bond angles of 120°

3.4

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Basic Electron Pair Repulsion of a Full Octet

• Consider CH4 – There are 4 shared electron pairs around the central

Carbon– Minimal electron repulsion when electrons are placed at

the four corners of a tetrahedron– Each H-C-H bond angle is 109.5°

• Tetrahedron is the primary structure of a full octet

3.4

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Basic Electron Pair Repulsion of a Full Octet with One Lone Pair

Consider NH3

• There are 4 electron pairs around the central Nitrogen• 3 pairs are shared electron pairs• 1 pair is a lone pair

– A lone pair is more electronegative with a greater electron repulsion– The lone pair takes one of the corners of the tetrahedron without being

visible, distorting the arrangement of electron pairs• Ammonia has a trigonal pyramidal structure with 107° angles

3.4

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Basic Electron Pair Repulsion of a Full Octet with Two Lone Pairs

Consider H2O• There are 4 electron pairs around the central Oxygen

• 2 pairs are shared electron pairs• 2 pairs are lone pairs

– All 4 electron pairs are approximately tetrahedral to each other– The lone pairs take two of the corners of the tetrahedron without being

visible, distorting the arrangement of electron pairs• Water has a bent or angular structure with 104.5° bond angles

Predicting Geometric Shape Using Electron Pairs

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Basic Procedure to Determine Molecular Shape

1. Write the Lewis structure

2. Count the number of shared electron pairs and lone pairs around the central atom

3. If no lone pairs are present, shape is:• 2 shared pairs - linear• 3 shared pairs - trigonal planar• 4 shared pairs - tetrahedral

4. Look at the arrangement and name the shape• Linear• Trigonal planar• Bent• Trigonal pyramid• Tetrahedral

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Consider dimethyl ether• Has 2 different central atoms:

• oxygen• carbon

– CH3 (methyl group) has tetrahedral geometry (like methane)

– Portion of the molecule linking the two methyl groups would bond angles similar to water

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Determine the Molecular Geometry

• PCl3

• SO2

• PH3

• SiH4

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Lewis Structures and Polarity• A molecule is polar if its centers of positive and

negative charges do not coincide

• Polar molecules when placed in an electric field will align themselves in the field• Molecules that are polar behave as a dipole (having

two “poles” or ends)

• One end is positively charged the other is negatively charged

• Nonpolar molecules will not align themselves in an electric field

Positive end of the bond, the less electronegative atom

Negative end of the bond, more electronegative atom attracts the electrons more strongly towards it3.

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To determine if a molecule is polar:

• Write the Lewis structure

• Draw the geometry

• Use the following symbol to denote the polarity of each bond

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Practice Determining PolarityDetermine whether the following bonds and molecules are polar:1. Si – Cl 1. O2

2. H – C 2. HF3. C – C 3. CH4

4. S – Cl 4. H2O

3.5 Properties Based on Electronic Structure and Molecular Geometry• Intramolecular forces – attractive forces

within molecules – Chemical bonds• Intermolecular forces – attractive forces

between molecules• Intermolecular forces determine many

physical properties– Intermolecular forces are a direct consequence

of the intramolecular forces in the molecules

Solubility - the maximum amount of solute that dissolves in a given amount of solvent at a specific temperature

• “Like dissolves like” – Polar molecules are most soluble in polar

solvents– Nonpolar molecules are most soluble in

nonpolar solvents• Does ammonia, NH3, dissolve in water?• Yes, both molecules are polar

3.5

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Geo

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Geo

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ry Interaction of Water and Ammonia

• The - end of ammonia, N, is attracted to the + end of the water molecule, H

• The + end of ammonia, H, is attracted to the - end of the water molecule, O

• The attractive forces, called hydrogen bonds, pull ammonia into water, distributing the ammonia molecules throughout the water, forming a homogeneous solution

3.5

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Geo

met

ry Interaction of Water and Oil• What do you know about oil

and water?– “They don’t mix”

• Why?– Because water is polar and

oil is nonpolar• Water molecules exert their

attractive forces on other water molecules

• Oil remains insoluble and floats on the surface of the water as it is less dense

3.5

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Geo

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ry Boiling Points of Liquidsand Melting Points of Solids

• Energy is used to overcome the intermolecular attractive forces in a substance, driving the molecules into a less associated phase

• The greater the intermolecular force, the more energy is required leading to – Higher melting point of a solid

– Higher boiling point of a liquid

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Geo

met

ry Factors Influencing Boiling and Melting Points

• Strength of the attractive force holding the substance in its current physical state

• Molecular mass• Larger molecules have higher m.p. and b.p. than

smaller molecules as it is more difficult to convert a larger mass to another phase

• Polarity• Polar molecules have higher m.p. and b.p. than

nonpolar molecules of similar molecular mass due to their stronger attractive force

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Geo

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ry Melting and Boiling Points –Selected Compounds by Bonding Type

Chapter 4

Calculations and the Chemical Equation

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4.1 The Mole Concept and Atoms• Atoms are exceedingly small

– Unit of measurement for mass of an atom is atomic mass unit (amu) – unit of measure for the mass of atoms

• carbon-12 assigned the mass of exactly 12 amu• 1 amu = 1.66 x 10-24 g

• Periodic table gives atomic weights in amu

• What is the atomic weight of one atom of fluorine? Answer: 19.00 amu

• What would be the mass of this one atom in grams?

• Chemists usually work with much larger quantities – It is more convenient to work with grams

than amu when using larger quantities

Mass of Atoms4.

1 Th

e M

ole

Con

cept

and

A

tom

s

atom FF g10156.3

Famu 1g101.661

atom FFamu 19.00 23-24 −×=××

• A practical unit for defining a collection of atoms is the mole

1 mole of atoms = 6.022 x 1023 atoms

• This is called Avogadro’s number– This has provided the basis for the concept

of the mole

The Mole and Avogadro’s Number4.

1 Th

e M

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Con

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and

A

tom

s

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The Mole

• To make this connection we must define the mole as a counting unit– The mole is abbreviated mol

• A mole is simply a unit that defines an amount of something– Dozen defines 12– Gross defines 144

F mol 1F atom10022.6

Famu 1F g1066.1

F atom 1Famu 00.19 2324 ××××

−

=19.00 g F/mol F or 19.00 g/mol F4.1

The

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Atomic Mass• The atomic mass of one atom of an element

corresponds to:– The average mass of a single atom in amu

– The mass of a mole of atoms in grams

– 1 atom of F is 19.00 amu 19.00 amu/atom F

– 1 mole of F is 19.00 g 19.00 g/mole F

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Calculating Atoms, Moles, and Mass

• We use the following conversion factors:

• Density converts grams – milliliters

• Atomic mass unit converts amu –grams

• Avogadro’s number converts moles –number of atoms

• Molar mass converts grams – moles

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Strategy for Calculations• Map out a pattern for the required

conversion

• Given a number of grams and asked for number of atoms

• Two conversions are required • Convert grams to moles

1 mol S/32.06 g S OR 32.06 g S/1 mol S• Convert moles to atoms

mol S x (6.022 x 1023 atoms S) / 1 mol S

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Practice Calculations1. Calculate the number of atoms in 1.7

moles of boron.

2. Find the mass in grams of 2.5 mol Na (sodium).

3. Calculate the number of atoms in 5.0 g aluminum.

4. Calculate the mass of 5,000,000 atoms of Au (gold)

Interconversion Between Moles, Particles, and Grams

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4.2 The Chemical Formula, Formula Weight, and Molar Mass

• Chemical formula - a combination of symbols of the various elements that make up the compound

• Formula unit - the smallest collection of atoms that provide two important pieces of information– The identity of the atoms– The relative number of each type of atom

4.2

The

Che

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Fo

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ass

Chemical FormulaConsider the following formulas:

• H2 – 2 atoms of hydrogen are chemically bonded forming diatomic hydrogen, subscript 2

• H2O – 2 atoms of hydrogen and 1 atom of oxygen, lack of subscript means one atom

• NaCl – 1 atom each of sodium and chlorine

• Ca(OH)2 – 1 atom of calcium and 2 atoms each of oxygen and hydrogen, subscript outside parentheses applies to all atoms inside

4.2

The

Che

mic

al F

orm

ula,

Fo

rmul

a W

eigh

t and

Mol

ar M

ass

Chemical FormulaConsider the following formulas:• (NH4)3SO4 – 2 ammonium ions and 1 sulfate ion

– Ammonium ion contains 1 nitrogen and 4 hydrogen

– Sulfate ion contains 1 sulfur and 4 oxygen

– Compound contains 2 N, 8 H, 1 S, and 4 O

• CuSO4.5H2O

– This is an example of a hydrate - compounds containing one or more water molecules as an integral part of their structure

– 5 units of water with 1 CuSO4

Comparison of Hydrated and Anhydrous Copper Sulfate

Hydrated copper sulfate Anhydrous copper sulfate

4.2

The

Che

mic

al F

orm

ula,

Fo

rmul

a W

eigh

t and

Mol

ar M

ass

Marked color difference illustrates the factthat these are different compounds

Formula Weight and Molar Mass• Formula weight - the sum of the atomic weights

of all atoms in the compound as represented by its correct formula– expressed in amu

• What is the formula weight of H2O?– 16.00 amu + 2(1.008 amu) = 18.02 amu

• Molar mass – mass of a mole of compound in grams / mole– Numerically equal to the formula weight in amu

• What is the molar mass of H2O?– 18.02 g/mol H2O

4.2

The

Che

mic

al F

orm

ula,

Form

ula

Wei

ghta

nd M

olar

Mas

s

Formula Unit• Formula unit – smallest

collection of atoms from which the formula of a compound can be established

• When calculating the formula weight (or molar mass) of an ionic compound, the smallest unit of the crystal is used

4.2

The

Che

mic

al F

orm

ula,

Form

ula

Wei

ght a

nd M

olar

Mas

s

What is the molar mass of (NH4)3PO4?

3(N amu) + 12(H amu) + P amu + 4(O amu)=3(14.01) + 12(1.008) + 30.97 + 4(16.00)=

149.10 g/mol (NH4)3PO4

Molar Mass• Molar mass - The mass in grams of 1 mole of

atoms• What is the molar mass of carbon?

12.01 g/mol C

• This means counting out a mole of Carbon atoms (i.e., 6.022 x 1023) they would have a mass of 12.01 g

• One mole of any element contains the same number of atoms, 6.022 x 1023, Avogadro’s number

4.2

The

Che

mic

al F

orm

ula,

Form

ula

Wei

ght a

nd M

olar

Mas

s

4.3 The Chemical Equation and the Information It Conveys

A Recipe For Chemical Change• Chemical equation - shorthand notation of a

chemical reaction– Describes all of the substances that react and all

the products that form, physical states, and experimental conditions

– Reactants – (starting materials) – the substances that undergo change in the reaction

– Products – substances produced by the reaction

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys Features of a Chemical Equation1. Identity of products and reactants must

be specified using chemical symbols2. Reactants are written to the left of the

reaction arrow and products are written to the right

3. Physical states of reactants and products may be shown in parentheses

4. Symbol over the reaction arrow means that energy is necessary for the reaction to occur

5. Equation must be balanced

)(O )2Hg( )2HgO( 2 gls +→∆

Products – written on the right

Reactants – written on the left of arrow

Products and reactants must be specified using chemical symbols

Physical states are shown in parentheses

– energy is needed

Features of a Chemical Equation4.

3 Th

e C

hem

ical

Equ

atio

nan

d th

e In

form

atio

n It

Con

veys

The Experimental Basis of a Chemical Equation

We know that a chemical equation represents a chemical change

• One or more substances changed into new substances

• Different chemical and physical properties

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Evidence of a Reaction OccurringThe following can be visual evidence of a reaction:•Release of a gas

– CO2 is released when acid is placed in a solution containing CO3

2- ions

•Formation of a solid (precipitate)– A solution containing Ag+ ions mixed with a solution

containing Cl- ions

•Heat is produced or absorbed – Acid and base are mixed together

•Color changes

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ysSubtle Indications of a Reaction

• Heat or light is absorbed or emitted

• Changes in the way the substances behave in an electrical or magnetic field before and after a reaction

• Changes in electrical properties

Writing Chemical Reactions• We will learn to identify the following

patterns of chemical reactions:– combination– decomposition– single-replacement– double-replacement

• Recognizing the pattern will help you write and understand reactions

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

A + B AB• Examples:

2Na(s) + Cl2(g) 2NaCl(s)

MgO(s) + CO2(g) MgCO3(s)

Combination Reactions

• The joining of two or more elements or compounds, producing a product of different composition

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Types of Combination Reactions

1. Combination of a metal and a nonmetal to form a salt

2. Combination of hydrogen and chlorine molecules to produce hydrogen chloride

3. Formation of water from hydrogen and oxygen molecules

4. Reaction of magnesium oxide and carbon dioxide to produce magnesium carbonate

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

AB A + B

• Examples:

2HgO(s) 2Hg(l) + O2(g)

CaCO3(s) CaO(s) + CO2(g)

Decomposition Reactions

• Produce two or more products from a single reactant

• Reverse of a combination reaction

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys Types of Decomposition Reactions

1. Heating calcium carbonate to produce calcium oxide and carbon dioxide

2. Removal of water from a hydrated material

1. Single-replacement• One atom replaces another in the

compound producing a new compound

• Examples:

Cu(s)+2AgNO3(aq) 2Ag(s)+Cu(NO3)2(aq)

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

A + BC B + AC

Replacement Reactions4.

3 Th

e C

hem

ical

Equ

atio

nan

d th

e In

form

atio

n It

Con

veys

1. Replacement of copper by zinc in copper sulfate

2. Replacement of aluminum by sodium in aluminum nitrate

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys Types of Replacement Reactions

2. Double-replacement• Two compounds undergo a “change

of partners”• Two compounds react by

exchanging atoms to produce two new compounds

Replacement Reactions

AB + CD AD + CB

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

AB + CD AD + CB

Types of Double-Replacement• Reaction of an acid with a base to

produce water and salt

HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)

• Formation of solid lead chloride from lead nitrate and sodium chloride

Pb(NO3)2(aq) + 2NaCl(aq)

PbCl2(s) + 2NaNO3(aq)

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Types of Chemical Reactions

Precipitation Reactions• Chemical change in a solution that

results in one or more insoluble products• To predict if a precipitation reaction can

occur it is helpful to know the solubilities of ionic compounds

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Solubilities of Some Common Ionic Compounds

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Predicting Whether Precipitation Will Occur

• Recombine the ionic compounds to have them exchange partners

• Examine the new compounds formed and determine if any are insoluble according to the rules in Table 4.1

• Any insoluble salt will be the precipitatePb(NO3)2(aq) + NaCl(aq)

PbCl2 (?) + NaNO3 ( ?)(s) (aq)

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Predict Whether These Reactions Form Precipitates

• Potassium chloride and silver nitrate

• Potassium acetate and silver nitrate

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Reactions with Oxygen• Reactions with oxygen generally release

energy

• Combustion of natural gas– Organic compounds CO2 and H2O are

usually the products

CH4+2O2CO2+2H2O

• Rusting or corrosion of iron

4Fe + 3O2 2Fe2O3

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

The H+ on HCl was transferred to the oxygen in OH-, giving H2O

Acid-Base Reactions• These reactions involve the transfer of a

hydrogen ion (H+) from one reactant (acid) to another (base)

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Two electrons are transferred from Zn to Cu2+

Oxidation-Reduction Reactions

• Reaction involves the transfer of one or more electrons from one reactant to another

Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

Writing Chemical Reactions

Consider the following reaction:hydrogen reacts with oxygen to produce water

• Write the above reaction as a chemical equation

H2 + O2 H2O• Don’t forget the diatomic elements

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

4.4 Balancing Chemical Equations

• A chemical equation shows the molar quantity of reactants needed to produce a particular molar quantity of products

• The relative number of moles of each product and reactant is indicated by placing a whole-number coefficient before the formula of each substance in the chemical equation

Law of Conservation of Mass

• Law of conservation of mass - matter cannot be either gained or lost in the process of a chemical reaction– The total mass of the products must equal

the total mass of the reactants

4.3

The

Che

mic

al E

quat

ion

and

the

Info

rmat

ion

It C

onve

ys

)(O )2Hg( )2HgO( 2 gls +→∆

Coefficient - how many of that substance are in the reaction

4.4

Bal

anci

ng C

hem

ical

Eq

uatio

nsBalancing

• The equation must be balanced – All the atoms of every reactant must also

appear in the products• Number of Hg on left? 2

– on right 2• Number of O on left? 2

– on right 2

4.4

Bal

anci

ng C

hem

ical

Eq

uatio

nsExamine the Equation

H2 + O2 H2O

• Is the law of conservation of mass obeyed as written? NO

• Balancing chemical equations uses coefficientsto ensure that the law of conservation of mass is obeyed

• You may never change subscripts!

• WRONG: H2 + O2 H2O2

Step 1. Count the number of moles of atoms of each element on both product and reactant sides

Reactants Products2 mol H 2 mol H2 mol O 1 mol O

4.4

Bal

anci

ng C

hem

ical

Eq

uatio

nsSteps in Equation Balancing

The steps to balancing:H2 + O2 H2O

H2 + O2 H2O

H2 + O2 2H2O

This balances oxygen, but is hydrogen still balanced?

4.4

Bal

anci

ng C

hem

ical

Eq

uatio

nsStep 2. Determine which elements are not

balanced – do not have same number on both sides of the equation– Oxygen is not balanced

Step 3. Balance one element at a time by changing the coefficients

Steps in Equation Balancing

Reactants Products4 mol H 4 mol H2 mol O 2 mol O4.

4 B

alan

cing

Che

mic

al

Equa

tions

Steps in Equation Balancing

H2 + O2 2H2O

How will we balance hydrogen?

2H2 + O2 2H2O

Step 4. Check! Make sure the law of conservation of mass is obeyed

Balancing an Equation4.

4 B

alan

cing

Che

mic

al

Equa

tions

4.4

Bal

anci

ng C

hem

ical

Eq

uatio

nsPractice Equation Balancing

Balance the following equations:

1. C2H2 + O2 CO2 + H2O

2. AgNO3 + FeCl3 Fe(NO3)3 + AgCl

3. C2H6 + O2 CO2 + H2O

4. N2 + H2 NH3

4.5 Calculations Using the Chemical Equation

• Calculation quantities of reactants and products in a chemical reaction has many applications

• Need a balanced chemical equation for the reaction of interest

• The coefficients represent the number of moles of each substance in the equation

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nGeneral Principles

1. Chemical formulas of all reactants and products must be known

2. Equation must be balanced to obey the law of conservation of mass

• Calculations of an unbalanced equation are meaningless

3. Calculations are performed in terms of moles

• Coefficients in the balanced equation represent the relative number of moles of products and reactants

Using the Chemical Equation

• Examine the reaction:

2H2 + O2 2H2O

• Coefficients tell us?

– 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O

• What if 4 moles of H2 reacts with 2 moles of O2?

– It yields 4 moles of H2O

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

n

2H2 + O2 2H2O

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nUsing the Chemical Equation

• The coefficients of the balanced equation are used to convert between moles of substances

• How many moles of O2 are needed to react with 4.26 moles of H2?

• Use the factor-label method to perform this calculation

2H2 + O2 2H2O

=×2

22 H mol__

O __molH mol 26.4 12

2.13 mol O2

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nUse of Conversion Factors

• Digits in the conversion factor come from the balanced equation

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nConversion Between Moles

and Grams• Requires only the formula weight• Convert 1.00 mol O2 to grams

– Plan the path – Find the molar mass of oxygen

• 32.0 g O2 = 1 mol O2

– Set up the equation– Cancel units 1.00 mol O2 x 32.0 g O2

1 mol O2– Solve equation 1.00 x 32.0 g O2 = 32.0 g O2

moles ofOxygen

grams ofOxygen

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nConversion of Mole Reactants to

Mole Products• Use a balanced equation• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)• 1 mol C3H8 results in:

– 5 mol O2 consumed 1 mol C3H8 /5 mol O2

– 3 mol CO2 formed 1 mol C3H8 /3 mol CO2

– 4 mol H2O formed 1 mol C3H8 /4 mol H2O

• This can be rewritten as conversion factors

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nCalculating Reacting Quantities

• Calculate grams O2 reacting with 1.00 mol C3H8

• Use 2 conversion factors– Moles C3H8 to moles O2

– Moles of O2 to grams O2

– Set up the equation and cancel units– 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 =

1 mol C3H8 1 mol O2

– 1.00 x 5 x 32.0 g O2 = 1.60 x 102 g O2

moles Oxygen

grams Oxygen

moles C3H8

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nCalculating Grams of Product

from Moles of Reactant• Calculate grams CO2 from combustion of 1.00

mol C3H8• Use 2 conversion factors

– Moles C3H8 to moles CO2– Moles of CO2 to grams CO2

– Set up the equation and cancel units– 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 =

1 mol C3H8 1 mol CO2

– 1.00 x 3 x 44.0 g CO2 = 1.32 x 102 g CO2

moles CO2

grams CO2

moles C3H8

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nRelating Masses of Reactants

and Products• Calculate grams C3H8 required to produce

36.0 grams of H2O• Use 3 conversion factors

– Grams H2O to moles H2O– Moles H2O to moles C3H8– Moles of C3H8 to grams C3H8

– Set up the equation and cancel units

36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H818.0 g H2O 4 mol H2O 1 mol C3H8

– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8

moles H2O

grams C3H8

moles C3H8

gramsH2O

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nCalculating a Quantity of Reactant

• Ca(OH)2 neutralizes HCl• Calculate grams HCl neutralized by 0.500 mol

Ca(OH)2– Write chemical equation and balance

• Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l)– Plan the path

– Set up the equation and cancel units0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl

1 mol Ca(OH)2 1 mol HClSolve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl

molesCa(OH)2

grams HCl

molesHCl

A Visual Example of the Law of Conservation of Mass

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

n

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nGeneral Problem-solving Strategy

Na + Cl2 NaCl

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nSample Calculation

1. Balance the equation

2. Calculate the moles Cl2 reacting with 5.00 mol Na

3. Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2

4. Calculate the grams Na reacting with 5.00 g Cl2

2Na + Cl2 2NaCl

%100yield ltheoretica

yield actual yield % ×=

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nTheoretical and Percent Yield

• Theoretical yield - the maximum amount of product that can be produced – Pencil and paper yield

• Actual yield - the amount produced when the reaction is performed– Laboratory yield

• Percent yield:

= 125 g CO2 actual x 100% = 97.4%132 g CO2 theoretical

4.5

Cal

cula

tions

Usi

ng th

e C

hem

ical

Equ

atio

nSample Calculation

If the theoretical yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield:

2 Al(s) + Fe2O3(s) Al2O3(aq) + 2Fe(aq)

• [25.0 g / 30.0 g] x 100% = 83.3%

• Calculate the % yield if 26.8 grams iron was collected in the same reaction

Chapter 5

States of Matter Gases, Liquids, and Solids

Denniston Topping Caret

6th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Changes in State• Changes in state are considered to be

physical changes• During a change of physical state many

other physical properties may also change• This chapter focuses on the important

differences in physical properties among– Gases– Liquids– Solids

Comparison of Physical Properties of Gases, Liquids, and Solids

5.1 The Gaseous State

Ideal Gas Concept• Ideal gas - a model of the way that particles

of a gas behave at the microscopic level• We can measure the following of a gas:

– temperature– volume– pressure – mass

We can systematically change one of the properties and see the effect on the others

5.1

The

Gas

eous

Sta

teMeasurement of Gases

• Gas laws involve the relationship between:– number of moles (n) of gas– volume (V)– temperature (T)– pressure (P)

• Pressure - force per unit area

• Gas pressure is a result of force exerted by the collision of particles with the walls of the container

5.1

The

Gas

eous

Sta

teBarometer

• Measures atmospheric pressure– Invented by Evangelista Torricelli

• Common units of pressure– atmosphere (atm)

– torr (in Torricelli’s honor)

– pascal (Pa) (in honor of Blaise Pascal)

• 1 atm is equal to:– 760 mmHg– 760 torr– 76 cmHg

5.1

The

Gas

eous

Sta

teKinetic Molecular Theory of Gases

1. Gases are made up of small atoms or molecules that are in constant, random motion

2. The distance of separation is very large compared to the size of the individual atoms or molecules

– Gas is mostly empty space3. All gas particles behave independently

– No attractive or repulsive forces exist between them

5.1

The

Gas

eous

Sta

te

4. Gas particles collide with each other and with the walls of the container without losing energy – The energy is transferred from one atom or

molecule to another

5. The average kinetic energy of the atoms or molecules increases or decreases in proportion to absolute temperature – As temperature goes up, particle speed goes up

Kinetic Molecular Theory of Gases

5.1

The

Gas

eous

Sta

te Kinetic Molecular Theory of Gases Explains the following statements:

• Gases are easily compressible – gas is mostly empty space, room for particles to be pushed together

• Gases will expand to fill any available volume– move freely with sufficient energy to overcome attractive forces

• Gases have low density – being mostly empty space; gases have low mass per unit volume

5.1

The

Gas

eous

Sta

te • Gases readily diffuse through each other – they are in continuous motion with paths readily available due to large space between adjacent particles

• Gases exert pressure on their containers – pressure results from collisions of gas particles with the container walls

• Gases behave most ideally at low pressure and high temperature – Low pressure, average distance of separation is

greatest, minimizing interactive forces– High temperature, rapid motion overcomes interactive

forces more easily

Gas Diffusion5.

1 Th

e G

aseo

us S

tate

Ammonia (17.0g/mol)

Ammonia diffused farther in same time, lighter moves faster

Hydrogen chloride (36.5g/mol)

Top:Start of expt

Bottom:End of expt

• Boyle’s law - volume of a gas varies inversely with the pressure exerted by the gas if the temperature and number of moles are held constant

• The product of pressure (P) and volume (V) is a constant

• Used to calculate– Volume resulting from pressure change– Pressure resulting from volume change

PV = k1

5.1

The

Gas

eous

Sta

teBoyle’s Law

PiVi = PfVf

5.1

The

Gas

eous

Sta

teApplication of Boyle’s Law

• Gas occupies 10.0 L at 1.00 atm pressure• Product, PV = (10.0 L) (1.00 atm) = k1• Double the pressure to 2.0 atm, decreases the

volume to 5.0 L– (2.0 atm)(Vx) = (10.0 L)(1.00 atm)– Vx = 5.0 L

5.1

The

Gas

eous

Sta

teBoyle’s Law Practice

1. A 5.0 L sample of a gas at 25oC and 3.0 atm is compressed at constant temperature to a volume of 1.0 L. What is the new pressure?

2. A 3.5 L sample of a gas at 1.0 atm is expanded at constant temperature until the pressure is 0.10 atm. What is the volume of the gas?

• It is possible to relate gas volume and temperature

• Charles’s law - volume of a gas varies directly with the absolute temperature (K) if pressure and number of moles of gas are constant

• Ratio of volume (V) and temperature (T) is a constant

2kTV =

f

f

i

i

TV

TV =5.

1 Th

e G

aseo

us S

tate

Charles’s Law

5.1

The

Gas

eous

Sta

teApplication of Charles’s Law

• If a gas occupies 10.0 L at 273 K with V/T = k2• Doubling temperature to 546 K, increasesvolume to 20.0 L10.0 L / 273 K = Vf / 546 K

5.1

The

Gas

eous

Sta

tePractice with Charles’s Law

1. A 2.5 L sample of gas at 25oC is heated to 50oC at constant pressure. Will the volume double?

2. What would be the volume?

3. What temperature would be required to double the volume?

• If a sample of gas undergoes change involving volume, pressure, and temperature simultaneously, use the combined gas law

• Derived from a combination of Boyle’s law and Charles’s law

f

ff

i

ii

TVP

TVP =

5.1

The

Gas

eous

Sta

teCombined Gas Law

• Calculate the volume of N2 resulting when 0.100 L of the gas is heated from 300. K to 350. K at 1.00 atm

• What do we know?– Pi = 1.00 atm Pf = 1.00 atm– Vi = 0.100 L Vf = ? L– Ti = 300. K Tf = 350. K

• Vf = ViTf / Ti this is valid as Pi = Pf

• Vf = (0.100 L)(350. K) / 300. K = 0.117 L• Note the decimal point in the temperature to indicate

significance

f

ff

i

ii

TVP

TVP =

5.1

The

Gas

eous

Sta

teUsing the Combined Gas Law

5.1

The

Gas

eous

Sta

tePractice With the Combined

Gas LawCalculate the temperature when a 0.50 L sample of gas at 1.0 atm and 25oC is compressed to 0.05 L of gas at 5.0 atm.

• Avogadro’s Law - equal volumes of any ideal gas contain the same number of moles if measured under the same conditions of temperature and pressure

• Changes in conditions can be calculated by rewriting the equation

3knV =

f

f

i

i

nV

nV =5.

1 Th

e G

aseo

us S

tate

Avogadro’s Law

5.1

The

Gas

eous

Sta

teUsing Avogadro’s Law

• If 5.50 mol of CO occupy 20.6 L, how many liters will 16.5 mol of CO occupy at the same temperature and pressure?

• What do we know?– Vi = 20.6 L Vf = ? L– ni = 5.50 mol nf = 16.5 mol– Vf = Vinf / ni = (20.6 L)(16.5 mol)

(5.50 mol)= 61.8 L CO

5.1

The

Gas

eous

Sta

te

• Molar volume - the volume occupied by 1 mol of any gas

• STP – Standard Temperature and Pressure– T = 273 K (or 0oC)– P = 1 atm

• At STP the molar volume of any gas is 22.4 L

Molar Volume of a Gas

5.1

The

Gas

eous

Sta

te Gas Densities• Density = mass / volume• Calculate the density of 4.00 g He

– What is the mass of 1 mol of H2? 4.00 g

DensityHe = 4.00g / 22.4L = 0.178 g/L at STP

• Combining: – Boyle’s law (relating volume and pressure)

– Charles’s law (relating volume and temperature)

– Avogadro’s law (relating volume to the number of moles)

gives the Ideal Gas Law

• R is a constant, ideal gas constant• R = 0.0821 L.Atm/mol.KIf units are P in atm, V in L, n in number of

moles, T in K

PV=nRT

5.1

The

Gas

eous

Sta

teThe Ideal Gas Law

=⋅⋅

==atm 1

K 273)Kmol

atmL6mol(0.08201

PnRTV 22.4 L

5.1

The

Gas

eous

Sta

teCalculating a Molar Volume

• Demonstrate molar volume of O2 gas

at STP

5.1

The

Gas

eous

Sta

te Practice Using the Ideal Gas Law1. What is the volume of gas occupied by

5.0 g CH4 at 25oC and 1 atm?

2. What is the mass of N2 required to occupy 3.0 L at 100oC and 700 mmHg?

• Dalton’s law – a mixture of gases exerts a pressure that is the sum of the pressures that each gas would exert if it were present alone under the same conditions

• Total pressure of our atmosphere is equal to the sum of the pressures of N2 and O2– (principal components of air)

Pt=p1+p2+p3+...

22 ONair ppP +=5.1

The

Gas

eous

Sta

teDalton’s Law of Partial Pressures

5.1

The

Gas

eous

Sta

te Ideal Gases vs. Real Gases

• In reality there is no such thing as an ideal gas– It is a useful model to explain gas behavior

• Nonpolar gases behave more ideally than polar gases because attractive forces are present in polar gases

5.2 The Liquid State• Liquids are practically incompressible

– Enables brake fluid to work in your car• Viscosity - a measure of a liquid’s

resistance to flow– A function of both attractive forces between

molecules and molecular geometry– Flow occurs because the molecules can easily

slide past each other• Glycerol - example of a very viscous liquid

– Viscosity decreases with increased temperature

5.2

The

Liqu

id S

tate

Surface Tension• Surface tension - a measure of the attractive forces

exerted among molecules at the surface of a liquid

– Surface molecules are surrounded and attracted by fewer liquid molecules than those below

– Net attractive forces on surface molecules pull them downward

• Results in “beading”

• Surfactant - substance added which decreases the surface tension, for example – soap

5.2

The

Liqu

id S

tate

Vapor Pressure of a Liquid

• Place water in a sealed container– Both liquid water and water vapor will exist in

the container• How does this happen below the boiling

point?– Temperature is too low for boiling conversion

• Kinetic theory - liquid molecules are in continuous motion, with their average kinetic energy directly proportional to the Kelvin temperature

energy + H2O(l) H2O(g)

5.2

The

Liqu

id S

tate

Temperature Dependence of Liquid Vapor Pressure

• Average molecular kinetic energy increases as does temperature

• Some high energy molecules have sufficient energy to escape from the liquid phase

• Even at cold temperatures, some molecules can be converted

H2O(g) H2O(l) + energy

5.2

The

Liqu

id S

tate

Movement From Gas Back to Liquid

• Molecules in the vapor phase can lose energy and be converted back to the liquid phase

• Evaporation - the process of conversion of liquid to gas at a temperature too low to boil

• Condensation - conversion of gas to the liquid state

5.2

The

Liqu

id S

tate

Liquid Water in Equilibrium With Water Vapor

• When the rate of evaporation equals the rate of condensation, the system is at equilibrium

• Vapor pressure of a liquid - the pressure exerted by the vapor at equilibrium

5.2

The

Liqu

id S

tate

Boiling Point• Boiling point - the temperature at which the vapor

pressure of the liquid becomes equal to the atmospheric pressure

• Normal boiling point - temperature at which the vapor pressure of the liquid is equal to 1 atm

• What happens when you go to a mountain where the atmospheric pressure is lower than 1 atm?– The boiling point lowers

• Boiling point is dependant on the intermolecular forces– Polar molecules have higher b.p. than nonpolar

molecules

5.2

The

Liqu

id S

tate

Van der Waals Forces

• Physical properties of liquids are explained in terms of their intermolecular forces

• Van der Waals forces are intermolecular forces having 2 subtypes– Dipole-dipole interactions

– Attractive forces between polar molecules – London forces

– As electrons are in continuous motion, a nonpolar molecule could have an instantaneous dipole

5.2

The

Liqu

id S

tate

London Forces• Exist between all molecules• The only attractive force between nonpolar

atoms or molecules• Electrons are in constant motion• Electrons can be, in an instant, arranged in

such a way that they have a dipole (Instantaneous dipole)

• The temporary dipole interacts with other temporary dipoles to cause attraction

5.2

The

Liqu

id S

tate

Hydrogen Bonding

• Hydrogen bonding:– not considered a Van der Waals force

– is a special type of dipole-dipole attraction

– is a very strong intermolecular attraction causing higher than expected b.p. and m.p.

• Requirement for hydrogen bonding:– molecules have hydrogen directly bonded to O,

N, or F

5.2

The

Liqu

id S

tate

Examples of Hydrogen Bonding• Hydrogen bonding has an extremely important

influence on the behavior of many biological systems

• H2O

• NH3

• HF

5.3 The Solid State

• Particles highly organized, in a defined fashion

• Fixed shape and volume• Properties of solids:

– incompressible– m.p. depends on strength of attractive force

between particles– crystalline solid - regular repeating structure– amorphous solid - no organized structure

5.3

The

Solid

Sta

teTypes of Crystalline Solids

1. Ionic solids• held together by electrostatic forces• high m.p. and b.p.• hard and brittle• if dissolves in water, electrolytes• NaCl

2. Covalent solids• held together entirely by covalent bonds• high m.p. and b.p.• extremely hard• diamond

5.3

The

Solid

Sta

te3.Molecular solids

• molecules are held together with intermolecular forces• often soft• low m.p.• often volatile• ice

4.Metallic solids• metal atoms held together with metal bonds• metal bonds

– overlap of orbitals of metal atoms– overlap causes regions of high electron density

where electrons are extremely mobile - conducts electricity

5.3

The

Solid

Sta

teFour Types of Crystalline Solids