chemistry 127 - styrke fitness · these temperature scales are called relative temperature scales....

Chemistry 127

General Chemistry I

(Fall 2011)

Darin J. Ulness

Department of Chemistry

Concordia College Moorhead, MN

Contents

1 Lecture L1: Introduction 1

2 Physical and Chemical Properties 3

2.1 Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.1 Pure Substances and Mixtures . . . . . . . . . . . . . . 4

2.2 Elements, Compounds and Molecules . . . . . . . . . . . . . . 4

2.3 Physical Properties of Matter . . . . . . . . . . . . . . . . . . 5

2.3.1 Extensive and Intensive Properties . . . . . . . . . . . 6

2.4 Physical and Chemical Changes . . . . . . . . . . . . . . . . . 6

2.5 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Measurement 10

3.1 Units of Measure . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.1.1 Reporting Numbers . . . . . . . . . . . . . . . . . . . . 12

3.2 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Anatomy of the Atom and the Periodic Table 18

4.1 Basic Structure of the Atom . . . . . . . . . . . . . . . . . . . 18

4.2 Atomic Number, Atomic Mass and Isotopes . . . . . . . . . . 19

4.2.1 Atomic number . . . . . . . . . . . . . . . . . . . . . . 19

1

4.2.2 Atomic mass and atomic mass number . . . . . . . . . 19

4.2.3 Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.3 The Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . 21

4.3.1 Group 1A . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.3.2 Group 2A . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.3.3 The B groups . . . . . . . . . . . . . . . . . . . . . . . 21

4.3.4 Group 3A . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.3.5 Group 4A . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3.6 Group 5A . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3.7 Group 6A . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3.8 Group 7A . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3.9 Group 8A . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.4 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5 The Mole 25

5.1 The Mole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.2 Molar Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.2.1 Converting between grams and moles . . . . . . . . . . 26

5.2.2 Percent Composition . . . . . . . . . . . . . . . . . . . 26

5.2.3 Empirical formula . . . . . . . . . . . . . . . . . . . . . 27

5.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2

6 Molecules and Compounds 30

6.1 Chemical Formula . . . . . . . . . . . . . . . . . . . . . . . . . 30

6.2 Naming Molecules . . . . . . . . . . . . . . . . . . . . . . . . . 31

6.3 Group Work: . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6.4 Problem Set: . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7 Ions and Ionic Compounds 34

7.1 Oxidation Number . . . . . . . . . . . . . . . . . . . . . . . . 34

7.2 Polyatomic Ions . . . . . . . . . . . . . . . . . . . . . . . . . . 35

7.3 Ionic Compounds . . . . . . . . . . . . . . . . . . . . . . . . . 36

7.3.1 Naming Ionic Compounds . . . . . . . . . . . . . . . . 36

7.4 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

7.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

8 Chemical Equations and Stoichiometry 40

8.1 Balancing Equations . . . . . . . . . . . . . . . . . . . . . . . 41

8.2 Quantitative Relations in Chemical Reactions . . . . . . . . . 42

8.3 Mass Relations in Chemical Reactions . . . . . . . . . . . . . 42

8.4 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

8.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

9 Limiting Reagents 47

9.1 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

9.2 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3

10 Net Ionic Equations 51

10.1 Ions in Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 51

10.1.1 Electrolytes . . . . . . . . . . . . . . . . . . . . . . . . 52

10.2 Precipitation Reactions . . . . . . . . . . . . . . . . . . . . . . 52

10.3 Net Ionic Equations . . . . . . . . . . . . . . . . . . . . . . . . 53

10.4 Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . 53

10.4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 53

10.4.2 A Word on the Hydrogen Ion . . . . . . . . . . . . . . 54

10.4.3 Reactions of Acids and Bases . . . . . . . . . . . . . . 55

10.5 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

10.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

11 Redox Reactions 58

11.0.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 58

11.0.2 Redox reactions . . . . . . . . . . . . . . . . . . . . . . 58

11.1 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

11.2 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

12 Concentration and Molarity 62

12.1 Concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

12.2 Molarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

12.3 pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

12.4 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

12.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4

13 Titration Reactions 68

13.1 Titrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

13.2 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

13.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

14 Energy, Heat and the First Law of Thermodynamics 72

14.1 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

14.1.1 Conservation of energy . . . . . . . . . . . . . . . . . . 72

14.1.2 Work and heat . . . . . . . . . . . . . . . . . . . . . . 73

14.2 The First Law of Thermodynamics . . . . . . . . . . . . . . . 73

14.3 Temperature and Heat . . . . . . . . . . . . . . . . . . . . . . 73

14.3.1 Heat capacity . . . . . . . . . . . . . . . . . . . . . . . 73

14.4 Phase Changes . . . . . . . . . . . . . . . . . . . . . . . . . . 74

14.5 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

14.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

15 Enthalpy 77

15.1 Work and the First Law . . . . . . . . . . . . . . . . . . . . . 77

15.2 Enthalpy Changes for Reactions . . . . . . . . . . . . . . . . . 79

15.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

15.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

16 Lecture L19: Calorimetry 82

16.1 Calorimetry and Heats of Combustions . . . . . . . . . . . . . 82

16.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 83

16.2 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

16.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5

17 Hess’ Law 86

17.1 State Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 86

17.2 Hess’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

17.2.1 Heats of formation . . . . . . . . . . . . . . . . . . . . 87

17.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

17.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

18 Light 92

18.1 Electromagnetic Radiation . . . . . . . . . . . . . . . . . . . . 92

18.2 The Photon Description . . . . . . . . . . . . . . . . . . . . . 93

18.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

18.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

19 The Fall of Classical Physics and the Bohr Model 97

19.1 Brief Overview . . . . . . . . . . . . . . . . . . . . . . . . . . 97

19.2 Bohr’s Atomic Theory . . . . . . . . . . . . . . . . . . . . . . 98

19.2.1 First attempts at the structure of the atom . . . . . . . 98

19.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

19.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

20 Atomic Orbitals 104

20.1 The Modern Theory of the Hydrogen Atom . . . . . . . . . . 104

20.2 The Quantum Numbers of the Hydrogen Atom . . . . . . . . . 105

20.3 Visualizing the Atomic Orbitals . . . . . . . . . . . . . . . . . 106

20.4 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

20.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

6

21 Multielectron Atoms and the Pauli Exclusion Principle 110

21.1 The Other Atoms . . . . . . . . . . . . . . . . . . . . . . . . . 110

21.2 The Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . 111

21.3 Electronic Configurations of Atoms . . . . . . . . . . . . . . . 111

21.4 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

21.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

22 Periodic Trends 115

22.1 Periodic Properties of the Atoms . . . . . . . . . . . . . . . . 115

22.2 Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

22.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

22.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

23 Lewis Dot Structures I 119

23.1 Fundamentals of Chemical Bonding . . . . . . . . . . . . . . . 119

23.2 Valence Electrons . . . . . . . . . . . . . . . . . . . . . . . . . 119

23.3 Lewis Dot Structures . . . . . . . . . . . . . . . . . . . . . . . 119

23.4 Chemical Bond Formation . . . . . . . . . . . . . . . . . . . . 120

23.5 Algorithm for Drawing Lewis Structures . . . . . . . . . . . . 123

23.6 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

23.7 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

24 Lewis Dot Structure II 126

24.1 Formal Charge . . . . . . . . . . . . . . . . . . . . . . . . . . 126

24.2 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

24.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

7

25 Properties of Chemical Bonds 129

25.1 Properties of the Chemical Bond . . . . . . . . . . . . . . . . 129

25.2 Charge Distribution in Covalent Compounds . . . . . . . . . . 130

25.2.1 Electronegativity and bond polarity . . . . . . . . . . . 130

25.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

25.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

26 The VSEPR Model 132

26.1 VSEPR Model . . . . . . . . . . . . . . . . . . . . . . . . . . 132

26.2 Molecular Dipole Moments . . . . . . . . . . . . . . . . . . . . 134

26.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

26.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

27 Valence Bond Theory 137

27.1 Valence Bond Theory . . . . . . . . . . . . . . . . . . . . . . . 137

27.2 Multiple Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . 138

27.3 Hybridization . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

27.4 Multiple Bonds Revisited . . . . . . . . . . . . . . . . . . . . . 140

27.5 Representing the 3D structure of molecules . . . . . . . . . . . 140

27.6 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

27.7 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

28 Symmetry 144

28.1 Symmetry Elements . . . . . . . . . . . . . . . . . . . . . . . 144

28.2 Symmetry Point Groups . . . . . . . . . . . . . . . . . . . . . 145

28.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

8

29 Intermolecular Forces 147

29.1 Polarizability and Induced Dipoles . . . . . . . . . . . . . . . 147

29.2 Intermolecular Forces . . . . . . . . . . . . . . . . . . . . . . . 148

29.3 Phase Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . 150

30 Hydrogen Bonding and Water 152

30.1 Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

30.1.1 The Grotthuss mechanism . . . . . . . . . . . . . . . . 154

30.2 Acids and Bases Revsited . . . . . . . . . . . . . . . . . . . . 154

30.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

30.4 Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

31 Ideal Gases I 157

31.1 Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

31.2 The Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . 157

31.2.1 A word on units . . . . . . . . . . . . . . . . . . . . . . 158

31.2.2 Ideal gas law in terms of density . . . . . . . . . . . . . 159

31.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

31.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

32 Ideal Gases II 162

32.1 Gas Mixtures and Partial Pressure . . . . . . . . . . . . . . . 162

32.2 Chemical Reactions Between Gases . . . . . . . . . . . . . . . 163

32.3 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

32.4 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

9

33 Kinetic Theory of Gases 167

33.1 The Kinetic Molecular Theory of Gases . . . . . . . . . . . . . 167

33.2 Group Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

33.3 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168

i

Lecture NotesChemistry 127: General Chemistry I

Darin J. Ulness

Fall 2011

ii

1. Lecture L1: Introduction

These notes are designed to cover what I will be writing on the board during

the video lectures. You should have these notes in front of you when you

watch. The idea is to give you more time to think about what is being said

rather than scrambling to write down the notes. One disadvantage is that it

may be easier to gap out for awhile. To get the most out of video lectures you

will want to avoid this. Also, try to find a dedicated place and time to watch

without disctraction. In the videos I will say things that are in addition to

the notes so you will want to write those things down as well. The notes are

organized such that each chapter is a lecture. Watching the video lectures as

part of your study time for this class will leave the class time for more active

learning with group work and other in-class activities. It can not be stressed

enough that the group work is as important or even more important than the

lecture itself. Some problems you work on in groups will introduce concepts

that we do not discuss in lecture. You will be expected to know the group

material at the same level as the lecture material when it comes time for the

exams. So, make sure that you as an individual understand what your group

is discussing. Finally, the problem sets for each lecture are given at the end

of each section. These problems are due at the beginning of the next lecture

period.

1

Part I: Material for Exam 1

2

2. Physical and Chemical Properties

The video lectures over this material are (*** these are optional

***)

• V010101• V010102• V010103

2.1 Matter

Matter is the “stuff” materials are made of–anything that takes up space

and is perceptible by the senses in some way.

Common states (or phases) of matter:

• Solid— rigid shape

— fixed volume

• Liquid— non-rigid shape–takes the form of the container

— fixed volume

• Gas— non-rigid shape–takes the form of the container

— variable volume–expands to fill the container

Uncommon states of matter:

• Plasma

3

— A “gas” of free charges–free electrons and bare nuclei.

— Exists at very high temperatures

• Dark matter— Proposed–not directly observed

— Exotic form of matter–unlike any of the above forms

— Amount determines the ultimate fate of the universe

2.1.1 Pure Substances and Mixtures

A pure substance is a piece of matter that cannot be separated into different

species by any physical technique.

A mixture is a piece of matter that is a physical combination of the at least

two different pure substances each of which maintains its own microscopic

identity.

Types of mixtures:

• Heterogeneous— A mixture which consists of regions of pure substances

• Homogeneous— Amixture in which the pure components are completely dispersed

on the microscopic level.

2.2 Elements, Compounds and Molecules

All matter is made up of atoms.

The word atom means indivisible. It was believed that atoms were the small-

est pieces of matter. We now know that atoms themselves have component

parts which we will learn about over the course of the year.

There are now over 112 different types of atoms which are listed in the

periodic table (only 90 of which occur naturally).

4

Each of these different types of atoms correspond to a different element.

An element is a pure substance consisting of only one type of atom.

If all there was were the 90 naturally occurring elements, the universe would

not be very exciting.

Fortunately these elements can be combined with each other via a chemical

bond in all sorts of different ways to form what are called compounds and

molecules.

A compound is a collection of at least two chemically combined (bonded)

atoms of different types.

Many compounds can also be called molecules; the choice of word to use

often depends on the type of chemical bond, but we have not learned enough

to make the distinction yet.

It is important to note, however, that a molecule is not a subclass of com-

pound. For example, oxygen normally exists as the molecule O2, but since

both atoms in the molecule are oxygen so, technically, it is not a compound.

2.3 Physical Properties of Matter

Substances are recognized and distinguished by their physical properties.

Some physical properties include

• color• density• conductivity• state (or phase) at a particular temperature

In general physical properties are temperature dependent.

A quick aside on temperature scales:

Temperature scales can be chosen completely arbitrarily.

One example is the familiar Fahrenheit scale. This scale was originally cho-

sen such that 0◦F was the freezing point of a saturated salt water solution

5

and 100◦F was normal body temperature (which turned out to actually be98.6◦F).

Another example is the Celsius scale. This scale is chosen such that 0◦C and100◦C are the freezing point and boiling point of pure water respectively.

These temperature scales are called relative temperature scales.

A better scale is the Kelvin scale which is an absolute temperature scale.

This scale is defines 0K as the lowest possible temperature–a point called

absolute zero. Kelvin units are also chosen to coincide with the Celsius units.

That is a change of one kelvin is equivalent to a change of one degree Celsius.

The conversion from Celsius to Kelvin is

Kelvin = Celsius + 27315

2.3.1 Extensive and Intensive Properties

What about, say, mass () or volume ( )?

These are in fact physical properties, but they are not useful for distinguishing

matter because they depend on the amount of material present.

Physical properties fall into two categories depending on whether or not they

depend in the amount of material.

Extensive properties depend on the amount of material

Intensive properties do not depend on the amount of material

Although extensive properties are not useful for distinguishing materials,

one can always define an intensive property as the ratio of two extensive

properties.

For example, density =

2.4 Physical and Chemical Changes

Any change of any physical property of a substance is called a physical change

provided the molecular make-up of the substance remains the same.

Any change that involves a change in the molecular make-up of a substance

to form one or more new substances is called a chemical change.

6

For a chemical change, a chemical reaction takes place.

For a physical change no chemical reactions take place.

2.5 Group Work

Get together with your group to work on the following questions.

1. Why are extensive properties not useful in distinguishing mater?

2. The following are a list of properties that we will encounter as we

go along. For those properties that you are familiar with, classify as

extensive or intensive: temperature, heat capacity, viscosity, electrical

conductivity, color, index of refraction.

Let us take a closer look at temperature.

3. Temperature is not heat (heat has units of energy), but it is related

to heat. One definition of temperature is that it gives the direction

of heat flow. Based on your intuition and daily experience answer the

following.

(a) If object A is placed in thermal contact with object B and heat en-

ergy flows from object A to object B, what must be true regarding

the relative temperatures of object A and B?

(b) What do you suppose it means to say object C and object D are

in thermal equilibrium?

4. Challenge question: The so-called zeroth law of thermodynamics

states that if substance 1 is in thermal equilibrium with substance 2

and substance 2 is in thermal equilibrium with substance 3, then sub-

stance 1 and substance 3 are in thermal equilibrium. Mathematically

this law is written as

if 1 = 2 and 2 = 3 then 1 = 3 (2.1)

This law allows us to build a thermometer. How?

7

2.6 Problem Set

Reading: Chapter 1

Exercises: Chapter 1: 1, 3, 7, 11, 21, 29.

8

3. Measurement

The video lectures over this material are

• V010201• V010202

3.1 Units of Measure

Much of chemistry is about taking and interpreting measurements.

One standard system of measurement is called the metric system and the set

of units are called SI units (SI stands for Système International).

Base SI Units

Property Name Symbol

Mass kilogram kg

Length meter m

Time second s

Temperature kelvin K

Amount mole mol

Current ampere A

Derived Units

10

Property Name Symbol Def. from base

Force Newton N m kg

s2

Pressure Pascal Pa kg

m s2

Energy Joule J m2 kg

s2

Power Watt W m2 kg

s3

Charge Coulomb C s A

Frequency Hertz Hz 1s

Prefixes

Factor Prefix Symbol Factor Prefix Symbol

10−1 deci d 101 deca da

10−2 centi c 102 hecto h

10−3 milli m 103 kilo k

10−6 micro 106 mega M

10−9 nano n 109 giga G

10−12 pico p 1012 tera T

10−15 femto f 1015 peta P

10−18 atto a 1018 exa E

Other units

Often the common unit used in the scientific literature is not an SI unit or

is an SI unit with a different unit. Some that will be important to us are:

• The liter (L) which is a unit of volume equal to a cubic decimeter.• The angstrom (Å) which is a unit of length equal to 1× 10−10 m• The torr (mmHg), the bar (bar), and the atmosphere (atm) whichare units of pressure equal to 133.322 Pa, 100,000 Pa and 101,325 Pa

respectively.

• The atomic mass unit (amu) which is a unit of mass that we will discusslater.

• The micron which is just another name for micrometer (1× 10−6 m)

11

• The fermi which is just another name for femtometer (1× 10−15 m)Bench marks

It is important to begin to get a feel for the sizes of the things we will be

talking about.

One way to do that without trying to memorize everything is to have a few

bench marks in your head.

A chemical bond is typically about 1.5 Å so molecules are on the order of a

few angstroms.

The wavelength of visible light is about 500 nm.

The nucleus of an atom is on the order of a fermi, whereas the atom itself

(including the elctrons) is on the order of an angstrom.

3.1.1 Reporting Numbers

The way scientists report numbers is important.

Several concepts go along with reporting numbers

• significant figures• average (or mean)• precision• accuracy• error

Significant Figures

The number of significant figures used to report a value is determined by the

number of digits provided by the measurement device.

For example some of the balances in the lab report mass to the tenth of a

milligram. So a value might be, say, 3.4175g. The number of significant

digits is five.

Other balances only report mass to the hundreth of a gram, this balance

would read 3.42g. Now the number of significant digits is three.

12

Of the significant digits all but the last digit (the right-most digit) are ‘exact;’

the last digit is rounded.

Notice in the above example the 3 and the 4 are the same for both balances

but the third digit is 1 in the first case and 2 in the second case.

The results of a calculation should be consistent with the number of signifi-

cant figures of its inputs.

∗ ∗ ∗ See “significant figure" section p35 of Kotz et.al. ∗ ∗∗

Average

It is usually the case that a measurement is repeated a number of times. A

scientist does not report all the values obtained, but only reports the average.

The average is determined by summing all the values obtained and then

dividing that result by the number of measurement, :

=

P

=1

(3.1)

whereP

=1 is shorthand notation for the act of summing numbers.

For exampleP3

=1 = (1+2+3) where 1 2 and 3 are three different

numbers.

Precision

Because instruments, techniques and scientists are not perfect, repeated mea-

surement of the same quantity gives slightly different values.

The precision of a particular method of measurement deals with the repro-

ducibility of the method.

One measure of precision that your book uses is the average deviation which

is defined as

4 =

P

=1 | − |

(3.2)

13

A more common measure of precision and the one that you will use in lab is

the standard deviation

=

sP

=1 ( − )2

− 1 (3.3)

This is a scary looking formula, but once you practice it with several examples

it will seem much less intimidating.

The scientist almost always reports a value as the average plus or minus some

measure of the precision (we will use average deviation). That is, ±

Accuracy

Accuracy deals with how a measurement deviates from the accepted value.

A measure of accuracy is the error,

error = − ac (3.4)

or better yet percent error,

%error =− ac

ac× 100% (3.5)

It is important to be clear that precision and accuracy are not the same.

14

3.2 Group Work


1. One mole of carbon should weigh exactly 12.011g. Libby went into the

lab and recorded the mass of one mole of carbon three times. She got

12.009 g, 12.008 g and 12.011 g.

(a) How many significant digits is Libby working with?

(b) What is the average value of her measurements? [ans: 12.009

g]

(c) What is the standard deviation of her measurements? [ans:

0.0015]

(d) What is her percent error? [ans: -0.016%]

2. Find the average and standard deviation of the following measurements.

15 Å, 13 Å, 17 Å, 19 Å. [ans: 16, 2.6]

3. Why do you suppose percent error is a better measure of error then

simply the error?

4. The shortest light pulse ever created is about 6 fs. What is that in

seconds? How about in minutes?

5. My computer at home has a 2.0 GHz processor. What is that in mega-

Hertz?

6. A regular light microscope can resolve objects only as small as about

two optical wavelengths. Using your bench marks, what are the typical

units used to describe the lengths of objects viewed under a microscope?

7. Use your bench marks to give a estimate of the radius of an atom

relative to the radius of its nucleus.

3.3 Problem Set

Reading: Chapter “Let’s Review”

Exercises: Chapter “Let’s Review”: 1, 5, 7, 13, 17, 21, 23, 25, 29, 31, 35, 39,

57, 61, 65

15

4. Anatomy of the Atom and the Periodic Table


• V020201• V020202

4.1 Basic Structure of the Atom

The atom was once thought to be indivisible, but it is now known that the

atom consists of a nucleus which contains two types of components: protons

and neutrons.

The atom also contains electrons which encompass the nucleus.

Electrons

• Electrons are extremely light weight objects which carry a negativeelectric charge.

• The electron is what is called an elementary particle since it is indivis-ible.

Protons

• Protons are relatively heavy objects compared to the electron (about2000 time heavier).

• They carry a positive electric charge which is exactly equal and oppositethat of the electron charge.

• It is the electrostatic force between electron and proton that holds theatom together

• Protons are not elementary particles, but are made up of even smallerobjects called quarks which are elementary particles.

18

Neutrons

• Neutrons are very similar to protons but they are electrically neutral.• Neutrons are also made up of quarks.

4.2 Atomic Number, Atomic Mass and Isotopes

4.2.1 Atomic number

All of the different types of atoms each corresponding to a different element

are listed on the periodic table.

The elements are arranged in columns and rows.

Each column contains a group of different types atoms which exhibit many

similar characteristics.

The feature that distinguishes atoms of different elements is the number of

protons contain in the nucleus.

This number is called the atomic number (labelled ).

So if an atom has, say, 6 protons ( = 6) for example it must be a carbon

atom regardless of how many neutrons it may have.

The atoms are placed into the periodic table according to increasing atomic

number.

4.2.2 Atomic mass and atomic mass number

Different atoms not only have different atomic numbers but they also have

different atomic masses.

The unit of mass for atoms is the amu (atomic mass unit). The amu is

defined such that a carbon atom with six protons and six neutrons weighs

exactly 12 amu.

The atomic mass of any particular atom is very close to (but not exactly) the

sum of the number of protons and neutrons. This sum is called the atomic

mass number. So the atomic mass is very close to, but not exactly the atomic

mass number.

19

For our purposes it will be sufficient to take the atomic mass and the atomic

mass number to be the same.

So, if a carbon had six protons and six neutrons its atomic mass number

would be 12 and written as 12C.

But if it had six protons and seven neutrons its atomic mass number would

13 (written as 13C) and its mass would be very close to 13 amu (so close that

we would take it to be 13).

4.2.3 Isotopes

The fact that the same type of atom can have different numbers of neutrons

is very important and in fact it has been exploited in many techniques.

Atoms of the same type, which must have the same number of protons, but

having different numbers of neutrons are called isotopes.

A natural sample of an element will contain all isotopes of that element.

The different isotopes will be present in their natural abundance. The percent

abundance of a given isotope is given by

% abundance =number of atoms of the given isotope

total number of atoms of all isotopes of that element×100%(4.1)

The mass given on the periodic table is the weighted average of the all the

naturally occurring isotopes.

So, for example, carbon naturally appears as 12C 98.9% of the time and 13C

1.1% of the time.

Thus the mass given on the periodic table for carbon is

0989| {z }989%

× 12000| {z }12C

+ 0011| {z }11%

× 13000| {z }13C

= 12011 amu| {z }periodic table

rather than 12.000 amu.

Note we did not use the exact mass of 13C which is actually 13.003 amu,

because this information will usually not be readily available to us.

20

4.3 The Periodic Table

As we have stated before, the elements are arranged in the periodic table

with each column listing a group of elements with similar properties.

We will now briefly mention some of the properties of each of the groups

4.3.1 Group 1A

Group 1A contains Hydrogen and the alkali metals.

The alkali metals are relatively soft and all very chemically reactive. In

particular they react violently with water to form hydrogen gas and produce

an alkaline (basic) solution.

All the elements in group 1A form compounds with oxygen (called oxides)

which have the general formula R2O where R is any of the element.

For example H2O, Li2O, Na2O etc.

4.3.2 Group 2A

The alkaline earth metals make up group 2A.

The alkaline earth metals are less reactive than the alkali metal, nonetheless

they too react with water to yield an alkaline solution (except for Be).

The group 2A elements form oxides of the from RO. For example, BeO, MgO,

CaO etc.

Many of these oxides themselves form alkaline solutions.

4.3.3 The B groups

The B groups are a collection of metals called transition metals.

Many of the familiar metals are transition metals (copper, nickel, gold, silver,

iron etc.)

4.3.4 Group 3A

Group 3A also consists mainly of metals except for Boron which is a metalloid–

it has properties intermediate between those of metals and nonmetals

21

The Group 3A oxides are of the form R2O3. For example B2O3 Al2O3 etc.

4.3.5 Group 4A

Group 4A has a real mix of elements: carbon is a nonmetal, silicon and

germanium are metalloids, and tin and lead are metals.

Carbon is the single most important element; mainly because it can form

extended bonding patterns which result in a enormous variety of molecules.

This property, unique to carbon, suggests that carbon based life may be the

only type of chemical-based life possible.

Silicon, because it is a metalloid, has found very important usage as the basis

for semiconductor devices.

The oxides of the Group 4A elements are of the form RO2 For example,

CO2 SiO2 etc.

4.3.6 Group 5A

Group 5A also contains a mix of elements with nitrogen and phosphorous

being nonmetals, arsenic and antimony being metalloids, and bismuth being

a metal.

4.3.7 Group 6A

Group 6A contains three nonmetals–oxygen, sulphur and selenium, a met-

alloid tellurium and a metal polonium

4.3.8 Group 7A

The Group 7A elements are called halogens. All but astatine exist naturally

as diatomic molecules (R2). For example, F2, Cl2 etc.

The halogens are very reactive. They react with many of the metals to form

halide salts. For example NaCl (sodium chloride), NaBr (sodium bromide),

etc.

4.3.9 Group 8A

The elements in group 8A are called the noble gases.

22

These elements are essentially unreactive–It is very difficult to form mole-

cules with these elements.

4.4 Group Work


1. Chlorine has two naturally occurring isotopes. The one with 18 neu-

trons occurs 75.8% of the time and the one with 20 neutrons occurs

the remainder of the time.

(a) Give the atomic mass number for each of chlorine’s two isotopes.

(b) Calculate the weighted average mass for chlorine. Does your num-

ber jive with that given on the periodic table? [ans: 35.5 amu]

2. A technique calledmass spectrometry, which you will use in lab, can dis-

tinguish masses that differ by only one amu. By comparing the counts

registered by the mass spectrometer at different masses one can deter-

mine the relative amounts of material present. In an experiment using

magnesium it is found that at mass 24 amu there was 3300 counts, at

25 amu there was 418 counts and at 26 amu 460 counts were measured.

Calculate the percent abundance for each of the isotopes of magnesium.

[ans: 24: 78.99%, 25: 10.00%, 26: 11.01%]

4.5 Problem Set

Reading: Chapter 2.1—2.5

Exercises: Chapter 2: 5, 7, 11, 15, 17, 19, 21, 23, 25, 27, 29, 101, 109

23

5. The Mole


• V030101• V030102

5.1 The Mole

In the lab, a chemist deals with, say, grams of chemicals. That is, a chemist

might mix two grams of one chemical with one gram of another to produce

1.5 grams each of two different products.

There needs to be a way to translate a convenient laboratory amount of

material to the number of atoms molecules or compounds used.

The unit of measure which does this is the mole.

A mole is defined as the number of atoms in exactly 12 grams of carbon—12.

That number is 6.022×1023 and is called Avogadro’s number.

1 mole = 6022× 1023 particles (5.1)

So the mole is just a count–just like a dozen is a count. It has no units.

5.2 Molar Mass

The mole is the simply the conversion factor from amu to grams: 1 g =

6022× 1023 amu.The mass of one mole of atoms, molecules, or compounds is called the molar

mass (given the symbol )

For atoms the molar mass in grams is numerically equal to the atomic mass

in amu. For example one mole of sodium atoms has a mass of 22.9898 g.

25

For molecules and compounds the molar mass in grams is numerically equal

to the sum of the atomic masses (in amu) of the atoms which make-up the

compound. For example on mole of sodium chloride has a mass of 229898+

354527 = 584425 g

5.2.1 Converting between grams and moles

One of the most important conversions that you will need to become very

good at is converting from grams to mole and converting frommoles to grams.

Example: grams to moles.

Let’s you have 100.00 g of sodium chloride (NaCl). How many moles do you

have.10000 g/ NaCl 1 mol NaCl

584425 g/ NaCl= 17111 mol NaCl (5.2)

Example: moles to grams.

Let’s you have 3.500 mol of sodium chloride (NaCl). How many grams do

you have.

3.500 mol/ NaCl 584425 g NaCl

1 mol/ NaCl= 2045 g NaCl (5.3)

5.2.2 Percent Composition

Since a particular type of molecule or compound always has the same number

of component atoms, the mole allows one to determine the percent composi-

tion by mass of each of the component elements of the molecule or compound.

For example, let us determine the percent composition by mass of carbon

and oxygen in carbon dioxide (CO2).

One mole of carbon dioxide has mass of 12011 + 2× 159994 = 44010 gFor every mole of carbon dioxide there is one mole of carbon atoms and two

moles of oxygen atoms

So of the 44.010g for carbon dioxide,

12011 g

44010 g× 100% = 2729% (5.4)

26

is carbon by mass and

2× 159994 g44010 g

× 100% = 7271% (5.5)

is oxygen by mass

5.2.3 Empirical formula

One can work this calculation backwards to determine the so-called empirical

formula for a substance.

Pretend for the moment that we do not know carbon dioxide has the molec-

ular formula CO2

Lets say we perform an experiment on 100 g of carbon dioxide and find three

results

• The only elements present in the sample were carbon and oxygen• Our experiment found that 27.29 g of carbon was present.• Our experiment found that 72.71 g of oxygen was present.

We can analyze our data as follows.

2729 g C× 1 mol C

12011 g C= 2272 mol C (5.6)

7271 g O× 1 mol O

159994 g O= 4544 mol O (5.7)

From the ratio of these two results we find the empirical formula

2272 mol C

4544 mol O=1 mol C

2 mol O(5.8)

That is the empirical formula of carbon dioxide is CO2 This happens to be

the molecular formula also.

The difference between the empirical formula and the molecular formula is

that the empirical formula only gives the molar ratio of the component atoms.

Whereas the molecular formula gives the total number of each atom.

For example, the molecular formula of benzene is C6H6 but the empirical

formula is CH since the ratio of C to H is 1 to 1.

27

5.3 Group Work


1. Why do we use the weighted averaged mass from the periodic table in

determining the molecular mass rather than the atomic mass numbers?

2. How many moles are in 50.00 g of benzene (C6H6). [ans: 0.640

mol]

3. How many grams are in 2.750 mol of carbon tetrachloride (CCl4)?

[ans: 423.0 g]

4. Hexane has the formula C6H14 and cyclohexane has the formula C6H12

Fred has a 100.0g sample of one or the other. In a elemental analysis

experiment he obtained the following data. 85.6 g of carbon present,

14.4 g of hydrogen present. Does Fred have hexane or cyclohexane?

5. Challenge problem: Make up a problem similar to the previous one

that involves chloroform (HCCl3) and methylene chloride (H2CCl2).

Make sure you problem has a solution.

5.4 Problem Set


Exercises: Chapter 2: 61, 63, 65, 67, 69, 71, 73, 75, 79, 81, 83, 85, 87, 89, 91,

93, 95.

28

6. Molecules and Compounds


• V020101• V020102

Compounds and molecules are collections of atoms.

• Compounds are collections of atoms in which there are at least twodifferent elements

• Molecules are collections of atoms that may be compounds or elements— For example a carbon dioxide molecule (CO2) is also a compound.

These type of molecules are called molecular compounds

— But, oxygen (O2) is a molecular element and not a compound

Note: some compounds are not typically called molecules. Table salt NaCl

is one example. It is to a little too early yet to understand the difference.

6.1 Chemical Formula

There are a number of ways chemists represent a molecule or compound. The

representation of a molecule or compound is called a chemical formula.

There are several different types of chemical formulae and the one you would

use depends on the amount of detail you want to convey.

• condensed formula— most concise way of representing a molecule

— more than one molecule can have the same condensed formula

• structural formula (without bonds)

30

— The atoms are listed roughly according to how they are connected

• structural formula (with bonds)— solid lines are drawn for each bond which connects two atoms of

the molecule together

Let us consider the example of a molecule called butanol.

Butanol has four carbon atoms, ten hydrogen atoms and one oxygen atom.

The figure shows the various formulae representation of butanol

6.2 Naming Molecules

There is a systematic way to name molecules, but many molecules also have

common names.

The systematic naming of molecules gets complicated for those having many

atoms.

You will spend a lot of time on this next year in organic chemistry.

We will focus only on binary molecular compounds (two different types of

atoms)

Naming binary molecular compounds

31

• For the first element listed use di-, tri-, tetra-, penta- etc. as appropri-ate

• For the second element listed use mono-, di-, tri-, tetra-, penta- etc asappropriate and ide ending

— For example, N2O is dinitrogen monoxide, NO2 is nitrogen dioxide.

6.3 Group Work:

worksheet

6.4 Problem Set:

*** Note problems are due with the next lectures set. ***

Reading: 2.6

Exercises: 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.

32

7. Ions and Ionic Compounds


• V020103• V020104

The bonds in molecules generally involve a sharing of electrons between the

two connected atoms. These types of bonds are called covalent bonds (more

on this latter)

In contrast, for a different class of compounds called ionic compounds the

bonds generally form ionic bonds which arise when one atom “steals” an

electron from the adjacent atom.

The atom which gains the electron now has an excess negative charge. It is

called an anion.

The atom which loses an electron now has a positive charge. It is called a

cation.

When speaking generically, that is, when one is not interested in the sign of

the charge, the charged atoms are called ions.

The positive and negative ions are attracted to on another via the electro-

static force between them.

7.1 Oxidation Number

More than one electron may be gained or lost by an atom.

The charge of the atom is counted in units of the charge of one electron and

it is called the oxidation number (or oxidation state) of the atom.

So, if the atom loses one electron to become a cation, it has a charge of +1

unit; its oxidation number is 1+

34

If the atom loses two electrons to become a cation, it has a charge of +2 unit;

its oxidation number is 2+

Likewise if the atom gains one electron, it has a charge of −1 and so itsoxidation number is 1− etc.As it turns out, most metals tend to give-up electrons to form cations and

most nonmetals accept electrons to from anions.

The alkali metals readily give-up one electron and so they all form cations

having a 1+ oxidation number.

The alkaline earth metals easily give-up two electrons to form cations with

2+ oxidation numbers.

The halogens readily gain one electron to form an anion with oxidation num-

ber 1−The group 6A elements tend to gain two electrons–oxidation number 2−

Ions are written in the form Xox.# (e.g., Na+ O2− Cl− etc.–the “1” isdropped for 1+ and 1−).Recall that the noble gases are particularly unreactive. This means that they

are stable as single atoms.

Notice that when an atom loses or gains electrons it now has exactly the

same number of electrons as the appropriate noble gas.

For example, Na+ has exactly as many electrons as Ne, Cl− has as manyelectrons as Ar, etc.

This allows us to predict the favorable oxidation numbers for many elements.

The story for the transition metals is a bit more complicated so, for now, we

will just have to get a feel for their oxidation numbers through practice.

7.2 Polyatomic Ions

Molecules may also gain or lose electrons to become polyatomic ions. Some

examples are carbonate CO2−3 and ammonium NH+4

For more see your book, Table 2.2, p 53.

35

7.3 Ionic Compounds

Although the ions lose or gain electrons to obtain a more stable number of

electrons, the ions carry an electric charge which is not a very stable situation

itself.

Consequently, cations combine with anions to form more stable neutral ionic

compounds.

An example is table salt NaCl which is an ionic compound formed by com-

bining Na+ and Cl− (Na++Cl− →NaCl). Note that the charge on NaCl iszero.

7.3.1 Naming Ionic Compounds

It is important that chemists use a precise and unambiguous naming system

for compounds, so that everyone understands each other.

Naming Cations

• For a monatomic (single atom) ion, give the name and add the word“ion”

— For example, Na+ is sodium ion.

• If an atom has more than one stable oxidation state, use roman numer-als to indicate the oxidation number.

— For example, Cu+ is copper(I) ion and Cu2+ is copper(II) ion.

• The only polyatomic cation that we will run into is NH+4 which is calledammonium ion.

Naming anions

• For monatomic anions add “ide” to the end of the name and add theword “ion.”

— For example, Cl− is chloride ion

• Polyatomic anions must simply be learned through experience.

36

• A class of polyatomic anions called oxoanions follow the series illus-

trated in the example below

— ClO−4 is perchlorate ion, ClO−3 is chlorate ion, ClO

−2 is chlorite ion

and ClO− is hypochlorite.

Naming ionic compounds

• Build the name of the ionic compound from the names of the ions withthe cation coming first.

— For example NaCl is sodium chloride

— NH4ClO−4 is ammonium perchlorate

7.4 Group Work


1. Why does 1+ make sense for the oxidation state of an ion that loses

an electron (instead of 1−)?2. Name the following compounds and give the oxidation number for the

cation and anion.

(a) Na2SO4

(b) CaO

(c) CuCl

(d) CuCl2

3. Give the formula for the following compounds and give the oxidation

number for the cation and anion

(a) cesium sulfide

(b) iron(II) chloride (chloride has oxidation number 1−)

37

7.5 Problem Set

Reading: 2.7—2.8

Exercises: 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.

38

8. Chemical Equations and Stoichiometry


• V030201• V030202

During a chemical change, or chemical reaction, a chemical or set of chemicals

react and produce a new chemical or set of chemicals.

The initial chemical(s) which react are called reactants.

The chemical(s) which are produced are called products.

A chemical reaction is generally written as the chemical equation

reactants→ products (8.1)

For example, the equation for a reaction between hydrogen and oxygen to

produce water is written as

2H2 +O2 → 2H2O (8.2)

Notice in this example that exactly twice as much H2 as O2 must react. This

reaction is said to be balanced

In general there is a precise relationship between the quantities of reactants

and the quantities of products. This is called stoichiometry.

In the water formation example the 2’s in front of the H2 and H2O ande

the 1 in front of the O2 (not shown explicitly) are called the stoichiometric

coefficients.

40

8.1 Balancing Equations

When one works with a chemical equation it must be balanced. That is, each

element must appear in equal quantities for the reactants and the product.

In the above balanced equation for the production of water, the reactants

contain four H’s and two O’s and so does the product.

Throughout the semester, we will be balancing many different types of equa-

tions. Here we will practice with a particular type of reaction called a com-

bustion reaction (we consider only the combustion of hydrocarbons).

A combustion reaction involves burning (reacting) a hydrocarbon in oxygen

to from water and carbon dioxide.

For example consider the combustion of benzene (C6H6).

• An unbalanced equation isC6H6 +O2 → CO2 +H2O (8.3)

• First balance the carbons

C6H6 +O2 →↓6CO2 +H2O (8.4)

• Next balance the hydrogens

C6H6 +O2 → 6CO2 +↓3H2O (8.5)

• Finally balance the oxygens

C6H6 +

↓

71

2O2 → 6CO2 + 3H2O (8.6)

• The above equation is balanced, but often one wants the stoichiometriccoefficients to be integers. To accomplish this just multiply every thing

by 2.

2C6H6 + 15O2 → 12CO2 + 6H2O (8.7)

• Always check your equation when you are done.

41

— 12 carbons√

— 12 hydrogens√

— 30 oxygens√

8.2 Quantitative Relations in Chemical Reactions

The stoichiometry of a balanced chemical equation gives a quantitative rela-

tion between the chemical speices in the reaction.

For concreteness let us consider the combustion of benzene,

2C6H6 + 15O2 → 12CO2 + 6H2O. (8.8)

This equation tells us that exactly 2 moles of benzene reacts with exactly 15

moles of oxygen to produce exactly 12 moles of carbon dioxide and exactly

6 moles of water.

So, if we had, say, 0.30 moles of benzene than we know that exactly 152×030 =

225 moles of oxygen reacts.

Also 122× 030 = 18 moles of carbon dioxide and 6

2× 030 = 090 moles of

water.

The stoichiometry of a reaction gives us a conversion factor for converting

moles of one species to moles of another.

For the above reaction we would write

030 mol/ C6H6/ 15 mol O22 mol/ C6H6/

= 225 mol O2

8.3 Mass Relations in Chemical Reactions

The balanced equations are in terms of moles. That is, the stoichiometric

coefficients correspond to the number of moles of each of the reactants and

products.

However, often times in the lab, the convenient measure of quantity is mass.

So it is important to know how to figure out from the stoichiometry how

much mass of each reactant is needed to produce a desired mass of product.

42

Again let us consider the combustion of benzene reaction from above,

2C6H6 + 15O2 → 12CO2 + 6H2O. (8.9)

Lets say that for some reason we wanted to generate 100g of water.

• First determine how many moles of water are in 100g

100 g/ H2O 1 mol H2O

18015 g/ H2O= 555 mol H2O (8.10)

• Next we use the stoichiometry to determine the moles of the otherproducts and the reactants

555 mol/ H2O 12 mol CO26 mol/ H2O

= 111 mol CO2 (8.11)

555 mol/ H2O 15 mol O26 mol/ H2O

= 139 mol O2 (8.12)

and

555 mol/ H2O 2 mol C6H66 mol/ H2O

= 185 mol C6H6 (8.13)

• Now convert the moles of each of chemicals in the previous step intograms

111 mol/ CO2 4401 g CO21 mol/ CO2

= 489 g CO2 (8.14)

139 mol/ CO2 31999 g O21 mol/ O2

= 445 g O2 (8.15)

and

185 mol/ C6H6 78113 g C6H61 mol/ C6H6

= 145 g C6H6 (8.16)

43

8.4 Group Work


1. Balance the following equations

(a) Sn + NaOH → Na2SnO2 + H2

(b) Al + Fe3O4 → Al2O3 +Fe

(c) H3COH + O2 → CO2 + H2O

2. 0.620 moles of ethanol (C2H5OH) is combusted in oxygen according to

C2H5OH + 3O2 → 2CO2 + 3H2O. (8.17)

(a) How many moles of O2 are used? [ans: 1.86 mol]

(b) How many moles of CO2 and H2O are produced? [ans: 1.24

mol CO2 and 1.86 mol H2O]

3. How many moles of oxygen are needed to produce 4.27 moles of Fe2O3?

The reaction is

4Fe + 3O2 → 2Fe2O3 (8.18)

[ans: 6.41 mol]

4. How many grams of nitrogen dioxide are needed to produce 10.3 g of

nitric acid (HNO3) via the reaction

3NO2 + H2O→ 2HNO3 + NO? (8.19)

[ans: 11.2 g]

5. 576 g of tungsten(VI) oxide is reacted with hydrogen to produce pure

tungsten and water:

WO3 + 3H2 → W + 3H2O. (8.20)

How many grams of pure tungsten are produced? [ans: 457 g]

44

8.5 Problem Set

*** Note: Problems are due after the next lecture ***

Reading: Chapter 3.1, 3.2, 4.1, 4.3

Exercises: Chapter 3: 1, 3, 5. Chapter 4: 1, 3, 5, 7, 9, 19, 21, 23, 25, 27, 29,

33, 35, 77, 87, 99.

45

9. Limiting Reagents

The video lecture over this material is

• V030301Consider again the combustion of benzene:

2C6H6 + 15O2 → 12CO2 + 6H2O. (9.1)

What would happen if we had 445g of O2 but we only had 100g of C6H6?

Which reagent would run out first?

To figure this out we need first find the number of moles of each reagent.

445 g/ O2 1 mol O231999 g/ O2

= 1390 mol O2 (9.2)

and100 g/ C6H6 1 mol C6H6

78113 g/ C6H6= 128 mol C6H6 (9.3)

Now we need to account for the stoichiometry of the reaction. Namely that

15 moles of oxygen react with every 2 moles of benzene.

So, we need to convert moles of oxygen to moles of benzene.

1390 mol/ O2 2 mol C6H615 mol/ O2

= 185 mol C6H6 (9.4)

This number is larger than the 1.28 moles available in the 100 g of benzene.

The reaction would proceed until all of the C6H6 was gone. There would still

be O2 left over and we would have generated less than 100g of water.

The C6H6 is said to be the limiting reagent (or limiting reactant) for this

reaction. That is, the extent of the reaction is limited to proceed only until

the C6H6 is all used up.

47

9.1 Group Work


1. Consider the reaction

CO + 2H2 → H3COH, (9.5)

with 36.4 g of CO and 11.2 g of H2 are present.

(a) Which species is the limiting reagent?

(b) How many grams of methanol (H3COH) are produced? [ans:

41.6 g]

2. In lab you will need to calculate theoretical yield and percent yield.

The theoretical yield is the number of grams of product that could be

produced based on the amount of starting material and assuming a

perfect experiment. The percent yield is the actual number of grams

of product divided by the theoretical yield time 100%. Consider the

reaction

C7H6O3 + CH3OH → C8H8O3 + H2O, (9.6)

with 1.50 g of C7H6O3 and 11.20 g of CH3OH present.

(a) What is the limiting reagent?

(b) What is the theoretical yield of C8H8O3? [ans: 1.65 g]

(c) Let’s say Liana got 1.31 g of C8H8O3 during her experiment. What

was Liana’s percent yield? [ans: 79.4%]

9.2 Problem Set

Reading: Chapter 4.2

Exercises: Chapter 4: 11, 12, 13, 14, 15, 16, 17, 18.

48

Part II: Material for Exam 2

49

10. Net Ionic Equations


• V040201• V040202

Many important chemical reactions take place in aqueous solutions.

Aqueous solutions are homogeneous mixtures in which the chemicals of in-

terest are dissolved in water.

In aqueous solutions the water is what is called the solvent and the other

chemicals are called solutes.

10.1 Ions in Solution

We stated earlier that ions formed compounds because the compounds were

more stable than the free ions.

In aqueous solutions, however, sometimes “free” (or solvated) ions are more

stable.

If this is the case, when one places a compound in water the compound

dissolves into solvated ions.

For example consider table salt (NaCl). This compound does in fact dissolve

in water to leave free Na+(aq)

and Cl−(aq)

ions in solution. (The subscript (aq)

reminds us that we are talking about an aqueous solution.)

NaCl→ Na+(aq) +Cl−(aq) (10.1)

51

10.1.1 Electrolytes

Compounds which dissolve in water are called electrolytes.

• If a compound fully dissolves in water it is called a strong electrolyte— example: NaCl

• If a compound does not fully dissolves in water it is called a weakelectrolyte

— example: Acetic acid (CH3CO2H)

• If a compound does not dissolve at all in water it is called a nonelec-trolyte.

— example: AgCl

One can not calculate if a compound will be soluble in water, but there are

some general empirical rules that can be used.

∗ ∗ ∗ See chart on p126 of Kotz et.al. ∗ ∗∗

10.2 Precipitation Reactions

Let us return to our dissolved table salt example.

NaCl→ Na+(aq) +Cl−(aq) (10.2)

Now let’s add another compound, silver nitrate (AgNO3)

AgNO3 → Ag+(aq) +NO−3(aq) (10.3)

But from our solubility rules AgCl is not soluble.

This means that AgCl is more stable as a compound than as solvated ions.

In the above example, the silver ions will combine with the chloride ions

to form what is called a solid precipitate (which in this case is the AgCl

compound)

Na+(aq) +Cl−(aq) +Ag

+(aq) +NO

−3(aq) → Na+(aq) +NO

−3(aq) +AgCl(s) (10.4)

52

(where the subscript (s) means solid)

The whole process discussed above is more simply written as

AgNO3(aq) +NaCl(aq) → AgCl(s) +NaNO3(aq) (10.5)

where is understood that the compounds with the subscript (aq) really means

that these compounds are dissolved into ions.

10.3 Net Ionic Equations

Many times a chemist is only interested in a change that took place during

a reaction.

Let us consider the previous example in this context

AgNO3(aq) +NaCl(aq) → AgCl(s) +NaNO3(aq) (10.6)

is, of course, really

Ag+(aq) +NO−3(aq) +Na

+(aq) +Cl

−(aq) → AgCl(s) +Na

+(aq) +NO

−3(aq) (10.7)

Notice that both before and after the reaction there are sodium and nitrate

ions in solution.

The sodium and nitrate ions are not really doing anything. They are said to

be spectator ions.

For brevity one cancels the spectator ions and writes the so-called net ionic

equation. Which for this example is

Ag+(aq)+NO/−3(aq)

+Na/+

(aq)+Cl−(aq) → AgCl(s) +Na/

+

(aq)+NO/

−3(aq)

(10.8)

Ag+(aq) +Cl−(aq) → AgCl(s) (10.9)

10.4 Acids and Bases

10.4.1 Definitions

There are special types of strong and weak electrolyte compounds which

contain either the hydrogen cation, H+ or the hydroxide anion, OH−

53

• if the electrolyte contains H+ it is an acid— example:HCl

— if it is a strong electrolyte the acid is a strong acid

— if it is a weak electrolyte the acid is a weak acid

• if the electrolyte contains OH− it is a base.— example: NaOH

— if it is a strong electrolyte the base is a strong base

— if it is a weak electrolyte the base is a weak base

Next semester we will encounter several technical definitions of acids and

bases, but for now we can work with the general definitions without worrying

about details.

The general definition of an acid is: any compound that increases the H+

concentration in the solution.

Likewise a base is any compound that increases the OH− concentration inthe solution.

We see immediately that the electrolyte definition of acids and bases are

consistent with these more general definitions.

As the compound is dissolved the H+ or OH− concentration increases.

To see that these are more general definitions of acids and bases than the

electrolyte definition consider adding ammonia to water,

NH3 +H2O→ NH+4 +OH− (10.10)

one hydrogen atom from the water combines with the ammonia molecule to

form the ammonium ion and leave behind the hydroxy ion thus increasing

the OH− concentration. So ammonia is a base.

10.4.2 A Word on the Hydrogen Ion

The hydrogen ion really does not exist as H+ in solution, in fact, it combines

with one or more water molecules for form H3O+, H5O

+2 etc.

54

We will always assume the form H3O+, but we will write H+ whenever it is

convenient to do so.

10.4.3 Reactions of Acids and Bases

Acids and bases react with one another in what are called neutralization

reactions.

A strong acid will react with a strong base to form water and a salt.

Let us represent a generic strong acid as HA and a generic strong base as

BOH

The strong acid—strong base neutralization reaction is written as

HA(aq) + BOH(aq) → BA(aq) +H2O. (10.11)

A specific example would be

HCl(aq) +NaOH(aq) → NaCl(aq) +H2O (10.12)

Notice that the net ionic equation for this reaction is

H+(aq) +OH−(aq) → H2O (10.13)

For now we will treat weak acids and bases in the same manner as strong

acids and bases.

Next semester when we discuss equilibrium we will be more careful to account

for the fact that weak acids and weak bases don’t fully dissociate.

10.5 Group Work


1. Using the solubility rules predict the solubility of the following

(a) PbS

(b) Na2CO3

(c) AgNO3

55

(d) CaI2

2. Write the net ionic equation for the following reactions.

(a) HBr(aq)+ KOH(aq) → KBr(aq) + H2O(l)

(b) AgNO3(aq) + NaBr(aq) → AgBr(s) + NaNO3(aq)

(c) CaS(aq) + 2HBr(aq) → CaBr2(aq) + H2S(g)

(d) Pb(NO3)2(aq) + 2NaBr(aq) → PbBr2(s) + 2NaNO3(aq)

(e) NaCl(aq) + KNO3(aq) → NaNO3(aq) + KCl(aq) (trick question)

3. In a few days we will be talking about the heat energy given off from

reaction. If you mix HCl and NaOH in a beaker, the beaker will get

very hot. I claim that this will happen for the reaction of any pair of

strong acid and strong base. Why can I say this without doing the

experiment?

4. Complete and balance each of the following neutralization reactions.

Identify the acid and the base. Include the phase labels and write the

net ionic equation for each.

(a) NaOH + HNO3 →(b) HCl + Ba(OH)2 →(c) HC2H3O2 + Ca(OH)2 →(d) NH3 + HNO3 →

10.6 Problem Set


Exercises: Chapter 3: 35, 36, 37, 38, 39, 40

56

11. Redox Reactions


• V040301A type of reaction that is fundamentally different from the above types is

the oxidation—reduction reaction. Note: these reactions are also called redox

reactions.

What distinguishes this type of reaction from the others is the fact that

electrons are transferred during the reaction which results in a change in the

oxidation numbers of the involved compounds.

11.0.1 Definitions

We first need a few definitions.

• When a species increases its oxidation number during a reaction it isoxidized.

• When a species decreases its oxidation number during a reaction it isreduced.

• A species which has a tendency to be oxidized is called a reducing agentbecause if it is oxidized the other reactant must be reduced.

• A species which has a tendency to be reduced is called an oxidizing

agent.

11.0.2 Redox reactions

A general form for the oxidation—reduction reaction is

Ahigh +Dlow → Alow +Dhigh (11.1)

58

Here a species A accepts an electron from a donor species D. A is reduced

since its oxidation number decreases and D is oxidized since its oxidation

number increases

For example an oxidation—reduction reaction between zinc sulfate and copper

metal is written as

CuSO4(aq) + Zn(s) → ZnSO4(aq) +Cu(s) (11.2)

It is more illustrative to write the net ionic equation by cancelling the spec-

tator sulfate ion

Cu2+(aq) + Zn0(s) → Zn2+(aq) +Cu

0(s) (11.3)

Here we can clearly see that the oxidation number of the copper is reduced

from 2+ to 0 whereas the oxidation number for zinc is increased from 0 to

2+.

For this reaction two electrons are transferred from each zinc atom to each

copper ion to create the zinc ion and the copper neutral atom.

Zinc is oxidized and copper is reduced.

Zinc acts as a reducing agent and copper acts as an oxidizing agent

11.1 Group Work


1. For the following reactions label the oxidizing agent and the reducing

agent. Also give the oxidation number for each species present (both

reactants and products). As an example:

4−C1+

H4 +0

O2 →4+

C2−O2+ 2

1+

H22−O

(a) P4 + 5O2 → P4O10

(b) Co + Cl2 → CoCl2

(c) 2HCl + Zn → H2 + ZnCl2

(d) ZnO + C → Zn + CO

(e) Fe2O3 + 3CO→ 2Fe + 3CO2

59

11.2 Problem Set

Reading: Chapter 3.7, 3.8

Exercises: Chapter 3: 41, 43, 45, 47, 49, 51, 53, 55, 57

60

12. Concentration and Molarity


• V040101• V040102

12.1 Concentration

An important piece of information about aqueous solutions is the concentra-

tion of each of the components.

Concentration can be measured in a variety of ways

• Molarity, = molesliter

is a measure of how many moles of a given solute

is in one liter of solution.

— We will almost always use molarity as our measure of concentra-

tion.

• Molality, = moles, is a measure of how many moles of a given solute

is in one kilogram of solution.

• Mole fraction, X = is a measure of the ratio of the moles of com-

ponent to the total number of moles in solution.

• Weight-weight percent, % =

× 100% is a measure of the ratioof the weight of species to the total weight of the solution.

12.2 Molarity

We will show here, by way of example, the general procedures for converting

to and from molarity.

The homework will give you practice with a wide variety problems involving

molarity

62

Example 1: Determine the molarity of a solution which is prepared by dis-

solving 30.00g of NaCl in water to make a 500mL solution.

• Molarity is moles per liter so we need to do is convert grams to molesand milliliters to liters

30.00 g/ NaCl 1 mol NaCl 1000 mL/ sol

500 mL/ sol 58.4425 g/ NaCl 1 L sol= 1027 M NaCl

(12.1)

Example 2: How many grams of NaCl are in 0.250 L of a 0.150 M NaCl

solution?

• Now we are given molarity and want grams.

0.250 L/ sol 0.15 mol/ NaCl 58.4425 g NaCl

1 L/ sol 1 mol/ NaCl= 219 g NaCl

(12.2)

Example 3: 0.100 L of pure water is added to 0.500 L of a 0.200 M solution

of NaCl. What is the new molarity after this dilution

• All we need to do here is maintain the same number of moles but changethe volume.

— Number of moles of NaCl:

0.500 L/ sol 0.200 mol NaCl

1 L/ sol= 0100 mol NaCl (12.3)

— New volume

0500 L+ 0100 L = 0600 L (12.4)

— New molarity

0100 mol NaCl

0600 L= 0167 M NaCl (12.5)

63

• There is a short-cut for dilution problems. One can set-up the relation11 =22 (12.6)

where the subscript 1 refers to the original solution and the subscript

2 refers to the final solution after dilution. For this example we have

0200 M× 0500 L = M× 0600 L (12.7)

solving for the unknown final concentration we get

M =0200 M

0600 L/× 0500 L/ = 0167 M NaCl (12.8)

12.3 pH

Water itself is a very weak electrolyte, but an electrolyte nonetheless:

H2O H+ + OH− (12.9)

where the means the reaction does not go to completion as written.

It turns out that for pure water the H+ and OH− concentrations are 1×10−7molar.

We have learned that acids increase the H+ concentration and bases increase

the OH− concentration.

The strength of an acidic or basic solution can thus be characterized by the

concentration H+ or OH−.

Since the concentrations of H+ and OH− are related through the stoichiom-etry of the above reaction, we need only look at one or the other.

The convention is to use the H+ concentration.

Because numbers like 1× 10−7 are ugly, one instead uses the negative log ofthe H+ concentration.

This is the pH of the solution

pH = − log ¡£H+¤¢ (12.10)

where the square brackets in£H+¤means concentration.

64

Thus for pure water

pH = − log ¡1× 10−7¢ = −(−7) = 7 (12.11)

Water is considered neutral.

So if the pH is less then 7 then the H+ concentration is higher than that of

pure water and the solution is acidic.

And if the pH is greater than 7 then the H+ concentration is lower (thus the

OH− concentration is higher) than pure water and the solution is basic.

The lower the pH the stronger the acid solution and the higher the pH the

stronger the base solution.

12.4 Group Work


1. A 75.00 ml volume solution contains 0.0285 moles of AgNO3 What is

the molarity? [ans: 0.3800 M]

2. Considering the above problem, how many grams of AgNO3 are there

in 10 ml of solution? [ans: 0.6456 g]

3. A solution is made by adding 0.984 g of potassium permanganate,

KMnO4 to 100.00 ml of water. What is the molarity. [ans: 0.0623

M]

4. The potassium permangante solution from the above problem is purple.

The darkness of the purple solution is directly related to the concen-

tration. Next semester in lab we will exploit this fact in what is called

a Beer’s law plot. One step in doing this is to make a series of dilutions

to the original solution. Let’s say 25.00 ml of the solution from the pre-

vious problem was removed and put into a 100.00 ml volumetric flask

and diluted to the mark. What is the molarity of the final solution.

[ans: 0.0156 M]

5. Show mathematically that a solution of pH 3 is more acidic than a

solution of pH 6.

6. Danny is a stubborn freshman who insists on using the OH− concen-tration concentration instead of the H+ concentration for deteriming

65

the acidity or basicity of a solution. How would Danny’s pOH scale

work?

7. What is the pH of a 0.125 M HCl solution? [ans: 0.90]

8. What is the pH of a 0.050 M KOH solution? [ans: 12.7]

12.5 Problem Set

Reading: Chapter 4.5, 4.6, 4.8

Exercises: Chapter 4: 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 59, 61, 113.

66

13. Titration Reactions


• V040103

13.1 Titrations

A titration is a very important technique for determining the unknown con-

centration of some compound in solution.

This procedure is also called volumetric analysis.

It employs a standard solution of known concentration, the stoichiometry of

a known reaction and some sort of indicator to indicate the end of a reaction.

For example lets say a solution of HCl with unknown concentration needs to

be analyzed.

The HCl is called the analyte in this case.

Since we know HCl will react fully with NaOH, we can prepare a standard

NaOH solution with known concentration, say, 0.1000M.

The NaOH is called the titrant in this case.

We can take a sample of the HCl solution, say, 25.00mL an put it in a flask

with an indicator which will change color when the solution is no longer

acidic.

Now we can us a piece of glassware called a buret to deliver just enough of

the standard base solution to neutralize the acid solution and change the

color of the indicator.

The point where the number of moles of base that has been added exactly

equals the number of moles of acid is called the equivalence point.

68

The point where you visually detect a change in the indicator is called the end

point of the titration. Technically the end point is not exactly the equivalence

point, but in practice they are very close to being equal.

So, for this example lets say it took 18.00mL to reach the end point of the

titration (which we will take to be the equivalence point)

In the 18.00mL of standard 0.1000M NaOH are

0.01800 L/ NaOH sol 0.1000 mol NaOH

1 L/ NaOH sol= 0001800 mol NaOH.

(13.1)

From the stoichiometry of the reaction

HCl(aq) +NaOH(aq) → H2O+NaCl(aq) (13.2)

we see that for every one mole of NaOH that is used, one mole of HCl is

used. Hence

0.001800 mol/ NaOH 1 mol HCL

1 mol/ NaOH= 0001800 mol HCl. (13.3)

This tells us that 0001800 moles of HCl were contain in the initial 25.00mL

sample. So, we have now determined the unknown concentration to be

0001800 mol HCl

002500 L HCl sol= 00720 M HCl. (13.4)

Titrations can exploit any type of reaction not just acid—base reactions.

In fact, in lab, you will do a titration based on a precipitation reaction.

13.2 Group Work


1. We will calculated what is called the titration curve for the reaction of

a strong acid with a strong base. There will be a number of calculations

so divide the work up amongst your group members. We will calculate

the pH of a 25.00 mL solution that originally starts at 0.1000 M HCl

after we have added different amounts of 0.1000 M NaOH. We will then

69

plot the results. Calculate and plot the pH after adding 0.00 mL, 5.00

mL, 10.00 mL, 15.00 mL, 20.00 mL, 22.00 mL, 23.00 mL 24.00 mL,

25.00 mL, 26.00 mL, 27.00 mL, 28.00 mL, 30.00 mL, 35.00 mL, 40.00

mL, 45.00 mL and 50.00 mL of NaOH. It seems like a lot of calculations,

but once you set-up the first one the rest is just a matter of changing

one number. Based on your plot, why can we say that the end point is

effectively the same as the equivalence point?

2. A flask containing an unknown amount of HCl is titrated with 0.2530 M

NaOH. It takes 15.45 mL of the standardized NaOH solution to reach

the end point. How many grams of HCl are present? [ans: 0.1425

g]

3. Challenge problem: When an analyte is titrated to the end point

with a standardized titrant, like we have be doing, it is called a direct

titration. Sometimes it is more convenient to add one standardized

reagent in excess and then do a back titration with a second standard.

Let’s say some piece of solid material has an unknown amount of cal-

cium carbonate, CaCO3. When the solid is placed in 50.00 ml of a

standardized 0.1000 M HCl solution, the following reaction takes place

2HCl + CaCO3 → CaCl2 + CO2 + H2O. (13.5)

After the reaction is complete the solution is titrated with a 0.1000 M

NaOH solution. It takes 22.00 mL to reach the end point. How many

grams of CaCO3 were present in the original solid. [ans: 0.1401 g]

13.3 Problem Set


Exercises: Chapter 4: 67, 68, 69, 70, 71, 72, 74, 137.

70

14. Energy, Heat and the First Law of

Thermodynamics


• V210101

14.1 Energy

The energy of an object comes in two basic forms

• Kinetic energy–energy due to motion— mechanical motion, thermal motion (heat), motion if electrons

(electric energy), sound

• Potential energy–energy due to position— Chemical energy, gravitational energy, electrostatic energy, mass

energy ( = 2).

The SI unit for Energy is the Joule. An older but commonly used unit is the

calorie. A calorie is equal to 4.18 Joules.

We are used to seeing calorie units on food labels, but these food calories are

actually kilocalories (1000cal).

14.1.1 Conservation of energy

Law of conservation of energy: Energy can not be created or destroyed.

Energy can be converted from one from to another.

For example, if we burn a piece of wood, some of the chemical energy stored

in the bonds is released as heat which may in turn boil some water to make

steam which drives a turbine (mechanical energy) that turns a generator to

produce electrical energy.

72

14.1.2 Work and heat

Above we classified energy as kinetic or potential.

It turns out that a sometimes better way to classify energy is as work or heat.

Energy that enters or leaves a system is classified as either work or heat.

As system here means any macroscopic piece of matter such as a beaker of

water or an iron bar.

Any work, done on the system is by convention taken to be a positive

quantity and any work done by the system is then negative.

Likewise any heat, taken in by the system is by convention a positive

quantity and any heat given off by the system is negative.

14.2 The First Law of Thermodynamics

The first law of thermodynamics is simply a statement of the conservation of

energy for a system.

In words the first law states that a change in the total energy of a system is

equal to the heat taken in by the system plus the work done on the system.

The first law is stated mathematically as

4 = + (14.1)

14.3 Temperature and Heat

We have said before that temperature and heat are not the same.

Temperature gives the direction of heat flow–heat energy flows from a hot

object to a cold object until thermal equilibrium is reach and the objects are

at the same temperature.

14.3.1 Heat capacity

Temperature is intimately related to heat through the heat capacity.

=

4(14.2)

73

This is an extensive property of the material that determines how much heat

energy, is needed to transferred between the system and its environment

in order to change the temperature by a given amount, 4 .

This can be easily rearranged as

= 4 (14.3)

If we want an intensive property which gives similar information we define

the specific heat capacity, as the heat capacity per gram of the material

= J

1 g · 1 K (14.4)

We see that specific heat capacity has units of energy per gram per Kelvin.

The specific heat capacity for most materials is known and can be found in

reference books.

In general the specific heat capacity of a material depends on temperature,

but, for freshmen chemistry, it is OK to ignore this dependence and treat it

as a constant.

14.4 Phase Changes

It always requires a certain amount of energy called latent heat to be ex-

changed between the system and its environment whenever a piece of matter

changes it phase.

• when changing between the solid and liquid phases this energy is calledthe heat of fusion

• when changing between the liquid and gas phases this energy is calledthe heat of vaporization

Consider heating water from 260 K to 400 K.

• at 260 K water is a solid (ice), as we add heat the temperature beginsto rise

• at 273.15 K the ice begins to melt, as we add heat all the energy goesinto melting the ice and so the temperature remains constant

74

• after the ice has melted, we now have liquid water at 273.15 K• addition of heat causes the temperature to rise• at 373.15 K the water begins to boil, as we add more heat all the energygoes into boiling the water and so the temperature remains constant.

• after all the water has boiled we heat the system to 400 K

14.5 Group Work


1. Drop your pencil off your desk. It seems like that violates conservation

of energy because it had more potential energy before you dropped it

than after. In both cases it was not moving so it has no kinetic energy.

Is it indeed violating conservation of energy?

2. Lead has a specific heat capacity of 0.12 J/(K g). How much energy

does it take to heat a 275 g block of lead 25 degrees?

3. 13 J of heat caused a 30 K temperature change in a block of lead. What

was the mass of the lead?

4. Sketch a graph of heat added, on the -axis versus temperature on

the -axis for the example of water going from 260 K to 400 K.

(a) What is characteristic of the slope of your graph at the points of

phase change?

14.6 Problem Set


Exercises: Chapter 5: 5, 7, 9, 11, 13, 15, 17, 19, 21, 23.

75

15. Enthalpy


• V210102

15.1 Work and the First Law

Let us consider applying the first law to a gas in a piston

If we apply heat to the gas it expands against atmospheric pressure and in

doing so it does work.

77

Since we are adding heat to the system, 0

Since the system is doing work, 0

The first law for this case says

4 = (positive)+ (negative) (15.1)

Therefore no matter what

4 (15.2)

So the change in total energy is less than the amount of heat energy we added

to the system. If we add 10 J of heat energy to the system the total energy

does not change by 10 J

This is somewhat inconvenient so chemists have defined a new measure of

energy called enthalpy ()

Enthalpy is defined such that at constant pressure it is equal to the amount

of heat energy taken in by the system

4 ≡ (15.3)

where the subscript reminds us that this is true only for constant pressure.

(We will always work at constant pressure in this course.)

Consider again the expanding gas example. Now if we add 10J of heat to the

gas then 4 = 10J

78

15.2 Enthalpy Changes for Reactions

During a chemical reaction it is often rather convenient to measure the heat

given off by a reaction.

In fact, you will do this in lab.

If the heat is measured at constant pressure then it is simply the change in

enthalpy for the reaction:

4 = products −reactants (15.4)

If 4 is positive that means that is positive and heat energy is taken in

by the system.

Reactions with positive 4 are called endothermic reactions.

If 4 is negative that means that is negative and heat energy is given

off by the system.

Reactions with negative 4 are called exothermic reactions.

15.3 Group Work


1. Based on your experience, is 4 positive or negative for combustion

reactions?

2. It seems simple now to identify whether a reaction is exothermic or

endothermic and indeed it is. However there is one point of confusion

and that is that sometimes one needs to add heat to get a reaction

started. At first glance this might appear that these reactions are

endothermic (because one adds heat.). In fact these reaction could well

be exothermic. Can you think of a reaction that is clearly exothermic,

but needs some heat to get it started?

15.4 Problem Set


Exercises: Chapter 5: 25, 26, 27, 28

79

16. Lecture L19: Calorimetry


• V210103

16.1 Calorimetry and Heats of Combustions

Last time we saw how heats of formation can be used to calculate 4rxn for

and reaction.

Although the heats of formation are very handy for computation, many of

these reactions are difficult to perform in the laboratory.

Much easier reactions in which enthalpy changes can be measured are com-

bustion reactions.

Combustion reactions are performed in devices called bomb calorimeters

Steps of a combustion experiment

1. Combust a standard sample with known4rxn and measure the change

in temperature of the bomb.

2. Since 4rxn = is known and4 is measured, one can determine the

heat capacity of the bomb C = −4

3. Next the substance with unknown 4rxn is combusted in the bomb

and the change in temperature is measured

4. From the previously determining heat capacity one can calculate4rxn =

= C4

Since these experiments are relatively easy to perform there are standard

tables of heats of combustion.

When applying Hess’s law, one can either use heats of formations or heats

of combustion (or any other reaction in which 4rxn is known)

82

16.1.1 Example

Let’s say we wanted to know how many “food Calories” (kilocalories) are in

a gram of some type of food.

Step 1: Standardization with benzoic acid (BA). 0.9987 g of BA is

combusted to give a temperature change of 2.59 K. Since BA is standard we

know that for each gram of BA combusted 6318 cal of heat are released.

C =0.9987 g/ BA −(−6318) cal2.59 K 1 g/ BA

= 2436 cal/K (16.1)

Step 2: Determination of unknown. 1.018 g of food is combusted to

give a 2.76 degree temperature change. Now that we know the heat capacity

of the bomb, we can determine the calorie content of the food.

food =−2436 cal 2.76 K/1 K/ 1.018 g food

= −6605 cal/g.

Now the food calories are really kilocalories so this piece of food offers 6.6

Cal/g where Cal means kcal and the values are always negative so the minus

sign is dropped.

16.2 Group Work


1. Martin was doing some calorimetry and got the following data: 1.023

g of BA caused a 2.64 K temperature change while 0.946 g of mandelic

acid caused a 2.14 K temperature change.

(a) What did Martin calculate for the heat capacity of the bomb?

(b) What did he get for the heat released per gram of mandelic acid?

(c) Mandelic acid has a molecular mass of 152.15 g/mol. What is the

molar heat of combustion for mandelic acid?

83

16.3 Problem Set



84

17. Hess’ Law


• V210104

17.1 State Functions

Both total energy, and enthalpy, , are what are called state functions

A state function is a function that depends only on the particular state that

the system is in. It is completely independent of how the system arrived at

that state.

For example consider heating in water from 300 K to 310 K.

You would calculate the same4 or4 regardless as to whether you heated

directly from 300 K to 310 K or if you first cooled to 290 K then heated to

360 K and then cooled to 310 K.

17.2 Hess’ Law

An important consequence of enthalpy being a state function is Hess’ law

which states that if a reaction is the sum of two or more reactions, 4 for

the overall reaction is the sum of the 4’s for the component reactions.

That is 4 is solely determined by

4 = products −reactants (17.1)

regardless of how the reactants became the products.

For example, 4 for the reaction

2C6H6 + 15O2 → 12CO2 + 6H2O (17.2)

86

is the same as the sum of the 4’s for the following reactions

2C6H6 → 12C+ 6H2 (17.3)

12C+ 6H2 + 15O2 → 12CO2 + 6H2O (17.4)

Hess’s law is very handy because it would be impossible to measure enthalpy

changes for every imaginable reaction were it not for Hess’ law.

17.2.1 Heats of formation

With Hess’ law one only needs to determine 4 for so-called formation

reactions. These are called molar heats of formation (or molar enthalpies

of formation) and are tabulated in reference books. (symbol 4° ) for a

extensive table see

∗ ∗ ∗ Appendix L of Kotz et.al. ∗ ∗∗

So, by the use of Hess’s law, the enthalpy change for any reaction can be

written as

4rxn =X

4° (products)−

X4

° (reactants). (17.5)

As a specific example consider the reaction for the combustion of butane gas

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) (17.6)

To find 4rxn we look up each of the reactants and products in the table of

standard heats of reactions. From Table 6.2 of Kotz & Treichel we find

C4H10(g) 4° = −1271 kJ/mol (17.7)

O2(g) 4° = 0 kJ/mol (17.8)

CO2(g) 4° = −3935 kJ/mol (17.9)

H2O(l) 4° = −2858 kJ/mol (17.10)

Why does 4° = 0 kJ/mol for O2(g)?

This is because the definition of heat of formation is that you start with the

elements (that make up the compound) in there natural state whose heat of

87

formation is defined to be zero. The element oxygen is naturally O2(g) so by

definition 4° = 0 kJ/mol

To complete the calculation we must consider the stoichiometry of the reac-

tion.

The stoichiometry tells us that two moles of butane react with 13 moles of

oxygen to make eight moles of carbon dioxide and ten moles of water. So,X4

° (products) = 8 mol×−3935 kJ/mol+ 10mol×−2858 kJ/mol

= −6006 kJ (17.11)

andX4

° (reactants) = 2 mol×−1271 kJ/mol+ 13 mol× 0 kJ/mol

= −2542 kJ (17.12)

Hence

4rxn =X

4° (products)−

X4

° (reactants) (17.13)

= −6006 kJ− (−254 kJ) = −5752 kJThis says the when we burn butane it gives of heat (as expected).

17.3 Group Work


1. Use Hess’ law to determine 4rxn for

4NH3 + 5O2 → 4NO + 6H2O (17.14)

given the following data

N2 + O2 → 2NO; 4 = 1806 kJ

N2 + 3H2 → 2NH3; 4 = −918 kJ2H2 +O2 → 2H2O; 4 = −4837 kJ

2. Based on the previous problem what is 4rxn for

2NH3 +5

2O2 → 2NO + 3H2O (17.15)

88

3. Use the data from Table 6.2 and that4° = −7598 kJ/mol for CaCl2

to calculate 4rxn for

2HCl(aq) + CaCO3 → CaCl2 + CO2+H2O (17.16)

4. Challenge problem: How much heat is given off when 2.234 g of

CaCO3 is dissolved in excess HCl(aq)?

17.4 Problem Set



89

Part III: Material for Exam 3

90

18. Light


• V050101

18.1 Electromagnetic Radiation

An important tool in the study of atomic (and molecular) structure is light.

Light is also called electromagnetic radiation because objects which emit light

radiate an oscillating electric and magnetic fields.

These fields oscillate in time with a given frequency. The number of times

that the electric (or magnetic) field returns to any given value during one

second gives the frequency in Hertz. (1Hz is one cycle per second: 1Hz =

1s−1)

When one says light, one tends to think of the light we see with our eyes–the

colors of the rainbow.

This visible light is just a small portion of the electromagnetic spectrum

which happens to be centered around 1015Hz.

∗ ∗ ∗ See Fig. 6.2, p271 of Kotz et.al ∗ ∗∗

Frequencies higher than the visible frequencies consist of the ultraviolet light

region, the X-ray region and the -ray region. (One typically does not use

the word ‘light’ when dealing with the X-ray and the -ray regions, but all

the regions are really the same aside from the frequency.)

Frequencies lower than the visible frequencies consist of the infrared light

region, the microwave region and the radio wave region. (Again one does not

typical use the word ‘light’ when dealing with the microwave and radio wave

regions)

92

The frequency, of light is directly related to its energy, . The two are

related by Planck’s constant,

= (18.1)

Let’s check units for this equation.

Frequency is in Hertz or one over seconds and Planck’s constant is 6626 ×10−34 J·s so,

J · s · s−1 = J √ (18.2)

Another way to characterize light is by its wavelength.

The wavelength is the distance the light travels in one cycle.

Since the wavelength, is a distance it has units of meter. If we multiply

by the frequency of the light which has units of seconds we get a product

which has units of meters per second (speed). That is,

· = (18.3)

The speed of light is a fundamental constant which is the same of light of

any frequency.

Hence,

=

(18.4)

This says that frequency and wavelength are inversely related

High frequencies mean short wavelengths

Consequently, short wavelengths mean high energy and long wavelengths

mean low energy.

18.2 The Photon Description

One can also describe light from a particle point of view.

Under certain experimental conditions light behaves like a stream of particles

called photons.

Each photon carries = amount of energy.

93

Unlike particles of matter photon are massless and not conserved.

18.3 Group Work


1. Early on we used as one of our bench marks 500 nm as a typical wave-

length for visible light. Light at this wavelength is green. What is the

frequency of this light? What is the energy carried by one photon of

this light? [ans: 6.0×1014 Hz, 3.97×10−19 J]2. The visible region is 400—700 nm. What is this range in frequency? How

about energy? [ans: 4.3×1014 Hz --- 7.5×1014 Hz, 2.8×10−19J -- 5.0×10−19 J]

3. Another way to describe light is in units of wavenumber (cm−1). Basi-cally this is the number of wavelengths that will fit into on centimeter.

What is the visible range in wavenumbers?[ans: 14,286 cm−1 --25,000 cm−1]

4. A very interesting and important property of light is that it can prop-

agate through a vacuum. That is, it does not need a material to host

it. Think about other types of waves that you are familiar with and

convince yourself that they are indeed tied to the material in which

they propagate.

5. Challenge question: How many photons are emitted per second from

a 100 W light bulb? [ans: on order of 1020 photons/s]

18.4 Problem Set



94

19. The Fall of Classical Physics and the Bohr Model


• V050201

19.1 Brief Overview

We learned previously that atoms are made up of component parts; namely

electrons, protons and neutrons.

The protons and neutrons make up the nucleus around which electrons en-

circle.

The nucleus does not participate in the chemistry of the atoms; it is the

electrons which are the primary players in chemistry.

In the lectures to come we will look closer at the structure of the atom and

in particular at just how the electrons behave.

As it turns out, Newton’s laws of classical mechanics only approximately

describe nature.

At very small scales, like that of atomic dimensions, Newton’s laws essentially

completely fail.

At these small scales, one must use a more accurate description of nature.

This is quantum mechanics.

The quantum mechanical description of nature leads to some unintuitive

results. Some of which appear quite strange at first.

We shall encounter some of these unintiutive results. The two most important

of these are

• the electron must be viewed as an object that has particle-like proper-ties and wave-like properties, but it is neither a particle nor a wave.

97

• the electrons exist in discrete (or quantized) energy states about theatom.

Unfortunately the mathematics which goes along with quantum mechanics

is significantly more complicated than the = of Newton’s classical

mechanics, so we can not present it in this course.

We will have to be satisfied with accepting the important results of quantum

mechanics on faith

19.2 Bohr’s Atomic Theory

19.2.1 First attempts at the structure of the atom

The “solar system” model.

• The electron orbits the nucleus with the attractive electrostatic forcebalances by the repulsive centrifugal force.

Flaws of the solar system model

• Newton: OK √

• Maxwell: problem √

— As the electron orbits the nucleus, the atom acts as an oscillating

dipole.

98

• — The classical theory of electromagnetism states that oscillating

dipoles emit radiation and thereby lose energy.

— The system is not stable and the electron spirals into the nucleus.

The atom collapses!

Bohr’s model: Niels Bohr (1885—1962)

• Atoms don’t collapse =⇒ what are the consequences.

Experimental clues

• Atomic gases have discrete spectral lines.

99

• If the orbital radius was continuous the gas would have a continuousspectrum.

Conclusion

• Atomic orbitals must be quantized. Only certain orbitals are allowed

The total energy of the Bohr atom is related to its quantum number

=−R2

(19.1)

where R is called the Rydberg constant. R = 2179× 10−18 J = 13606 eV(the eV is another unit of energy called the electron volt). The allowed (or

quantized) values of are 1,2,3,· · · Tests of the Bohr atom

• Ionization energy of Hydrogen atoms— The Ionization energy for Hydrogen atoms is the minium energy

required to completely remove an electron form it ground state,

i.e., = 1→ =∞

ionize = ∞ −1 =

µ −R(∞)2

¶−µ−R(1)2

¶= R (19.2)

— ionize = R = 13606 eV.

100

— ionize experimentally observed from spectroscopy is 13.605 eV

(very good agreement)

• Spectroscopic lines from Hydrogen represent the difference in energy

between the quantum states

— Bohr theory: Difference energies

4 = − = −Rµ1

2− 1

2

¶(19.3)

— Since the orbitals are quantized, the atom may only change its

orbital radius by discrete amounts.

— Doing this results in the emission or absorption of a photon with

energy 4

Failure of the Bohr model

• No fine structure predicted (electron-electron coupling)• No hyperfine structure predicted (electron-nucleus coupling)• No Zeeman effect predicted (response of spectrum to magnetic field)

19.3 Group Work


1. Have your group pick three different quantum states.

(a) Calculate the energy of each of these states.

(b) Calculate 4 for all possible combinations of your three states.

2. Use the graph paper provide to plot an energy level diagram for the

Bohr model of hydrogen by drawing a vertical energy axes and then

drawing a horizontal line at the appropriate energies to represent the

quantum state.

101

19.4 Problem Set


Exercises: Chapter 6: 13, 15, 17, 19, 21, 23, 25..

102

20. Atomic Orbitals


• V050301• V050302• V050303• V050304• V050305

20.1 The Modern Theory of the Hydrogen Atom

The Bohr atom was an important step towards the formulation of quantum

mechanics (developed in 1926)

• Erwin Schrödinger (1887—1961): Wave mechanics• Werner Heisenberg (1902—1976): Matrix mechanics• Paul Dirac (1902—1984): Abstract approach

From its inception in 1926 to the present, quantum mechanics has proven to

be an extraordinarily successful theory.

The mathematics of quantum mechanics and in particular the application of

quantum mechanics to the hydrogen atom problem is beyond the scope of

general chemistry.

We shall have to be satisfied with the results coming from solving the for the

hydrogen atom and it will be our job to study these results.

Results from quantum mechanics regarding the hydrogen atom

1. Five quantum numbers arise unlike only one for the Bohr model. We

will discuss each of these quantum numbers in detail below.

104

• The five quantum numbers are and

2. The solution of the problem yields what are called wavefunctions (Ψ).

• Each wavefunction is associated with a unique set of quantumnumbers Ψ

.

• The square of the wavefunction ¡Ψ2

¢gives the probability

density of finding the electron in a particular spot in the three

dimensional space around the nucleus.

3. At any given time the electron is described by one of the wavefunctions

and hence by the five quantum numbers.

20.2 The Quantum Numbers of the Hydrogen Atom

We will be most concerned with understanding the quantum numbers of the

hydrogen atom and how they are eventually are applied to the rest of the

atoms on the periodic table.

The quantum numbers of the hydrogenic system

• The principle quantum number, : determines the total energy of the

atom and the atomic orbitals (or shells).

— The principle quantum number, can take on values of 1,2,3

• The angular momentum quantum number, : determines the total an-

gular momentum of the system. It also determines the shape of the

atomic sub-orbitals (sub-shells)

— The angular momentum quantum number, can take on values

of 0, 1 (− 1)— For historical reasons = 0 is called , = 1 is called = 2 is

called = 3 is called etc.

• The orientation quantum number, : determines the orientation in

space of the atomic sub-orbitals.

— The magnetic quantum number, can take on values of 0,

±1 ±

105

• The spin quantum number, : determines the total spin angular mo-

mentum.

— Spin is an intrinsic property of the electron. There is no analogy

for it.

— Unfortunately the word “spin” implies rotational motion. But the

electron is not spinning like a top.

— For electrons = 12 (Other fundamental particles may have

different values.)

— Since is always the same this quantum number never changes so

it is not really important for what we want to do with the quantum

numbers

— Your book ignores this quantum number.

• The spin orientation quantum number, : determines the orientation

of the spin called “spin-up” or “spin-down”).

— For electrons = +12 (spin-up) or −12 (spin-down).

20.3 Visualizing the Atomic Orbitals

As stated the mathematics are too difficult present in this course. You will

have to wait for physical chemistry for that.

We can view the graphs of the square of the wavefunctions for hydrogen.

This gives us a three dimensional view of the probability of the electron being

in a given position in space.

The natural solution of the Schrödinger equation for hydrogen gives what is

called the “physicist picture” of the atomic orbitals.

∗ ∗ ∗ See overhead ∗ ∗∗

The advantage of this picture is that each graph corresponds to a single set

of quantum numbers.

The disadvantage is that the wavefunctions are complex functions rather

than real functions.

106

So, chemists use a different but equivalent picture of the atomic orbitals.

This is the “chemist picture”

∗ ∗ ∗ See overhead ∗ ∗∗

The disadvantage here is that the quantum number cannot be associated

with a particular atomic orbital.

For example the 2p orbital is given by quantum number = 2 = 1 But

we can not say = 1 0 or −1

20.4 Group Work


1. One important thing to look for when viewing the atomic orbitals are

nodes. Nodes are surfaces (either planes, cones or spheres) where the

electron density is zero. On either side of the node the wavefunction

changes sign from positive to negative or vice versa. There are always

− 1 nodes. Can you find them for each of the atomic orbitals shown

on the overhead?

2. A node is classified as radial if the surface it traces out is a sphere and

angular if the surface is a plane or a cone. See if you can figure out

a relation between the number of each type of node and the and

quantum numbers.

3. Interestingly the energy levels for the hydrogen atom in quantum me-

chanics have the same form as for the Bohr model,

=−R2

The difference from the Bohr model comes in from the other quantum

numbers. The value of the principle quantum number, determines

the values of the other quantum numbers according to the rules given

above in lecture. The set of distinct quantum numbers determines a

single state for the hydrogen atom. Since the energy depends only

on then there must be several states of equivalent energy. State of

equivalent energy are said to be degenerate and the number of states

107

of equivalent energy is the degeneracy of the energy level. With this

in mind draw the energy level diagram for the hydrogen atom and

explicitly show the degeneracy of each state.

20.5 Problem Set



108

21. Multielectron Atoms and the Pauli Exclusion

Principle


• V050401

21.1 The Other Atoms

Despite the tremendous success of quantum mechanics in describing the re-

sults of numerous experiments, we can only solve the hydrogen atom system

exactly.

Even helium which only has one more electron than hydrogen is impossible

to solve exactly.

So we do not have the exact helium wavefunctions and helium quantum

numbers. The same is true for the remainder of the atoms.

To handle the other atoms beyond hydrogen wemust make the approximation

that the each of the electrons in the atom behaves independently of the other

electrons.

Clearly this is an approximation since electrons are negatively charged and

negative charge repel one another.

With this approximation each electron in the atom behaves as if it were in a

hydrogen atom by itself (with one exception which we will discuss shortly).

In this way we may say that the wavefunction for the other atoms are built-

up of hydrogen wavefunctions. This is called the aufbau principle (aufbau is

German for build-up).

110

21.2 The Pauli Exclusion Principle

We noted above that there is one exception to our approximation that the

electrons behave as though they are in a hydrogen atoms

This exception is that once one of the electron in the atom is described by a

given set of five quantum numbers none of the other electrons can have the

same five quantum numbers.

This is a consequence of the Pauli exclusion principle.

The precise statement of the Pauli exclusion principle is at a technical level

beyond the scope of general chemistry.

Most books, including your book, simply give the Pauli exclusion principle

as: “no two electrons can have the same set of quantum numbers”

Another consequence of the Pauli exclusion principle is that no atomic sub-

orbital can contain more than two electrons.

21.3 Electronic Configurations of Atoms

So, in accordance with the Pauli exclusion principle, we fill the atomic orbitals

with the appropriate number of electrons.

Recall that the ground state hydrogen has one electron in the 1s subshell

(the electron may be either spin up or spin-down).

For helium which has two electrons we see that the ground state would be

the case when there are two electrons in the 1s subshell (one spin-up and one

spin-down).

For the three electrons of lithium, the ground state would be one in which

two electrons are in the 1s subshell and one electron is in the 2s subshell.

We put the third electron in the 2s subshell rather than the 2p subshell

because the electrons fill the subshell according to the following rule: the

subshell with the lowest value of + is the next to fill.

So, for the case of lithium the third electron goes into the 2s subshell (+ =

2 + 0 = 2) instead of the 2p subshell since (+ = 2 + 1 = 3).

111

∗ ∗ ∗ See Figure 2.7 on p307 of Kotz et.al ∗ ∗∗

One final rule for filling electrons is Hund’s rule which states that when

possible electrons fill different subshells with the same spin orientation.

For example consider the case when two electrons are to fill the 2p subshell.

We could put the electrons into the same 2p subshell, but then they would

have be spin-up and spin-down.

Alternatively we put the two electrons into two different 2p subshells. Then

the electrons could have the same spin orientation. So, according to Hund’s

rule this latter choice is the appropriate one.

There is a shorthand way of denoting how the electrons fill the subshells for

the ground state of each of the atoms.

This is called the electronic configuration.

∗ ∗ ∗ See Handout ∗ ∗∗

Example: Silicon.

Silicon has 14 electrons so we simply fill according to the rules: 2 electrons

in 1s, 2 in 2s, 6 in 2p, 2 in 3s and the last 2 in two different 3p orbitals both

spin up (or both spin down).

In electron configuration notation it is

Si: 1s22s22p63s23p2 (21.1)

In noble gas notation it is

Si :

same as Nez }| {1s22s22p63s23p2 (21.2)

Si : [Ne]3s23p2

21.4 Group Work


112

1. Practice writing the noble gas configuration for 8 of your favorite el-

ements. Compare your result with Table 8.3 or the handout. Note:

there are several exceptions to the rules.

2. Practice writing the noble gas configuration for 4 of your favorite monatomic

ions.

3. Chromium is one example of an exception to the filling rules. Inspect

the noble gas configurations of chromium and the elements next to it

(Ti, V, Mn, and Fe). Can you think of any reason why Cr might be

the way it is?

4. Challenge problem: Term symbols. There is even a shorter notation

for a atomic states called term symbols. To produce the term symbol

from the electron configuration, you look only at the unfilled subshells

and give the maximum sum of the values as the base symbol. For the

example of Si, the 3p orbital is the only one unfilled. The maximum

value we can get here is for the case when one electron is in the = 1

orbital and one is in the = 0 orbital. Hence = 1 + 0 = 1 We give

the capital letter for the corresponding value as the base symbol, so

here the base term symbol is P. We decorate the base term symbol with

a superscript on the left side that gives the spin degeneracy–that is

the number of ways the spins can be oriented. Here we have 2 electrons

in the unfilled shell and so the spin degeneracy is 3: both spin up, both

spin down, one spin up and one spin down. Thus the final term symbol

is 3P. Try and see if you can get the term symbols for several different

atoms.

21.5 Problem Set

Reading: Chapter 6.7, 7.1—7.4

Exercises: Chapter 7: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21.

113

22. Periodic Trends

The videos lectures over this material are

• V050501• V050502

22.1 Periodic Properties of the Atoms

There are many properties of the atoms which are intimately tied to the

electronic configuration of the atoms.

Chapter 6 of your book discuss many of these. Please read these sections

carefully.

Here we will only discuss oxidation states of the atomic ions.

We are well familiar with the common oxidation states for the groups of

atoms. For example the alkali metals tend to lose an electron to form cations

having a 1+ oxidation state.

Earlier in the semester we simply had to memorize the common oxidation

states for the groups. Now we can predict them from the electronic configu-

rations.

First we need to know one more important fact about the filled atomic sub-

shells. That is, completely filled subshells are more stable then partially filled

subshells.

Furthermore the most stable case is when the and subshells in a given

shell are completely filled

Let us see how this determines the oxidation states for the groups.

We can start with the noble gases.

• These elements have completely filled and subshells in the th shell.

115

• It is unstablizing to either lose or gain and electron, since either casewould result in a partially filled and subshell.

• Consequently the noble gases do not readily from ions.

Turning to the alkali metals

• These elements have 1 electron in the subshell.• Either gaining an electron or losing an electron results in fully filledsubshells

• However, losing an electron to become a cation of +1 charge not onlyresults in completely filled and subshells but the electronic con-

figuration of the cation becomes identical to the previous noble gas

atom.

• Consequently, the alkali ions tend to have a 1+ oxidation state as ex-pected.

The alkaline earth metals

• These elements have filled subshells, but the subshells in the outershell are empty.

• If these atoms give-up two electrons they obtain the electronic config-uration of the previous noble gas.

• Hence the cations of the alkaline earth metals have an oxidation stateof 2+.

Similar arguments can be made to explain the common oxidation states for

the other A groups.

The same rules about stability apply to the B groups (the transition metals).

Many of the transition metals have two or more common oxidation states

because there is more than one way to obtain filled subshells and one is not

dominant over the other as is the case for the A groups.

Also, the half-filled d orbital is relatively stable. This explains the electron

configuration of Cr.

116

22.2 Shielding

Electrons carry a negative charge that is equal and opposite to a proton.

As a result each individual electron feels a positive charge that is much less

than the total nuclear charge.

The electrons shield each other from the full effect of the nucleus.

Rules of thumb for shielding

• Electron in inner shells shield a full charge from the valance electrons.

• For electrons in the same shell the shielding effectiveness goes as s & p d f.

22.3 Group Work


1. The ionization energy is the energy required to remove an electron.

Without looking at your book,

(a) how does the ionization energy go from left to right across a row?

(b) how does the ionization energy go down a column?

2. Electron affinity is a measure of how much an atom wants to gain an

electron. Without looking at your book,

(a) how does the electron affinity go across a row?

(b) how does the electron affinity go down a column?

3. Use the idea of shielding to explain the fact that the atomic radius

decreases from left to right across the row.

4. Shielding is responsible the so called lanthanide contraction which re-

sults in hafnium having a smaller atomic radius than zirconium in spite

of the fact that it is below zirconium. Why?

117

22.4 Problem Set


Exercises: Chapter 7: 23, 25, 27, 29, 31.

118

23. Lewis Dot Structures I


• V060101• V060102

23.1 Fundamentals of Chemical Bonding

There would not be much to the universe if all there was were the elements.

The variety of the universe comes from the interaction of atoms of different

elements to form bonded compounds and molecules.

We have already been discussing molecules through most of this semester,

but now after the previous chapter we are in position to discuss chemical

bonding on a more fundamental level.

23.2 Valence Electrons

The electrons of any given atom can be divided into two groups.

• The valence electrons which are those electrons in the outermost shell.• The core electrons which are the remaining electrons.

It is the valence electrons which are the primary participants in bonding.

23.3 Lewis Dot Structures

A very useful way to display an atom symbolically is to include the valence

electrons as dots around the symbol. This is called the Lewis dot structure

of the atom.

We shall soon see that the Lewis dot structures will lead us to our first

approach to bonding.

119

As an example let us consider the second row of the period table. The lewis

dot structures for these elements are

Notice the noble gas neon. It has eight valence electrons. We know that the

noble gases are extremely stable and rarely react with anything.

This points to a general rule regarding atoms and valence electrons. The rule

is the octet rule, which states that atoms form bonds which provides them

with and octet of valence electrons.

The octet rule is intimately related to the idea from the previous chapter

that the most stable configuration is one in which the and subshells are

filled.

There are exceptions to this rule; most notably hydrogen forms bond to

obtain two valence electrons.

Boron usually only forms bonds to obtain six valence electrons

Heavier atoms (those in row three and higher) can sometimes form bonds in

which the number of valence electrons exceeds eight.

23.4 Chemical Bond Formation

Based on what was said above we can use the Lewis dot structures to form

our first theory of chemical bond formation.

120

Simply obeying the octet rule, correctly predicts a very large number of

compounds and molecules.

One way the octet rule can be satisfied is if one atom which needs to gain

a few electrons in order to obtain the desirable eight can pair with an atom

that needs to lose a few electrons in order to have an outer shell with eight

electrons.

In this case the electrons are simply transferred from one atom to the other.

This is exactly the ionic bond which we know from earlier in the course.

For example consider the formation of NaCl.

An alternative way atoms can pair to obtain eight valence electrons is to

share valence electrons.

The shared electrons are counted for each atom. So in a sense you get to

count some electrons twice.

This type of pairing is called a covalent bond.

For example consider the formation of Br2

121

So far we have only been talking about pairs of atoms. These ideas can be

extended to many-atom molecules, compounds and polyatomic ions.

For example consider the formation of H2O

and CCl4

and NH+4

122

23.5 Algorithm for Drawing Lewis Structures

It can be a little more complicated, so it is best the have a general algorithm

to create the Lewis structures.

Lewis Dot Structure

• Determine the total number of electrons.• Draw the skeleton structure by connecting the atoms with single bonds.• Add lone pairs to all the atoms to obtain an octet around each atom.

— if there are now too many electrons create double or triple bonds

to reduce the total number of electrons.

— if there is not enough electrons add lone pair electrons to the

central atom.

23.6 Group Work


1. Draw the lewis dot structures for H2CCl2 PH3 F2 SiH4

23.7 Problem Set

**NOTE** These problems are not due until after the next two lectures but

you can get started on them now.

Reading:: Chapter 8.1, 8.2, 8.5

123

Exercises: Chapter 8: 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16.

124

24. Lewis Dot Structure II


• V060103• V060104

24.1 Formal Charge

Sometimes there will be more than one valid Lewis structure.

To determine the appropriate one, one must consider the formal charge.

The formal charge on an atom in a compound or polyatomic ion is defined

as

formal charge = group #− (# of lone pair e− +1

2# of bonding e−)

(24.1)

Note: the sum of the formal charges for all the atoms in a compound or

polyatomic ion must equal the overall charge.

The Lewis structure that has the least number of atoms with a non-zero

formal charge, or the one that has the formal charge on appropriate atoms

is the most correct representation of the molecule.

We will get practice with this on our handout.

24.2 Group Work


1. Draw the Lewis dot structures for the molecules on the handout. Try

to do this individually first then compare with the group.

126

24.3 Problem Set

Reading:: Chapter 8.3, 8.4

Exercises: Chapter 8: 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16.

127

25. Properties of Chemical Bonds


• V060201

25.1 Properties of the Chemical Bond

A step towards understanding molecules is to understand some of the general

properties of the chemical bond.

The bond order of a chemical bond is defined as

B.O. =1

2(number of shared electrons between the bonded pair) (25.1)

The bond length is the distance between the nuclei for the bonded pair.

Bond lengths are typically on the order of 1Å (100 pm).

The bond lengths general decrease with increasing bond order.

The bond dissociation energy, is the energy required to break a bond and

it is often giving in an energy per mole of bonds.

Bond dissociation energies vary significantly depending on the particular

molecule, but typically the values are several hundreds of kilojoules per mole.

Bond dissociation energies tend to increase with increasing bond orders.

During a chemical reaction bonds are broken to release energy and new bonds

are formed which require energy.

129

25.2 Charge Distribution in Covalent Compounds

We know that in ionic compounds electrons move entirely from one atom to

another and the bond is formed due to the mutual attraction of the anion

and the cation.

For covalent compounds the electron density is shared. However, it is not

necessarily shared evenly between all the atoms in the compound.

As a result some atoms in the compound carry what are called partial charges.

25.2.1 Electronegativity and bond polarity

One way to characterize the charge distribution in molecules is by formal

charge.

Another way to characterize the charge distribution in compounds and poly-

atomic ions is to introduce the idea of electronegativity,

Electronegativity is a measure of the ability of an atom in a compound to

attract electron density to itself.

For any given bonded pair the difference in electronegativities will determine

just how equally the atoms share the bonding electrons.

As a rule of thumb,

• if the electronegativities for the bonding pair differ by less than 0.5then they essentially equally share the electrons. This type of bond is

called a nonpolar covalent bond.

• if the electronegativity difference is between 0.5 and 2.0 then one atomattracts more electron density. This type of bond is called a polar

covalent bond. The atom with the higher electronegativity carries a

partial negative charge while its partner with lower electron density

carries a positive charge.

• if the electronegativity exceeds 2.0 the bond is an ionic bond.

25.3 Group Work


130

1. Give the bond polarity and dipole moments for the molecules in the

hand out.

25.4 Problem Set

Reading:: Chapter 8.9

Exercises: Chapter 8: 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 43, 45, 47,

49, 51, 55

131

26. The VSEPR Model


• V010201• V010202

There are two common theoretical treatments of the chemical bond: valence

bond theory and molecular orbital theory. We will only discuss valence bond

theory.

26.1 VSEPR Model

Prior to discussing valence bond theory we need to learn something about

the shapes of molecules. A good approach to this is given by the valence

shell electron repulsion (VSEPR) model.

The primary principle of the VSEPR model is that bonds and/or lone pairs

of the valence electrons on each element of a compound repel each other and

try to be as far away from one another as possible.

Under the VSEPR model molecules can be classified according to there elec-

tronic geometry. The electronic geometry is the three dimensional structure

that the electron density–either a bond or lone pair–adopts.

Electronic Geometries

• Linear: Two bonds or lone pairs• Trigonal planar: Three bonds or lone pairs• Tetrahedral: Four bonds or lone pairs• Trigonal-bipyramidal: Five bonds or lone pairs• Octahedral: Six bonds or lone pairs

132

Once the molecule has been assigned its electronic geometry, it then gets

assigned its molecular geometry.

Molecular Geometries (electronic geometry in bold, molecular in italic)

• Linear— Linear : Two bonds, no lone pairs.

• Trigonal planar— Tringonal planar : Three bonds, no lone pairs.

— Bent: Two bonds, one lone pair.

• Tetrahedal— Tetrahedral : Four bonds, no lone pairs.

— Trigonal Pyramidal : Three bonds, one lone pair.

— Bent: Two bonds, two lone pairs.

• Trigonal-bipyramidal— Trigonal-bipyramidal : Five bonds, no lone pairs.

— Seesaw : Four bonds, one lone pair.

— T-shaped : Three bonds, two lone pairs.

— Linear : Two bonds, Three lone pairs.

• Octahedral— Octahedral : Six bonds, no lone pairs.

— Square-pryamidal : Five bonds, one lone pair.

— Square-planar : Four bonds, two lone pairs.

133

26.2 Molecular Dipole Moments

Polar covalent bonds with in molecule can potentially lead to a net dipole

moment for the molecule.

If all the polar bonds in a molecule cooperate in such a way that one side

of the overall molecule is partially negative and the other side is partially

positive then the molecule has a net dipole moment.

If all the polar bonds in a molecule cancel each other out then the molecule

has no net dipole moment.

26.3 Group Work


1. Give the electronic and molecular geometries for the molecules on the

handout. Try to do this individually, then compare with the group.

26.4 Problem Set

*** Note: These problems are due along with the problem from next lecture

***

Reading: Chapter 8.6—8.8, 8.10

Exercises: Chapter 8: 39, 41.

134

27. Valence Bond Theory


• V060401• V060402• V060403

27.1 Valence Bond Theory

The basic idea of valence bond theory is that a bond can form if there is

significant overlap of the electron densities of the original atoms

The overlap of electron densities are not limited to like subshells as is shown

in the Fig. 10.1 of Koltz & Treichel. An orbital may overlap with a

orbital or a orbital etc.

The orbital can form only one type of bond and that is one in which the

overlap region is directly between the nuclei.

These type of bonds are called -bonds.

The (and etc.) can form -bonds, but they can also form more different

types of bonds called -bonds.

In -bonds the overlap of the electron densities are off the internuclear axis

as shown in the figure

137

27.2 Multiple Bonds

We are already familiar with single, double and triple bonds.

All single bonds are -bonds

A double bond is a -bond and a -bond

A triple bond is a -bond and two -bonds

27.3 Hybridization

According to the VSEPRmodel bonding electrons try to orientate themselves

such that are as far away from one another as possible.

For example when the carbon atom bonds to four other atoms via single

bonds, the molecular geometry about the carbon atom is tetrahedral.

Now, the orbital is spherical and the three orbitals are orientated along

the and axes. There is no way to get the desired tetrahedral orientation

of bonds from the orbitals.

138

In order to achieve the appropriate molecular geometry, the and the three

orbitals must “mix” together in what is called the hybridization of atomic

orbitals.

When this happen we may view the atomic orbitals as corresponding to four

equivalent states called the 3 hybrid orbitals.

+ + + ⇒ 43 (27.1)

Note this is not a chemical reaction.

The four 3 hybrid orbitals are orientated in the tetrahedral geometry.

It is the overlap of these orbitals with the orbitals of other atoms which leads

to bonds and molecules with the proper geometry.

The 3 hybridization scheme is not the only scheme possible. It can happen

that the orbital and only two orbitals hybridize to form three equivalent

2 orbitals.

mixz }| {+ + + ⇒ 32 + (27.2)

The 2 orbitals have a trigonal planar geometry

Finally, it may happen that the orbital and only one of the orbitals are

involved in hybridization.

mixz }| {+ + + ⇒ 2+ + (27.3)

For this case two equivalent hybrid orbitals are formed. They have a linear

geometry.

For the heavier atoms (third row atoms and higher), the hybridization schemes

are not limited to and orbital, but they can also involve several orbitals.

The 3 scheme has a trigonal bipyramidal geometry.

The 32 scheme has an octahedral geometry.

139

27.4 Multiple Bonds Revisited

We saw eariler that a double bond is a and bond.

The bond is formed by the overlap of the hybrid orbitals.

The bond is formed by the overlap of the free orbitals.

27.5 Representing the 3D structure of molecules

Formulae at best show only the connectivity of the atoms which make up the

molecule.

Real molecules are three dimensional, so in some cases it is nice to go beyond

the formula to what is called a model of the molecule (or molecular model).

The three dimensional shape of a molecule can be represented on a flat piece

of paper using the line—wedge—dash convention.

• One draws a picture of the molecule by choosing a convenient set ofthree atoms. These three atoms form a plane which coincides with the

plane of the paper.

• Any bonds which lie in the plane of the paper are drawn as lines.• Any bonds which lie in back of the plane are drawn as dashes• Any bonds which lie in front of the plane are drawn as wedges

The figure illustrates some examples of this type of model.

The line-wedge-dash is useful for relatively small molecules.

140

Other molecular model techniques involve building a three dimensional model

using a model set or using computer graphics to show the representation of

the molecule “in 3D”

27.6 Group Work


1. Give the hybridizations for the atoms in the molecules on the handout.

Try to do this individually, then compare with the group.

27.7 Problem Set


Exercises: Chapter 9: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,

18, 19, 20.

141

Part IV: Material for Exam 4

142

28. Symmetry


• V070101The physical symmetry of molecules upon rotations, inversions and reflec-

tions is a very important property of molecules.

Molecules that have similar symmetries often have similar characteristics.

Advanced theoretical treatment of molecules often is greatly aided by knowl-

edge symmetry.

28.1 Symmetry Elements

The key symmetry elements are:

• C rotations

— Is the molecule that same after rotation by 360 degrees?

— The axes of highest C is the principal axes.

• Mirror symmetry and

— Is the molecule the same after mirror reflection.

— If the mirror plane contains the principle axes, it is called a vertical

mirror plane ()

— If the mirror plane is perpendicular to the principle axes, it is

called a horizontal mirror plane ()

• Inversion .

— Is the molecule the same after reflecting through the origin?

• Improper rotation, S

144

— This is just a C followed by a .

Example: Water.

• Principle axis is C2• Has two vertical mirror planes• No or S

28.2 Symmetry Point Groups

Believe it or not there are only 32 distinct groups of symmetry that any given

molecule will fall into.

And, in fact, most molecules fall in an ever smaller subset of this.

Mathematicians have worked out the properties of all 32 of these groups.

So, when a chemist categorizes a molecule in a particular point group, he or

she automatically knows a number of properties of that molecule.

Since some symmetry elements are hard to find, a streamlined flow chart for

assigning molecules to point groups has been developed.

We will spend the rest of today and the next period practicing assigning

point groups.

The goal for this is to develop our abilities to think about molecules as three

dimensional objects.

∗ ∗ ∗ practice with flow chart ∗ ∗∗

28.3 Problem Set

Practice assigning molecules to point groups

145

29. Intermolecular Forces


• V080101• V080201• V080301• V080401

29.1 Polarizability and Induced Dipoles

Last time we reminded ourselves of the molecular dipole moment.

The dipole moment is important in determined intermolecular forces and the

physical properties of substances.

Another equally important and somewhat related property of the molecule

is its polarizability.

The polarizabilty, is a measure of how the electron density of the molecules

responds to an external electric field.

An external electric field causes the electron cloud of a molecule to become

distorted.

This distortion means that, even for neutral molecules, there can exits a

dipole moment.

Unlike the permanent dipole moment that comes from appropriate molecular

geometry, this induced dipole is caused by the presence of the electric field,

= (29.1)

147

29.2 Intermolecular Forces

There are several types of intermolecular forces that differ in their make-up

and in their interaction length.

The Coulombic Forces

Ion—Ion:

The force between two ions is the strongest molecular force and it has the

longest interaction length.

The force, , falls off relatively slowly with distance, ,

∝ 1

2(29.2)

Ion—dipole:

148

This force is weaker than ion—ion interaction and it falls off with distance

quicker,

∝ 1

3(29.3)

The Van der Waals Forces

Even when ions are not present one still sees intermolecular forces.

These forces are called Van der Waals forces.

The Van der Waals forces are weaker than the coulombic forces. They also

have a shorter interaction length

dipole—dipole:

The strongest of the Van der Waals forces with the longest interaction length,

∝ 1

4(29.4)

dipole—induced dipole:

When a molecule has a permanent dipole, that dipole creates an electric field

that neighboring molecule feel.

Consequently, an induced dipole can be created in the neighboring dipole.

As you might expect these forces are weaker then dipole—dipole forces, having

149

the form of

∝ 1

5

induced dipole—induced dipole (called Loudon or dispersive forces):

One might think that there would be no interaction between molecules that

have no permanent dipole.

In fact there is an attractive force that does arise in these situations.

The electron cloud and hence the polarization of the molecule is always fluc-

tuating.

This random fluctuation spontaneously creates a transient dipole moment

which can then induce a dipole in a neighboring molecule.

This induced dipole then strengthens the transient dipole by back induction.

The force is very weak and drops off very rapidly,

∝ 1

6

29.3 Phase Diagrams

The phase change behavior of a substance can be expressed conveniently with

a pressure versus temperature graph called a phase diagram.

∗ ∗ ∗ HANDOUT ∗ ∗∗

The lines on the phase diagrams represent the precise temperatures and

pressures where an equilibrium between two phases occurs.

The solid—liquid line:

• crossing the line from solid to liquid represents melting

150

• crossing the line from liquid to solid represents freezing

The gas—liquid line:

• crossing the line from gas to liquid represents condensation

• crossing the line from liquid to gas represents boiling

The solid—gas line:

• crossing the line from solid to gas represents sublimation

• crossing the line from gas to solid represents deposition

There are two special points on the phase diagram.

• The triple point— The point where the solid-liquid line, solid-gas line and gas-liquid

line converge.

— The single value of temperature and pressure in which all three

phases can co-exist.

• The critical point— The termination of the gas-liquid line.

151

30. Hydrogen Bonding and Water

Last time we discussed intermolecular forces.

There is one very important and special type of intermolecular force that we

did not mention: hydrogen bonding.

Hydrogen bonding can occur in between some molecules that contain hydro-

gen.

If a hydrogen is bonded to an oxygen, nitrogen or fluorine atom in one mole-

cule, it can hydrogen bond to an oxygen, nitrogen or fluorine atom in another

molecule.

A hydrogen bond can be thought of as either an exceptionally strong physical

interaction or a very weak chemical bond.

Molecules that have a hydrogen bonded to an oxygen, nitrogen or fluorine

atom can be hydrogen bond donors and acceptors.

152

Molecules that have an oxygen, nitrogen or fluorine atom but no hydrogen

attached to it can only be hydrogen bond acceptors.

30.1 Water

One of the most interesting, important and complicated substances know is

water.

We don’t always appreciate how unique water is because we are so familiar

with it.

Almost all of the unique aspects of water are do to hydrogen bonds.

Some of the unique properties of water are

• An abnormally high melting point• An abnormally high boiling point

153

• The fact that ice floats• The Grotthuss mechanism• A high heat capacity• A high dielectric constant.

30.1.1 The Grotthuss mechanism

The Grotthuss mechanism accounts for the abnormally high mobility of H+

and OH− ions compared to other ions.

Disruption of the Grotthuss mechansim allow aquaporins in our bodies to

transmit water in and out of the cell while blocking proton flow.

30.2 Acids and Bases Revsited

We now consider two ways to define acids and bases.

1. Brønsted—Lowry

• Acid: Donates a proton. For example,HCl+H2O→ Cl− +H3O

+

154

• Base: Accepts a proton. For example,

NH3 +H2O→ NH+4 +OH−

2. Lewis

• Acid: Accepts electrons. For example,

Fe3+ + 3H2O→ Fe(OH)3 + 3H+

• Base: Donates electrons. For example,

NH3 +Ni2+ → Fe(NH3)

2+

6

30.3 Group Work

1. Give the principle type of intermolecular force for each of the following

examples

(a) NaCl in water

(b) pure water

(c) “Salt bridges” in proteins

(d) methane dissolved in water

(e) pure CCl4

2. How would the slope of the solid liquid line for the phase diagram for

water compare to the generic one shown in the handout.

3. Suggest a way to go from a liquid to a gas without boiling.

4. Is there a way to go from a solid to a liquid without melting?

5. Restate in your own words why hydrogen bonding is the reason for the

unusual properties of water that we discussed in lecture.

155

30.4 Homework

Reading: 12, 13.6, 14.1-14.4

Exercises: In-class work.

156

31. Ideal Gases I


• V090101

31.1 Gases

We have learned that matter exists in three common states: gases, liquids

and solids.

Of these, gases are the easiest to understand from a theoretical standpoint.

The reason for this is that in a gas the constituent atoms or molecules are

very far apart such that there is little or no interaction between them.

We now study some of the important physical properties of gases.

31.2 The Ideal Gas Law

We stated above that the particles making up the gas have minimal interac-

tion with one another.

This serves the basis for an important simplifying approximation. For gases

at “low” pressures (“low” here can still mean pressures ten or twenty times

greater than atmospheric pressures), we may approximate the gas as a so-

called ideal gas

An ideal gas is one in which the particles do not interact at all. Furthermore,

the particles of an ideal gas have no volume. (note: the ideal gas of course

has volume, but the particles themselves don’t)

With these very good approximations, we can describe any real gas under

most circumstances using the ideal gas law:

= (31.1)

157

where is pressure, is the volume occupied by the gas, is the number of

moles of ideal gas particles, is the gas constant and is the temperature.

Let us solve the ideal gas equation for volume and check to see of the equation

is consistent with our common sense. The equation becomes

=

(31.2)

This equation says that the volume of a gas is directly proportional to both

the amount of gas particles and the temperature. It is inversely proportional

to pressure.

These relations are consistent with common sense.

31.2.1 A word on units

Let’s look at the ideal gas law in terms of units.

We see that the right hand side is which has units of energy

⇒ mol/ es energy tem/ ptem/ p mo/ le

= energy. (31.3)

If we are using SI units then the energy units would be joules.

The left hand side of the ideal gas equation must then also have units of

energy.

Indeed a pressure times a volume is energy.

If we were in SI units pressure would be in pascals and volume in cubic

meters: Pa m3 = J.

These units of pressure and volume are not very common so instead when

working with gases one uses atmospheres and liters repsectively.

The energy unit then is the liter-atmosphere: L atm.

Because we are in different units takes on a different value.

In units of L atm, = 0082 L atm/(K mol).

158

31.2.2 Ideal gas law in terms of density

We can express the ideal gas law in terms of density by observing that

moles of a gas whose constituents have a molar mass of will have a have

mass of

= (31.4)

From this relation we have

=

(31.5)

which we can substitute into the ideal gas law to get

= (31.6)

=

We can rearrange this equation to the form

=

z}|{

(31.7)

=

Solving for density gives

=

(31.8)

Notice that this version of the ideal gas law contains only intensive variables.

31.3 Group Work


1. Solve the ideal gas law for pressure and see if the relation between

pressure and the other variables are consistent with your experience.

2. Solve the ideal gas law for temperature and see if the relation between

temperature and the other variables are consistent with your experi-

ence.

3. How much volume does 1.234 g of oxygen take up at 300 K and 1 atm.

[ans: 0.95 L]

159

4. The intensive variable molar volume is volume per mole.

(a) Write the ideal gas law in terms of molar volume.

(b) What is the molar volume of gases at what are called standard

conditions 1 atm and 298 K. [ans: 24.4 L/mol]

5. What is the density of nitrogen at 1 atm and 298 K? [ans: 1.15

g/L]

6. Challenge problem: Let’s say gas A is twice the molar mass as gas

B and both gases are at the same pressure and density. Gas A is at

300 K. What is the temperature of gas B? [ans: 150]

31.4 Problem Set



160

32. Ideal Gases II


• V090102• V090103

32.1 Gas Mixtures and Partial Pressure

To understand a mixture of ideal gases we need the concept of partial pres-

sure.

Consider a mixture of two different types of gases and

The partial pressure of component (denoted as ) is the pressure that

would exist if all of component was absent.

Likewise the partial pressure of (denoted as ) is the pressure that would

exist if all of component was absent.

As it turns out the total pressure of the mixture is simply the sum of the

partial pressures. So, in this case = +

This is known as Dalton’s law of partial pressures, which applies more gen-

erally to mixtures of an arbitrary number of components:

= 1 + 2 + · · · (32.1)

Partial pressures can be calculated using separate ideal gas equations. That

is,

1 = 1 2 = 2 etc. (32.2)

162

So the ideal gas law for mixtures can be written as

= (1 + 2 + · · · ) (32.3)

=

µ1

+

2

+ · · ·

¶

= 1 + 2 + · · · = (1 + 2 + · · · )| {z }

total

Dalton’s law gives us the total pressure in terms of the constituent pressures.

Often it is useful to have an expression for one of the components in terms

of the total pressure.

To obtain this relation we start with the ratio of the component pressure to

the total pressure,

=

tot

=

tot

¡

¢¡

¢ =

tot (32.4)

Now, the relation tot

is the definition of the mole fraction of component

( =tot) So,

= =⇒ = (32.5)

The partial pressure of component is equal to the total pressure times the

mole fraction of component

32.2 Chemical Reactions Between Gases

Reactions between gases can change the total number of moles of gas in a

reaction vessel.

Since the number of moles of gas appears in the ideal gas law, the pressure

or volume of a reaction vessel may change during reaction.

Consider the combustion of methane

CH4() + 2O2() → CO2() + 2H2O() (32.6)

in a closed container of fixed volume (10.0 L).

163

Let us determine the change in pressure for the case of the reactants at 300.0

K and the products at 300.0 K.

Say we start with 1 mole of methane and 2 moles of oxygen in our reaction

vessel. The partial pressure of methane is

CH4 =CH4

=1 mol (0082057 L·atm

K·mol)3000 K

100 L= 246 atm. (32.7)

The partial pressure of oxygen is

O2 =O2

=2 mol(0082057 L·atm

K·mol)300 K

100 L= 492 atm. (32.8)

So the total pressure of the reactants is

= CH4 + O2 = 738 atm. (32.9)

After the reaction only one mole of CO2 is present so the total pressure of

the products is

= CO2 =CO2

=1mol(0082057 L·atm

K·mol)300 K

100 L= 246 atm. (32.10)

Thus the change in pressure due to the reaction is 246 atm − 738 atm

= −492 atm.

32.3 Group Work


1. Brenda placed 0.500 g of oxygen and 0.500 g of argon in a 10.0 L flask

at 298 K.

(a) What is the partial pressure of each gas? [ans: PAr = 0.0305

atm, PO2 = 0.0381 atm]

(b) What is the total pressure of the flask? [ans: Ptot = 0.0686

atm]

(c) What is the mole fraction of each gas? [ans: XO2 = 0.556, XAr= 0.445]

164

2. Jasmine placed 0.231 g of methane and 0.250 g of oxygen are placed in

a 10.0 L at 300 K and ignited the gases. After the reaction was com-

plete, Jasmine returned the flask back to 300 K. What is the partital

pressure the species present at the end? Hint: be careful to determine

the limiting reagent. [ans: PCO2 = 9.61×10−3 atm, PCH4 = 0.0256atm]

32.4 Problem Set



165

33. Kinetic Theory of Gases


• V090201

33.1 The Kinetic Molecular Theory of Gases

A view of gases from a molecular perspective is given by the kinetic molecular

theory of gases

We will not discuss the theory in lecture other than to point out several key

results.

1. The probability of finding a molecule at a give speed is given by the

Maxwell-Boltzmann distribution.

() = 4

µ

2

¶32

−2

2 2 (33.1)

167

2. The average kinetic energy of the gas particles is

=3

2 (33.2)

3. The average speed or root mean square (rms) speed of the gas particles

is p2 =

r3

(33.3)

where is the molar mass of the particles.

Read section 12.13 of your book for a more elaborate discussion of the theory.

33.2 Group Work


1. Calculate the rms of Br2 at 298 K. [ans: 216 m/s]

2. Show that the ratio of rms for two gases at the same temperature is

simply the inverse of the ratio of the square roots of their molecular

masses.

3. Wendy measured the rms of an unknown gas in a container at 300 K

to be 498 m/s. What would she measure for the rms at 400 K? [ans:

575 m/s]

33.3 Problem Set


Exercises: Chapter 11: 41, 43, 45, 47, 49.

168

Index

accuracy 12, 14

acid

definition of 54

acid/base reactions 55

alkali metals 21

alkaline earth metals 21

analyte 68

angstrom 11

anion 34

atmosphere 11

atomic mass 19

atomic mass number 19

atomic mass unit (amu) 11

atomic number 19

atomic orbitals

chemist picture of 107

physicist picture of 106

aufbau principle 110

average 12

formula for 13

average deviation

formula for 13

back titration 70

balancing chemical equations 41

bar 11

base

definition of 54

Beer’s law plot 65

benchmarks

important ones 12

Bohr model

failure of 101

Bohr model of the atom 99

bond

covalent 34, 121

ionic 121

bond dissociation energy 129

bond length 129

bond order 129

bond polarity 130

cation 34

chemical bond 5

chemical change 6

chemical energy 72

chemical equation 40

chemical equations

balancing 41

chemical formula 30

chemical reaction 40

classical mechanics 97

combustion experiment

steps of 82

combustion reaction 41

compound 5

compounds 30

concentration 62

condensed formula 30

converting

grams to moles 26

moles to grams 26

169

core electrons 119

covalent bond 34, 121

Dalton’s law of partial pressures 162

dark matter 4

degeneracy 108

dipole moment 134

direct titration 70

electrolytes 52

electromagnetic radiation 92

electron affinity 117

electron configuration 112

noble gas notation 112

electronegativity 130

electronic geometry 132

electrons 18

element 5

emperical formula 27

end point

for a titration 69

endothermic reaction 79

energy

conservation of 72

energy level diagram

for the Bohr model 101

for the hydrogen atom 108

energy levels


energy states

of the atom 98

enthalpy 78

equvalence point

for a titration 68

error 12

formula for 14

exothermic reaction 79

extensive property 6

fermi 12

first law of thermodynamics 73

food calories 83

formal charge 126

formation reactions 87

halogens 22

heat 73

heat capcity 73

heat of formation 87

heat of fusion 74

heat of vaporization 74

Hess’ law 86

heterogenous mixture 4

homogenous mixture 4

Hund’s rule 112

hybridization 139

ideal gas 157

ideal gas law 157

indicator 68

intensive property 6

inversion symmetry 144

ion 34

ionic bond 121

ionization energy 100, 117

isotope 20

percent aboundance 20

kinetic energy 72

kinetic theory of gases 167

lanthanide contration 117

latent heat 74

Lewis dot structure 119

general algorithm 123

light

energy of 93

frequency of 93

speed of 93

limiting reagent 47

liter 11

mass spectrometry 23

matter

170

physical properties of 5

states of 3

Maxwell-Boltzmann distribution 167

mean 12

metalloid 21

micron 11

mirror symmetry 144

mixture 4

types of 4

molality 62

molar mass 25

molarity 62

mole 25

mole fraction 62

molecular geometry 133

molecular orbital theory 132

molecule 5

molecules 30

naming anions 36

naming binary compounds 31

naming cations 36

naming ionic compounds 37

net ionic equation 53

neutron 19

noble gases 35

nodes 107

octet rule 120

odixation-reduction reactions 58

oxidation number 34, 58

oxidation state 34

oxides 21

oxidizing agent 58

partial pressure 162

Pauli exclusion principle 111

percent composition 26

percent error

formula for 14

percent yield 48

pH 64

physcal change 6

piston 77

plasma 3

point group 145

potential energy 72

precipitate 52

precipitation reaction 52

precision 12

prefixes

for SI units 11

products 40

proton 18

pure substance 4

quantum mechanics 97, 104

quantum number 100

quantum numbers


reactants 40

redox reactions 58

reducing agent 58

rotation symmetry 144

Rydberg constant 100

Schrodinger equation 104

shielding 117

SI units 10

derived units 10

prefixes for 11

significant figures 12

solar system model of the atom 98

solute 51

solvated ions 51

solvent 51

specific heat capacity 74

spectator ions 53

spectroscopic lines 101

standard conditions 160

standard deviation

171

formula for 14

state function 86

stoichiometric coefficients 40, 42

stoichiometry 40

structural formula 30

symmetry elements 144

temperature 73

172

chemistry 127 - styrke fitness · these temperature scales are called relative temperature scales....

Documents