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General Chemistry I

By: John Hutchinson

Online: < http://cnx.org/content/col10263/1.3> This selection and arrangement of content as a collection is copyrighted by John Hutchinson.

It is licensed under the Creative Commons Attribution License: http://creativecommons.org/licenses/by/2.0/

Collection structure revised: 2007/07/18

For copyright and attribution information for the modules contained in this collection, see the " Attributions" section at the end of the collection.

General Chemistry I

Table of Contents

Chapter 1. The Atomic Molecular Theory

1.1.

Foundation

Goals

Observation 1: Mass relationships during chemical reactions

Observation 2: Multiple Mass Ratios

Review and Discussion Questions

Chapter 2. Relative Atomic Masses and Empirical Formulae

2.1.

Foundation

Goals

Observation 1: Volume Relationships in Chemical Reactions

Determination of Atomic Weights for Gaseous Elements

Determination of Atomic Weights for Non-Gaseous Elements

Moles, Molecular Formulae and Stoichiometric Calculations

Review and Discussion Questions

Chapter 3. The Structure of an Atom

3.1.

Foundation

Goals

Observation 1: Scattering of α particles by atoms

Observation 2: X-ray emission

Observation 3: Ionization energies of the atoms

Review and Discussion Questions

Chapter 4. Quantum Energy Levels In Atoms

4.1.

Foundation

Goals

Observation 1: The Spectrum of Hydrogen

Observation 2: The Photoelectric Effect

Quantized Energy Levels in Hydrogen Atoms

Observation 3: Photoelectron Spectroscopy of Multi-Electron Atoms

Electron Waves, the Uncertainty Principle, and Electron Energies

Electron Orbitals and Subshell Energies

Review and Discussion Questions

Chapter 5. Covalent Bonding and Electron Pair Sharing

5.1.

Foundation

Goals

Observation 1: Valence and the Periodic Table

Observation 2: Compounds of Carbon and Hydrogen

Observation 3: Compounds of Nitrogen, Oxygen, and the Halogens

Interpretation of Lewis Structures

Extensions of the Lewis Structure Model

Resonance Structures

Review and Discussion Questions

Chapter 6. Molecular Geometry and Electron Domain Theory

6.1.

Foundation

Goals

Observation 1: Geometries of molecules

Observation 2: Molecules with Double or Triple Bonds

Observation 3: Distortions from Expected Geometries

Review and Discussion Questions

Chapter 7. Molecular Structure and Physical Properties

7.1.

Foundation

Goals

Observation 1: Compounds of Groups I and II

Observation 2: Molecular Dipole Moments

Observation 3: Dipole Moments in Polyatomic Molecules

Review and Discussion Questions

Chapter 8. Chemical Bonding and Molecular Energy Levels

8.1.

Foundation

Goals

Observation 1: Bonding with a Single Electron

Observation 2: Bonding and Non-Bonding in Diatomic Molecules

Observation 3: Ionization energies of diatomic molecule

Review and Discussion Questions

Chapter 9. Energetics of Chemical Reactions

9.1.

The Foundation

Goals

Observation 1: Measurement of Heat by Temperature

Observation 2: Hess' Law of Reaction Energies

Observation 3: Bond Energies in Polyatomic Molecules

Review and Discussion Questions

Index

Chapter 1. The Atomic Molecular Theory

Foundation

There are over 18 million known substances in our world. We will begin by assuming that all

materials are made from elements, materials which cannot be decomposed into simpler

substances. We will assume that we have identified all of these elements, and that there a very

small number of them. All other pure substances, which we call compounds, are made up from

these elements and can be decomposed into these elements. For example, metallic iron and

gaseous oxygen are both elements and cannot be reduced into simpler substances, but iron rust, or

ferrous oxide, is a compound which can be reduced to elemental iron and oxygen. The elements

are not transmutable: one element cannot be converted into another. Finally, we will assume that

we have demonstrated the Law of Conservation of Mass.

Law 1.1.

The total mass of all products of a chemical reaction is equal to the total mass of all reactants of

that reaction.

These statements are summaries of many observations, which required a tremendous amount of

experimentation to achieve and even more creative thinking to systematize as we have written

them here. By making these assumptions, we can proceed directly with the experiments which led

to the development of the atomic-molecular theory.

Goals

The statements above, though correct, are actually more vague than they might first appear. For

example, exactly what do we mean when we say that all materials are made from elements? Why

is it that the elements cannot be decomposed? What does it mean to combine elements into a

compound? We want to understand more about the nature of elements and compounds so we can

describe the processes by which elements combine to form compounds, by which compounds are

decomposed into elements, and by which compounds are converted from one to another during

chemical reactions.

One possibility for answering these questions is to assume that a compound is formed when

indestructible elements are simply mixed together, as for example, if we imagine stirring together

a mixture of sugar and sand. Neither the sand nor the sugar is decomposed in the process. And the

mixture can be decomposed back into the original components. In this case, though, the resultant

mixture exhibits the properties of both components: for example, the mixture would taste sweet,

owing to the sugar component, but gritty, characteristic of the sand component.

In contrast, the compound we call iron rust bears little resemblance to elemental iron: iron rust

does not exhibit elemental iron's color, density, hardness, magnetism, etc. Since the properties of

the elements are not maintained by the compound, then the compound must not be a simple

mixture of the elements.

We could, of course, jump directly to the answers to these questions by stating that the elements

themselves are comprised of atoms: indivisible, identical particles distinctive of that element.

Then a compound is formed by combining the atoms of the composite elements. Certainly, the

Law of Conservation of Mass would be easily explained by the existence of immutable atoms of

fixed mass.

However, if we do decide to jump to conclusions and assume the existence of atoms without

further evidence (as did the leading chemists of the seventeenth and eighteenth centuries), it does

not lead us anywhere. What happens to iron when, after prolonged heating in air, it converts to

iron rust? Why is it that the resultant combination of iron and air does not maintain the properties

of either, as we would expect if the atoms of each are mixed together? An atomic view of nature

would not yet provide any understanding of how the air and the iron have interacted or combined

to form the new compound, and we can't make any predictions about how much iron will produce

how much iron rust. There is no basis for making any statements about the properties of these

atoms. We need further observations.

Observation 1: Mass relationships during chemical reactions

The Law of Conservation of Mass, by itself alone, does not require an atomic view of the

elements. Mass could be conserved even if matter were not atomic. The importance of the Law of

Conservation of Mass is that it reveals that we can usefully measure the masses of the elements

which are contained in a fixed mass of a compound. As an example, we can decompose copper

carbonate into its constituent elements, copper, oxygen, and carbon, weighing each and taking the

ratios of these masses. The result is that every sample of copper carbonate is 51.5% copper, 38.8%

oxygen, and 9.7% carbon. Stated differently, the masses of copper, oxygen, and carbon are in the

ratio of 5.3 : 4 : 1, for every measurement of every sample of copper carbonate. Similarly, lead

sulfide is 86.7% lead and 13.3% sulfur, so that the mass ratio for lead to sulfur in lead sulfide is

always 6.5 : 1. Every sample of copper carbonate and every sample of lead sulfide will produce

these elemental proportions, regardless of how much material we decompose or where the

material came from. These results are examples of a general principle known as the Law of

Definite Proportions.

Law 1.2.

When two or more elements combine to form a compound, their masses in that compound are in a

fixed and definite ratio.

These data help justify an atomic view of matter. We can simply argue that, for example, lead

sulfide is formed by taking one lead atom and combining it with one sulfur atom. If this were true,

then we also must conclude that the ratio of the mass of a lead atom to that of a sulfur atom is the same as the 6.5 : 1 lead to sulfur mass ratio we found for the bulk lead sulfide. This atomic

explanation looks like the definitive answer to the question of what it means to combine two

elements to make a compound, and it should even permit prediction of what quantity of lead

sulfide will be produced by a given amount of lead. For example, 6.5g of lead will produce exactly

7.5g of lead sulfide, 50g of lead will produce 57.7g of lead sulfide, etc.

There is a problem, however. We can illustrate with three compounds formed from hydrogen,

oxygen, and nitrogen. The three mass proportion measurements are given in the following table.

First we examine nitric oxide, to find that the mass proportion is 8 : 7 oxygen to nitrogen. If this is

one nitrogen atom combined with one oxygen atom, we would expect that the mass of an oxygen

atom is 8/7=1.14 times that of a nitrogen atom. Second we examine ammonia, which is a

combination of nitrogen and hydrogen with the mass proportion of 7 : 1.5 nitrogen to hydrogen. If

this is one nitrogen combined with one hydrogen, we would expect that a nitrogen atom mass is

4.67 times that of a hydrogen atom mass. These two expectations predict a relationship between

the mass of an oxygen atom and the mass of a hydrogen atom. If the mass of an oxygen atom is

1.14 times the mass of a nitrogen atom and if the mass of a nitrogen atom is 4.67 times the mass

of a hydrogen atom, then we must conclude that an oxygen atom has a mass which is 1.14 × 4.67

= 5.34 times that of a hydrogen atom.

But there is a problem with this calculation. The third line of the following table shows that the compound formed from hydrogen and oxygen is water, which is found to have mass proportion

8:1 oxygen to hydrogen. Our expectation should then be that an oxygen atom mass is 8.0 times a

hydrogen atom mass. Thus the three measurements in the following table appear to lead to contradictory expectations of atomic mass ratios. How are we to reconcile these results?

Table 1.1. Mass Relationships for Hydrogen, Nitrogen, Oxygen Compounds

"Expected" "Expected" "Expected"

Mass

Relative

Relative

Relative

Total

Mass of

Mass of

Compound

of

Atomic

Atomic

Atomic

Mass Hydrogen Nitrogen Oxygen Mass of

Mass of

Mass of

Hydrogen

Nitrogen

Oxygen

Nitric

15.0

-

7.0 g

8.0 g

-

7.0

8.0

Oxide

g

Ammonia

8.5 g

1.5 g

7.0 g

-

1.5

7.0

-

Water

9.0 g

1.0 g

-

8.0 g

1.0

-

8.0

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One possibility is that we were mistaken in assuming that there are atoms of the elements which

combine to form the different compounds. If so, then we would not be surprised to see variations

in relative masses of materials which combine.

Another possibility is that we have erred in our reasoning. Looking back, we see that we have to

assume how many atoms of each type are contained in each compound to find the relative masses

of the atoms. In each of the above examples, we assumed the ratio of atoms to be 1:1 in each

compound. If there are atoms of the elements, then this assumption must be wrong, since it gives

relative atomic masses which differ from compound to compound. How could we find the correct

atomic ratios? It would help if we knew the ratio of the atomic masses: for example, if we knew

that the oxygen to hydrogen mass ratio were 8:1, then we could conclude that the atomic ratio in

water would be 1 oxygen and 1 hydrogen. Our reasoning seems to circular: to know the atomic

masses, we must know the formula of the compound (the numbers of atoms of each type), but to

know the formula we must know the masses.

Which of these possibilities is correct? Without further observations, we cannot say for certain

whether matter is composed of atoms or not.

Observation 2: Multiple Mass Ratios

Significant insight into the above problem is found by studying different compounds formed from

the same elements. For example, there are actually three oxides of nitrogen, that is, compounds

composed only of nitrogen and oxygen. For now, we will call them oxide A, oxide B, and oxide C.

Oxide A has oxygen to nitrogen mass ratio 2.28 : 1. Oxide B has oxygen to nitrogen mass ratio

1.14 : 1, and oxide C has oxygen to nitrogen mass ratio 0.57 : 1.

The fact that there are three mass ratios might seem to contradict the Law of Definite Proportions,

which on the surface seems to say that there should be just one ratio. However, each mass

combination gives rise to a completely unique chemical compound with very different chemical

properties. For example, oxide A is very toxic, whereas oxide C is used as an anesthesia. It is also

true that the mass ratio is not arbitrary or continuously variable: we cannot pick just any

combination of masses in combining oxygen and nitrogen, rather we must obey one of only three.

So there is no contradiction: we simply need to be careful with the Law of Definite Proportions to

say that each unique compound has a definite mass ratio of combining elements.

These new mass ratio numbers are highly suggestive in the following way. Notice that, in each

case, we took the ratio of oxygen mass to a nitrogen mass of 1, and that the resultant ratios have a

very simple relationship:

()

The masses of oxygen appearing in these compounds are in simple whole number ratios when we

take a fixed amount of nitrogen. The appearance of these simple whole numbers is very

significant. These integers imply that the compounds contain a multiple of a fixed unit of mass of

oxygen. The simplest explanation for this fixed unit of mass is that oxygen is particulate. We call the fixed unit of mass an atom. We now assume that the compounds have been formed from

combinations of atoms with fixed masses, and that different compounds have differing numbers of

atoms. The mass ratios make it clear that oxide B contains twice as many oxygen atoms (per

nitrogen atom) as does oxide C and half as many oxygen atoms (per nitrogen atom) as does oxide

A. The simple mass ratios must be the result of the simple ratios in which atoms combine into

molecules. If, for example, oxide C has the molecular formula NO , then oxide B has the formula N O 2 , and oxide A has the formula N O 4 . There are other possibilities: if oxide B has molecular formula NO , then oxide A has formula N O 2 , and oxide C has formula N 2 O . Or if oxide A has formula NO , then oxide B has formula N 2 O and oxide C has formula N 4 O . These three possibilities are listed in the following table.

Table 1.2. Possible Molecular Formulae for Nitrogen Oxides

Oxide C is

Oxide B is

Oxide A is

Assuming that:

NO

NO

NO

Oxide A is

N O 4

N O 2

NO

Oxide B is

N O 2

NO

N 2 O

Oxide C is

NO

N 2 O

N 4 O

We don't have a way (from these data) to know which of these sets of molecular formulae are

right. But we can assert that either one of them or one analogous to them is right.

Similar data are found for any set of compounds formed from common elements. For example,

there are two oxides of carbon, one with oxygen to carbon mass ratio 1.33:1 and the other with

mass ratio 2.66:1. The second oxide must have twice as many oxygen atoms, per carbon atom, as

does the first. The general statement of this observation is the Law of Multiple Proportions.

Law 1.3.

When two elements combine to form more than one compound, the mass of element A which

combines in the first compound with a given amount of element B has a simple whole number

ratio with the mass of element A which combines in the second compound with the same given

mass of element B.

This sounds confusing, but an example clarifies this statement. Consider the carbon oxides, and let

carbon be element B and oxygen be element A. Take a fixed given mass of carbon (element B),

say 1 gram. The mass of oxygen which combines with 1 gram of carbon to form the first oxide is

1.33 grams. The mass of oxygen which combines with 1 gram of carbon to form the second oxide

is 2.66. These masses are in ratio 2.66 : 1.33 = 2 : 1 , a simple whole number ratio.

In explaining our observations of the Law of Multiple Proportions for the carbon oxides and the

nitrogen oxides, we have concluded that the simple mass ratio arises from the simple ratio of

atoms contained in the individual molecules. Thus, we have established the following postulates

of the Atomic Molecular Theory.

Theory

the elements are comprised of identical atoms

all atoms of a single element have the same characteristic mass

these number and masses of these atoms do not change during a chemical transformation

compounds consist of identical molecules formed of atoms combined in simple whole number

ratios

Review and Discussion Questions

Exercise 1.

Assume that matter does not consist of atoms. Show by example how this assumption leads to

hypothetical predictions which contradict the Law of Multiple Proportions. Do these hypothetical

examples contradict the Law of Definite Proportions? Are both observations required for

confirmation of the atomic theory?

Exercise 2.

Two compounds, A and B, are formed entirely from hydrogen and carbon. Compound A is 80.0%

carbon by mass, and 20.0% hydrogen, whereas Compound B is 83.3% carbon by mass and 16.7%

hydrogen. Demonstrate that these two compounds obey the Law of Multiple Proportions. Explain

why these results strongly indicate that the elements carbon and hydrogen are composed of atoms.

Exercise 3.

In many chemical reactions, mass does not appear to be a conserved quantity. For example, when

a tin can rusts, the resultant rusty tin can has a greater mass than before rusting. When a candle

burns, the remaining candle has invariably less mass than before it was burned. Provide an

explanation of these observations, and describe an experiment which would demonstrate that mass

is actually conserved in these chemical reactions.

Exercise 4.

The following question was posed on an exam:

An unknown non-metal element (Q) forms two gaseous fluorides of unknown molecular

formula. A 3.2 g sample of Q reacts with fluorine to form 10.8 g of the unknown fluoride A.

A 6.4 g sample of Q reacts with fluorine to form 29.2 g of unknown fluoride B. Using these

data only, demonstrate by calculation and explanation that these unknown compounds obey

the Law of Multiple Proport