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
Observation 1: Mass relationships during chemical reactions
Observation 2: Multiple Mass Ratios
Review and Discussion Questions
Chapter 2. Relative Atomic Masses and Empirical Formulae
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
Observation 1: Scattering of α particles by atoms
Observation 3: Ionization energies of the atoms
Review and Discussion Questions
Chapter 4. Quantum Energy Levels In Atoms
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
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
Review and Discussion Questions
Chapter 6. Molecular Geometry and Electron Domain Theory
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
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
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
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
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
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