Download The science of chemistry is concerned with the

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Chapter two
ATOMS, MOLECULES
AND CHEMICAL REACTIONS
In Chap. 1 we described two very important things that chemists (and scientists in general)
do. They make quantitative measurements, and they communicate the results of such
experiments as clearly and unambiguously as possible. In this chapter we will deal with
another important activity of chemists-the use of their imaginations to devise theories or
models to interpret their observations and measurements. Such theories or models are
useful in suggesting new observations or experiments which yield additional data. They
also serve to summarize existing information and aid in its recall.
The atomic theory, first proposed in modern form by John Dalton, is extremely important
to chemists. It interprets observations of the every-day world in terms of particles called
atoms and molecules. Macroscopic events—those which humans can observe or
experience with their unaided senses—are interpreted by means of microscopic objects—
those so small that a special instrument or apparatus must be used to detect them. (Perhaps
the term submicroscopic really ought to be used, because most atoms and molecules are
much too small to be seen even under a
microscope.) In any event, chemists continually try to explain the macroscopic world in
microscopic terms.
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2.1 MACROSCOPIC PROPERTIES AND MICROSCOPIC MODELS
As a simple example of how the macroscopic properties of a substance can be explained on
a microscopic level, consider the liquid mercury. Macroscopically, mercury at ordinary
temperatures is a silvery liquid which can be poured much like water—rather unusual for a
metal. Mercury is also the heaviest known liquid. Its density is 13.6 g cm–3, as compared
with only 1.0 g cm–3 for water. When cooled below –38.9°C mercury solidifies and behaves
very much like more familiar solid metals such as copper and iron. Mercury frozen around
the end of a wooden stick can be used to hammer nails, as long as it is kept sufficiently
cold. Solid mercury has a density of 14.1 g cm–3 slightly greater than that of the liquid.
When mercury is heated, it remains a liquid until quite a high temperature, finally boiling at
356.6°C to give an invisible vapor. Even at low concentrations gaseous mercury is
extremely toxic if breathed into the lungs. It has been responsible for many cases of human
poisoning. In other respects mercury vapor behaves much like any gas. It is easily
compressible. Even when quite modest pressures are applied, the volume decreases
noticeably. It is also much less dense than the liquid or the solid. At 400°C and ordinary
pressures, its density is 3.6 × 10–3 g cm–3 about one four-thousandth that of solid or liquid
mercury.
A modem chemist would interpret these macroscopic properties in terms of a microscopic
model involving atoms of mercury. As shown in Fig. 2.1, the atoms may be thought of as
small, hard spheres. Like billiard balls they can move around and bounce off one another.
In solid mercury the centers of adjacent atoms are separated by only 300 pm (300 × 10–12 m
or 3.00Å). Although each atom can move around a little, the others surround it so closely
that it cannot escape its allotted position. Hence the solid is rigid. Very few atoms move out
of position even when it strikes a nail. As temperature increases, the atoms vibrate more
violently, and eventually the solid melts. In liquid mercury, the regular, geometrically rigid
structure is gone and the atoms are free to move about, but they are still rather close
together and difficult to separate. This ability of the atoms to move past each other accounts
for the fact that liquid mercury can flow and take the shape of its container. Note that the
structure of the liquid is not as compact as that of the solid; a few gaps are present. These
gaps explain why liquid mercury is less dense than the solid.
In gaseous mercury, also called mercury vapor, the atoms are very much farther apart than
in the liquid and they move around quite freely and rapidly. Since there are very few atoms
per unit volume, the density is considerably lower than for the liquid and solid. By moving
rapidly in all directions, the atoms of mercury (or any other gas for that matter) are able to
fill any container in which they are placed. When the atoms hit a wall of the container, they
bounce off. This constant bombardment by atoms on the microscopic level accounts for the
pressure exerted by the gas on the macroscopic level. The gas can be easily compressed
because there is plenty of open space between the atoms. Reducing the volume merely
reduces that
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Figure 2.1 Microscopic view of the element mercury (a) in the gaseous state (above 356.6°C); (b) as a
liquid (between –38.9 and 356.6°C) and (c) in solid form (below –38.9°C)
empty space. The liquid and the solid are not nearly so easy to compress because there is
little or no empty space between the atoms.
You may have noticed that although our microscopic model can explain many of the
properties of solid, liquid, and gaseous mercury, it cannot explain all of them. Mercury’s
silvery color and why the vapor is poisonous remain a mystery, for example. There are two
approaches to such a situation. We might discard the idea of atoms in favor of a different
theory which can explain more macroscopic properties. On the other hand it may be
reasonable to extend the atomic theory so that it can account for more facts. The second
approach has been followed by chemists. In this and the next chapter we shall discuss in
more detail those facts which require only a
33
simple atomic theory for their interpretation. Many of the subsequent chapters will describe
extensions of the atomic theory which allow interpretations of far more observations.
2.2 HISTORICAL DEVELOPMENT OF THE ATOMIC THEORY
The atomic, microscopic way of looking at matter is actually a fairly new development. The
United States has already celebrated its two-hundredth birthday, whereas the atomic theory
is only about 175 years old. None of the Founding Fathers—not even Benjamin Franklin,
the most scientific of them—thought about matter in terms of atoms.
Neither, for that matter, did the man whose experiments and ideas led directly to the theory
itself. Antoine Lavoisier was born in 1743, the same year as Thomas Jefferson. The son of a
wealthy French lawyer, he was well educated and became a successful businessman,
gentleman farmer, economist, and social reformer, as well as the leading chemist of his day.
It was Lavoisier’s position as a tax collector, not his chemical research, which led to his
death by guillotine in 1794, at the height of the French Revolution. Much of Lavoisier’s
work as a chemist was devoted to the study of combustion. He became convinced that when
a substance is burned in air, it combines with some component of the air. Eventually he
realized that this component was the dephlogisticated air which had been discovered by
Joseph Priestly (1733 to 1804) a few years earlier. Lavoisier renamed this substance
oxygen. In an important series of experiments he showed that when mercury is heated in
oxygen at a moderate temperature, a red substance, calx of mercury, is obtained. (A calx is
the ash left when a substance burns in air.) At a higher temperature this calx decomposes
into mercury and oxygen. Lavoisier’s careful experiments also revealed that the combined
masses of mercury and oxygen were exactly equal to the mass of calx of mercury. That is,
there was no change in mass upon formation or decomposition of the calx. Lavoisier
hypothesized that this should be true of all chemical changes, and further experiments
showed that he was right. This principle is now called the law of conservation of mass.
As Lavoisier continued his experiments with oxygen, he noticed something else. Although
oxygen combined with many other substances, it never behaved as though it were itself
a combination of other substances. Lavoisier was able to decompose the red calx into
mercury and oxygen, but he could find no way to break down oxygen into two or more new
substances. Because of this he suggested that oxygen must be an element—an ultimately
simple substance which could not be decomposed by chemical changes.
Lavoisier did not originate the idea that certain substances (elements) were fundamental
and all others could be derived from them. This had first been proposed in Greece during
the fifth century B.C. by Empedocles, who speculated that all matter consisted of
combinations of earth, air, fire, and water. These ideas were further developed and taught
by Aristotle and re-
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Figure 2.2 Lavoisier’s table of elements.
(From Lavoisier, “Traite de chemie,” 1789.
English translation, 1790.)
mained influential for 200 years. Lavoisier did, however, produce the first table of the
elements which contained a large number of substances that modern chemists would agree
should be classifies as elements. The list in Fig. 2.2 is taken from the first English edition of
Lavoisier’s “Textbook of Chemistry” (1790). He published it with the knowledge that
further research might succeed decomposing some of the substances listed, thus showing
them not to be elements. One of his objectives was to prod his contemporaries into just that
kind of research. Sure enough the “earth substances” listed at the bottom were eventually
shown to be combinations of certain metals with oxygen. It is also interesting to note that
not even Lavoisier could entirely escape from Aristotle’s influence. The second element in
his list is Aristotle’s “fire,” which Lavoisier called “caloric,” and which we now call “heat.”
Both heat and light, the first two items in the table, are now regarded as forms of energy
rather than of matter.
Although his table of elements was incomplete, and even incorrect in some instances,
Lavoisier’s work represented a major step forward. By
classifying certain substances as elements, he stimulated much additional chemical
research and brought order and structure to the subject where none had existed before. His
contemporaries accepted his ideas very readily, and he became known as the father of
chemistry.
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2.3 THE ATOMIC THEORY
John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from
him in almost every respect. Dalton came from a working class family and only attended
elementary school. Apart from this, he was entirely self-taught. Even after he became
famous, he never aspired beyond a modest bachelor’s existence in which he supported
himself by teaching mathematics to private pupils. Dalton made many contributions to
science, and he seems not to have realized that his atomic theory was the most important of
them. In his “New System of Chemical Philosophy” published in 1808, only the last seven
pages out of a total of 168 are devoted to it!
The postulates of the atomic theory are given in Table 2.1. The first is no advance on the
ancient Greek philosopher Democritus who had theorized
TABLE 2-1 The Postulates of Dalton’s Atomic Theory.
1
2
3
4
All matter is composed of a very large number of very small particles called atoms.
For a given element, all atoms are identical in all respects. In particular all atoms of
the same element have the same constant mass, while atoms of different elements
have different masses.
The atoms are the units of chemical changes. Chemical reactions involve the
combination, separation, or rearrangement of atoms, but atoms are neither created,
destroyed, divided into parts, or converted into atoms of any other kind.
Atoms combine to form molecules in fixed ratios of small whole numbers.
almost 2000 years earlier that matter consists of very small particles. The second postulate,
however, shows the mark of an original genius; here Dalton links the idea of atom to the
idea of element. Lavoisier’s criterion for an element had been essentially a macroscopic,
experimental one. If a substance could not be decomposed chemically, then it was probably
an element. By contrast, Dalton defines an element in theoretical, microscopic terms. An
element is an element because all its atoms are the same. Different elements have different
atoms. There are just as many different kinds of elements as there are different kinds of
atoms.
Now look back a moment to Fig. 2.1, where microscopic pictures of solid, liquid, and
gaseous mercury were given. Applying Dalton’s second postulate to this figure, you can
immediately conclude that mercury is an element, because only one kind of atom appears.
Although mercury atoms are drawn as black spheres in Fig. 2.1, it would be more common
today to represent them using chemical symbols. The chemical symbol for an element (or
an atom of that element) is a one- or two-letter abbreviation of its name. Usually, but not
always, the first two letters are used. To complicate matters further, chemical symbols are
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TABLE 2.2 Names, Chemical Symbols, and Atomic Weights of the Element
37
TABLE 2.2 (Continued)
sometimes derived from a language other than English. For example the symbol for Hg for
mercury comes from the first and seventh letters of the element’s Latin name, hydrargyrum.
The chemical symbols for all the currently known elements are listed in Table 2.2 (the
atomic weights in the table will be discussed later in this chapter.) These symbols are the
basic vocabulary of chemistry because the atoms they represent make up all matter. You
will see symbols for the more important elements over and over again, and the sooner you
know what element they stand for, the easier it will be for you to learn chemistry. These
more important element have been indicated in Table 2.2 by colored shading around their
names.
Dalton’s forth postulate states that atoms my combine to form molecules. An example of
this is provided by bromine, the only element other than mercury which is a liquid at
ordinary room temperature (20°C). Macroscopically, bromine consists of dark-colored
crystals below –7.2°C and a
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Figure 2.3 Microscopic view of the element bromine (a) in the gaseous state (above 58°C); (b) in liquid
form (between -7.2 and 58.8°C); and (c) in solid form (below -7.2°C).
reddish brown gas above 58.8°C. The liquid is dark red-brown and has a pungent odor
similar to the chlorine used in swimming pools. It can cause severe burns on human skin
and should not be handled without the protection of rubber gloves.
The microscopic view of bromine in Fig. 2.3 is in agreement with its designation as an
element—only one kind of atom is present. Except at very high temperatures, though,
bromine atoms always double up. Whether in solid, liquid, or gas, they go around in pairs.
Such a tightly held combination of two or more atoms is called a molecule.
The composition of a molecule is indicated by a chemical formula. A subscript to the right
of the symbol for each element tells how many atoms of that element are in the molecule.
For example, Table 2.2 gives the chemical symbol Br for bromine, but each molecule
contains two bromine atoms, and so the chemical formula is Br2. According to Dalton’s
fourth postulate in Table 2.1, atoms combine in the ratio of small whole numbers, and so
the subscripts in a formula should be small whole numbers.
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2.4 MACROSCOPIC AND MICROSCOPIC VIEWS
OF A CHEMICAL REACTION
Dalton’s third postulate states that atoms are the units of chemical changes. What this
means can be seen in the macroscopic and microscopic views of a chemical change in Plate
3. (Color plates will be found in Chap. 5). When macroscopic quantities of mercury and
bromine are mixed at room temperature, a chemical reaction occurs, and a new substance,
mercuric bromide, is produced. It is a white solid, quite different in appearance from the
two elements from which it was formed.
A chemist’s microscopic theory about what is going on is shown in Plate 3 along with the
photographs of the macroscopic reaction. Soon after the two liquids are mixed together, a
rearrangement of atoms begins. The two bromine atoms of each Br2 molecule become
separated and combine instead with mercury atoms. When the chemical reaction is
complete, all that remains is a collection of mercuric bromide molecules, each of which
contains one mercury atom and two bromine atoms. Notice that there are just as many
mercury atoms after the reaction as there were before the reaction. The same applies to
bromine atoms. Atoms were neither created, destroyed, divided into parts, or changed into
other kinds of atoms during the chemical reaction.
The view of solid mercuric bromide shown in Plate 3f is our first microscopic example of a
compound. Each molecule of a compound is made up of two (or more) different kinds of
atoms. Since these atoms may be rearranged during a chemical reaction, the molecules may
be broken apart and the compound can be decomposed into two (or more) different
elements.
The formula for a compound involves at least two chemical symbols—one for each
element, present. In the case of mercuric bromide each molecule contains one mercury and
two bromine atoms, and so the formula is HgBr2. Both Plate 3f and the formula tell you that
any sample of pure mercuric bromide contains twice as many bromine atoms as mercury
atoms. This 2:1 ratio agrees with Dalton’s fourth postulate that atoms combine in the ratio
of small whole numbers.
Although John Dalton originally used circular symbols like those in Plate 3 to represent
atoms in chemical reactions, a modern chemist would use chemical symbols and a chemical
equation like
Hg + Br2 → HgBr2
Reactant(s)
(2.1)
Product(s)
This equation may be interpreted microscopically to mean that 1 mercury atom and 1
bromine molecule react to form 1 mercuric bromide molecule. It should also call to mind
the macroscopic change shown in the photographs in Plate 3. This macroscopic
interpretation is often strengthened by specifying physical states of the reactants and
products:
Hg(l) + Br2(l) → HgBr2(s)
(2.2)
Thus liquid mercury and liquid bromine react to form solid mercuric bromide. [If the
bromine had been in gaseous form, Br2(g) might have been
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used. Occasionally (c) may be used instead of (s) to indicate a crystalline solid.] Chemical
equations such as (2.1) and (2.2) summarize a great deal of information on both
macroscopic and microscopic levels, for those who know how to interpret them.
2.5 TESTING THE ATOMIC THEORY
Two criteria are usually applied to any theory. First, does it agree with facts which are
already known? Second, does it predict new relationships and stimulate additional
observation and experimentation? Dalton’s atomic theory was able to do both of these
things. It was especially useful in dealing with data regarding the masses of different
elements which were involved in chemical compounds or chemical reactions.
To test a theory, we first use it to make a prediction about the macroscopic world. If the
prediction agrees with existing data, the theory passes the test. If it does not, the theory
must be discarded or modified. If data are not available, then more research must be done.
Eventually the results of new experiments can be compared with the predictions of the
theory.
Several examples of this process of testing a theory against the facts are afforded by
Dalton’s work. For example, postulate 3 in Table 2.1 states that atoms are not created,
destroyed, or changed in a chemical reaction. Postulate 2 says that atoms of a given element
have a characteristic mass: By logical deduction, then, equal numbers of each type of atom
must appear on left and right sides of chemical equations such as (2.1) and (2.2), and the
total mass of reactants must equal the total mass of products. Dalton’s atomic theory
predicts Lavoisier’s experimental law of conservation of mass.
A second prediction of the atomic theory is a bit more complex. A compound is made up of
molecules, each of which contains a certain number of each type of atom. No matter how,
when, or where a compound is made, its molecules will always be the same. Thus mercuric
bromide molecules always have the formula HgBr2. No matter how much we have or where
the compound came from, there will always be twice as many bromine atoms as mercury
atoms. Since each type of atom has a characteristic mass, the mass of one element which
combines with a fixed mass of the other should always be the same. In mercuric bromide,
for example, if each mercury atom is 2.510 times as heavy as a bromine atom, the ratio of
masses would be
Mass of 1 mercury atom
Mass of 2 bromine atoms
=
2.510 × mass of 1 bromine atom
2 × mass of 1 bromine atom
= 1.255
No matter how many mercuric bromide molecules we have, each has the same proportion
of mercury, and so any sample of mercuric bromide must have that same proportion of
mercury.
We have just derived the law of constant composition, sometimes called the law of
definite proportions. When elements combine to form a compound, they always do so in
exactly the same ratio of masses. This law had been postulated in 1799 by the French
chemist Proust (1754 to 1826)
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4 years before Dalton proposed the atomic theory, and its logical derivation from the theory
contributed to the latter’s acceptance.
The law of constant composition makes the important point that the composition and other
properties of a pure compound are independent of who prepared it or where it came from.
The carbon dioxide found on Mars, for instance, can be expected to have the same
composition as that on Earth, while the natural vitamin C extracted and purified from rose
hips has exactly the same composition as the synthetic vitamin C prepared by a drug
company. Absolute purity is, however, an ideal limit which we can only approach, and the
properties of many substances may be affected by the presence of very small quantities of
impurities.
A third law of chemical composition may be deduced from the atomic theory. It involves
the situation where two elements can combine in more than one way, forming more than
one compound. For example, if mercuric bromide is ground and thoroughly mixed with
liquid mercury, a new compound, mercurous bromide, is formed. Mercurous bromide is a
white solid which is distinguishable from mercuric bromide because of its insolubility in
hot or cold water. Mercurous bromide also changes directly from a solid to a gas at 345°C.
From the microscopic view of solid mercurous bromide in Fig. 2.4, you can readily
determine that its chemical formula is Hg2Br2. (Since there are two atoms of each kind in
the molecule, it would be incorrect to write the formula as HgBr.) The chemical equation
for synthesis of mercurous bromide is
Hg(l) + HgBr2(s) → Hg2Br2(s)
(2.3)
From the formulas HgBr2 and Hg2Br2 we can see that mercuric bromide has only 1 mercury
atom for every 2 bromines, while mercurous bromide has 2 mercury atoms for every 2
bromines. Thus, for a given number of bromine atoms, mercurous bromide will always
have twice as many mercury atoms as mercuric bromide. Again using postulate 2 in Table
2.1, the atoms have characteristic masses, and so a given number of bromine atoms
corresponds to a fixed mass of bromine. Twice as many mercury atoms correspond to twice
the mass of mercury. Therefore we can say that for a given
Figure 2.4 The arrangement of atoms and molecules in
a crystal of mercurous bromide, Hg2Br2.
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mass of bromine, mercurous bromide will contain twice the mass of mercury that mercuric
bromide will. [The doubled mass of mercury was provided by adding liquid mercury to
mercuric bromide in Eq. (2.3).]
EXAMPLE 2.1 Given that the mass of a mercury atom is 2.510 times the mass of a
bromine atom, calculate the mass ratio of mercury to bromine in mercurous bromide.
Solution The formula Hg2Br2 tells us that there are 2 mercury atoms and 2 bromine atoms
in each molecule. Thus the mass ratio is
Mass of 2 mercury atoms
Mass of 2 bromide atoms
=
2 × (2.510 × mass of 1 bromide atom)
2 × mass of 1 bromide atoms
= 2.510
Note that the mass of mercury per unit mass of bromine is double that calculated earlier for
mercuric bromide.
The reasoning and calculations above illustrate the law of multiple proportions. When two
elements form several compounds, the mass ratio in one compound will be a small wholenumber multiple of the mass ratio in another. In the case of mercuric bromide and
mercurous bromide the mass ratios of mercury to bromine are 1.255 and 2.510,
respectively. The second value is a small whole-number multiple of (2 times) the first.
Until the atomic theory was proposed, no one had expected any relationship to exist
between mass ratios in two or more compounds containing the same elements. Because the
theory predicted such relationships, Dalton and other chemists began to look for them.
Before long, a great deal of experimental evidence was accumulated to show that the law of
multiple proportions was valid. Thus the atomic theory was able to account for previously
known facts and laws, and it also predicted a new law. In the process of verifying that
prediction, Dalton and his contemporaries did many additional quantitative experiments.
These led onward to more facts, more laws, and, eventually, new or modified theories.
This characteristic of stimulating more research and thought put Dalton’s postulates in the
distinguished company of other good scientific theories.
2.6 ATOMIC WEIGHTS
Our discussion of the atomic theory has indicated that mass is a very important
characteristic of atoms—it does not change as chemical reactions occur. Volume, on the
other hand, often does change, because atoms or molecules pack together more tightly in
liquids and solids or become more widely separated in gases when a reaction takes place.
From the time Dalton’s theory
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was first proposed, chemists realized the importance of the masses of atoms, and they spent
much time and effort on experiments to determine how much heavier one kind of atom is
than another.
Dalton, for example, studied a compound of carbon and oxygen which he called carbonic
oxide. He found that a 100-g sample contained 42.9 g C and 57.1 g O. In Dalton’s day there
were no simple ways to determine the microscopic nature of a compound, and so he did not
know the composition of the molecules (and hence the formula) of carbonic oxide. Faced
with this difficulty, he did what most scientists would do—make the simplest possible
assumption. This was that the molecules of carbonic oxide contained the minimum number
of atoms: one of carbon and one of oxygen. Carbonic oxide was the compound we now
know as carbon monoxide, CO, and so in this case Dalton was right. However, erroneous
assumptions about the formulas for other compounds led to half a century of confusion
about atomic weights.
Since the formula was CO, Dalton argued that the ratio of the mass of carbon to the mass of
oxygen in the compound must be the same as the ratio of the mass of 1 carbon atom to the
mass of 1 oxygen atom:
Mass of 1 C atom
Mass of 1 O atom
=
mass of C in CO
mass of O in CO
=
42.9 g
57.1 g
=
0.751
1
= 0.751
In other words the mass of a carbon atom is about three-quarters (0.75) as great as the mass
of an oxygen atom.
Notice that this method involves a ratio of masses and that the units grams cancel, yielding
a pure number. That number (0.751, or approximately ¾) is the relative mass of a carbon
atom compared with an oxygen atom. It tells nothing about the actual mass of a carbon
atom or of an oxygen atom–only that carbon is three-quarters as heavy as oxygen.
The relative masses of the atoms are usually referred to as atomic weights. Their most
recently determined values were given in Table 2.2, along with the names and symbols for
the elements. The atomic-weight scale was originally based on a relative mass of 1 for the
lightest atom, hydrogen. As more accurate methods for determining atomic weight were devised, it proved convenient to shift to oxygen and then carbon, but the scale was adjusted so
that hydrogen’s relative mass remained close to 1. Thus nitrogen’s atomic weight of
14.0067 tells us that a nitrogen atom has about 14 times the mass of a hydrogen atom.
The fact that atomic weights are ratios of masses and have no units does not detract at all
from their usefulness. It is very easy to determine how much heavier one kind of atom is
than another.
EXAMPLE 2.2 Use the atomic weights in Table 2.2 to show that the mass of a mercury
atom is 2.510 times the mass of a bromine atom, as stated earlier in this chapter.
Solution The actual masses of the atoms will be in the same proportion as their relative
masses. Table 2.2 gives atomic weights of 200.59 for
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mercury and 79.904 for bromine. Therefore
Mass of a Hg atom
Mass of a Br atom
or
=
relative mass of a Hg atom
relative mass of a Br atom
=
200.59
79.904
= 2.5105
Mass of a Hg atom = 2.5104 × mass of a Br atom
The atomic-weight table also permits us to obtain the relative masses of molecules. These
are called molecular weights and are calculated by summing the atomic weights of all
atoms in the molecule.
EXAMPLE 2.3 How heavy would a mercurous bromide molecule be in comparison to a
single bromine atom?
Solution First, obtain the relative mass of an Hg2Br2 molecule (the molecular weight):
2Hg atoms: relative mass = 2 × 200.59 = 401.18
2Br atoms: relative mass = 2 × 79.904 = 159.808
1Hg2Br2 molecule:
relative mass = 560.99
Therefore
Mass of a Hg 2 Br2 molecule
Mass of a Br atom
=
560.99
79.904
= 7.0208
The Hg2Br2 molecule is about 7 times heavier than a bromine atom.
2.7 THE AMOUNT OF SUBSTANCE: MOLES
According to the atomic theory, atoms are the units of chemical reactions. The formula
HgBr2 indicates that each molecule of this substance contains one mercury and two
bromine atoms. Therefore, if we ask how much bromine is required to make a given
quantity of mercuric bromide, the answer is two bromine atoms for each mercury atom or
two bromine atoms per molecule. In other words, how much substance we have depends in
a very important way on how many atoms or molecules are present.
How much in the above sense of how many atoms or molecules are present is not the same
thing as how much in terms of volume or mass. It takes 3.47 cm3 Br2(l) to react with a 1cm3 sample of Hg(l). That same 1 cm3 Hg(l) would weigh 13.59g, but only 10.83 g Br2(l)
would be needed to react with it. In terms of volume, more bromine than mercury is
needed, while in terms of mass, less bromine than mercury is required. In the atomic sense,
however, there are exactly twice as many bromine atoms as mercury atoms and twice as
much bromine as mercury.
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The Mole
Because atoms and molecules are extremely small, there are a great many of them in any
macroscopic sample. The 1 cm3 of mercury referred to previously would contain 4.080 ×
1022 mercury atoms, for example, and the 3.47 cm3 of bromine would contain twice as
many (8.160 × 1022) bromine atoms. The very large numbers involved in counting
microscopic particles are inconvenient to think about or to write down. Therefore chemists
have chosen to count atoms and molecules using a unit called the mole. One mole
(abbreviated mol) is 6.022 × 1023 of the microscopic particles which make up the substance
in question. Thus 6.022 × 1023 Br atoms is referred to as 1 mol Br. The 8.160 × 1022 atoms
in the sample we have been discussing would be (8.160 × 1022)/(6.022 × 1022) mol Br, or
0.1355 mol Br.
The idea of using a large number as a unit with which to measure how many objects we
have not unique to chemists. Eggs, doughnuts, and many other things are sold by the
dozen—a unit of twelve items. Smaller objects, such as pencils, may be ordered in units of
144, that is, by the gross, and paper is packaged in reams, each of which contains 500
sheets. A chemist who refers to 0.1355 mol Br is very much like a bookstore manager who
orders 2½ dozen sweat shirts, 20 gross of pencils, or 62 reams of paper.
There is a difference in degree, however, because the chemist’s unit, 6.022 × 1023, is so
large. A stack of paper containing a mole of sheets would extend more than a million times
the distance from the earth to the sun, and 6.022 × 1023 grains of sand would cover all the
land in the world to a depth of nearly 2 ft. Obviously there are a great many particles in a
mole of anything.
Why have chemists chosen such an unusual number as 6.022 × 1023 as the unit with which
to count the number of atoms or molecules? Surely some nice round number would be
easier to remember. The answer is that the number of grams in the mass of 1 mol of atoms
of any element is the atomic weight of that element. For example, 1 mol of mercury atoms
not only contains 6.022 × 1023 atoms, but its mass of 200.59 g is conveniently obtained by
adding the unit gram to the atomic weight in Table 2.2. Some other examples are
1mol H contains 6.022 × 1023 H atoms; its mass is 1.008 g.
1 mol C contains 6.022 × 1023 C atoms; its mass is 12.01 g.
1 mol O contains 6.022 × 1023 O atoms; its mass is 16.00 g.
1 mol Br contains 6.022 × 1023 Br atoms; its mass is 79.90 g.
(Here and in subsequent calculations atomic weights are rounded to two decimal places,
unless, as in the case of H, fewer than four significant figures would remain.)
The mass of a mole of molecules can also be obtained from atomic weights. Just as a dozen
eggs will have a dozen whites and a dozen yolks, a mole of CO molecules will contain a
mole of C atoms and a mole of O atoms.
46
The mass of a mole of CO is thus
Mass of 1 mol C + mass of 1 mol O = mass of 1 mol CO
12.01 g +
16.00 g = 28.01 g
The molecular weight of CO(28.01) expressed in grams is the mass of a mole of CO. Some
other examples are
Molecule
Molecular Weight
Mass of 1 Mol
of Molecules
It is important to specify what kind of particle a mole refers to. A mole of Br atoms, for
example, has only half as many atoms (and half as great a mass) as a mole of Br2
molecules. It is best not to talk about a mole of bromine without specifying whether you
mean 1 mol Br or 1 mol Br2.
The Amount of Substance
Chemists use the mole so often to measure how much of a substance is present that it is
convenient to have a name for the quantity which this unit measures. In the International
System this quantity is called the amount of substance and is given the symbol n. Here
again a common English word has been given a very specific scientific meaning. Although
amount might refer to volume or mass in everyday speech, in chemistry it does not. When a
chemist asks what amount of Br2 was added to a test tube, an answer like “0.0678 mol Br2”
is expected. This would indicate that 0.0678 × 6.022 × 1023 or 4.08 × 1022, Br2 molecules
had been added to the test tube.
The Avogadro Constant
Although chemists usually work with moles as units, occasionally it is helpful to refer to
the actual number of atoms or molecules involved. When this is done, the symbol N is
used. For example, in referring to 1 mol of mercury atoms, we could write
nHg = 1mol
and
NHg = 6.022 × 1023
Notice that NHg is a pure number, rather than a quantity. To obtain such a pure number,
we need a conversion factor which involves the number of particles per unit amount of
substance. The appropriate factor is given the symbol NA and is called the Avogadro
constant. It is defined by the equation
NA =
N
n
47
Since for any substance there are 6.022 × 1023 particles per mole, NA =6.022 × 1023/1 mol =
6.022 × 1023 mol–1.
EXAMPLE 2.4 Calculate the number of O2 molecules in 0.189 mol O2.
Solution Rearranging Eq. (2.4), we obtain
N = n × NA = 0.189 mol–1 × 6.022 × 1023 mol–1 = 1.14 × 1023
Alternatively, we might include the identity of the particles involved:
N = 0.189 mol O 2 ×
6.022 × 10 23 O 2 molecules
1 mol O 2
= 1.14 × 10 O 2 molecules
23
Notice that Eq. (2.4), which defines the Avogadro constant, has the same form as Eq. (1.1),
which defined density. The preceding example used the Avogadro constant as a conversion
factor in the same way that density was used in Examples 1.10 and 1.11. As in those
previous examples, all that is necessary is to remember that number of particles and amount
of substance are related by a conversion factor, the Avogadro constant.
Avogadro constant
Number of ←⎯⎯⎯⎯⎯
→ amount of
particles
substance
N ←⎯A⎯
→n
N
As long as the units mole cancel, NA is being used correctly.
2.8 THE MOLAR MASS
It is often convenient to express physical quantities per unit amount of substance (per
mole), because in this way equal numbers of atoms or molecules are being compared. Such
molar quantities often tell us something about the atoms or molecules themselves. For
example, if the molar volume of one solid is larger than that of another, it is reasonable to
assume that the molecules of the first substance are larger than those of the second.
(Comparing the molar volumes of liquids, and especially gases, would not necessarily give
the same information since the molecules would not be as tightly packed.)
A molar quantity is one which has been divided by the amount of substance. For example,
an extremely useful molar quantity is the molar mass M:
Molar mass =
mass
amount of substance
M=
m
n
(2.5)
48
It is almost trivial to obtain the molar mass, since atomic and molecular weights expressed
in grams give us the masses of 1 mol of substance.
EXAMPLE 2.5 Obtain the molar mass of (a) Hg and (b) Hg2Br2.
Solution
a) The atomic weight of mercury is 200.59, and so 1 mol Hg weighs 200.59 g.
M Hg =
m Hg
nHg
=
200.59 g
1 mol
-1
= 200.59 g mol −1
b) Similarly, for Hg2Br2 the molecular weight is 560.98, and so
M Hg Br =
2
2
m Hg 2 Br2
nHg 2 Br2
= 560.98 g mol −1
The molar mass is numerically the same as the atomic or molecular weight, but it has units
of grams per mole.
Equation (2.5), which defines the molar mass, has the same form as Eq. (1.1), which
defined density, and Eq. (2.4), which defined the Avogadro constant. As in the case of
density or the Avogadro constant, it is not necessary to memorize or manipulate a formula.
Simply remember that mass and amount of substance are related via molar mass.
Molar mass
M
Mass ←⎯⎯⎯
→ amount of substance m ←⎯
→n
The molar mass is easily obtained from atomic weights and may be used as a conversion
factor, provided the units cancel.
EXAMPLE 2.6 Calculate the amount of octane (C8H18) in 500 g of this liquid.
Solution Any problem involving interconversion of mass and amount of substance requires
molar mass
M = (8 × 12.01 + 18 × 1.008) g mol–1 = 114.2 g mol–1
The amount of substance will be the mass times a conversion factor which permits
cancellation of units:
n = m × conversion factor
1
1 mol
= 500 g ×
= 4.38 mol
=m×
114.2 g
M
49
In this case the reciprocal of the molar mass was the appropriate conversion factor.
The Avogadro constant, molar mass, and density may be used in combination to solve more
complicated problems.
EXAMPLE 2.7 How many molecules would be present in 25.0 ml of pure carbon
tetrachloride (CCl4)?
Solution In Example 2.4 we showed that the number of molecules may be obtained from
the amount of substance by using the Avogadro constant. The amount of substance may be
obtained from mass by using the molar mass (Example 2.6), and mass from volume by
means of density (Example 1.12). A road map to the solution of this problem is
density
Molar mass
Avogadro constant
Volume ⎯⎯⎯
→ mass ←⎯⎯⎯
→ amount ←⎯⎯⎯⎯⎯
→ number of molecules
or in shorthand notation
ρ
M
A
V ⎯⎯
→ m ⎯⎯
→ n ⎯⎯→
N
N
The road map tells us that we must look up the density of CCl4 in Table 1.4:
ρ = 1.595 g cm–3
The molar mass must be calculated from atomic weights:
M = (12.01 + 4 × 35.45) g mol–1 = 153.81 g mol–1
and we recall that the Avogadro constant is
NA = 6.022 × 1023 mol–1
The last quantity (N) in the road map can then be obtained by starting with the first (V) and
applying successive conversion factors:
N = 25.0 cm 3 ×
1.595 g
1 cm 3
×
1 mol
153.81 g
×
6.022 × 1023 molecules
1 mol
= 1.56 × 1023 molecules
Notice that in this problem we had to combine techniques from three previous examples. To
do this you must remember relationships among quantities. For example, a volume was
given, and we knew it could be converted to the corresponding mass by means of density,
and so we looked up the density in a table. By writing a road map, or at least seeing it in
your mind’s eye, you can keep track of such relationships, determine what conversion
factors are needed, and then use them to solve the problem.
50
2.9 FORMULAS AND COMPOSITION
In Plate 3 we present a microscopic view of the chemical reaction between mercury and
bromine. Equations (2.1) and (2.2) represent the same event in terms of chemical symbols
and formulas. But how does a practicing chemist find out what is occurring on the
microscopic scale? When a reaction is carried out for the first time, little is known about the
microscopic nature of the products. It is therefore necessary to determine experimentally the
composition and formula of a newly synthesized substance.
The first step in such a procedure is usually to separate and purify the products of a
reaction. For example, although the combination of mercury with bromine produces mainly
mercuric bromide, a little mercurous bromide is often formed as well. A mixture of
mercurous bromide with mercuric bromide has properties which are different from a pure
sample of HgBr2, and so the Hg2Br2 must be removed. The low solubility of Hg2Br2 in
water would permit purification by recrystallization. The product could be dissolved in a
small quantity of hot water and filtered to remove undissolved Hg2Br2. Upon cooling and
partial evaporation of the water, crystals of relatively pure HgBr2 would form.
Once a pure product has been obtained, it may be possible to identify the substance by
means of its physical and chemical properties. The reaction of mercury with bromine yields
white crystals which melt at 236°C. The liquid which is formed boils at 322°C. Since it is
made by combining two elements, the product is a compound. Comparing its properties
with a handbook or table of data leads to the conclusion that it is mercuric bromide.
But suppose you were the first person who ever prepared mercuric bromide. There were no
tables which listed its properties then, and so how could you determine that the formula
should be HgBr2? One answer involves quantitative analysis—the determination of the
percentage by mass of each element in the compound. Such data are usually reported as the
percent composition.
EXAMPLE 2.8 When 10.0 g mercury reacts with sufficient bromine, 18.0 g of a pure
compound is formed. Calculate the percent composition from these experimental data.
Solution The percentage of mercury is the mass of mercury divided by the total mass of
compound times 100 percent:
% Hg =
mHg
mcompound
× 100% =
10.0 g
18.0 g
× 100% = 55.6%
The remainder of the compound (18.0 g – 10 g = 8.0 g) is bromine:
% Br =
mBr
mcompound
× 100% =
8.0 g
18.0 g
As a check, verify that the percentages add to 100:
55.6% + 44.4% = 100%
× 100% = 44.4%
51
To obtain the formula from percent-composition data, we must find how many bromine
atoms there are per mercury atom. On a macroscopic scale this corresponds to the ratio of
the amount of bromine to the amount of mercury. If the formula is HgBr2, it not only
indicates that there are two bromine atoms per mercury atom, it also says that there are 2
mol of bromine atoms for each 1 mol of mercury atoms. That is, the amount of bromine is
twice the amount of mercury. The numbers in the ratio of the amount of bromine to the
amount of mercury (2:1) are the subscripts of bromine and mercury in the formula.
EXAMPLE 2.9 Determine the formula for the compound whose percent composition was
calculated in the previous example.
Solution For convenience, assume that we have 100 g of the compound. Of this, 55.6 g
(55.6%) is mercury and 44.4 g is bromine. Each mass can be converted to an amount of
substance
1 mol Hg
nHg = 55.6 g ×
= 0.277 mol Hg
200.59 g
nHg = 44.4 g ×
1 mol Br
79.90 g
= 0.556 mol Br
Dividing the larger amount by the smaller, we have
nBr
nHg
=
0.556 mol Br
0.227 mol Hg
=
2.01 mol Br
1 mol Hg
The ratio 2.01 mol Br to 1 mol Hg also implies that there are 2.01 Br atoms per 1 Hg atom.
If the atomic theory is correct, there is no such thing as 0.01 Br atom; furthermore, our
numbers are only good to three significant figures. Therefore we round 2.01 to 2 and write
the formula as HgBr2.
EXAMPLE 2.10 A bromide of mercury has the composition 71.5% Hg, 28.5% Br. Find its
formula.
Solution Again assume a 100-g sample and calculate the amount of each element:
nHg = 71.5 g ×
nHg = 28.5 g ×
The ratio is
nBr
nHg
=
1 mol Hg
200.59 g
1 mol Br
79.90 g
0.357 mol Br
0.356 mol Hg
=
= 0.356 mol Hg
= 0.357 mol Br
1.00 mol Br
1 mol Hg
We would therefore assign the formula HgBr.
52
The formula obtained in Example 2.10 does not correspond to either of the two mercury
bromides we have already discussed. Is it a third one? The answer is no because our method
can only determine the ratio of Br to Hg. The ratio 1:1 is the same as 2:2, and so our
method will give the same result for HgBr or Hg2Br2 (or Hg7Br7, for that matter, should it
exist). The formula determined by this method is called the empirical formula or simplest
formula. Occasionally, as in the case of mercurous bromide, the empirical formula differs
from the actual molecular composition, or the molecular formula. Experimental
determination of the molecular weight in addition to percent composition permits
calculation of the molecular formula.
EXAMPLE 2.11 A compound whose molecular weight is 28 contains 85.6% C and 14.4%
H. Determine its empirical and molecular formulas.
Solution
nC = 85.6 g ×
nH = 14.4 g ×
nH
nC
=
1 mol C
12.01 g
1 mol H
1.008 g
14.3 mol H
7.13 mol C
=
= 7.13 mol C
= 14.3 mol H
2.01 mol H
1 mol C
The empirical formula is therefore CH 2 . The molecular weight corresponding to the
empirical formula is
12.01 + 2 × 1.008 = 14.03
Since the experimental molecular weight is twice as great, all subscripts must be doubled
and the molecular formula is C2H4.
Occasionally the ratio of amounts is not a whole number.
EXAMPLE 2.12 Aspirin contains 60.0% C, 4.48% H, and 35.5% O. What is its empirical
formula?
Solution
nH = 14.4 g ×
nC = 85.6 g ×
nO = 35.5 g ×
1 mol H
1.008 g
1 mol C
12.01 g
1 mol O
16.00 g
= 14.3 mol H
= 7.13 mol C
= 2.22 mol O
53
Divide all three by the smallest amount of substance
nC
nO
nH
nO
=
=
5.00 mol C
=
2.22 mol O
4.44 mol H
2.22 mol O
=
2.25 mol H
1 mol O
2.00 mol H
1 mol O
Clearly there are twice as many H atoms as O atoms, but the ratio of C to O is not so
obvious. We must convert 2.25 to a ratio of small whole numbers. This can be done by
changing the figures after the decimal point to a fraction. In this case, .25 becomes ¼. Thus
2.25 = 2¼ = 94 , and
nC
nO
=
2.25 mol C
1 mol O
=
9 mol C
4 mol O
We can also write
nH
nO
=
2 mol H
1 mol O
=
8 mol H
4 mol O
Thus the empirical formula is C9H8O4.
Once someone has determined a formula–empirical or molecular—it is possible for
someone else to do the reverse calculation. Finding the weight-percent composition from
the formula often proves quite informative, as the following example shows.
EXAMPLE 2.13 In order to replenish nitrogen removed from the soil when plants are
harvested, the compounds NaNO3 (sodium nitrate), NH4NO3 (ammonium nitrate), and NH3
(ammonia) are used as fertilizers. If a farmer could buy each at the same cost per gram,
which would be the best bargain? In other words, which compound contains the largest percentage of nitrogen?
Solution We will show the detailed calculation only for the case of NH4NO3.
1 mol NH4NO3 contains 2 mol N, 4 mol H, and 3 mol O. The molar mass is thus
M = (2 × 14.01 + 4 × 1.008 + 3 × 16.00) g mol–1 = 80.05 g mol–1
A 1-mol sample would weigh 80.05 g. The mass of 2 mol N it contains is
Therefore the percentage of N is
54
The percentages of H and O are easily calculated as
Though not strictly needed to answer the problem, the latter two percentages provide a
check of the results. The total 35.00 + 5.04% + 59.96% = 100.00% as it should.
Similar calculations for NaNO3 and NH3 yield 16.48% and 82.24% nitrogen, respectively.
The farmer who knows chemistry chooses ammonia!
2.10 BALANCING CHEMICAL EQUATIONS
We have now determined symbols and formulas for all the ingredients of chemical
equations, but one important step remains. We must be sure that the law of conservation of
mass is obeyed. The same number of atoms (or moles of atoms) of a given type must appear
on each side of the equation. This reflects our belief in Dalton’s third postulate that atoms
are neither created, destroyed, nor changed from one kind to another during a chemical
process. When the law of conservation of mass is obeyed, the equation is said to be
balanced.
As a simple example of how to balance an equation, let us take the reaction which occurs
when a large excess of mercury combines with bromine. In this case the product is a white
solid which does not melt but instead changes to a gas when heated above 345°C. It is
insoluble in water. From these properties it can be identified as mercurous bromide, Hg2Br2.
The equation for the reaction would look like this:
Hg + Br2 → Hg2Br2
(2.6)
but it is not balanced because there are 2 mercury atoms (in Hg2Br2) on the right side of the
equation and only 1 on the left.
An incorrect way of obtaining a balanced equation is to change this to
Hg + Br2 → HgBr2
(2.7)
This equation is wrong because we had already determined from the properties of the
product that the product was Hg2Br2. Equation (2.7) is balanced, but it refers to a different
reaction which produces a different product. The equation might also be incorrectly written
as
Hg 2 + Br2 → Hg2Br2
The formula Hg2 suggests that molecules containing 2 mercury atoms each
(2.8)
55
were involved, but our previous microscopic experience with this element (Figure 2.1)
indicates that such molecules do not occur.
In balancing an equation you must remember that the subscripts in the formulas have been
determined experimentally. Changing them indicates a change in the nature of the reactants
or products. It is permissible, however, to change the amounts of reactants or products
involved. For example, the equation in question is correctly balanced as follows:
2Hg + Br2 → Hg2Br2
(2.9)
The 2 written before the symbol Hg is called a coefficient. It indicates that on the
microscopic level 2Hg atoms are required to react with the molecule. On a macroscopic
scale the coefficient 2 means that 2 mol Hg atoms are required to react with 1 mol Br2
molecules. Twice the amount of Hg is required to make Hg2Br2 as we needed for HgBr2
[see Eq. (2.1)].
To summarize: Once the formulas (subscripts) have been determined, an equation is
balanced by adjusting coefficients. Nothing else may be changed.
EXAMPLE 2.14 Balance the equation
Hg2Br2 + Cl2 → HgCl2 + Br2
Solution Although Br and Cl are balanced, Hg is not. A coefficient of 2 with HgCl2 is
needed:
Hg2Br2 + Cl2 → 2HgCl2 + Br2
Now C1 is not balanced. We need 2 Cl2 molecules on the left:
Hg2Br2 + 2Cl2 → 2HgCl2 + Br2
We now have 2Hg atoms, 2Br atoms, and 4C1 atoms on each side, and so balancing is
complete.
Most chemists use several techniques for balancing equations. 1 For example, it helps to
know which element you should balance first. When each chemical symbol appears in a
single formula on each side of the equation (as Example 2.14), you can start wherever you
want and the process will work. When a symbol appears in three or more formulas,
however, that particular element will be more difficult to balance and should usually be left
until last.
EXAMPLE 2.15 When butane (C4H10) is burned in oxygen gas (O2), the only products are
carbon dioxide and (CO2) water. Write a balanced equation to describe this reaction.
1
Laurence E. Strong, Balancing Chemical Equations, Chemistry, vol. 47, no. 1, pp. 13-16, January 1974,
discusses some techniques in more detail.
56
Solution First write an unbalanced equation showing the correct formulas of all the
reactants and products:
C4H10 + O2 → CO2 + H2O
We note that O atoms appear in three formulas, one on the left and two on the right.
Therefore we balance C and H first. The formula C4H10 determines how many C and H
atoms must remain after the reaction, and so we write coefficients of 4 for CO2 and 5 for
H2O:
C4H10 + O2 → 4CO2 + 5H2O
We now have a total of 13 O atoms on the right-hand side, and the equation can be
balanced by using a coefficient of 132 in front of O2:
C4H10 +
13
2
O2 → 4CO2 + 5H2O
Usually it is preferable to remove fractional coefficients since they might be interpreted to
mean a fraction of a molecule. (One-half of an O2 molecule would be an O atom, which has
quite different chemical reactivity.) Therefore we multiply all coefficients on both sides of
the equation by two to obtain the final result:
2C4H10 + 13O2 → 8CO2 + 10H2O
(Sometimes, when we are interested in moles rather than individual molecules, it may be
useful to omit this last step. Obviously the idea of half a mole of O2 molecules, that is,
3.011 × 1023 molecules, is much more tenable than the idea of half a molecule.)
Another useful technique is illustrated in Example 2.15. When an element (such as O2)
appears by itself, it is usually best to choose its coefficient last. Furthermore, groups such as
NO3, SO4, etc., often remain unchanged in a reaction and can be treated as if they consisted
of a single atom. When such a group of atoms is enclosed in parentheses followed by a
subscript, the subscript applies to all of them. That is, the formula involves Ca(NO3)2
involves 1Ca, 2N and 2 × 3 = 6 O atoms.
EXAMPLE 2.16 Balance the equation
NaMnO4 + H2O2 + H2SO4 → MnSO2 + Na2SO4 + O2 + H2O
Solution We note that oxygen atoms are found in every one of the seven formulas in the
equation, making it especially hard to balance. However, Na appears only in two formulas:
2NaMnO4 + H2O2 + H2SO4 → MnSO2 + Na2SO4 + O2 + H2O
as does manganese, Mn:
2NaMnO4 + H2O2 + H2SO4 → 2MnSO2 + Na2SO4 + O2 + H2O
57
We now note that the element S always appears with 4 O atoms, and so we balance the SO 4
groups:
2NaMnO4 + H2O2 + 3H2SO4 → 2MnSO2 + Na2SO4 + O2 + H2O
Now we are in a position to balance hydrogen:
2NaMnO4 + H2O2 + 3H2SO4 → 2MnSO2 + Na2SO4 + O2 + 4H2O
and finally oxygen. (We are aided by the fact that it appears as the element.)
2NaMnO4 + H2O2 + 3H2SO4 → 2MnSO2 + Na2SO4 + 3O2 + 4H2O
Notice that in this example we followed the rule of balancing first those elements whose
symbols appeared in the smallest number of formulas: Na and Mn in two each, S (or SO4)
and H in three each, and finally O. Even using this rule, however, equations in which one or
more elements appear in four or more formulas are difficult to balance without some
additional techniques which we will develop in Chap. 11.
The balancing of chemical equations has an important environmental message for us. If
atoms are conserved in a chemical reaction, then we cannot get rid of them. In other words
we cannot throw anything away. There are only two things we can do with atoms: Move
them from place to place or from compound to compound. Thus when we "dispose" of
something by burning it, dumping it, or washing it down the sink, we have not really gotten
rid of it at all. The atoms which constituted it are still around someplace, and it is just as
well to know where they are and what kind of molecule they are in. Discarded atoms in
places where we do not want them and in undesirable molecules are known as pollution.
SUMMARY
The atomic theory is one of the most important and useful ideas of chemistry. Thinking
about matter in terms of microscopic particles such as atoms and molecules permits us to
interpret numerous macroscopic observations. The contrasting properties of solids, liquids,
and gases, for example, may be ascribed to differences in spacing between and speed of
motion of the constituent atoms or molecules.
In the form originally proposed by John Dalton, the atomic theory distinguished elements
from compounds and was used to explain the law of constant composition. The theory also
agreed with Lavoisier's law of conservation of mass, and it was able to predict the law of
multiple proportions. An important aspect of the atomic theory is the assignment of relative
masses (atomic weights) to the elements.
Atoms and molecules are extremely small. Therefore, when calculating how much of one
substance is required to react with another, chemists use a unit called the mole. One mole
contains 6.022 × 1023 of whatever kind of microscopic particles one wishes to consider.
Referring to 2 mol Br2 specifies a certain number of Br2 molecules in the same way that
referring to 10 gross of pencils specifies a certain
58
number of pencils. The quantity which is measured in the units called moles is known as
the amount of substance.
The somewhat unusual number 6.022 × 1023, which specifies how many particles are in a
mole, has been chosen so that the mass of 1 mol of atoms of any element is the atomic
weight of that element expressed in grams. Similarly, the mass of a mole of molecules is
the molecular weight expressed in grams. The molecular weight is obtained by summing
atomic weights of all atoms in the molecule. This choice for the mole makes it very
convenient to obtain molar masses–simply add the units grams per mole to the atomic or
molecular weight. Using molar mass and the Avogadro constant, it is possible to determine
the masses of individual atoms or molecules and to find how many atoms or molecules are
present in a macroscopic sample of matter.
A table of atomic weights and the molar masses which can be obtained from it can also be
used to obtain the empirical formula of a substance if we know the percentage by weight of
each element present. The opposite calculation, determination of weight percent from a
chemical formula, is also possible. Once formulas for reactants and products are known, a
balanced chemical equation can be written to describe any chemical change. Balancing an
equation by adjusting the coefficients applied to each formula depends on the postulate of
the atomic theory which states that atoms are neither created, destroyed, nor changed into
atoms of another kind during a chemical reaction.