Copyright (c) 1996 by James H. Burness
Chemical reactions represent the heart of chemistry: they
describe the myriad of ways that substances can combine with each
other to form new substances and the energy which accompanies
these changes. In essence, they describe what chemistry really is
-- the study of matter and its transformations. Chemical
reactions can be as large in scope as the industrial production
of millions of pounds of sulfuric acid or as small in scope as
the production of a few millionths of a gram of a hormone in your
body. Irregardless of scope, they are extremely important. It is
because of chemical reactions that you are able to read this
paragraph, breathe without even realizing it, and process the
information you learn in chemistry class. Indeed, it is because
of chemical reactions that you are alive.
The purpose of this chapter is to describe the concepts of stoichiometry
[pronunciation: stoy-key-ahm'-eh-tree; from the Greek stoicheion
(element) and metron (to measure).] Stoichiometry is the
area of chemistry which deals with quantitative relationships in
chemical reactions. It is the area which allows chemists to know
how much starting material is needed to produce a million pounds
of sulfuric acid or how many molecules are in a microgram of a
particular hormone.
The calculations described in this chapter are, in a word,
trivial. They are based on simple ratios and proportions, as are
most chemical calculations. This shouldn't be surprising; if we
burn 20 grams of propane gas, we would expect to get twice the
mass of products (and twice the energy) that would have been
obtained if we had burned 10 grams.
Your favorite candy costs $2.00 per pound. You have 50 cents.
How much candy can you buy? How much can you buy with $1.00?
If you answered that you can buy a quarter of a pound with 50
cents and a half pound with one dollar, you will easily be able
to do the calculations which chemists do every day.
In order to make it easier to understand what a chemical reaction really means and to help you to visualize the stoichiometric relationships between the starting materials and products in a chemical reaction, we'll use an analogy throughout this chapter. Imagine a factory which assembles Teddy Bear heads. The heads will be attached to the body in another factory; we are only concerned here with the construction of the head itself. The "elements" we'll use to put the head together are
Have you made a mental picture of this situation? If not, be
sure to do so now. To successfully solve any chemistry problem,
the first step is to understand it. Making a mental picture of
the situation helps you to understand what is happening. We
intentionally haven't supplied a figure of the Teddy Bear parts
-- you should be picturing three elements which will be used to
construct the final head: a face, a pair of eyes which will fit
into sockets in the front of the face, and a pair of ears -- each
of which has tabs for insertion into the top of the face.
If we wanted to describe the construction of the face, we
could verbally state that "one face is combined with one
pair of eyes and a pair of ears to produce a complete head",
or we could simplify the description by using symbols:
FA + EYE2 + EAR2
-------> FA(EYE)2(EAR)2 (Eq. 1)
Likewise, chemists usually find it easier to describe chemical
changes with a chemical equation, which lists the starting
materials (reactants), an arrow which is read as
"yields", "produces" or "gives",
followed by the products of the reaction.
Unfortunately, the symbolism of chemistry isn't always
unambiguous. For example, the subscript in the notation EYE2
on the left of the arrow in Eq. 1 indicates
that the two eyes are joined together as a pair. The same can be
said for the pair of ears. But the "formula" on the
right side, even though it contains the same two subscripts, has
a totally different meaning. Neither the two eyes nor the two
ears are connected to each other in the completed head; rather,
there are four separate pieces which are all attached to the
central face. When ethylene reacts with chlorine, the reaction
is:
C2 H4 + Cl2
-------> C2 H4 Cl2 .
Although the two chlorine atoms are bonded together in the
chlorine molecule on the left side, they are not bonded to each
other in the molecule shown on the right side.
Earlier we discussed the mass of products from the burning of
propane gas. Verbally, this could be stated, "Propane reacts
with oxygen to produce carbon dioxide and water." The
chemical equation representing the same process is:
C3H8 + O2
-----> CO2 + H2O (Eq. 2)
You already learned in chapter 2 that
atoms are never created or destroyed in a chemical reaction.
Thus, the number of atoms which were present before the reaction
must be present after the reaction. An equation which meets this
criterion is said to be balanced. Eq. 1
above is balanced, since there is a face, two eyes, and two ears
before and after assembly. Eq. 2, on the
other hand, is not balanced, because we have appeared to
"lose" carbon and hydrogen atoms, and "gain"
oxygen atoms, in going from reactants to products. Since the
subscripts in the formulas above represent the number of atoms in
a molecule of the substance, we cannot adjust them to balance an
equation. It would be incorrect to change oxygen's subscript from
a 2 to a 3 in order to balance oxygen atoms; this would result in
an equation which describes the reaction between propane and ozone
(O3) rather than oxygen (O2),
thereby completely changing the meaning of the reaction. The
first fact you should remember about chemical equations is
Never change the subscripts in a chemical
formula in order to balance an equation.
The proper way to balance an equation is to change the stoichiometric
coefficients (the numbers in front of the formulas, sometimes
simply called coefficients) in the equation. By convention,
coefficients of 1 are understood, but not shown, as in the
equation above. One form of the balanced equation for the burning
of propane is:
C3H8 +
5O2 -----> 3CO2 + 4H2O (Eq.
3)
For the time being, let's interpret the coefficient as simply a "unit". Thus, we can read Eq. 3 as, "One unit of propane reacts with five units of oxygen to produce three units of carbon dioxide and four units of water." We'll return to a more useful interpretation of the meaning of a coefficient later in the chapter. If we interpret the coefficients as units, then they really represent ratios, in much the same way that a recipe calls for "one part of ingredient A mixed with two parts of ingredient B." Since they represent ratios, the coefficients in Eq. 3 can all be multiplied or divided by the same number, giving in infinite number of balanced equations. Instead of the coefficients 1, 5, 3, and 4, we could have (respectively),
2, 10, 6, and 8 OR 0.5, 2.5, 1.5, and 2 OR 3, 15, 9, and 12, and so on...
By the way, without even knowing what a "unit" might represent, try to answer the following question:
If one unit of water contains 3,000 atoms, how many atoms are
there in five units?
If you answered 15,000 atoms, you understand the basics of stoichiometry. Congratulations!
* * * * * * * * *
The second thing to remember about chemical equations is that
The coefficients in a
chemical equation do not represent the actual
amounts or reactants and/or products present. They represent ratios.
For example, just because the coefficients in front of each of
the "reactants" in Eq. 1 is a
number one doesn't mean that there must be one unit of each. It
simply means that one unit of each is used up when the head is
constructed. There is no reason why I couldn't start with five
dozen faces, one dozen pairs of eyes, and three dozen pairs of
ears. But at the end of the process, I'll have a dozen complete
heads, four dozen left over faces, and two dozen pairs of ears
which I didn't use. Likewise, the coefficients of 3 and 4 in
front of the carbon dioxide and water, respectively, in Eq. 3 don't say anything at all about the
actual amounts of water and carbon dioxide. What they do
mean is that for each four units (dozen, gross, million, etc.) of
water molecules formed in this reaction, three units (dozen,
gross, million, etc.) of carbon dioxide molecules are formed.
Chemical equations are used so frequently that chemists have
developed a few conventions for writing them. Some of these
conventions are shown in the following table:
Let's return to our Teddy Bear analogy to see how a chemical
equation should be interpreted. At first glance, it might appear
most logical to interpret Eq. 1 as
"One face combines with one pair of eyes and a pair of ears
to produce a head." But the simple fact is that this does not
correctly describe how the head would be constructed. For one
thing, it is clear that the pair of eyes, as well as the pair of
ears, must be separated before they are placed in the face. The
face does not "combine" at all with a pair of anything.
Let's consider some possible ways that the head could be
constructed if we looked at the process step-by-step. This is,
after all, how things would have to happen in the
"reality" of our Teddy Bear factory. Each construction
scenario, and the steps involved for each scenario, will be shown
symbolically:
step 1: EYE2 -------> 2 EYE [separate the two
eyes]
step 2: EAR2 --------> 2 EAR [separate the two
ears]
step 3: FA + EYE ------> partial head(1) [place 1st eye in its
socket]
step 4: partial head(1) + EYE --------> partial head(2) [place
2nd eye in its socket]
step 5: partial head(2) + EAR --------> partial head (3) [put
1st ear in place]
step 6: partial head (3) + EAR --------> FA(EYE)2(EAR)2
[put 2nd ear in place]
-------------------------------------------------------------------------------------------------
When the above steps are added, note that common terms on both
sides of the arrow (the separate eyes, ears, and all of the
partial heads) cancel, giving the following net equation:
FA + EYE 2 + EAR 2 --------> FA(EYE)2(EAR)2.
step 1: EYE 2 -------> 2 EYE [separate the two
eyes]
step 2: FA + EYE ------> partial head(1) [place 1st eye in its
socket]
step 3: partial head(1) + EYE --------> partial head(2) [place
2nd eye in its socket]
step 4: EAR 2 --------> 2 EAR [separate the two
ears]
step 5: partial head(2) + EAR --------> partial head (3) [put
1st ear in place]
step 6: partial head (3) + EAR --------> FA(EYE)2(EAR)2
[put 2nd ear in place]
-------------------------------------------------------------------------------------------------
Net equation: FA + EYE 2 + EAR 2
--------> FA(EYE)2(EAR)2.
Note that the net equation is the same as that for Scenario 1.
Devise a third scenario which gives the same net equation as that above.
Assume you know with high certainty that at the beginning of a
particular week, 10,000 faces, 10,000 pairs of eyes, and 10,000
pairs of ears were shipped to the factory. At the end of the
week, the complete heads were shipped out. Although you were
never inside the factory, someone asks you to list the steps used
to manufacture the heads. Think about this question for a few
seconds before continuing to read...........
Do you agree that it is impossible to list the steps? There
are a number of scenarios which are possible, all of which give
rise to the same net conversion. The only way you could figure
out how the heads were constructed would be to either watch them
being put together or to garner some clues which might help you
to rule out some scenarios. For example, suppose you examined the
10,000 heads and found a few incomplete pieces. None of these
incomplete pieces have any ears. Suppose there were three heads
with only one eye in a socket, two heads with both eyes in the
face, three single eyes, and a few pairs of ears with the rubber
band still around them. Wouldn't this evidence allow you to
confidently rule out Scenario 1 , which
requires that the ears be separated before any eyes are placed in
the sockets? The same sort or reasoning can be used for a
chemical reaction.
Chemists are interested in how chemical reactions happen on a
molecular level. This information is valuable for understanding
how atoms and molecules behave and for designing new reactions or
changing the rates of reactions. The steps which describe the
molecular events (collisions) between individual atoms and
molecules, are called elementary reactions. They are
analogous to the steps in the scenarios above.
An important example of an elementary reaction is one of the
steps in the depletion of ozone from our stratosphere:
Cl + O3 ------> O2 +
ClO
This reaction means that a single chlorine atom collides with
an ozone molecule, forming a transient species in which one of
the bonds in the ozone is breaking while a chlorine-oxygen bond
is being formed. This transient species then breaks apart into a
single oxygen molecule and a chlorine monoxide molecule.
The series of steps which describe the overall (net)
conversion of reactants to products is called a reaction
mechanism. Mechanisms are analogous to the scenarios
above. Since the elementary reactions (steps) represent molecular
events, and the net reaction is the overall combination of these
events, you should keep these rules in mind:
Let's consider an actual chemical example. The overall reaction for the decomposition of dinitrogen pentoxide is
2 N2O5 ------> 4 NO2
+ O2 .
The mechanism for this reaction has been determined to
be the following:
2 x ( N2O5 -----> NO2 + NO3 )
NO2 + NO3 -----> NO2 + O2 + NO
NO + NO3 ------> 2 NO2
The "2 x" in front of the first step means that this step occurs twice in the mechanism for every occurrence of the second and third steps. You should be able to identify the following intermediates in the mechanism: some of the NO2 molecules (how many?), NO3, and NO. You should also recognize that the sum of the elementary reactions gives the net reaction and that all stoichiometric coefficients in the steps of the mechanism are integers.
and
2NO2 + 2NO3
-----> 2NO2 + 2O2 + 2NO?
by subtracting NO2
from both sides?
N2O5
------>2 NO2 + ½O2 ?
Answers:
You will learn much more about reaction mechanisms in
chapter 12. For now, however, it is important to
understand what information you can and cannot get
out of a chemical equation. The vast majority of
reactions you'll consider in this course, and all of the
reactions we'll look at in the remainder of this chapter,
are overall reactions. This means that you'll get
the same results whether you balance them with
whole-number or fractional coefficients, but you
shouldn't expect them to provide information about what
is happening on a molecular scale.
Here is a summary of what we know about chemical
equations at this point:
It might seem that we spent a lot of time talking
about chemical equations, but the chemical equation
really is an important part of the foundation for your
eventual understanding of chemistry. It doesn't do much
good to perform calculations with an equation if you
don't know what the equation means.
It is easy to determine the mass of the completed
Teddy Bear head, even without weighing it directly. This
is because we know the masses of each of the component
parts and simply need to add them together. But weighing
atoms is another story altogether. Atoms are so small
that it is impossible to weigh a single atom or even a
few of the heaviest atoms known. We'll discuss how to
determine the masses of atoms and will use our Teddy Bear
head analogy to enhance your understanding of the
microscopic world of atoms.
Recall that the eyes for the head are distributed in
pairs. Since some of the eyes have different masses than
others, we need to think about the possible masses for
each pair of eyes. There are three possible ways to put
together EYE2:
| Composition of EYE2 | Mass of EYE2 |
| (EYE-35)(EYE-35) | 70 grams |
| (EYE-35)(EYE-37) | 72 grams |
| (EYE-37)(EYE-37) | 74 grams |
Let's consider now how frequently we would expect to
encounter each of these combinations. Recall that 75% of
the eyes are EYE-35 and 25% of them are EYE-37.
| Composition of EYE2 | Expected Frequency |
| (EYE-35)(EYE-35) | (0.75)2 = 0.5625 |
| (EYE-35)(EYE-37) or (EYE-37)(EYE-35) | (0.75)(0.25)(2) = 0.375* |
| (EYE-37)(EYE-37) | (0.25)2 = 0.0625 |
*We needed to multiply by
two because of the two ways of obtaining this
combination.
Fig. 1 on page 11 shows the
expected distribution, normalized at 100% for the most
probable frequency (by dividing each of the above numbers
by 0.5625):
Figure 1. Expected Masses of Teddy Bear Eye Pairs
Suppose now that we are asked to calculate the mass of
a thousand pairs of eyes. We'll consider two ways to
obtain this value.
One method is to use the statistical distribution to
determine how many of each variety of pairs we have, and
then multiply by the mass of each variety:
(1000 pairs)(56.25%) = 562.5 pairs of (EYE-35)(EYE-35). Then,
562.5 pairs x 70 grams/pair = 39375 grams
(1000 pairs)(37.5%) = 375 pairs of (EYE-35)(EYE-37)
375 pairs x 72 grams/pair = 27000 grams
(1000 pairs)(6.25%) = 62.5 pairs of (EYE-37)(EYE-37)
62.5 pairs x 74 grams/pair = 4625 grams.
Thus, the total mass of a thousand pairs of eyes is
39375 + 27000 + 4625 = 71000 grams.
Another way to do the calculation is to calculate the
"average" weight of a pair of eyes, then
multiply by the number of pairs. To calculate the weighted
average, we take into account the frequency of
occurrences:
(70 grams)(.5625) + (72 grams)(.375) + (74 grams)(.0625) = 71 grams. Then,
(71 grams/ "average" pair)(1000
"average" pairs) = 71000 grams, which is the
same answer as above. The advantage of using the latter
approach is that once the "average" weight is
determined, the weight of any number of units can be
determined very easily by multiplying the average weight
by the number of units. Naturally, none of the
pairs of eyes have a mass of 71 grams -- they weigh
either 70, 72, or 74 grams. Nevertheless, in a large
sample where the statistical distribution shown above is
valid, we can use this artificial number to calculate the
mass of the entire sample.
Although chemists aren't able to weigh single atoms or
molecules directly on a balance, they are able to
determine atomic masses by using an instrument called a mass
spectrometer. The operation of a mass spectrometer
consists of the following schematic steps:
The mass spectrometer, then, is capable of providing
two pieces of information about the atoms of an element:
the masses of the isotopes making up the element
(measured by the position of the peak) and the abundances
of the isotopes (measured by the intensity of the peak,
since the more abundant isotopes will produce more ions,
resulting in a stronger detector signal.) Fig.
2 shows the mass spectrum of neon. Since Ne is a
monatomic gas, a sample of neon is a mixture of single
atoms of the naturally-occurring isotopes of neon.
Figure 2. Mass Spectrum of Neon
| Mass | Relative Abundance | Actual Abundance |
| 20 | 100.0% | 90.92% |
| 21 | 0.3% | 0.257% |
| 22 | 9.7% | 8.82% |
It is conventional to display mass spectra with the most intense peak (called the base peak) scaled to 100%, as was done for the distribution shown in Figure 1 at the top of page 11. You can see from the table above that neon consists predominantly of two isotopes (Ne-20 and Ne-22) which collectively make up 99.74% of naturally-occurring neon. The peak for the Ne-21 isotope is so small that it isn't visible in the figure.
Notice that the mass unit shown in the figure above is
abbreviated as u. This represents a mass unit
known as the atomic mass unit. You may see this
abbreviated in some texts as amu. The atomic mass unit is
an extremely small mass unit, as shown by the following
equivalence:
1 atomic mass unit = 1.6605 x 10-24
g
The reason for defining this unit is much the same as
the reason for defining any unit; it makes it easier to
express the value. Can you imagine expressing the mass of
an automobile in units of ounces? It would be a very
large value. Using the mass unit "ton" makes
the mass of the car much more manageable. Likewise,
expressing the mass of a single atom in units of grams
gives a very small number. Using atomic mass units makes
it more manageable.
Actually, atomic masses are relative masses, based on the mass of the primary isotope of carbon, 12C. This atom has been assigned a mass of exactly 12u, and all other masses are measured relative to this standard. The following table shows the masses of the fundamental subatomic particles in terms of the atomic mass unit scale:
| Particle | Mass, u | Mass, g |
| Proton | 1.0073 | 1.6725 x 10-24 |
| Neutron | 1.0087 | 1.6748 x 10-24 |
| Electron | 0.0005486 | 9.109 x 10-28 |
Notice that the masses of the proton
and neutron are very close to 1u. Notice also that
expressing the masses in atomic mass units more clearly
shows how much more massive the proton and neutron are
than the electron, reiterating that the vast majority of
the mass of an atom comes from the nucleus.
Although the masses of the neon isotopes in Fig. 2 are shown as integers, mass spectral measurements actually provide very accurate values for the masses of a particular isotope. For example, the actual masses of the naturally-occurring neon isotopes are shown in the following table:
| Neon Isotope | Mass, u | % Natural Abundance |
| Ne-20 | 19.99244 | 90.92 |
| Ne-21 | 20.99395 | 0.257 |
| Ne-22 | 21.99138 | 8.82 |
By definition, the mass number
of an isotope is the total number of protons and neutrons
in the atom's nucleus. Since the masses of the proton and
neutron are a little greater than 1u, we would expect the
mass of each of the neon isotopes to have a mass a little
greater than the mass number. But the masses are actually
a little less than the mass number! (Refer to the
table above.) Have we lost mass somewhere? The answer is,
perhaps surprisingly, yes. It turns out that some
of the mass of the separate nucleons is converted to
energy (the binding energy) when the nucleus is
formed because the neutrons and protons are more stable
when they are together in the nucleus than when they are
separate particles. This phenomenon will be discussed in
more detail in chapter 21, Nuclear Chemistry.
We can calculate the mass of an "average" Ne
atom by finding the weighted average of the masses of
neon's isotopes:
(19.99244 u)(0.9092) + (20.99395
u)(0.00257) + (21.99138 u)(0.0882) = 20.17 u
This number is called the atomic mass, and is
shown underneath the atomic symbol for Ne in the periodic
table. Most importantly, it can be used to calculate the
mass of any given number of neon atoms, provided that
there are enough atoms that the frequencies of occurrence
of the isotopes correspond to the values we used in the
calculation above. But atoms are so incredibly small that
even a sample of atoms with a minuscule mass contains so
many atoms that the statistical distribution above is
valid. In other words, atomic masses can be used to
represent the average weight of an atom and will
always provide accurate results in any macroscopic sample
of the atoms. If we could close our eyes and grab a single
atom of neon, however, it would never have a mass
of 20.17 u. It would have a mass of 19.99244u, 20.99395u,
or 21.99138u, depending on which isotope we happened to
grab. The formal definition of atomic mass is:
The atomic mass of an
atom is the weighted average of the masses of the
naturally-occurring isotopes of the atom.
It turns out that our Teddy Bear
analogy is closer to reality than you might have
imagined. Chlorine exists naturally as two isotopes with
the following properties:
| Chlorine Isotope | Mass, u | % Natural Abundance |
| Cl-35 | 34.96885 | 75.53 |
| Cl-37 | 36.96590 | 24.47 |
If you compare the masses and
abundances above with the description of the EYEs of the Teddy Bear (page
2), you'll see that the values are very similar, with the
exception that the masses of the eyes have units of grams
and the chlorine atoms have units of atomic mass units.
The atomic mass of chlorine is easily calculated:
(34.96885 u)(0.7553) + (36.96590
u)(0.2447) = 35.457 u
Look at your periodic table and check this value.
Copper exists naturally as two isotopes: 63Cu
(62.9298 u, 69.09%) and 65Cu (64.9278 u,
30.91%). Calculate the atomic mass of copper.
Answer: Find the answer in a periodic table and
compare it to your calculated value. The problem could
have been solved even if the abundance of only one of the
isotopes had been specified, since the sum of the
isotopic abundances must equal 100%.
CONTINUED IN PART 2
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Last modified June 16, 1997