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:
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:
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:
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
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:
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
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:
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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:
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
The mechanism for this reaction has been determined to
be the following:
2 x ( N2O5 -----> NO2 + NO3 )
2 x ( NO2 + NO3 -----> NO2 + O2 + NO)
NO + NO3 ------> 2 NO2
The "2 x" in front of the first and second steps means that each of these steps occurs twice in the mechanism for every occurrence of the third step. 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.
by subtracting NO2 from both sides?
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:
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.
*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
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:
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:
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:
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:
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:
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:
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:
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%.
Since a molecule is made up of two or more atoms, chemists use
the term molecular mass to refer to the mass of a molecule.
The molecular mass of Cl2 is 2(35.457 u) = 70.91 u.
Just as a given pair of Teddy Bear eyes never have a mass of 71g
(the "average" mass), a single chlorine molecule never
has a mass of 70.91 u. Nevertheless, using average molecular masses
will always give accurate results for any macroscopic-sized collection
of molecules.
Determine the molecular mass, to two decimal places, for each of the following (refer to a periodic table for the atomic masses):
Answers:
1. 2(1.008 u) + 16.00 u = 18.02 u
2. 12(12.011 u) + 22(1.008 u) + 11(16.00 u) = 342.31 u
3. 22.99 u + 35.45 u = 58.44 u
4. 39.10 u + 26.98 u + 2(32.07 u) + 8(16.00 u) + 12(18.02u) =
474.46 u
Note: Strictly speaking, since the compounds in 3 and 4 are ionic,
they don't exist as molecules (see p. 38.) It is better to use
the term formula mass, rather than molecular mass, for
ionic substances.
When molecules, rather than atoms, are placed in a mass spectrometer,
it usually happens that some of the molecular ions formed by electron
bombardment are unstable and break apart into fragment ions.
Some of the molecular ions, however, make it to the detector without
fragmenting. These ions have the same mass as the molecule (the
mass of the electron which is removed to form the ion is negligible,
as shown in the first table
on page 13), and they are therefore
called parent molecular ions. For example, the parent molecular
ion in the mass spectrum of chlorine gas is either (35Cl-35Cl)+,
(35Cl-37Cl)+, or (37Cl-37Cl)+.
The cluster of peaks for the parent molecular ion in the mass
spectrum of Cl2 is shown in Fig. 3
on page 17. Compare
this with Fig. 1 on page 11. The similarity of the two figures
shouldn't be surprising at all, now that we have discussed the
concept of atomic and molecular masses.
Figure
3. Mass Spectrum of Chlorine in the Region of the Parent Molecular
Ion
Since a mass spectrometer detects the fragments formed when a
molecule breaks apart, and since these fragments are determined
by the molecule's structure, the mass spectrum of a compound is
a fingerprint. In other words, no two substances have the same
mass spectrum. This is true even if they have the same formula.
Consider the mass spectrum shown below
for acetone, which has
a molecular formula C3H6O (and thus a molecular
mass of 58 u):
Figure 4. Mass Spectrum
of Acetone
The peak at m/e=58 represents the parent molecular ion,
(CH3COCH3)+.
It is due to molecules of acetone, minus an electron, which reached
the mass spectrometer detector. But not all molecules strike the
detector intact; some of the molecules hit by the electrons are
fragmented, resulting in smaller mass pieces which reach the detector.
For example, the base peak, at m/e=43, has a mass which is 15
units less than that of the parent molecular ion.
Which species is giving rise to the peak at m/e=43?
Answer: There is a mass difference of 15 between 58 and
43. This corresponds to the mass of a methyl group, CH3
. Thus, the fragment at 43u corresponds to loss of a methyl group
from the parent molecule. The ion which causes the peak at
m/e=43, then, is (CH3CO)+. Since the
parent molecule can lose CH3 from either side of the
molecule, it isn't surprising that the peak at 43 is the base
peak.
The peak at m/e=15 is due, of course, to a methyl fragment ion.
The fragmentation of a molecule is analogous to hitting the Teddy
Bear heads with a baseball bat and looking at the pieces which
are left over. The "parent" species would be a complete
head which managed to stay together after the impact and would
have a mass equal to the sum of the masses of the components.
On the other hand, if an ear is knocked out of its slot, the resulting
"fragment" would have a mass 19 grams less than that
of a complete head, since each ear weighs 19g.
Consider now the molecule propanal, which has the same overall
formula (and thus the same molecular mass) as acetone. The arrangement
of atoms in the two molecules is, however, different (compare
the structures shown in the two figures of the mass spectra).
Compounds with the same formula but a different arrangement of
atoms are called isomers (don't confuse this term with
isotopes, which are atoms of the same type with different
numbers of neutrons).
The mass spectrum of propanal is shown in Fig. 5:
Figure 5. Mass Spectrum
of Propanal
Which species are giving rise to the peaks at m/e=58, 57, and
29?
Answer: The peak at m/e=58 is due to the parent molecular
ion, (CH3CH2COH)+. The peak at
m/e=57 is due to the (CH3CH2CO)+
ion, and the base peak at m/e=29 arises from two ions: (CH3CH2)+
and COH+, both of which happen to have the same mass.
As you can see, the mass spectra of the two isomers are indeed
quite different. Chemists can examine the mass spectrum of a substance
and identify the fragment ions. Then they can see how the fragments
must have originally fit together, like the pieces of a jigsaw
puzzle. This means that the structure of an unknown molecule can
be deduced. Since mass spectra are like molecular fingerprints,
they can also be used to unambiguously identify countless known
substances, such as illegal drugs and gasoline additives.
Molecules, as well as the atoms which make them up, the mechanistic steps
by which they undergo reactions with other molecules, and the
fragments which are left over when we blast them apart, are real.
Unlike our hypothetical Teddy Bear head factory, they exist.
Now that we have discussed the topics of atomic and molecular
mass, we can move on the the central concept of the chapter --
the mole. The mole is, actually, central to most of the concepts
and calculations in chemistry, including stoichiometry, gases,
equilibrium and electrochemistry, just to name a few. Yet although
the mole is so critically important, it is extremely simple to
understand. If you understand the concept of a dozen, you will
grasp what a mole represents.
In order to appreciate why chemists use this concept, it is necessary
to remember that the atom is so tiny. In fact, the atom's minuscule
size was also the reason scientists invented the atomic mass unit.
Let's ask the following question: "How much does a carbon-12
atom weigh in grams?"
Recall that the mass of 12C is defined as exactly
12 u. It is logical that if one atomic mass unit is equivalent
to 1.6605 x 10-24 g, then 12 u has a mass twelve times
as much:
Now let's ask the question, "How many carbon-12 atoms would
we need to have for a total mass of exactly 12 grams?" Before
we determine this number, think about it for a moment. Atoms have
an extremely small mass. Thus, we would expect an incredibly huge
number of atoms to be present in a 12-gram sample, which would
fit in the palm of your hand. Algebraically, we can rephrase our
question as
Solving for x, we get x = 12g1.9925 x 10-23 g/atom
= 6.022 x 1023 atoms. Since the 12 is an exact
[defined] value, it doesn't limit the significant figures allowed
in the answer. The number we just calculated is, indeed, a very
large number. We'll discuss just how large it is shortly, but
first let's find the answer to another question:
"How many 'average' chlorine atoms are present in a sample
of chlorine atoms with a mass of 35.45 grams?"
First, we obtain the mass of an "average" chlorine atom
using chlorine's atomic mass:
Notice that this is the same answer as the number of atoms in
12 grams of 12C.
We are now in a position to define the mole:
The abbreviation for mole isn't much of one; we can drop the 'e'
and abbreviate it as "mol". Never use "m"
as the abbreviation, since this represents molality, a unit for
measuring the concentration of a solution.
Note that the definition of a mole refers to particles,
not to specific species like atoms or molecules. Thus, it is proper
to talk about a mole of atoms, a mole of molecules, a mole of
ions, a mole of electrons, etc. It is not proper, however,
to use the term mole for a heterogeneous or homogeneous mixture.
For example, it is incorrect to refer to a mole of air, since
air consists of a mixture of gases (one can, however, refer to
a mole of nitrogen or a mole of oxygen, since these are pure substances.)
A chemist will never say "one mole of a sodium chloride solution",
and the term "mole of concrete" doesn't make any sense
because concrete is a heterogeneous mixture.
The mole, then can be interpreted in two ways:
If you examine the calculation which we carried out above for
the carbon-12 and the chlorine atoms, you'll see that it is no
coincidence that the number of atoms worked out to be the same.
For example, if we substitute the left-hand side of Eq. 6
for
the value 5.886 x 10-23 in Eq. 7
, we get:
(x)(35.45)( 1.6605 x 10-24) = 35.45 The atomic weight
cancels, and this is why we got the same answer for carbon as
for chlorine. If we solve for x, we see that x is equal to the
reciprocal of 1.6605 x 10-24, which is 6.022 x 1023.
The latter number is so important that it is given a special name,
in honor of the Italian scientist Amadeo Avogadro.
It is interesting that the monumental contributions made by Avogadro
were made as a result of his interest in chemistry as a hobby.
He was a lawyer by profession.
The mole, then, simply represents a counting unit. There are 6.022
x 1023 of something in a mole, just as there are 12
of something in a dozen and 144 of something in a gross. The mole
is often referred to as the chemist's dozen. The only reason
that Avogadro's number is so big is that atoms are so small.
Here are some analogies to help you visualize just how tremendously
large this number really is:
In terms of doing stoichiometry calculations, the most useful
result of the foregoing discussion is that
One mole of "average" atoms (or molecules) has a
mass in grams equal to the atomic (or molecular) mass.
This mass is known as the substance's Molar Mass.
In other textbooks or chemical literature sources, you may see
molar mass referred to as atomic weight, molecular weight, or
formula weight, depending on whether the substance exists as single
atoms, molecules, or formula units of an ionic compound, respectively.
The term molar mass is preferable because it can be used for any
one of these three types of substances. Chemists use the adjective
"molar" to specify that the value of a particular property
is for one mole of a substance. Thus, molar volume is the volume
of a mole; molar conductivity is the conductivity of a mole of
a metal; and molar mass is simply the mass of a mole. The abbreviation
for molar mass is MM. The units of molar mass are grams/mole if
the value is shown alone, e.g. Sucrose, MM = 342.31 g/mole (see
the Concept Check on page 15.) If molar mass is stated verbally,
however, the units are grams, e.g. "The molar mass of sucrose
is 342.31 g", since the adjective "molar" already
specifies "per mole."
Refer to a periodic table as you examine each of the following
examples. Make sure that you understand each and every sentence.
Notice that the word "average" didn't need to be used for the second example above. This is because fluorine exists monoisotopically in nature as the 19F isotope.
How many peaks would you expect to find for the parent molecular
ion in the mass spectrum of F2? At which m/e value
would it (they) be found?
Answer: One peak would be expected, at m/e = 38 u.
In two moles of ethane, C2H6,
Answers:
As it turns out, our Teddy Bear analogy has another connection
to real life. The entire Teddy Bear scenario can actually be used
as a model for the formation of a chemical compound:
FA + EYE2 + EAR2 -------> FA(EYE)2(EAR)2
100 g + 71g (average) + 38 g
-------> 209 g (average)
C + Cl2 + F2 ----------> CCl2F2
12.011 g (average) + 70.9 g (average) + 38 g --------->
120.9 g (average)
Remember that the face had four points of attachment; likewise,
carbon always forms four bonds in its compounds. In the same way
that the eyes and ears were always supplied as pairs, the elements
chlorine and fluorine always exist in nature as diatomic molecules.
There are two differences between the analogy and the real chemical
equation: 1) the mass of carbon is not the same as the mass of
the face (a 12-g face wouldn't be very heavy!) and 2) we assumed
that all faces were identical, whereas carbon actually exists
as a mixture of two isotopes (C-12 and C-13, with abundances of
98.89% and 1.11%, respectively).
Notice that the equation above for the formation of CCl2F2
is already balanced. What is the best way to interpret the stoichiometric
coefficients in the balanced equation? We already learned that
a generic interpretation of a coefficient is a "unit",
but which unit should we use? A dozen? A gross? A billion molecules?
The best "unit" to use for the stoichiometric coefficient
is the chemist's dozen, the mole:
Stoichiometric coefficients in balanced chemical
equations represent ratios between the number of moles of reactants
and products in the equation.
The best interpretation of the equation above is "One mole
of carbon will react completely with one mole of chlorine and
one mole of fluorine to yield one mole of dichlorodifluoromethane."
If we had chosen to balance the equation with fractional coefficients,
e.g.
then we would read the equation as "One-half mole of carbon
reacts completely with one-half mole of chlorine and one-half
mole of fluorine to produce one-half mole of dichlorodifluoromethane.
Remember that the coefficients themselves say nothing about how
much of the reactants or products is actually present; we could
have 100 grams each of carbon, chlorine, and fluorine, but the
two interpretations above would not be any different. The ratios
of moles which react are always one-to-one whether we use integral
or fractional coefficients.
It is important to remember the distinction between a subscript
in a formula and a stoichiometric coefficient. For example, the
notations Cl2 and 2Cl each represents a total of 70.9
grams of chlorine and a total of two moles of chlorine atoms,
but the meanings are entirely different. The former means one
mole of diatomic chlorine molecules, for a total of two
moles of chlorine atoms, whereas the latter means two moles of
monatomic chlorine atoms. The important difference is that
the two moles of chlorine atoms in the former example are bonded
to each other (like a pair of Teddy Bear eyes) while the two moles
of chlorine atoms in the latter example are separate atoms. Refer
again to scenario 1
on page 5. The Cl2 would be analogous
to the reactant in step 1, while 2Cl would be analogous to the
product in the same step.
Now that you realize that stoichiometric coefficients represent
moles, you will be able to perform stoichiometric calculations.
Common examples of the sorts of stoichiometric questions which
chemists might ask are
The two most important things to remember about stoichiometric
calculations are that
We'll look at step 1 first, then use balanced equations to perform
a variety of stoichiometric calculations. Refer back to the concept
check on the very first page of this chapter. Remember that if
you were able to get the answer to this, then you can easily do
stoichiometric calculations.
Except for reactions known as redox (an abbreviation for oxidation-reduction) reactions, which can sometimes be complex, chemical equations are balanced by a trial-and-error procedure. Redox reactions are balanced by a special set of rules which will be considered in chapter 4. Some reactions are easy to balance by trial-and-error, while others are a little more time-consuming. The following guidelines may help you to balance equations more easily:
We'll look at two examples of common types of reactions and will
refer to each of the rules above, when necessary, as we balance
them.
Combustion is the reaction of a substance with oxygen. When metals react with oxygen, the corresponding metal oxides are usually formed (common exceptions are sodium, which reacts with oxygen to form sodium peroxide, Na2O2, instead of Na2O, and potassium, which reacts with oxygen to form potassium superoxide, KO2, instead of K2O.)
When hydrocarbons (compounds containing only carbon and
hydrogen) or compounds containing carbon, hydrogen, and oxygen
react with oxygen, the products are always carbon dioxide and
water.
Complete and balance the equation for the combustion of aluminum.
Answer:
The unbalanced equation is
Following rules 1 and 2, we begin with the product and save oxygen
until last. Thus, we balance aluminum, obtaining
We don't want to change the coefficient in front of the product,
since that will force us to change the "2" we just placed
in front of Al. Thus, we want to change the coefficient in front
of O2 so that we have three oxygen atoms on the left.
A coefficient of 3/2 will do the trick:
If we wanted to use simplest whole-number coefficients, we would have:
Complete and balance the equation for the reaction between lithium
and oxygen.
Complete and balance the equation for the combustion of ethyl
alcohol, C2H5OH.
Answer:
The unbalanced equation is
Notice that it is easier to collect the hydrogens together before
beginning. Following rules 1 and 2, we begin with the ethyl alcohol
and balance carbon:
Balancing hydrogen next gives:
The total number of oxygen atoms on the right side is seven, so
the final balanced equation is
Complete and balance the equation for the combustion of methane,
CH4. This reaction occurs when a gas furnace or a gas
stove is used for heating.
Answer:
Methane is a hydrocarbon. The unbalanced equation is
Following rules 1 and 2, we begin with the methane and notice
that carbon atoms are already balanced. Since hydrogen shows up
in only one product and oxygen shows up in two, it might be easier
to balance hydrogens next:
Oxygen is easily balanced:
When an acid and base react, the products of the reaction are
a salt and water. The water is formed as a result of the H+
from the acid reacting with the OH- from the base.
The salt, then, is composed of the cation from the base and the
anion from the acid.
Complete and balance the equation for the reaction between hypochlorous
acid and magnesium hydroxide.
Answer:
From the nomenclature rules you learned in chapter 2, you first
translate the names into chemical formulas and complete the reaction:
acid + base ------> salt
+ water
According to rule 1, we should begin with either the base or the
salt. Let's begin with the salt. Magnesium is already balanced.
We note that the hypochlorite ion stays intact from reactants
to products, so rule 3 allows us to balance the ion as a unit.
Two moles of hypochlorite ions on the right side require two moles
of hypochlorous acid on the left side:
Since the stoichiometric coefficients for the reactants are now
fixed, we can balance H and O. There are 4 hydrogens on the left,
which requires a 2 in front of water:
Since oxygens balance, we are finished. Note that we could have
balanced oxygen before hydrogen. But we shouldn't count the O
atom in the hypochlorite ion, since it was included when we balanced
the ion as a unit. Thus, going back to the equation before the
previous one, we would count two oxygens on the left (from the
magnesium hydroxide), requiring a 2 in front of water. This balances
hydrogen, giving us the same final equation as before.
It is easier to balance acid/base reactions if we realize that
the formula for water can be written HOH, showing that one mole
of hydroxide ion from the base is needed for each mole of hydrogen
ion from the acid. Refer again to the example above. Since one
mole of magnesium hydroxide contains two moles of hydroxide ion
and one mole of the hypochlorous acid contains only one mole of
hydrogen ion, it will require two moles of acid for every mole
of base, and two moles of water will form.
Complete and balance the equation for the reaction between phosphoric
acid and potassium hydroxide. The product of this reaction is
a common ingredient in commercial fertilizers.
Answer:
The formulas for the reactants are H3PO4
and KOH. The unbalanced reaction is
Since the acid has three H+ ions and the base has only
one OH- ion, three moles of base will be required for
each mole of acid, producing three moles of water:
Now that you know how to balance a reaction, let's discuss the
procedure for using a balanced reaction to perform calculations.
Review the steps for solving stoichiometry problems, on page 25.
We'll use a visual algorithm to implement each step and to help
you think about the chemical concepts as you solve the problem.
We'll first solve the mole-mole problem on page 24:
"How many moles of water are formed if five moles of hydrogen
gas react with excess oxygen gas?"
Step 1: Write the balanced equation for the reaction.
Step 2: Write known quantities above or below the formula
for the substance; moles always go below the formula, and
quantities which will be used to determine moles go above the
formula. Use a question mark to represent the unknown (desired)
quantity.
Step 3: Draw an arrow from the quantity above the formula
for the known to the number of moles shown below the formula.
Show the factor you'll use to convert from the former to the latter.
[Note: this step does not apply here, since we were given moles
of H2.]
Step 4: Draw an arrow from the number of moles of the
known substance to moles of the unknown, and write the mole ratio
above the arrow. Use this ratio to calculate moles of unknown
substance from moles of the known substance.
Since the number of moles of water formed is equal to the moles
of hydrogen which react, five moles of hydrogen will produce five
moles of water.
Step 5: Draw an arrow from moles of unknown to desired
quantity of unknown (above the formula). Show the factor to be
used to convert from moles to the desired quantity. [Note: this
step does not apply here, since we were asked to determine moles
of H2O.]
The answer to the problem, then, is that 5 moles of water
will be formed.
Let's use these same steps to perform the mass-mass calculation
on page 24:
"How many grams of oxygen are needed to completely convert
50 g of mercury to mercuric oxide (HgO)?"
Step 1: The balanced equation is
Steps 2,3: We list grams of mercury above the atomic symbol,
a question mark above the O2, then draw our conversion
arrow:
If there are 200 grams in a mole of mercury, then 50 grams is
obviously one-quarter of a mole. Recall the concept check at the
beginning of this document:
Your favorite candy costs $2.00 per pound. You have 50 cents.
How much candy can you buy?
Do you see a connection between the candy example and the mercury?
Chemistry really is common sense.
Step 4: Draw the conversion arrow between moles of mercury and
moles of oxygen, showing the mole ratio:
The stoichiometric coefficients show that only one mole of oxygen is needed for every two moles of mercury. Thus, to find moles of oxygen we simply divide moles of mercury by 2. Since the mole ratio is 2:1, you know that you'll be either multiplying or dividing moles of mercury by the number 2. How do you know which to do? The answer is common sense. If you had multiplied by two, you would have obtained a larger number of moles of oxygen than the number of moles of mercury you had, clearly contradicting the meaning of the stoichiometric coefficients. Our stoichiometry map now looks like this:
Step 5: Finally, we draw the conversion arrow for changing moles
of oxygen to grams, and complete the calculation:
Thus, it will require 4.0 grams of oxygen to completely convert
50 g of mercury to HgO. It might seem surprising that 50 grams
of mercury only requires 4 grams of oxygen, but remember that
chemical equations represent mole ratios. Since the molar
mass of Hg is so large (200g/mol), we still only have a fraction
of a mole even with 50 grams. In other words, most of the mass
of any sample of HgO is due to mercury. We'll explore the idea
of percentage composition in a section later in the chapter.
The examples we have considered so far have assumed that one of the reactants, the one on which we based the stoichiometric calculations, has been consumed completely. This will only be true if two conditions are met:
For example, the reaction
is complete; if we start with an equal number of moles of each
reactant, at the end of the reaction there is no reactant remaining;
the hydrogen and chlorine have been converted to HCl. On the other
hand, the reaction between hydrogen and nitrogen to form ammonia,
is not complete; even if we start with three moles of hydrogen
for every mole of nitrogen, there will still be some unreacted
nitrogen and hydrogen at the end of the reaction, and fewer than
two moles of ammonia will have been formed; in other words, not
all of the reactants would have been stoichiometrically
converted to products. It turns out that the type of stoichiometric
calculations which we have considered in this chapter are only
valid for complete reactions. Reactions which do not go to completion
are called equilibrium reactions and will be considered
in more detail in chapter 13.
The reactant which is used up completely in a reaction limits
how much of the products are formed and is therefore called
the limiting reactant.
In a sense, then, both of the problems which we solved above were
limiting reactant problems; what made them easier was that we
knew which reactant was limiting, since we were told that there
was an excess of the other reactant. Sometimes, however, we do
not know which reactant is present in excess, and we must first
determine which is the limiting reactant before we continue with
the problem. Therefore, we should modify step 3 for solving stoichiometry
problems (see p. 25). The modified step is:
3. Determine which reactant is the limiting reactant, if necessary.
Usually, the signal for a limiting reactant problem is that the
quantities of two or more reactants are given in the problem.
Base all subsequent stoichiometric calculations on the actual
number of moles of the limiting reactant.
Let's consider the solution to the limiting reactant problem shown
on p. 24:
"How many grams of water are formed if 12.0 grams of hydrogen
gas react with 50.0 grams of oxygen gas?"
Step 1: Write the balanced equation for the reaction.
Step 2: Write known quantities above or below the formula
for the substance; moles always go below the formula, and
quantities which will be used to determine moles go above the
formula. Use a question mark to represent the unknown (desired)
quantity.
Step 3: Draw an arrow from the quantity above the formula
for the known(s) to the number of moles shown below the formula.
Show the factors you'll use to convert from the former to the
latter
Step 4: Determine the limiting reactant. We'll consider two ways to do this: a systematic way and a shortcut way.
Systematic Method:
Arbitrarily choose one of the reactants and ask how many moles
of all other reactants would be needed to react completely with
the actual moles of the reactant you chose. If there is enough
of all other reactants, then the one you chose is the limiting
reactant because it is, indeed, used up completely. If there isn't
enough moles of one of the other reactant to use up the one you
chose, then that other reactant is the limiting reactant.
For example, let's arbitrarily choose hydrogen in this problem.
We know from the hydrogen-to-oxygen mole ratio of 2-to-1, it will
take one-half of 5.95, or about 3 moles of oxygen to react with
all of the hydrogen. We don't have this much oxygen, since there
are only 1.56 mol O2. Thus, there is not enough oxygen
to use up all of the hydrogen; oxygen will be used up completely
instead, and therefore oxygen is the limiting reactant.
Choose oxygen, instead of hydrogen, and make sure that you reach
the same conclusion that oxygen is the limiting reactant.
Shortcut Method:
The shortcut method is based on a recipe analogy. Suppose a recipe
calls for 2 eggs, one cup of milk, and half a cup of flour. How
many complete recipes can be made if you have 6 eggs on hand,
4 cups of milk, and 2 cups of flour? It is easy to obtain the
answer if we divide what we have on hand for each ingredient by
what one recipe requires:
6 eggs / (2 eggs/recipe) = 3 recipes
4 C milk / (1 C/recipe) = 4 recipes
2 C flour / (½C/recipe) = 4 recipes
Thus, we conclude that although we could theoretically make four
recipes based on the milk and flour, the eggs we have on hand
"limits" the number of recipes which are possible to
three, and we'll have leftover milk and flour at the end.
In a similar way, if we divide the actual number of moles of each
reactant by the corresponding stoichiometric coefficient from
the balanced equation, the smallest quotient identifies the limiting
reactant. Using the moles of hydrogen and oxygen in the problem
above, we get:
5.95 mol hydrogen / 2 mol (stoichiometric coefficient) = 2.98
1.56 mol oxygen / 1 mol (stoichiometric coefficient) = 1.56
The smaller quotient, 1.56, indicates that oxygen is the limiting
reactant. This is the same conclusion we reached by using the
systematic method. Use whatever method you feel more comfortable
with to determine the limiting reactant in a reaction.
Step 5: Base all subsequent calculations on the quantity of limiting
reactant present.
Calculate the number of grams of carbon dioxide which will be
formed in the reaction between 10.0 grams of propane (C3H8)
with 10.0 grams of oxygen. Do a "back of the envelope"
estimate before picking up your calculator to do the calculation.
Answer:
The ability to estimate answers is very important in chemistry.
The estimate not only serves as check on your calculation, but
also can be very fast. You'd be surprised how close the estimate
often is to the actual answer. There is one cardinal rule for
"back of the envelope" estimating: NO CALCULATORS.
Often, you can do the estimate in your head, without an envelope!
The nice thing about this skill is that you get faster and faster
at it with practice.
Before beginning the estimate, we'll write the balanced equation.
You'll need to picture the arrows and conversion factors which
we have used in the problems so far.
Here's what your estimation process might sound like if you thought
out loud: "The molar mass of propane is 3(12)+8, or about
44g/mol. If I have 10 grams of propane, that's 10/44, or about
1/4 of a mole. Now for oxygen... The molar mass is 32g/mol, and
thus 10 grams is about 1/3 of a mole of oxygen. If I wanted to
use all of the 1/4 of a mole of propane, it would take five times
as much, or 5/4 of a mole of oxygen. This is more than a mole
of O2, and I only have 1/3 of a mole. Thus, there isn't
enough O2 to use up the propane, and propane is left
over, making O2 the limiting reactant. The moles of
CO2 formed are equal to 3/5 times the moles of O2,
thus (3/5)( 1/3) = 1/5 of a mole of CO2 should be formed.
The molar mass is 12+2(16)=44g/mol, so 1/5 of a mole is 44/5,
or a little less than 9 grams of CO2 should be
formed.
Now let's do the problem with a calculator, following the same
reasoning:
Calculation of initial moles:
(10.0 g propane) / (44.1 g/mol) = 0.227 mol C3H8
(10.0 g oxygen) / (32.0 g/mol) = 0.312 mol O2
Determination of limiting reactant:
0.227/1 [stoichiometric coefficient] = 0.227
0.312/5 = .0624. This smaller quotient identifies O2
as the limiting reactant.
Use of mole ratio:
(0.312 mol O2)(3 moles CO2/5 moles O2)
= 0.187 mol CO2
Conversion from moles to grams:
(0.187 mol CO2)(44.0 g/mol) = 8.23 g CO2.
This answer is consistent with our ballpark estimate.
Calculate the number of grams of propane which are left over after
the reaction above. Calculate the number of grams of water which
are formed. In each case, do a "back of the envelope"
estimate before picking up your calculator to do the calculations,
then compare your answer to your estimate.
Answer:
7.26 grams of propane are left over; 4.50 g H2O are
formed.
In summary, you see that stoichiometric calculations are very
easy to perform. In a typical problem, it is usually necessary
to divide the mass of a substance by its molar mass to get moles,
use the mole ratios from the balanced equation to get moles of
another substance, then multiply by the molar mass of the other
substance to get grams. This is the procedure for solving any
mass-mass stoichiometry problem. Limiting reactant problems only
add one more wrinkle: it is necessary to identify the limiting
reactant before using the stoichiometric relationships. Chemists
use stoichiometric relationships on a daily basis, because they
study chemical reactions. And, as you now know, chemical reactions
are based on a well-defined and easy-to-use scheme: atoms combine
with each other based on simple whole-number ratios. These ratios
are still the same when we scale up the quantities to moles of
atoms, and we can measure these incredibly large numbers of atoms
easily using a substance's molar mass.
As mentioned above, you have learned that atoms combine with each
other in small whole-number ratios. These small numbers are represented,
in fact, by the subscripts in chemical formulas. But you also
know, as a result of what you learned in this chapter, that a
constant atom ratio must also mean a constant mass ratio, since
the molar mass of a given atom is constant. Thus, the formula
CO2 not only means that there are two oxygen atoms
for every carbon atom; it also means that there are two moles
of oxygen atoms per mole of carbon atoms, and thus 32.00 grams
of oxygen for every 12.01 grams of carbon.
Chemical Formulas and Percentage Composition
From the figures above, it is easy to calculate the percentage
by mass of each element in carbon dioxide.
By definition, the percentage by mass (sometimes called the mass
percentage) of any component is given by the expression
Notice that the mass units aren't important; they simply have
to be the same for the component and the sample which contains
the component. Since the units will cancel, mass percentage has
no units.
It is important to understand that the mass percentage of a component
doesn't depend on the size of the sample; it is an intensive property.
The percentage by mass of eyes in the "average"
Teddy
Bear head (see p. 22) is
This is true for 100, 1000, or 6.02 x 1023 heads. We
can calculate the mass percentage of an element in a compound
in a similar fashion. Although the mass percentage is the same
irregardless of sample size, it is more convenient to assume we
have a mole of the substance. This allows us to simply use molar
masses for the calculation. Thus, the mass percentage of carbon
in carbon dioxide is given by
The percentage by mass of oxygen could be calculated in a similar
fashion, but it is easier to realize that since the sum of the
percentages must equal 100%, the percentage oxygen is 100%(exact)
- 27.29% = 72.71% O.
Calculate the mass percentage of each element in potassium phosphate,
K3PO4.
Answer:
The molar mass of K3PO4 is 212.3 g/mol.
One mole of the compound contains 3 mol K, 1 mol P, and 4 mol
O. Thus,
% K = (3)(39.10)/212.3 x 100 = 55.25%
% P = 30.97/212.3 x 100 = 14.59%
% O = (4)(16.00)/212.3 x 100 = 30.15%
The sum of the percentages is 99.99%. The difference is a result
of round-off error.
Calculate the mass percentage of each element in aluminum sulfate,
Al2(SO4)3.
Answer:
% Al = 15.77%; % S = 28.12%; %O = 56.11%
Chemists are not only interested in reactions between substances;
they also want to know the composition of a particular
sample. The branch of chemistry which is concerned with the determination
of the qualitative and quantitative composition of a sample is
known as analytical chemistry. There are techniques in
analytical chemistry which allow chemists to determine the percentages
of elements in a compound. It is then possible to use these percentages
to determine the formula of the compound. This is the reverse
of what we accomplished above, i.e. determining percentage composition
from known formulas. Let's consider the following problem:
An unidentified compound is known to have the formula MSO4,
where M represents an unknown metal. The compound is soluble in
water. When a solution of 5.00 g of the compound is reacted with
an excess of aqueous barium nitrate solution, all of the sulfate
ion in the unknown compound is precipitated as insoluble BaSO4.
The dry precipitate has a mass of 9.68 g. What is the formula
of the compound?
Let's consider the steps we could use to solve this problem. If
we could find moles of sulfate ion in the BaSO4, this
would also be the moles of sulfate in the compound. Since the
formula shows that there is one mole of metal M for each mole
of sulfate, we would also then know how many moles of metal were
in the unknown compound. In addition, if we know moles of sulfate
in the sample, we can convert to grams of sulfate ion; if we then
subtract this number from the total mass of the compound (5.00
grams, as given) we can find the number of grams of metal in the
unknown. Finally, if we know the mass of metal and the corresponding
number of moles, we can find the molar mass and identify the metal
from the periodic table:
Moles of sulfate ion in BaSO4 = moles BaSO4 = 9.68 g BaSO4/MM BaSO4, so
moles of sulfate ion = 9.68 g / (233.4 g/mol) = 0.0415 mol SO42-.
0.0415 moles of sulfate in sample means that there is 0.0415 moles
of metal in sample
(0.0415 moles sulfate in sample)(96.07 g/mol sulfate) = 3.99 g
sulfate in sample
5.00 grams of compound minus 3.99 grams of sulfate = 1.01 grams
of metal in sample
Molar mass of metal = 1.01g / 0.0415 mol = 24.3 grams/mol, which
means that
The problem can also be solved using mass percentages. Try to
figure out how to do the problem this way and verify that your
answer is the same as the one above.
In the above problem, the mole ratio between the metal and sulfate
ions was known and the molar mass of the metal was the unknown.
It is more common for chemists to know the identities of all elements
in a compound and for them to know the mass percentages of each
element as well. This information allows a determination of the
relative number of each type of atom in the compound. In other
words, it allows the determination of the empirical formula.
The empirical formula for a compound
gives the relative number of each type of atom in the compound.
The molecular formula gives the actual number of
each type of atom in one molecule of the compound.
Examine the table below to understand the difference between these
two types of formulas.
| Water | ||
| Benzene | ||
| Acetylene | ||
| Sodium Sulfate | ||
| Glucose |
Ionic compounds do not exist as discrete molecules or even as
a small number of well-defined units. They always exist in a large
(on a molecular scale) crystal which has a very large number of
ions. Thus, ionic formulas are always empirical formulas. The
formulas for substances which exist as molecules, however, can
be either empirical or molecular formulas. You can see that it
is possible for the empirical and molecular formulas for a substance
to be the same (e.g., water, above). Notice also that it is possible
for two or more completely different substances, with different
molecular formulas, to have the same empirical formula (e.g.,
benzene and acetylene, above). Naturally, since isomers have the
same molecular formula, they also have the same empirical formula.
Although benzene and acetylene have different molecular formulas,
they will have the same mass percentage of carbon and hydrogen,
since they have the same empirical formula. This can easily be
demonstrated:
Benzene: 
Acetylene: 
It should be clear from the calculations above that the mass percentages
of the elements in a compound are determined entirely by the empirical
formula. Conversely, if we were told that we had a compound which
consisted of 92.26% carbon and 7.74% hydrogen, we would only know
that the empirical formula was CH; we wouldn't be able to tell
from this information whether the compound was benzene or acetylene.
In other words,
If the mass percentages of the elements in a compound
are known, then the empirical formula of the
compound can be determined.
Determining Empirical Formulas from Elemental Composition Data
The procedure for determining the empirical formula from the composition
will be illustrated by reversing the above problem. That is, if
we know that a compound is made up of 92.26% carbon and 7.74%
hydrogen, then what is the empirical formula?
The steps for solving this type of problem are listed below. Don't
memorize them; rather, examine each step and make sure you understand
it.
Applying these steps to the specific problem above gives:
An analytical chemist determines that a compound of phosphorus
and oxygen contains 43.64% P. Help her to deduce the empirical
formula of the compound.
Answer:
The compound contains 100%-43.64% = 56.36% oxygen.
On April 19, 1995, a compound which is normally used for fertilizer
was mixed with fuel oil, producing a powerful explosive which
destroyed 169 lives and half of the Alfred P. Murrah Federal Building
in Oklahoma City. The compound has the composition 35.00% nitrogen,
59.96% oxygen, and 5.04% hydrogen. What is its empirical formula?
Answer:
The empirical formula is N2O3H4.
The compound is ammonium nitrate, NH4NO3.
You know now that empirical formulas can be determined from the
mass percentages of the elements in a compound. But these percentages
don't just magically appear during a chemist's dream. How is it
possible to determine the composition of a compound? The answers
to this question vary depending on the particular substance which
is being analyzed. In fact, the sole purpose of many analytical
chemistry texts is to describe these methods in detail. However,
we can describe how it is done for two groups of compounds: those
which are made up of only carbon and hydrogen (the hydrocarbons)
and those which also contain oxygen.
The strategy for determining the percentage composition of these
compounds is simple, but elegant. Chemists rely on the fact that
combustion of a substance in either class of compounds results
in the production of carbon dioxide and water (see the section
on combustion reactions, p. 25.) Since the only two reactants
are CxHyOz and O2,
all of the carbon in the CO2 produced must have originally
been in the organic reactant. Likewise, this reactant must have
originally contained all of the hydrogen atoms which eventually
ended up in the form of water.
In practice, a weighed sample of the substance is reacted with excess oxygen and the product gases (carbon dioxide and water) are forced to pass through two reaction vessels: the first one absorbs all of the carbon dioxide which was produced in the reaction, and the second one absorbs all of the water. The mass of each product is simply equal to the mass increase of the materials which absorbed each of the products. Let's see how this technique can be used to determine the empirical formula of a compound which contains at least carbon and hydrogen, and may also contain oxygen. Consider the following problem:
3.00 g of a compound with the general formula CxHy
or CxHyOz is burned in excess
oxygen. Analysis of the combustion products shows that 4.25 g
of CO2 and 2.61 g of H2O were formed. What
is the empirical formula for the compound?
Answer:
Let's see which general scheme we might use to solve this problem.
We know that we can get the empirical formula if we determine
the percentage composition of the compound. But is it necessary
to do this? Since the empirical formula represents the relative
number of moles of each type of atom, we could also get the empirical
formula directly from the number of moles of C and H (and O, if
it contains oxygen). Let's consider the following reasoning scheme:
mass CO2 ----> moles CO2 ----> moles C ----> mass C,
mass H2O ----> moles H2O ----> moles
H----> mass H.
Then, we can add the masses of C and H. If they total the mass
of the compound, 3.00g in this case, there is no oxygen in the
compound. If the total is less than 3.00 grams, then the difference
in mass must be due to the oxygen.
Finally, we take the ratio of the number of moles of C and H (and
O, if present), determine the smallest whole-number ratio, and
we have the empirical formula. Let's put our plan into action:
4.25 g CO2 / 44.0 g/mol = 0.0966 mol CO2.
Since there is one mole of carbon atoms per mole of CO2,
the total moles of carbon in the product is 0.0966 mol. If there
are 12.011 grams of carbon per mole, then we have (12.011g/mol)(0.0966
mol) = 1.16 grams of carbon. This is also the amount of carbon
which was originally in the organic reactant. We do the same for
hydrogen:
2.61 g H2O / 18.016 g/mol = 0.145 mol H2O.
Since there are two moles of hydrogen atoms per mole of water
molecules, we have 2(0.145) = 0.290 moles of hydrogen. The mass
of hydrogen originally in the organic reactant is therefore (1.008
g/mol)(0.290 mol) = 0.292 gram.
The total mass of the C and H is 1.16 g + 0.292 g = 1.45 g. Since
this is less than the mass of the original reactant, there was
also oxygen present in the compound. The mass of oxygen is 3.00g
- 1.45 g = 1.55 g O. Since the molar mass of oxygen atoms is 16.00
g/mol, 1.55 g oxygen corresponds to 1.55 g / (16.00 g/mol) = 0.0969
mol O.
We have the number of moles of C, H, and O in the same 3.00 gram
sample of the original compound, so we can determine the mole
ratios by dividing through by the smallest number:
0.0966 mol C / 0.0966 = 1
0.290 mol H / 0.0969 = 2.99 3
0.0969 mol O / 0.0966 = 1.00 (to 3 s.f.)
Thus, the empirical formula is CH3O.
Notice that we cannot write a balanced equation for the combustion
of this substance, since we still don't know the molecular
formula. After you solve the next Concept Check, we'll discuss
how to obtain the molecular formula for a substance if we know
the empirical formula.
A small amount of an organic compound is found in an unlabeled
bottle in the lab by your chemistry professor, who sends a 600
mg sample of it to a laboratory for analysis. After one week,
the professor is informed that the analysis is complete, but that
the chemist who performed the analysis is ill and won't be able
to complete the calculations for another two days. Anxious to
know the identity of the compound, the professor tells the lab
to provide the data so that one of the chemistry students in the
class can work up the results. The lab sends the following results:
The 600mg sample was burned in excess oxygen, resulting in
the formation of 0.879 grams of carbon dioxide. Furthermore,
the lab quotes a notation in the chemist's notebook: "I
noticed that the same number of moles of water and carbon dioxide
were produced."
Can you help your chemistry professor by determining the empirical
formula for the compound?
Answer:
The empirical formula of the compound is CH2O.
Determining Molecular Formulas
A molecular formula can always be obtained by multiplying the
subscripts in the empirical formula by an integer: 1 (if the empirical
and molecular formulas are the same), 2, 3, etc. Thus, a compound
with an empirical formula of NO2 could have molecular
formulas of NO2, N2O4, N3O6,
or any molecular formula of the form NxO2x.
Although any of these formulas are possible from a compositional
point of view, it is usually the case that only one or, sometimes,
a small number of formulas actually represent real chemical species.
For example, only the first two nitrogen-containing compounds
listed above actually exist. It is also possible, of course, that
many different compounds share a particular empirical formula,
and each of these compounds can have its own unique molecular
formula. For example, the empirical formula CH2 could
represent C2H4 (ethene), C3H6
(cyclopropane or propene, two different compounds), C6H12
(cyclohexane), and so on. Finally, remember that the term molecular
formula cannot be applied to ionic compounds.
You will have a better feel for which of the many possible molecular
formulas for a compound might actually exist after you have studied
more about chemical bonding. For now, we have a simple goal: to
discover how a molecular formula can be obtained from an empirical
formula. The answer is easy:
To determine the molecular formula, we need to know how many "empirical
formula units" (efu) make up the molecular formula. For example,
each of the "possible" molecular formulas for the nitrogen
oxides shown on the previous page can be written in the form (NO2)x,
i.e. (NO2), (NO2)2, (NO2)3,
and so on. This notation does not suggest anything about the bonding
which might exist in each of the possible molecular formulas.
For example, the third formula doesn't necessarily mean that there
would be three NO2 units bonded to each other sequentially.
The notations simply mean that there is one, two, or three empirical
formula units in a molecule, respectively.
To find the number of empirical formula units in a molecule, simply
divide the molar mass by the empirical formula mass:
Let's look at an example of how this procedure will allow us to
deduce the molecular formula from the empirical formula. Refer
again to Concept Check 3.11d above. If the analytical lab also
reported that the molar mass of a compound was determined to be
60.0 g/mol, then we have:
The empirical formula is CH2O, so the efu mass is 12.011
+ 2(1.008) + 16.00 = 30.0 g/mol of efu.
60 g/mol of molecules / (30.0 g/ mol of efu) = 2.00 mol of efu
/ mol of molecules.
In other words, there are two empirical formula units per molecule,
and the subscripts in the empirical formula can be multiplied
by the factor of 2 to obtain the molecular formula:
The molar mass of a substance can be obtained in a variety of
ways: by knowing the mass of a specified number of molecules (and
using Avogadro's Number), by identifying the parent molecular
ion in the mass spectrum of the substance, or by knowing the mass
of a sample and the corresponding number of moles. In chapter
5, you'll learn that the molar mass of a gas can be determined
in a number of ways. Irregardless of the method used to determine
it, the molar mass can be used in conjunction with the empirical
formula to obtain the molecular formula of the compound.
The compound described in Example Problem 3.11c
is used as antifreeze
in automobile radiators. The most intense peak of the parent ion
cluster in the mass spectrum shows up at m/e = 62 u. Determine
the molecular formula of the compound.
Answer:
The molecular formula is C2H6O2.
The common name of the compound is ethylene glycol.
We started out with Teddy Bears in this chapter and ended up with
molecular formulas. Although our path was at times circuitous,
and we took a few side trips, hopefully you enjoyed the journey.
In the next chapter, we'll continue the excursion by looking at
chemical reactions in solution.
The formulas of compounds, the chemical equations which chemists
use to describe the countless reactions between them, and the
molecular-scale hustle and bustle which occur as these reactions
take place are all the result of a simple fact: chemical compounds
are made up of atoms, and chemical reactions result in a shuffling
around of these atoms, with no loss or gain of a single atom.
The masses of these atoms, miniscule as they are, can be measured
very accurately relative to the arbitrarily-defined mass of the
12C standard. Since atoms combine and get shuffled
around in small whole-number ratios, the masses of atoms which
combine are also in these same ratios. The mole is a chemist's
way of counting huge numbers of molecules and is convenient because
the mass of a mole of atoms is simply its molecular mass expressed
in grams. Mass relationships in chemical formulas and chemical
reactions are a big part of the foundation of chemical knowledge
which you will need to understand as you explore even more of
this wonderful chemical world around you.
The next time you see a Teddy Bear, think of the atoms which make
it up or of the chemical reactions which might have been used
to synthesize its fur, eyes, and ears. Understanding nature on
a more intimate level doesn't take away any of its beauty and
poetry; instead, it allows a deeper appreciation of these qualities.
Last modified 12/07/96