Copyright (c) 1996 by James H. Burness
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 168 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.
RETURN TO PART 2
Last modified June 16, 1997