More Complex Stoichiometric Calculations
When you’re asked to do stoichiometric calculations on
the SAT II Chemistry exam, make sure that if you need to write out
the chemical formulas, you do this correctly. No matter how good
you are at math and how well you understand the stoichiometric rules
that follow, you won’t get the right answer if your chemical formulas
are wrong! If you feel that you’re weak in this area, see the review
(in Appendix II) of chemical formula naming and writing.
Perhaps the easiest way to approach problems that ask
you to calculate the amounts of reactants consumed or products produced
during the course of a reaction is to start by creating a table
or chart. Let’s work through a typical example. Say the SAT II Chemistry
test asks you what mass of oxygen will react completely with 96.1
grams of propane. Notice that for this question, you’ll need to
start by writing the chemical formulas. Now follow these steps:
-
Write the chemical equation.
- Calculate
the molar masses and put them in parentheses above the formulas; soon
you’ll figure out you don’t have to do this for every reactant and product,
just those you’re specifically asked about.
- Balance
the equation.
- Next
put any amounts that you were given into the table. In this example, you
were told that the reaction started with 96.1 g of propane.
- Find
the number of moles of any compounds for which you were given masses.
Here you’d start with propane: you divide 96.1 grams by the molar mass
of propane (44.11 g/mol) to get the number of moles of propane (2.18 mol).
- Use
the mole:mole ratio expressed in the coefficients of each of the compounds
to find moles of all of the necessary compounds involved. The only
one you really need to know is oxygen, but let’s run through all
of them for practice. If the coefficient for propane, which is 1,
is equal to 2.18 moles of propane, then the number of moles of oxygen
must be 5
2.18 = 10.9, the moles of CO2 is
3
2.18 = 6.54, and the moles of H2O
= 4
2.18 = 8.72.
| Molar mass |
(44.11) |
(32.00) |
(44.01) |
(18.02) |
| Balanced equation |
C3H8 + |
5O2  |
3CO2 + |
4H2O |
| No. of moles |
2.18 |
10.9 |
6.54 |
8.72 |
| Amount |
96.1 g |
|
|
|
-
Reread the problem to determine which amount was asked
for. The question asks for the mass of oxygen, so convert moles
of oxygen to grams and you have the answer:
10.9 mol
44.01 g/mol = 349 g oxygen
| Molar mass |
(44.11) |
(32.00) |
(44.01) |
(18.02) |
| Balanced equation |
C3H8 + |
5O2  |
3CO2 + |
4H2O |
| Mole:mole |
1 |
5 |
3 |
4 |
| No. of moles |
2.18 |
10.9 |
6.54 |
8.72 |
| Amount |
96.1 g |
349 g |
|
|
But what if this question had asked you to determine the
liters of CO
2 consumed in this reaction at
STP (273K, 1 atm)? You would take the number of moles of CO
2 that
we calculated from the table and use the standard molar volume for
a gas, or 22.4 L/mol. So, 6.54 mol
22.4 L/mol = 146 L.
Finally, what if the question had asked how many water
molecules are produced? You would take the number of moles of water
and multiply it by Avogadro’s number, 6.02
10
23,
to get 5.25
10
24 molecules
of water.
| Molar mass |
(44.11) |
(32.00) |
(44.01) |
(18.02) |
| Balanced equation |
C3H8 + |
5O2  |
3CO2 + |
4H2O |
| Mole:mole |
1 |
5 |
3 |
4 |
| No. of moles |
2.18 |
10.9 |
6.54 |
8.72 |
| Amount |
96.1 g |
349 g |
146 L |
5.25 1024 |
You certainly don’t have to write out several tables;
we just did that to make the method clearer to you. Once you practice,
you won’t need to write the categories to the left, either. They
will become second nature. In essence, you can work the problems
faster and set yourself up nicely for a clear understanding
of equilibrium problems in your future.
Perhaps your teacher at school taught you to solve this
type of problem using dimensional analysis. If so, and if you feel
comfortable solving problems using that method, don’t learn to do
it our way: stick to the method with which you feel comfortable.
We’ll go through this problem using dimensional analysis: What
mass of oxygen will react with 96.1 grams of propane? Again,
you would first write the chemical formula and make sure your equation
is correctly balanced:
C3H8 +
5O2
3CO2 +
4H2O
The amount to start with, when setting up the dimensional
analysis, is 96.1 g, and your goal is to calculate the number of
grams of oxygen produced:
Not too hard, right? Now, how many liters of CO2 would
be produced at STP?
And how many water molecules are produced?
Some people prefer the table method, while others are
more comfortable with dimensional analysis. Use whatever method
you feel more comfortable using, but just be consistent—if you always
do the same types of problems the same way, you’ll feel much more
confident on test day. Now try the method you prefer on some problems.
Examples
-
Solid lithium hydroxide is used in space vehicles
to remove exhaled carbon dioxide from the air; it reacts with carbon
dioxide to form solid lithium carbonate and liquid water. What mass
of gaseous carbon dioxide will be consumed in a reaction with 1.00
kg of lithium hydroxide?
Explanation
First write the reaction, then create and fill in your
chart, and when you’re done filling in the necessary blocks, it
should look like this:
| Molar mass |
(23.95) |
(44.01) |
|
|
| Balanced equation |
2LiOH + |
CO2  |
Li2CO3 + |
H2O |
| No. moles |
1000 g/23.95
g/mol = 41.8 moles |
20.9 |
20.9 |
20.9 |
| Amount |
1.00 kg |
20.9 44.01
= 920 g |
|
|
-
Baking soda (NaHCO3) is often
used as an antacid. It neutralizes excess hydrochloric acid secreted
by the stomach in the reaction below:
NaHCO3(s) + HCl(aq)
NaCl(aq) + H2O(l ) +
CO2(aq)
How many grams of NaHCO3 would
be needed to completely react with 10.0 g of HCl?
Explanation
According to the balanced equation above, 1 mole of NaHCO3 reacts
with 1 mole of HCl. First you need to calculate how many moles of
HCl are in 10 grams of HCl. The formula weight of HCl is about 36
g/mol, so you set up the equation:
= 0.27 moles HCl. Now, knowing that 1 mole of baking soda reacts with 1
mole of NaCl, we next need to figure out how many grams of baking
soda would react with 0.27 moles of baking soda: 0.27 moles
134 g/mol baking soda = 37.2 grams/mole
NaHCO
3
| Molar mass |
(85.32) |
(36.46) |
|
|
|
| Balanced equation |
NaHCO3 + |
HCl  |
NaCl + |
H2O + |
CO2 |
| No. moles |
0.274 |
10.0/36.46 = 0.274 |
|
|
|
| Amount |
0.274 85.32 = 23.4 g |
10.0 g |
|
|
|
