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.
