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 7.1 The Mole 7.2 Percent Composition of Compounds 7.3 More Complex Stoichiometric Calculations 7.4 Limiting Reagents

 7.5 Chemical Yields 7.6 Practice Questions 7.7 Explanations
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:
1. Write the chemical equation.
2. 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.
3. Balance the equation.
4. 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.
5. 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).
6. 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 52.18 = 10.9, the moles of CO2 is 32.18 = 6.54, and the moles of H2O = 42.18 = 8.72.
Molar mass Balanced equation No. of moles Amount (44.11) (32.00) (44.01) (18.02) C3H8 + 5O2 3CO2 + 4H2O 2.18 10.9 6.54 8.72 96.1 g
1. 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 mol44.01 g/mol = 349 g oxygen
Molar mass Balanced equation Mole:mole No. of moles (44.11) (32.00) (44.01) (18.02) C3H8 + 5O2 3CO2 + 4H2O 1 5 3 4 2.18 10.9 6.54 8.72 96.1 g 349 g
But what if this question had asked you to determine the liters of CO2 consumed in this reaction at STP (273K, 1 atm)? You would take the number of moles of CO2 that we calculated from the table and use the standard molar volume for a gas, or 22.4 L/mol. So, 6.54 mol22.4 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, 6.021023, to get 5.251024 molecules of water.
Molar mass Balanced equation Mole:mole No. of moles (44.11) (32.00) (44.01) (18.02) C3H8 + 5O2 3CO2 + 4H2O 1 5 3 4 2.18 10.9 6.54 8.72 96.1 g 349 g 146 L 5.251024
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 + 5O23CO2 + 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
1. 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 Balanced equation No. moles Amount (23.95) (44.01) 2LiOH + CO2 Li2CO3 + H2O 1000 g/23.95 g/mol = 41.8 moles 20.9 20.9 20.9 1.00 kg 20.944.01 = 920 g
1. 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 moles134 g/mol baking soda = 37.2 grams/mole NaHCO3
 Molar mass Balanced equation No. moles Amount (85.32) (36.46) NaHCO3 + HCl NaCl + H2O + CO2 0.274 10.0/36.46 = 0.274 0.27485.32 = 23.4 g 10.0 g
 Jump to a New ChapterIntroduction to the SAT IIIntroduction to the SAT II Chemistry TestStrategies for Taking the SAT II Chemistry TestThe Structure of MatterThe States of MatterReaction TypesStoichiometryEquilibrium and Reaction RatesThermodynamicsDescriptive ChemistryLaboratoryBasic Measurement and Calculation ReviewChemical Formulas Review: Nomenclature and Formula WritingPractice Tests Are Your Best Friends
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