6.4 Stoichiometry
Learning Objectives
 Define stoichiometry.
 Relate quantities in a balanced chemical reaction on a molecular basis.
Consider a classic recipe for pound cake: 1 pound of eggs, 1 pound of butter, 1 pound of flour, and 1 pound of sugar. (That’s why it’s called “pound cake.”) If you have 4 pounds of butter, how many pounds of sugar, flour, and eggs do you need? You would need 4 pounds each of sugar, flour, and eggs.
Now suppose you have 1.00 g H_{2}. If the chemical reaction follows the balanced chemical equation 2H_{2}(g) + O_{2}(g) → 2H_{2}O(ℓ), then what mass of oxygen do you need to make water?
Curiously, this chemical reaction question is very similar to the pound cake question. Both of them involve relating a quantity of one substance to a quantity of another substance or substances. The relating of one chemical substance to another using a balanced chemical reaction is called stoichiometry. Using stoichiometry is a fundamental skill in chemistry; it greatly broadens your ability to predict what will occur and, more importantly, how much is produced.
Let us consider a more complicated example. A recipe for pancakes calls for 2 cups (c) of pancake mix, 1 egg, and ½ c of milk. We can write this in the form of a chemical equation:
2 c mix + 1 egg + ½ c milk → 1 batch of pancakes
If you have 9 c of pancake mix, how many eggs and how much milk do you need? It might take a little bit of work, but eventually you will find you need 4½ eggs and 2¼ c milk.
How can we formalize this? We can make a conversion factor using our original recipe and use that conversion factor to convert from a quantity of one substance to a quantity of another substance. Because one recipe’s worth of pancakes requires 2 c of pancake mix, 1 egg, and ½ c of milk, we actually have the following mathematical relationships that relate these quantities:
2 c pancake mix ⇔ 1 egg ⇔ ½ c milk
where ⇔ is the mathematical symbol for “is equivalent to.” This does not mean that 2 c of pancake mix equals 1 egg. However, as far as this recipe is concerned, these are the equivalent quantities needed for a single recipe of pancakes. So, any possible quantities of two or more ingredients must have the same numerical ratio as the ratios in the equivalence.
We can deal with these equivalences in the same way we deal with equalities in unit conversions: we can make conversion factors that essentially equal 1. For example, to determine how many eggs we need for 9 c of pancake mix, we construct the conversion factor:
[latex]\dfrac{1\text{ egg}}{2\text{ c pancake mix}}[/latex]
This conversion factor is, in a strange way, equivalent to 1 because the recipe relates the two quantities. Starting with our initial quantity and multiplying by our conversion factor, we get:
Note how the units cups pancake mix cancelled, leaving us with units of eggs. This is the formal, mathematical way of getting our amounts to mix with 9 c of pancake mix. We can use a similar conversion factor for the amount of milk:
Again, units cancel, and new units are introduced.
A balanced chemical equation is nothing more than a recipe for a chemical reaction. The difference is that a balanced chemical equation is written in terms of atoms and molecules, not cups, pounds, and eggs.
For example, consider the following chemical equation:
2H_{2}(g) + O_{2}(g) → 2H_{2}O(ℓ)
We can interpret this as, literally, “two hydrogen molecules react with one oxygen molecule to make two water molecules.” That interpretation leads us directly to some equivalences, just as our pancake recipe did:
2 H_{2} molecules ⇔ 1 O_{2} molecule ⇔ 2 H_{2}O molecules
These equivalences allow us to construct conversion factors:
[latex]\dfrac{2\text{ molecules }\ce{H2}}{1\text{ molecule }\ce{O2}}\Leftrightarrow \dfrac{2\text{ molecules }\ce{H2}}{2\text{ molecules }\ce{H2O}} \Leftrightarrow \dfrac{1\text{ molecule }\ce{O2}}{2\text{ molecules }\ce{H2O}}[/latex]
and so forth. These conversions can be used to relate quantities of one substance to quantities of another. For example, suppose we need to know how many molecules of oxygen are needed to react with 16 molecules of H_{2}. As we did with converting units, we start with our given quantity and use the appropriate conversion factor:
[latex]16\text{ }\cancel{\text{molecules }\ce{H2}}\times \dfrac{1\text{ molecule }\ce{O2}}{2\text{ }\cancel{\text{molecules }\ce{H2}}}=8\text{ molecules }\ce{O2}[/latex]
Note how the unit molecules H_{2} cancels algebraically, just as any unit does in a conversion like this. The conversion factor came directly from the coefficients in the balanced chemical equation. This is another reason why a properly balanced chemical equation is important.
Example 6.9
Problem
How many molecules of SO_{3} are needed to react with 144 molecules of Fe_{2}O_{3} given the balanced chemical equation Fe_{2}O_{3}(s) + 3SO_{3}(g) → Fe_{2}(SO_{4})_{3}?
Solution
We use the balanced chemical equation to construct a conversion factor between Fe_{2}O_{3} and SO_{3}. The number of molecules of Fe_{2}O_{3} goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO_{3} go on the top. Thus, the appropriate conversion factor is:
[latex]\dfrac{3\text{ molecules }\ce{SO3}}{1\text{ molecule }\ce{Fe2O3}}[/latex]
Starting with our given amount and applying the conversion factor, the result is:
[latex]144\text{ molecules }\ce{Fe2O3}\times \dfrac{3\text{ molecules }\ce{SO3}}{1\cancel{\text{ molecule }\ce{Fe2O3}}}=432\text{ molecules }\ce{SO3}[/latex]
We need 432 molecules of SO_{3} to react with 144 molecules of Fe_{2}O_{3}.
Test Yourself
How many molecules of H_{2} are needed to react with 29 molecules of N_{2} to make ammonia if the balanced chemical equation is N_{2} + 3H_{2} → 2NH_{3}?
Answer
87 molecules
Chemical equations also allow us to make conversions regarding the number of atoms in a chemical reaction because a chemical formula lists the number of atoms of each element in a compound. The formula H_{2}O indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors:
[latex]\dfrac{2\text{ atoms H}}{1\text{ molecule }\ce{H2O}} \Leftrightarrow \dfrac{1\text{ molecule }\ce{H2O}}{1\text{ atom O}}[/latex]
Conversion factors like this can also be used in stoichiometry calculations.
Example 6.10
Problem
How many molecules of NH_{3} can you make if you have 228 atoms of H_{2}?
Solution
From the formula, we know that one molecule of NH_{3} has three H atoms. Use that fact as a conversion factor:
[latex]228\text{ atoms H}\times \dfrac{1\text{ molecule }\ce{NH3}}{3\cancel{\text{ atoms H}}}=76\text{ molecules }\ce{NH3}[/latex]
Test Yourself
How many molecules of Fe_{2}(SO_{4})_{3} can you make from 777 atoms of S?
Answer
259 molecules
For a video lecture on stoichiometry, view this video on stoichiometry by Dr. Jessie A. Key.
Key Takeaways
 Quantities of substances can be related to each other using balanced chemical equations.
Exercises
Questions
 Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe?
 Think back to the pancake recipe. What possible conversion factors can you construct relating the components of the recipe?
 What are all the conversion factors that can be constructed from the balanced chemical reaction 2H_{2}(g) + O_{2}(g) → 2H_{2}O(ℓ)?
 What are all the conversion factors that can be constructed from the balanced chemical reaction N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)?
 Given the chemical equation Na(s) + H_{2}O(ℓ) → NaOH(aq) + H_{2}(g)
 Balance the equation.
 How many molecules of H_{2} are produced when 332 atoms of Na react?
 Given the chemical equation S(s) + O_{2}(g) → SO_{3}(g)
 Balance the equation.
 How many molecules of O_{2} are needed when 38 atoms of S react?
 For the balanced chemical equation 6H^{+}(aq) + 2MnO_{4}^{−}(aq) + 5H_{2}O_{2}(ℓ) → 2Mn^{2+}(aq) + 5O_{2}(g) + 8H_{2}O(ℓ), how many molecules of H_{2}O are produced when 75 molecules of H_{2}O_{2} react?
 For the balanced chemical reaction 2C_{6}H_{6}(ℓ) + 15O_{2}(g) → 12CO_{2}(g) + 6H_{2}O(ℓ), how many molecules of CO_{2} are produced when 56 molecules of C_{6}H_{6} react?
 Given the balanced chemical equation Fe_{2}O_{3}(s) + 3SO_{3}(g) → Fe_{2}(SO_{4})_{3}, how many molecules of Fe_{2}(SO_{4})_{3} are produced if 321 atoms of S react?
 For the balanced chemical equation CuO(s) + H_{2}S(g) → CuS + H_{2}O(ℓ), how many molecules of CuS are formed if 9,044 atoms of H react?
 For the balanced chemical equation Fe_{2}O_{3}(s) + 3SO_{3}(g) → Fe_{2}(SO_{4})_{3}, suppose we need to make 145,000 molecules of Fe_{2}(SO_{4})_{3}. How many molecules of SO_{3} do we need?
 One way to make sulfur hexafluoride is to react thioformaldehyde, CH_{2}S, with elemental fluorine, is described by CH_{2}S + 6F_{2} → CF_{4} + 2HF + SF_{6}. If 45,750 molecules of SF_{6} are needed, how many molecules of F_{2} are required?
 Construct the three independent conversion factors possible for these two reactions: 2H_{2} + O_{2} → 2H_{2}O and H_{2} + O_{2} → H_{2}O_{2}. Why are the ratios between H_{2} and O_{2} different? The conversion factors are different because the stoichiometries of the balanced chemical reactions are different.
 Construct the three independent conversion factors possible for these two reactions
2Na + Cl_{2} → 2NaCl and 4Na + 2Cl_{2} → 4NaCl. What similarities, if any, exist in the conversion factors from these two reactions?
Answers
 [latex]\dfrac{1\text{ pound butter}}{1\text{ pound flour}}[/latex] or [latex]\dfrac{1\text{ pound sugar}}{1\text{ pound eggs}}[/latex] are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible.
 [latex]\dfrac{2\text{ molecules }\ce{H2}}{1\text{ molecule }\ce{O2}}\Leftrightarrow \dfrac{1\text{ molecule }\ce{O2}}{2\text{ molecules }\ce{H2O}}\Leftrightarrow \dfrac{2\text{ molecules }\ce{H2}}{2\text{ molecules }\ce{H2O}}[/latex] and their reciprocals are the conversion factors that can be constructed.
 2Na(s) + 2H_{2}O(ℓ) → 2NaOH(aq) + H_{2}(g) and 166 molecules
 120 molecules
 107 molecules
 435,000 molecules

 [latex]\dfrac{2\text{ molecules }\ce{H2}}{1\text{ molecule }\ce{O2}}\Leftrightarrow \dfrac{1\text{ molecule }\ce{O2}}{2\text{ molecules }\ce{H2O}}\Leftrightarrow \dfrac{2\text{ molecules }\ce{H2}}{2\text{ molecules }\ce{H2O}}[/latex]
 [latex]\dfrac{1\text{ molecule }\ce{H2}}{1\text{ molecule }\ce{O2}}\Leftrightarrow \dfrac{1\text{ molecule }\ce{O2}}{1\text{ molecule }\ce{H2O2}}\Leftrightarrow \dfrac{1\text{ molecule }\ce{H2}}{1\text{ molecule }\ce{H2O2}}[/latex]