Example \(\PageIndex{1}\): Calculating the Theoretical Yield and the Percent Yield
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below.
\[2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)\nonumber \]
In a certain experiment, \(40.0 \: \text{g} \: \ce{KClO_3}\) is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed, the oxygen gas is collected, and its mass is found to be \(14.9 \: \text{g}\). What is the percent yield for the reaction?
Solution
First, we will calculate the theoretical yield based on the stoichiometry.
Step 1: List the known quantities and plan the problem.
Known
- Given: Mass of \(\ce{KClO_3} = 40.0 \: \text{g}\)
- Molar mass \(\ce{KClO_3} = 122.55 \: \text{g/mol}\)
- Molar mass \(\ce{O_2} = 32.00 \: \text{g/mol}\)
Unknown
- theoretical yield O2 = ? g
Apply stoichiometry to convert from the mass of a reactant to the mass of a product:
\[\text{g} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{KClO_3} \rightarrow \text{mol} \: \ce{O_2} \rightarrow \text{g} \: \ce{O_2} \nonumber\nonumber \]
Step 2: Solve.
\[40.0 \: \text{g} \: \ce{KClO_3} \times \frac{1 \: \text{mol} \: \ce{KClO_3}}{122.55 \: \text{g} \: \ce{KClO_3}} \times \frac{3 \: \text{mol} \: \ce{O_2}}{2 \: \text{mol} \: \ce{KClO_3}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \text{mol} \: \ce{O_2}} = 15.7 \: \text{g} \: \ce{O_2} \nonumber\nonumber \]
The theoretical yield of \(\ce{O_2}\) is \(15.7 \: \text{g}\).
Step 3: Think about your result.
The mass of oxygen gas must be less than the \(40.0 \: \text{g}\) of potassium chlorate that was decomposed.
Now we will use the actual yield and the theoretical yield to calculate the percent yield.
Step 1: List the known quantities and plan the problem.
Known
- Actual yield \(= 14.9 \: \text{g}\)
- Theoretical yield \(= 15.7 \: \text{g}\)
Unknown
\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \nonumber\nonumber \]
Use the percent yield equation above.
Step 2: Solve.
\[\text{Percent Yield} = \frac{14.9 \: \text{g}}{15.7 \: \text{g}} \times 100\% = 94.9\% \nonumber\nonumber \]
Step 3: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under \(100\%\).