Stoichiometry of Chemical Reactions

Percent Yield Calculations

The theoretical yield of a reaction is the calculated amount of product that is expected to be produced. The amount of the product determined experimentally is the actual yield. The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

In chemistry, the amount of product obtained in a chemical reaction is called the yield. The theoretical yield is the amount of product predicted from a reaction using the mole ratios from the balanced equation. The actual yield (or experimental yield) is the amount of product from a reaction determined experimentally. In practice, the actual yield of most reactions is not equal to the theoretical yield for various reasons. Some reactions may be reversible so that the products change back to reactants, or some may not give optimum products because of a lack of proper experimental conditions. There may also be two or more competing reactions that affect the amount of product. Thus, the actual yield is always less than the theoretical yield.

The percent yield is the actual yield divided by the theoretical yield, multiplied by 100 percent.
Percent yield=Actual yieldTheoretical yield×100%{\text{Percent yield}} = \frac{{{\text{Actual yield}}}}{{{\text{Theoretical yield}}}} \times 100\%
The percent yield is important because it indicates whether a required or optimal goal of product has been achieved during a reaction.

The steps for calculating the percent yield are:

1. Write the balanced chemical equation.
2. Determine the mole ratio between the reactants and the products.
3. Use the mole ratio to calculate the mass of the products.
4. Use the masses of the reactants and the products to calculate the percent yield.

Step-By-Step Example
Determining Percent Yield in the Reaction of Copper and Silver Nitrate
Silver nitrate (AgNO3) reacts with copper (Cu) to produce silver (Ag) and copper nitrate, Cu(NO3)2. Calculate the percent yield when 80.0 grams (g) of silver is obtained from 140.0 g of silver nitrate.
Step 1

To determine the percent yield, both the actual yield and the theoretical yield of Ag are required.

The actual yield, 80.0 g, is given.

The theoretical yield can be calculated using the mass-to-mass method. The given mass of silver nitrate can be used to find the number of moles of silver nitrate. This can be used to find the number of moles of silver. The number of moles of silver can be used to find the theoretical yield of silver.

Step 2
Write the balanced equation.
Cu+2AgNO3Cu(NO3)2+2Ag\rm{Cu}+2{\rm{AgNO}}_3\rightarrow\rm{Cu}({\rm{NO}}_3)_2+2\rm{Ag}
Step 3
Calculate the molar mass of silver nitrate. Multiply the number of each atom in the formula by its molar mass.
Molar mass of AgNO3=(1)(107.87g/mol)+(1)(14.01g/mol)+(3)(16.00g/mol)=169.88g/mol\begin{aligned}\text{Molar mass of }{\rm AgNO_{3}}&=(1)(107.87\rm{\; g/mol})+(1)(14.01\rm{\; g/mol})+(3)(16.00\rm{\; g/mol})\\&=169.88\rm{\; g/mol}\end{aligned}
Step 4
Use the molar mass to convert the mass of AgNO3 to the number of moles.
Number of moles of AgNO3=140.0g169.88g/mol=0.8241mol\begin{aligned}\text{Number of moles of }{\rm AgNO_{3}}&=\frac{140.0\rm{\; g}}{169.88\rm{\; g/mol}}\\&=0.8241\rm{\; mol}\end{aligned}
Step 5
The balanced equation shows that the mole ratio of silver nitrate to silver is 2:2. Therefore, 0.8241 mol of silver is produced.
Step 6
Use the molar mass of silver to convert the number of moles of silver to mass.
Mass of Ag=(0.8241mol)(107.87g/mol)=88.90g\begin{aligned}\text{Mass of }{\rm Ag}&=(0.8241\rm{\; mol})(107.87\rm{\; g/mol})\\&=88.90\rm{\; g}\end{aligned}
To the correct number of significant figures, the theoretical yield is therefore 88.9 g Ag.
Solution
Use the actual yield and the theoretical yield to calculate the percent yield of silver.
Percent yield=Actual yieldTheoretical yield×100%=80.0g88.9g×100%=90.0%\begin{aligned}\text{Percent yield}&=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%\\&=\frac{80.0\rm{\; g}}{88.9\rm{\; g}}\times100\%\\&=90.0\%\end{aligned}
Step-By-Step Example
Determining Percent Yield in the Combustion of Sulfur Dioxide
When 815 grams (g) of sulfur dioxide (SO2) reacts with excess oxygen, 835 g of sulfur trioxide (SO3) is produced. Calculate the percent yield.
Step 1

To find the percent yield, both the actual yield and the theoretical yield of SO3 are required.

The actual yield, 835 g, is given.

The theoretical yield can be calculated using the mass-to-mass method. The given mass of sulfur dioxide can be used to find the number of moles of sulfur dioxide. This can be used to find the number of moles of sulfur trioxide. The number of moles of sulfur trioxide can be used to find the theoretical yield of sulfur trioxide.

Step 2
Write the balanced equation.
2SO2+O22SO32{\rm{SO}}_2+{\rm O}_2\rightarrow2{\rm{SO}}_3
Step 3
Calculate the molar mass of sulfur dioxide. Multiply the number of each atom in the formula by its molar mass.
Molar mass of SO2=(1)(32.06g/mol)+(2)(16.00g/mol)=64.06g/mol\begin{aligned}\text{Molar mass of }{\rm SO_{2}}&=(1)(32.06\rm{\; g/mol})+(2)(16.00\rm{\; g/mol})\\&=64.06\rm{\; g/mol}\end{aligned}
Step 4
Use the molar mass to convert the mass of SO2 to the number of moles.
Moles of SO2=Mass of SO2Molar mass of SO2=815g64.06g/mol=12.72mol\begin{aligned}\text{Moles of }{\rm SO_{2}}&=\frac{\text{Mass of }{\rm SO_{2}}}{\text{Molar mass of }{\rm SO_{2}}}\\&=\frac{815\rm{\; g}}{64.06\rm{\; g/mol}}\\&=12.72\rm{\; mol}\end{aligned}
To the correct number of significant figures, there are 12.7 mol SO2.
Step 5
The balanced equation shows that the mole ratio of sulfur dioxide to sulfur trioxide is 2:2. Therefore, 12.72 mol of SO3 is produced.
Step 6
Calculate the molar mass of sulfur trioxide. Multiply the number of each atom in the formula by its molar mass.
Molar mass of SO2=(1)(32.06g/mol)+(3)(16.00g/mol)=80.06g/mol\begin{aligned}\text{Molar mass of }{\rm SO_{2}}&=(1)(32.06\rm{\; g/mol})+(3)(16.00\rm{\; g/mol})\\&=80.06\rm{\; g/mol}\end{aligned}
Step 7
Use the molar mass of sulfur trioxide to convert the number of moles of sulfur trioxide to mass.
Mass of SO3=(12.72mol)(80.06g/mol)=1,018g\begin{aligned}\text{Mass of }{\rm SO_{3}}&=(12.72\rm{\; mol})(80.06\rm{\; g/mol})\\&=1\rm{,}018{\;g}\end{aligned}
The theoretical yield is therefore 1,018 g SO3.
Solution
Use the actual yield and the theoretical yield to calculate the percent yield of sulfur trioxide.
Percent yield=Actual yieldTheoretical yield×100%=835g1018g×100%=82.0%\begin{aligned}\text{Percent yield}&=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%\\&=\frac{835\rm{\; g}}{1018\rm{\; g}}\times100\%\\&=82.0\%\end{aligned}