Can percent yield be greater than 100%?

Theoretically no. If an experiment yields over 100%, it indicates the product contains impurities, unreacted starting materials, or remaining solvent (such as water).

What if the actual yield is given in moles instead of grams?

You can calculate the theoretical yield in moles and perform the division directly. Percent yield is a dimensionless ratio, so as long as the units of actual and theoretical yield match, the calculation is valid.

Do I need to calculate the theoretical yield for all products in the reaction?

No. You only calculate the theoretical yield for the specific product whose actual yield is provided.

How to calculate percent yield

Percent yield compares the actual amount of product obtained from a reaction to the maximum possible amount calculated from stoichiometry. It applies when evaluating the efficiency of a chemical reaction where side reactions, incomplete conversion, or physical losses occur.

The setup

Identify the balanced chemical equation for the reaction. Obtain the actual yield of the product from experimental data or the problem statement, and identify the limiting reactant if multiple reactant masses are given.

The steps

Calculate the moles of the limiting reactant. 2. Use the stoichiometric ratio from the balanced equation to find the theoretical moles of the target product. 3. Convert the theoretical moles of the product to mass (grams) using its molar mass to obtain the theoretical yield. 4. Divide the actual yield by the theoretical yield and multiply by 100 to get the percent yield.

Checking the result

Verify that both yields are in the same units before dividing. The final percent yield must logically fall between 0% and 100% for a pure, dry product.

Common errors

Failing to identify the limiting reactant will result in an incorrectly high theoretical yield. Another frequent error is inverting the formula by dividing the theoretical yield by the actual yield.

Worked example

10.0 g of hydrogen gas ( $H_2$ ) reacts with excess oxygen gas ( $O_2$ ) to produce 75.0 g of water ( $H_2O$ ). Calculate the percent yield. The balanced equation is $2H_2 + O_2 \rightarrow 2H_2O$ .

Actual yield = 75.0 g $H_2O$ . Molar mass of $H_2$ = 2.016 g/mol. Moles of $H_2$ = 10.0 g / 2.016 g/mol = 4.96 mol $H_2$ . The stoichiometric ratio of $H_2$ to $H_2O$ is 2:2, or 1:1. Theoretical moles of $H_2O$ = 4.96 mol. Molar mass of $H_2O$ = 18.015 g/mol. Theoretical yield = 4.96 mol * 18.015 g/mol = 89.35 g $H_2O$ . Percent yield = (75.0 g / 89.35 g) * 100 = 83.9%.

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References: OpenStax Chemistry 2e, Chapter 4: Stoichiometry of Chemical Reactions · Khan Academy, Stoichiometry and molecular composition