What is percent yield and how do you calculate it?

Percent yield is a measure of how efficient a chemical reaction is. It compares the amount of product you actually made in the lab to the maximum amount you could have made under perfect conditions.

Think of it like baking cookies. If a recipe says it makes 24 cookies (your theoretical yield), but you drop some dough on the floor and end up with 20 cookies (your actual yield), your percent yield is the ratio of those two numbers. In chemistry, calculating percent yield tells us how much product was lost to side reactions, spills, or incomplete reactions.

The Theoretical Yield (The Perfect Scenario)

Before you can find the percent yield, you need to know the theoretical yield. This is the absolute maximum amount of product you can make, calculated using stoichiometry and the limiting reactant. It assumes every single molecule reacts perfectly with no mistakes, spills, or losses.

The Actual Yield (Reality)

The actual yield is the amount of product you actually recover when you do the experiment. This number cannot be calculated from a balanced equation; it must be measured on a scale in the lab or given to you outright in a word problem. It is almost always less than the theoretical yield.

The Formula

To calculate the percent yield, divide the actual yield by the theoretical yield, then multiply by 100. The formula looks like this: Percent Yield=(Actual YieldTheoretical Yield)×100%\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%.

Where Students Slip Up

The most common mistake is mixing up the actual and theoretical yields. Remember that the actual yield is what really happened (the measured mass), while the theoretical yield is the perfect math prediction. If your calculation gives you a percent yield over 100%100\%, double-check your math! If the math is right, it usually means your physical product is wet or contains impurities.

Worked through

You react 10.0 g10.0\text{ g} of hydrogen gas (H2H_2) with excess oxygen gas (O2O_2) to produce water (H2OH_2O). The balanced equation is 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O. In the lab, you collect 80.0 g80.0\text{ g} of water. What is the percent yield?

First, find the theoretical yield of water. Convert the mass of H2H_2 to moles: 10.0 g×(1 mol/2.02 g)=4.95 moles of H210.0\text{ g} \times (1\text{ mol} / 2.02\text{ g}) = 4.95\text{ moles of } H_2. Use the molar ratio to find moles of water: 4.95 moles of H2×(2 moles H2O/2 moles H2)=4.95 moles of H2O4.95\text{ moles of } H_2 \times (2\text{ moles } H_2O / 2\text{ moles } H_2) = 4.95\text{ moles of } H_2O. Convert moles of water to grams: 4.95 moles×18.02 g/mol=89.2 g of H2O4.95\text{ moles} \times 18.02\text{ g/mol} = 89.2\text{ g of } H_2O. This is the theoretical yield. Now, calculate the percent yield: (80.0 g/89.2 g)×100=89.7%(80.0\text{ g} / 89.2\text{ g}) \times 100 = 89.7\%. The percent yield is 89.7%89.7\%.

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Where this comes from: OpenStax Chemistry 2e, Chapter 4: Stoichiometry of Chemical Reactions · Khan Academy: Stoichiometry and molecular weight

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