Percent Yield and Limiting Reactant Calculator

Enter a balanced reaction and the amounts you started with to find the limiting reactant, the theoretical yield, how much excess reactant is left, and your percent yield. Type formulas such as Al, Cl2, or Ca(NO3)2 and the calculator finds each molar mass for you.

Reactant 1
Reactant 2
Product

Coefficients come from the balanced equation, here 2Al + 3Cl2 → 2AlCl3. Leave the actual yield blank to get only the theoretical yield.

For 2Al + 3Cl2 → 2AlCl3 with 10.0 g Al and 35.0 g Cl2, chlorine is limiting and the theoretical yield is 43.9 g AlCl3.

How to Use This Percent Yield Calculator

Formulas can include parentheses and hydrates, such as Ca(NO3)2 or CuSO4·5H2O. If you already know a molar mass, type the number instead of a formula. The equation has to be balanced first; if it is not, start with balancing the equation.

The Percent Yield Formula

Percent yield = (actual yield ÷ theoretical yield) × 100

How the Calculator Finds the Limiting Reactant

  1. Convert each reactant’s mass to moles using its molar mass.
  2. Use the coefficients to find how many moles of product each reactant could make on its own.
  3. The reactant that makes less product is limiting, and that amount is the theoretical yield.
  4. The other reactant is in excess; the leftover amount is what remains after the limiting reactant is used up.

Comparing grams directly gives the wrong answer, because reactants combine by moles. The full method, including the leftover excess and how it is scored, is worked through in AP Chemistry limiting reactant and percent yield.

Worked Example: Aluminum and Chlorine

2Al + 3Cl2 → 2AlCl3, starting with 10.0 g of Al and 35.0 g of Cl2. The lab collects 38.5 g of AlCl3.

StepAluminumChlorine
Moles10.0 ÷ 26.98 = 0.3706 mol35.0 ÷ 70.90 = 0.4937 mol
AlCl3 it could make0.3706 × 2/2 = 0.3706 mol0.4937 × 2/3 = 0.3291 mol
ResultIn excessLimiting

Theoretical yield: 0.3291 mol × 133.33 g/mol = 43.9 g AlCl3.
Aluminum left over: 0.3291 mol of Al is used (8.88 g), so 10.0 − 8.88 = 1.1 g remains.
Percent yield: 38.5 ÷ 43.9 × 100 = 87.7%.

Aluminum had the smaller mass but was not limiting, which is exactly why the comparison has to be made in moles.

Why Percent Yield Is Usually Below 100%

Comparing a measured value with an expected one is the same idea behind AP Chemistry percent error, which measures accuracy rather than how much product you recovered.

How the Calculator Handles Molar Masses and Sig Figs

Molar masses come from standard atomic masses, rounded to two decimal places for most elements as on the AP exam periodic table, so H2O is 18.02 g/mol. The theoretical yield is reported with the significant figures of the limiting reactant’s amount; coefficients are exact and never limit the answer. Leftover excess follows the subtraction rule, and percent yield uses the fewer significant figures of the two yields. To check any of these by hand, use the sig fig calculator, and for the mole conversions themselves see AP Chemistry molar mass. To see how stronger free-response answers could move your overall estimate, try the AP Chem Score Calculator.

Frequently Asked Questions

What is the formula for percent yield?

Percent yield = (actual yield ÷ theoretical yield) × 100. Both yields must be in the same unit, usually grams. If 38.5 g of product forms when 43.9 g was possible, the percent yield is 87.7%.

How do you find the limiting reactant?

Convert each reactant to moles, then use the mole ratio from the balanced equation to find how much product each one could make. The reactant that makes less product is the limiting reactant, and that smaller amount is the theoretical yield.

Can percent yield be over 100%?

Not for a pure, dry product. A percent yield above 100% usually means the product still contained water or another impurity when it was weighed, or that a measurement was wrong.

What is the difference between theoretical yield and actual yield?

The theoretical yield is the maximum amount of product the limiting reactant can make, calculated from the balanced equation. The actual yield is the amount you really collect and measure in the lab, which is usually less.

Is the limiting reactant the one with the smallest mass?

Not necessarily. What matters is moles compared through the coefficients of the balanced equation, not grams. A reactant with a larger mass can still run out first if its molar mass is large or the equation needs more of it.

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