Percent Yield Calculator
Compare what you actually recovered with what the equation says you could have made. Solve for percent yield, actual yield or theoretical yield.
yield % = actual ÷ theoretical × 100
Solve for
%
Percent yield
%
3 sig figs: 82.2
- Rearrange: % = actual ÷ theoretical × 100
- Substitute: act = 15.2 g, th = 18.5 g (converted to SI units first).
- Result: % = 82.1622 %, rounded to 3 significant figures because the least precise input has 3.
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Worked example
A student reacts enough reactant to make a theoretical 18.5 g of product and recovers 15.2 g after drying. Percent yield = 15.2 ÷ 18.5 × 100 = 82.2 %.
If you know the percent yield you expect (from a published procedure, say 75 %) and need 10.0 g of product, work backwards: theoretical = 10.0 × 100 ÷ 75 = 13.3 g, then use the stoichiometry calculator to find the starting mass for 13.3 g.
Actual, theoretical and percent yield
- Theoretical yield: calculated, the most product possible from the limiting reactant.
- Actual yield: measured, the dry mass you weigh at the end.
- Percent yield: the ratio of the two, a measure of how efficient the procedure was.
Frequently asked questions
What is the percent yield formula?
Percent yield = (actual yield ÷ theoretical yield) × 100. Both yields must be in the same unit, usually grams.
How do I get the theoretical yield?
From stoichiometry: convert the limiting reactant to moles, apply the mole ratio to the product and convert to grams. The limiting reactant calculator does this for you.
Why is my percent yield less than 100 %?
Product is lost on filter paper and glassware, reactions may not go to completion or reach equilibrium, side reactions make other products, and some product stays dissolved in the solvent.
What does a percent yield over 100 % mean?
Almost always that the product is not pure or not dry: trapped water or solvent, or unreacted starting material, adds mass. Dry to constant mass and recalculate.
What is a good percent yield?
It depends on the reaction. In teaching labs 70–90 % for a simple precipitation is typical; multi-step organic syntheses multiply losses, so each step’s yield matters a lot.