What is my percent yield?
Type the mass of product you got and the mass the equation predicts. The percent yield calculator divides them. If you do not know the theoretical yield, it finds it from the limiting reactant and the balanced equation.
- Percent yield
- 77.29%
Your percent yield is 77.29%: 0.392 g of 0.5072 g.
- Theoretical yield (g)
- 0.5072
- Actual yield (g)
- 0.392
Percent yield: 77.29%. Your percent yield is 77.29%: 0.392 g of 0.5072 g.
How to calculate
Finds the percent yield of a reaction from the actual and theoretical yield, or works out the theoretical yield from the limiting reactant first.
Example with the default inputs (Theoretical yield I know it, Actual yield 0.392 g, Theoretical yield 0.5072 g): Your percent yield is 77.29%: 0.392 g of 0.5072 g.
Method: Percent yield = actual yield ÷ theoretical yield × 100. From the limiting reactant: theoretical yield = reactant mass ÷ reactant molar mass × (product coefficient ÷ reactant coefficient) × product molar mass.
- The reactant you type is the limiting reactant: the one that runs out first. Check it with the mole ratios before you use this mode.
- Molar masses are typed in g/mol; masses can be typed in g, mg, kg, oz or lb and are worked in grams.
- Each typed mass and molar mass is used as the exact decimal you typed; the result is rounded once.
Worked examples
Each example is checked against the calculator on every build.
- Theoretical yield I know it, Actual yield 0.392 g, Theoretical yield 0.5072 g gives Percent yield 77.287066%, Theoretical yield (g) 0.5072, Actual yield (g) 0.392.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields, https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields (Example 4.13: 0.392 g ÷ 0.5072 g × 100% = 77.3%)
- Theoretical yield From the limiting reactant, Actual yield 0.392 g, Mass of limiting reactant 1.274 g, Reactant molar mass 159.62, Reactant coefficient 1, Product coefficient 1, Product molar mass 63.55 gives Percent yield 77.283786%, Theoretical yield (g) 0.507222, Theoretical product (mol) 0.007981.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields, https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields (Example 4.13: 1.274 g CuSO₄ ÷ 159.62 g/mol × 63.55 g/mol = 0.5072 g Cu)
- Theoretical yield From the limiting reactant, Actual yield 12.5 g, Mass of limiting reactant 32.9 g, Reactant molar mass 153.81, Reactant coefficient 1, Product coefficient 1, Product molar mass 120.91 gives Percent yield 48.332189%, Theoretical yield (g) 25.862681.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields, https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields (Check Your Learning after Example 4.13: 48.3%)
- Theoretical yield From the limiting reactant, Actual yield 10 g, Mass of limiting reactant 4 g, Reactant molar mass 2, Reactant coefficient 2, Product coefficient 2, Product molar mass 18 gives Percent yield 27.777778%, Theoretical yield (g) 36, Theoretical product (mol) 2.Source: OpenStax, Chemistry 2e, §4.4 Reaction Yields, https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields
How it works
Percent yield = actual yield ÷ theoretical yield × 100.
When you choose From the limiting reactant, the page first finds the theoretical yield:
- Moles of product = reactant mass (g) ÷ reactant molar mass (g/mol) × product coefficient ÷ reactant coefficient.
- Theoretical yield (g) = moles of product × product molar mass (g/mol).
Units: masses typed in mg, kg, oz or lb are worked in grams (1 oz = 28.349523125 g, 1 lb = 453.59237 g, exactly).
Exact arithmetic. Each mass and molar mass is used as the exact decimal you typed, in the unit you typed it in, and every step is an exact fraction; the result is rounded once at the end.
Rules
- The actual yield can be 0 (0%); the theoretical yield, the reactant mass and the reactant mass must be more than 0. Masses are at most 1,000 kg; molar masses are from 0.1 to 100,000 g/mol; coefficients are whole numbers from 1 to 1,000.
- A percent yield over 100% is shown, with a note that a pure product cannot exceed its theoretical yield.
- The page has no answer when the result is too large to show as a number (a theoretical yield billions of times smaller than the actual yield).
Output format. Percent yield shows with at most 2 decimals; the theoretical yield, actual yield and moles of product with 6 significant figures. Each is rounded half up from its exact value.
Worked examples by hand
OpenStax Example 4.13. 0.392 g of copper collected, 0.5072 g expected: 0.392 ÷ 0.5072 × 100 = 77.29% (77.3% in the book).
The same reaction from the reactant. CuSO₄ + Zn → Cu + ZnSO₄, 1 : 1. 1.274 g ÷ 159.62 g/mol = 0.0079815 mol CuSO₄ = 0.0079815 mol Cu; × 63.55 g/mol = 0.50722 g. 0.392 ÷ 0.50722 × 100 = 77.28%.
OpenStax Check Your Learning. CCl₄ + 2HF → CF₂Cl₂ + 2HCl. 32.9 g ÷ 153.81 g/mol = 0.21390 mol CCl₄ → 0.21390 mol CF₂Cl₂ × 120.91 g/mol = 25.863 g. 12.5 ÷ 25.863 × 100 = 48.33% (48.3% in the book).
Water. 2H₂ + O₂ → 2H₂O with 4 g of H₂ (2 g/mol) limiting: 2 mol H₂ × 2/2 = 2 mol H₂O × 18 g/mol = 36 g. Collecting 10 g gives 10 ÷ 36 × 100 = 27.78%.
Other questions people ask
What is the percent yield formula?
Percent yield = actual yield ÷ theoretical yield × 100%. If a reaction should give 0.5072 g of copper and you collect 0.392 g, the percent yield is 0.392 ÷ 0.5072 × 100 = 77.3%.
How do I find the theoretical yield?
Convert the mass of the limiting reactant to moles (divide by its molar mass), multiply by the mole ratio from the balanced equation (product coefficient ÷ reactant coefficient), and convert to grams (multiply by the product’s molar mass).
Can percent yield be over 100%?
Not for a pure, dry product. A result over 100% usually means the product still holds water or solvent, has an impurity, or a mass was weighed or typed wrong. The page still shows the number and adds a note.
Why is percent yield usually below 100%?
Side reactions make other products, reactions may not go to completion, and some product is lost when it is filtered, moved or purified.
What is the limiting reactant?
The reactant that runs out first. It sets how much product can form. To find it, convert each reactant to moles and divide by its coefficient; the smallest result is the limiting reactant.
Which units can I use?
Masses in grams, milligrams, kilograms, ounces or pounds; molar masses in g/mol. Both yields must be the same substance, so the ratio has no unit.