Percentage Yield Calculations

Percentage yield tells you how efficient a reaction was in practice compared to the theoretical maximum predicted by the balanced equation. It is a key concept for the Extended tier and connects directly to reacting masses calculations.

The method

1. Calculate the theoretical yield: the maximum mass of product possible from the given mass of reactant, using a reacting masses calculation.
2. Note the actual yield: the mass of product actually obtained (given in the question).
3. Apply the percentage yield formula.

The formula

percentage yield = (actual yield / theoretical yield) x 100

Where:

• actual yield is the mass of product actually obtained (g)
• theoretical yield is the maximum mass of product calculated from the equation (g)

Worked examples

Example 1: Simple percentage yield

Question: A student reacted 10 g of calcium carbonate with excess hydrochloric acid. The theoretical yield of calcium chloride is 11.1 g. The student obtained 8.9 g. Calculate the percentage yield.

Working:

• Percentage yield = (8.9 / 11.1) x 100 = 80.2%

Answer: 80.2%

Examiner note: When the theoretical yield is given in the question, the calculation is a single step. Show the substitution clearly.

Example 2: Calculating theoretical yield first

Question: 4.8 g of magnesium was burned in excess oxygen. The student collected 7.2 g of magnesium oxide. Calculate the percentage yield. (Ar: Mg = 24, O = 16)

Working:

• Balanced equation: 2Mg + O2 -> 2MgO
• Mr of Mg = 24, Mr of MgO = 40
• Moles of Mg = 4.8 / 24 = 0.2 mol
• Mole ratio 1:1, so moles of MgO = 0.2 mol
• Theoretical yield of MgO = 0.2 x 40 = 8 g
• Percentage yield = (7.2 / 8) x 100 = 90%

Answer: 90%

Examiner note: The theoretical yield must be calculated first using the reacting masses method. Show all steps.

Example 3: Iron from iron oxide

Question: 32 g of iron(III) oxide was reduced with excess carbon monoxide. 15.7 g of iron was obtained. Calculate the percentage yield. (Ar: Fe = 56, O = 16)

Working:

• Balanced equation: Fe2O3 + 3CO -> 2Fe + 3CO2
• Mr of Fe2O3 = (2 x 56) + (3 x 16) = 160
• Moles of Fe2O3 = 32 / 160 = 0.2 mol
• Mole ratio: 1 Fe2O3 : 2 Fe
• Moles of Fe = 0.2 x 2 = 0.4 mol
• Theoretical yield of Fe = 0.4 x 56 = 22.4 g
• Percentage yield = (15.7 / 22.4) x 100 = 70.1%

Answer: 70.1%

Examiner note: The mole ratio is 1:2 here. Make sure you double the moles when going from Fe2O3 to Fe.

Example 4: Ethanol production

Question: 18 g of glucose (C6H12O6) was fermented. 7.36 g of ethanol (C2H5OH) was produced. Calculate the percentage yield. (Ar: C = 12, H = 1, O = 16)

Working:

• Balanced equation: C6H12O6 -> 2C2H5OH + 2CO2
• Mr of C6H12O6 = (6 x 12) + (12 x 1) + (6 x 16) = 180
• Mr of C2H5OH = (2 x 12) + (6 x 1) + 16 = 46
• Moles of glucose = 18 / 180 = 0.1 mol
• Mole ratio: 1 glucose : 2 ethanol
• Moles of ethanol = 0.1 x 2 = 0.2 mol
• Theoretical yield = 0.2 x 46 = 9.2 g
• Percentage yield = (7.36 / 9.2) x 100 = 80%

Answer: 80%

Examiner note: Fermentation is a common context for percentage yield. The 1:2 ratio means one mole of glucose produces two moles of ethanol.

Example 5: Reverse question - finding actual yield

Question: The theoretical yield of copper from a reaction is 12.8 g. If the percentage yield is 75%, what mass of copper was actually obtained?

Working:

• Rearrange: actual yield = (percentage yield / 100) x theoretical yield
• Actual yield = (75 / 100) x 12.8 = 9.6 g

Answer: 9.6 g

Examiner note: The formula can be rearranged to find the actual yield when the percentage yield and theoretical yield are both given.

Common mistakes

1. Putting the numbers the wrong way round. Actual yield goes on top (numerator), theoretical yield on the bottom (denominator). If you get a percentage greater than 100%, you have them swapped.
2. Forgetting to calculate the theoretical yield. The theoretical yield is not the mass of reactant; it is the maximum mass of product from a reacting masses calculation.
3. Mixing up actual and theoretical. The actual yield is always the one measured in the experiment. The theoretical yield is always the one calculated from the equation.
4. Not using the mole ratio. The mole ratio is needed to calculate the theoretical yield. Skipping this step leads to an incorrect denominator.

When this appears

Percentage yield is Extended tier content and appears on Paper 4. It builds directly on reacting masses and often appears alongside percentage purity. See the stoichiometry topic for the underlying theory.