Theoretical Yield – IGCSE Chemistry Definition

Theoretical yield is the maximum mass of product that can be obtained from a reaction, calculated from the stoichiometry of the balanced equation. It assumes that the reaction goes to completion, that the limiting reagent is entirely converted to product, and that no product is lost during purification or transfer. In practice, the actual yield is always less than the theoretical yield.

How to calculate theoretical yield

The calculation follows the standard mole method. First, identify the limiting reagent and calculate its number of moles. Second, use the molar ratio from the balanced equation to find the number of moles of the desired product. Third, multiply by the molar mass (Mr) of the product to obtain the theoretical yield in grams.

For example, in the reaction 2Mg + O₂ → 2MgO, if 4.8 g of magnesium reacts with excess oxygen, then moles of Mg = 4.8/24 = 0.20 mol. The molar ratio of Mg to MgO is 1:1, so 0.20 mol of MgO is expected. Theoretical yield = 0.20 x 40 = 8.0 g.

Relationship to percentage yield

Theoretical yield is the denominator in the percentage yield formula. Percentage yield = (actual yield / theoretical yield) x 100. A percentage yield of 100% would mean the actual yield equals the theoretical yield, which rarely happens in practice.

Exam context

On Paper 4, candidates are often given a mass of reactant and asked to calculate the mass of product formed. This calculated mass is the theoretical yield. A follow-up part then gives the actual yield and asks for the percentage yield. Setting out each step clearly is essential for earning all available marks.

Worked exam question

Calculate the theoretical yield of iron from 320 g of iron(III) oxide. Fe₂O₃ + 3CO → 2Fe + 3CO₂. (Ar: Fe = 56, O = 16) (3 marks)

Mr of Fe₂O₃ = (2 x 56) + (3 x 16) = 160 [1]. Moles of Fe₂O₃ = 320/160 = 2.0 mol. Molar ratio Fe₂O₃ to Fe = 1 : 2, so moles of Fe = 4.0 mol [1]. Theoretical yield = 4.0 x 56 = 224 g [1].

Common mistakes

Candidates sometimes forget to identify the limiting reagent when both reactant masses are given. The theoretical yield must be based on the limiting reagent, not the reagent in excess. Another common error is using the wrong molar ratio, which produces an incorrect theoretical yield and cascades into a wrong percentage yield.