Percentage yield compares what you actually made with the theoretical yield from the equation: percentage yield = (actual yield / theoretical yield) x 100 %.
This page walks through percentage yield for SPM Chemistry, step by step: the formula you need, the units to watch, and a worked example.
When you use this
In real reactions you rarely collect all the product the equation promises. Percentage yield measures how close you came, and it appears in SPM Chemistry both as a calculation and as a reason why industrial processes are run the way they are.
The formula and units
percentage yield = (actual yield / theoretical yield) x 100 %
- The actual yield is the mass (or moles) you really obtained, given in the question.
- The theoretical yield is the maximum possible, worked out from the balanced equation assuming everything reacts perfectly.
Units. Both yields must be in the same unit (both in grams, or both in moles) so that the fraction cancels to a pure percentage. The answer is a percentage and can never sensibly be more than 100 %.
Worked example 1 (easy)
A reaction has a theoretical yield of 10.0 g of product, but only 8.0 g is collected. Find the percentage yield.
- percentage yield = (8.0 / 10.0) x 100 % = 80 %.
Worked example 2 (medium)
Calcium carbonate decomposes: CaCO3 → CaO + CO2. Heating 50 g of calcium carbonate gives 21 g of calcium oxide. Find the percentage yield. (Ca = 40, C = 12, O = 16)
- M(CaCO3) = 100 g mol−1, so n(CaCO3) = 50 / 100 = 0.50 mol.
- Ratio CaCO3 : CaO = 1 : 1, so theoretical n(CaO) = 0.50 mol.
- Theoretical mass = 0.50 x 56 = 28 g.
- percentage yield = (21 / 28) x 100 % = 75 %.
Worked example 3 (SPM level)
Ammonia is made by the Haber process: N2 + 3H2 → 2NH3. Reacting 28 g of nitrogen with excess hydrogen produces 23.8 g of ammonia. Find the percentage yield. (N = 14, H = 1)
- M(N2) = 28 g mol−1, so n(N2) = 28 / 28 = 1.0 mol.
- Ratio N2 : NH3 = 1 : 2, so theoretical n(NH3) = 2.0 mol.
- Theoretical mass = 2.0 x 17 = 34 g.
- percentage yield = (23.8 / 34) x 100 % = 70 %.
Why the yield is below the maximum
A yield below 100 % is normal, and the exam often asks you to explain it. The reaction may be reversible and reach equilibrium before finishing, as in the Haber process; some reactant may be lost in side reactions; and product is always lost during transfer, filtering or purifying. Knowing these reasons turns a bare number into a full-mark explanation, and it links directly to why industry adjusts temperature, pressure and catalysts to push the yield as high as is economic.
Yield in industry
Percentage yield is not only an exam sum; it decides whether a process makes money. A plant that converts more reactant into sellable product wastes less raw material and energy, so chemists raise the yield by recycling unreacted gases, removing product to shift an equilibrium, and choosing conditions that favour the forward reaction. When a Paper 2 question asks you to suggest how to improve a yield, these are the ideas the marking scheme is looking for, which is why percentage yield often appears alongside the manufactured-substances chapter rather than as a bare number exercise.
Common traps
- Mixing units. Compare mass with mass, or moles with moles, never one against the other.
- Forgetting the theoretical step. You must first calculate the theoretical yield from the equation; it is not simply the starting mass.
- A result above 100 %. This signals an error or an impure, still-wet product, not a brilliant reaction.
- Using actual yield as the denominator. The theoretical yield goes on the bottom.
Our teachers have students always write the theoretical yield calculation in full before the final line, because that is where the marks sit and where most yield questions are quietly won or lost.
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