Percentage yield GCSE chemistry questions compare the actual mass of product you collect against the maximum mass a reaction could theoretically produce, calculated as (actual ÷ theoretical) × 100. Atom economy is a related but different calculation, measuring what proportion of the reactants' total mass ends up in the useful product rather than in waste.
What is percentage yield?
In any real reaction, some product is always lost, through spillages, incomplete reactions, or side reactions producing other substances. Percentage yield compares the mass actually collected (the actual yield) with the maximum mass predicted by the balanced equation (the theoretical yield):
$$\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$
A percentage yield of 100% would mean every particle of reactant was converted perfectly into the desired product, with none lost — something that essentially never happens outside a calculation.
How do you calculate percentage yield, step by step?
- Write the balanced equation for the reaction and identify the desired product.
- Convert the mass of the limiting reactant into moles using $n = m \div M_r$.
- Use the mole ratio from the balanced equation to find the moles of product expected.
- Convert those moles of product into a theoretical mass using $m = n \times M_r$.
- Divide the actual mass obtained by the theoretical mass, then multiply by 100.
Worked example: 50g of calcium carbonate ($M_r$ = 100) is heated until it fully decomposes: $CaCO_3 \rightarrow CaO + CO_2$. A student collects 22g of calcium oxide ($M_r$ = 56). Calculate the percentage yield.
Moles of CaCO₃: $50 \div 100 = 0.5$ mol. The mole ratio is 1:1, so moles of CaO expected = 0.5 mol.
Theoretical mass of CaO: $0.5 \times 56 = 28$ g.
$$\text{Percentage yield} = \frac{22}{28} \times 100 = 78.6% \text{ (to 1 decimal place)}$$
Why is percentage yield almost never 100% in practice?
Real reactions lose product for several reasons: some of the mixture is left behind on equipment during filtering or transferring, the reaction may be reversible and not go to completion, unwanted side reactions can consume some of the reactants, and impure starting materials introduce substances that do not react as expected. GCSE mark schemes reward answers that name a specific practical cause of yield loss, rather than a vague statement that "some was lost."
What is atom economy?
Atom economy measures how much of the total mass of the reactants ends up in the useful, desired product, as opposed to unwanted by-products. It is calculated using relative formula masses from the balanced equation, not from an actual experiment:
$$\text{Atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all reactants}} \times 100$$
A reaction with a high atom economy wastes very little of its starting material as by-product, which matters for both cost and environmental impact in industrial chemistry.
How do you calculate atom economy, step by step?
- Write the balanced equation and identify which product is the desired one.
- Calculate the relative formula mass ($M_r$) of every reactant.
- Add the $M_r$ values of all reactants together to find the total starting mass.
- Calculate the $M_r$ of the desired product only.
- Divide the desired product's $M_r$ by the total reactant $M_r$, then multiply by 100.
Worked example: For the same reaction, $CaCO_3 \rightarrow CaO + CO_2$, calculate the atom economy if calcium oxide is the desired product.
There is only one reactant, CaCO₃, with $M_r = 100$. The desired product, CaO, has $M_r = 56$.
$$\text{Atom economy} = \frac{56}{100} \times 100 = 56%$$
This means 56% of the reactant's mass becomes the useful product, while the remaining 44% (as CO₂ gas) is a by-product, even though the percentage yield of that 56% portion could separately be anywhere up to 100%.
Why does atom economy matter for sustainable chemistry?
Industrial chemists prefer reactions with a high atom economy because a low atom economy means more raw material is wasted as by-product for every tonne of useful product made. This increases raw material costs, produces more waste that must be disposed of safely, and uses more energy overall. Choosing a reaction pathway with higher atom economy is one of the twelve principles of green chemistry, and GCSE questions often ask you to evaluate two possible routes to the same product on this basis.
Frequently asked questions
What is the key difference between percentage yield and atom economy?
Percentage yield is based on an actual experiment and compares the mass you really collected with the maximum mass the equation predicts, so it depends on practical losses. Atom economy is a purely theoretical calculation based only on the balanced equation and relative formula masses, and it does not depend on any experimental result at all. A reaction can have a high atom economy but a low percentage yield, or the reverse, because they measure different things.
Can percentage yield ever be more than 100%?
In a correctly performed calculation, percentage yield should not exceed 100%, since you cannot collect more product than the reaction can theoretically produce. A result above 100% in a real experiment usually means the collected product was impure or still wet, adding extra mass that was not the pure substance being measured. Exam questions sometimes ask you to explain this kind of anomalous result.
Why do some reactions have more than one possible desired product?
Some reactions produce two or more different products from the same reactants, and manufacturers choose which one to treat as the "desired" product based on which is more commercially useful. The atom economy calculation changes depending on which product is chosen as desired, because only that product's $M_r$ goes in the numerator. This is why atom economy questions always specify which product is wanted.
Do you need to know the mole ratio to calculate atom economy?
You need the mole ratio from the balanced equation only to make sure you are using the correct number of each reactant when adding up the total $M_r$, and to correctly identify the $M_r$ of the desired product. Unlike percentage yield, atom economy does not require converting between mass and moles for an actual sample — it works directly with the relative formula masses from the equation itself.
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