5.11C–5.12CPercentage yield
Chemistry only — not on the Combined Science specification
Even though no atoms are gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of a product because:
• the reaction may not go to completion because it is reversible • some of the product may be lost when it is separated from the reaction mixture • some of the reactants may react in ways different to the expected reaction.
Those three reasons are the answer to almost every "suggest why the yield was less than 100%" question, and they are worth learning as a list. Note what is not on it: atoms are never lost, so "some of the product disappeared" is not a reason.
The amount of a product obtained is known as the yield. When compared with the maximum theoretical amount as a percentage, it is called the percentage yield.
percentage yield = mass of product actually made ÷ maximum theoretical mass of product × 100
You should be able to calculate the percentage yield of a product from the actual yield of a reaction.
Worked example
A reaction has a maximum theoretical mass of 8.0 g of product. A student obtains 6.4 g. Calculate the percentage yield.
- Identify which mass is which: actual 6.4 g, theoretical 8.0 g.
- Divide the actual by the theoretical: 6.4 ÷ 8.0 = 0.8.
- Multiply by 100.
- Check the answer is below 100%, as it must be.
Answer: 80%
Common mistake
A percentage yield above 100% is impossible and means the calculation is upside down, or that the product was still wet when it was weighed. Check which mass you divided by.
- Percentage yield
- the mass of product actually obtained as a percentage of the maximum theoretical mass, which is less than 100% because reactions may be reversible, product is lost on separation, or there are side reactions
5.11CCalculating the theoretical massHigher tier
Chemistry only — not on the Combined Science specification
Higher tier students should also be able to calculate the theoretical mass of a product from a given mass of reactant and the balanced equation for the reaction — which means the percentage yield question may not hand you the theoretical mass at all.
The method is the mole calculation for amounts of substances in equations: convert the reactant mass to moles, apply the mole ratio from the balanced equation, and convert back to a mass.
Worked example
CaCO₃ → CaO + CO₂. A student heats 25.0 g of calcium carbonate and obtains 11.2 g of calcium oxide. Calculate the percentage yield. (Ar: Ca = 40, C = 12, O = 16.)
- Mr of CaCO₃ = 40 + 12 + 48 = 100; Mr of CaO = 40 + 16 = 56.
- Moles of CaCO₃ = 25.0 ÷ 100 = 0.25 mol.
- The ratio in the equation is 1 : 1, so the theoretical amount of CaO is 0.25 mol.
- Theoretical mass = 0.25 × 56 = 14.0 g.
- Percentage yield = 11.2 ÷ 14.0 × 100.
Answer: 80%
5.13C–5.14CAtom economy
Chemistry only — not on the Combined Science specification
The atom economy, or atom utilisation, is a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and for economic reasons to use reactions with high atom economy.
percentage atom economy = relative formula mass of the desired product from the equation ÷ sum of the relative formula masses of all reactants from the equation × 100
You should be able to calculate the atom economy of a reaction to form a desired product from the balanced equation.
- Percentage yield
- How much of the possible product was actually obtained, measured by experiment.
- Atom economy
- How much of the mass of the reactants ends up in the desired product, fixed by the equation alone.
The two are easy to confuse, so keep the distinction sharp. Atom economy is decided the moment you choose the reaction — it depends only on the balanced equation, and you could calculate it before doing any chemistry at all. Percentage yield depends on how the reaction is carried out, and you cannot know it until the product has been made and weighed.
A reaction with only one product has an atom economy of 100%, because every atom of the reactants ends up in the desired product. Any reaction that produces by-products has an atom economy below 100%, however carefully it is performed.
Worked example
Iron is extracted by Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the atom economy for producing iron. (Ar: Fe = 56, O = 16, C = 12.)
- Mr of the desired product, as it appears in the equation: 2Fe = 2 × 56 = 112.
- Mr of Fe₂O₃ = (2 × 56) + (3 × 16) = 160.
- Mr of 3CO = 3 × (12 + 16) = 84.
- Sum of all reactants = 160 + 84 = 244.
- Atom economy = 112 ÷ 244 × 100 = 45.9%.
Answer: 45.9% — less than half the mass of the reactants ends up as iron, the rest leaving as carbon dioxide.
Common mistake
Include the balancing numbers on both the top and the bottom. Using 56 for iron instead of 112, or the Mr of CO instead of 3CO, changes the answer completely. The masses must be the ones the equation actually specifies.
5.15CChoosing a reaction pathwayHigher tier
Chemistry only — not on the Combined Science specification
Higher tier students should be able to explain why a particular reaction pathway is chosen to produce a specified product, given appropriate data such as atom economy (if not calculated), yield, rate, equilibrium position and usefulness of by-products.
This is an evaluation question, not a calculation, and industry rarely picks the pathway that wins on every measure. Work through the factors you are given:
Worked example
Pathway A has 90% atom economy, a 40% yield and is very slow. Pathway B has 55% atom economy, a 95% yield, is fast, and its by-product is sold as a fertiliser. Which would industry choose?
- Compare on atom economy: A wastes less of the starting material, so A wins on that alone.
- Compare on yield and rate: B converts far more of what is put in, and does it faster, so B produces much more product per day.
- Consider the by-product: B's lower atom economy matters much less because the atoms that do not become the product are sold rather than discarded.
- Weigh them and commit to an answer, saying which factor decided it.
Answer: Pathway B. Its much higher yield and faster rate produce far more product for the same plant, and because the by-product is sold, the lower atom economy costs little — the waste is not really waste.
Before you practise
- percentage yield = (actual yield ÷ theoretical yield) × 100.
- Three reasons the actual yield falls short: the reaction is reversible and does not finish, some product is lost in transfer or filtering, and side reactions make something else.
- atom economy = (relative formula mass of the desired product ÷ total relative formula mass of all reactants) × 100.
- Atom economy is a property of the equation and never changes; percentage yield depends on how the reaction was run.
- A reaction with only one product has an atom economy of 100%.
- Choosing a pathway weighs atom economy, yield, rate, the usefulness of the by-products, and the cost and availability of the reactants.
- A high atom economy matters because it means less waste and less raw material used.