Whiteboard Chemistry with Joe White

Yield & Atom Economy

Chemistry-only: how much product you actually get versus the theoretical maximum, and how much of your starting material ends up as useful product.

AQA Specification Paper 1

Percentage Yield Triple

Even though no atoms are gained or lost, you can almost never collect the full calculated amount of product from a real reaction. The yield is the amount of product you actually obtain.

✅ Why yield is never 100%
  • the reaction may be reversible and not go to completion;
  • some product is lost when it is separated and purified (e.g. during filtration);
  • some reactants react in unexpected (side) reactions to give different products.

The percentage yield compares the actual yield with the maximum theoretical yield (the amount calculated from the balanced equation, assuming perfect conditions):

percentage yield = mass of product actually mademaximum theoretical mass of product× 100

H  Higher Tier students must also be able to calculate the theoretical mass of a product from a given mass of reactant and the balanced equation (using the reacting-mass method from section 8).

✅ Worked example — percentage yield

(a) A reaction has a maximum theoretical yield of 2.0 g, but only 1.6 g of product is actually collected. Find the percentage yield.

percentage yield = 1.62.0 × 100 = 80%

(b) H  25 g of calcium carbonate is heated until it fully decomposes. 12.6 g of calcium oxide is collected. Calculate the percentage yield. (Mr: CaCO₃ = 100, CaO = 56.)

CaCO₃ → CaO + CO₂

  1. Theoretical mass first. moles CaCO₃ = 25 ÷ 100 = 0.25 mol. Ratio is 1 : 1, so 0.25 mol of CaO should form: mass = 0.25 × 56 = 14 g (the theoretical yield).
  2. Then compare with the actual yield. percentage yield = 12.614 × 100 = 90%.
⚠️ Common mistake — don’t divide by 100 twice

To turn a fraction into a percentage you multiply by 100 — once. A frequent slip is to use the calculator’s % key (which already divides by 100) and then divide by 100 again, giving an answer 100× too small (e.g. 0.9% instead of 90%). Always sense-check: a percentage yield should sit between 0 and 100%.

🧪 Exam-style questions
Q1 [2 marks]

A reaction has a maximum theoretical yield of 6.0 g. A student actually collects 4.2 g of product. Calculate the percentage yield.

%
Show answer
  • percentage yield = actual yieldtheoretical yield × 100 = 4.26.0 × 100 1 mark = 70% 1 mark

The actual yield always goes on top, so the answer can never be more than 100%.

Q2 H [4 marks]

4.8 g of magnesium is burned completely in oxygen and 7.0 g of magnesium oxide is collected. Calculate the percentage yield of magnesium oxide.

2Mg + O₂ → 2MgO  ·  Ar: O = 16, Mg = 24
%
Show answer
  • moles of Mg = 4.8 ÷ 24 = 0.2 mol 1 mark
  • ratio Mg : MgO = 1 : 1, so theoretical mass of MgO = 0.2 × 40 = 8.0 g 1 mark
  • percentage yield = (7.0 ÷ 8.0) × 100 1 mark = 87.5% 1 mark

Atom Economy Triple

📖 Definition

Atom economy (atom utilisation) — a measure of the amount of the starting materials that end up as useful products. A high atom economy means little is wasted, which matters for sustainable development and for economic reasons.

percentage atom economy = Mr of the desired productsum of the Mr of all reactants× 100

Use the relative formula masses taken from the balanced equation, remembering to include the big numbers. A reaction that makes only the desired product has an atom economy of 100%; one that also makes by-products has a lower atom economy — although those by-products can sometimes be sold or used, which keeps the process economically worthwhile.

H  Higher Tier students should also be able to explain why a particular reaction pathway is chosen, weighing up atom economy alongside percentage yield, rate, equilibrium position and the usefulness of any by-products.

✅ Worked example — atom economy

Hydrogen is manufactured by reacting methane with steam (steam reforming). The desired product is the hydrogen. Calculate the atom economy. (Mr: CH₄ = 16, H₂O = 18, H₂ = 2.)

CH₄ + H₂O → CO + 3H₂

  1. Mr of the desired product: 3H₂ = 3 × 2 = 6.
  2. Sum of the Mr of all reactants: 16 + 18 = 34.
  3. Divide and × 100:

atom economy = 634 × 100 = 17.6%

This is a low atom economy — most of the starting mass ends up in the carbon monoxide by-product. Industrially this can still be worthwhile because the CO is itself useful and can be sold or used in other processes.

⚠️ Watch out — the big number counts

The most common mistake here is using Mr = 2 for the hydrogen instead of 3 × 2 = 6. The big number in 3H₂ tells you that three whole molecules of hydrogen are made, and every one of them counts towards the mass of useful product. Picture the actual particles on each side of the equation:

REACTANTS PRODUCTS C H H H H + O H H C O H H H H H H CH₄ H₂O CO 3H₂ Mr 16 Mr 18 Mr 28 2 + 2 + 2 = 6 Three H₂ molecules are made — the desired product is all of them, not one. Mr of desired product = 3 × 2 = 6
🧪 Exam-style questions
Q1 [2 marks]

Quicklime (calcium oxide) is made by heating limestone. Calculate the atom economy for producing calcium oxide.

CaCO₃ → CaO + CO₂  ·  Mr: CaCO₃ = 100, CaO = 56, CO₂ = 44
%
Show answer
  • atom economy = Mr of desired productsum of Mr of reactants × 100 = 56100 × 100 1 mark = 56% 1 mark

There is only one reactant here (CaCO₃), so the denominator is just its Mr.

Q2 [3 marks]

Iron is extracted from iron(III) oxide in the blast furnace. Calculate the atom economy for producing iron, where iron is the desired product.

Fe₂O₃ + 3CO → 2Fe + 3CO₂  ·  Mr: Fe₂O₃ = 160, CO = 28, Fe = 56, CO₂ = 44
%
Show answer
  • Mr of desired product = 2 × 56 = 112 1 mark
  • sum of Mr of reactants = 160 + (3 × 28) = 160 + 84 = 244 1 mark
  • atom economy = (112 ÷ 244) × 100 = 45.9% 1 mark

Remember to multiply each formula mass by its big number before adding — here 3 × CO.

Q3 [6 marks] T H

A company can manufacture the same product using one of two reactions:

Route A: high percentage yield, but a low atom economy (about half the starting atoms end up in a by-product)  ·  Route B: high atom economy and no useful by-product, but a low percentage yield and a slow reaction.

Evaluate the choice of reaction pathway, and decide which route the company should use. This is a levels-of-response question — weigh the factors against each other and reach a justified conclusion. Plan your answer, then compare it with the model answer.

Show a model answer

How it is marked (levels of response):

  • Level 3 (5–6): considers atom economy and yield (and at least one of rate or by-product use), then reaches a justified conclusion.
  • Level 2 (3–4): describes the trade-off using both atom economy and yield, but the conclusion is missing or not justified.
  • Level 1 (1–2): one or two simple relevant points about one factor.

In favour of Route B (high atom economy):

  • A high atom economy means less of the starting material is wasted as by-product, so it is more sustainable and uses raw materials efficiently.
  • With no by-product, there is less to separate and dispose of, which can lower costs and environmental impact.

In favour of Route A (high yield):

  • A high percentage yield means more product is made from each batch, so less reactant is needed for the same output.
  • Route A is also faster; a faster reaction makes more product per day, which matters for profit.
  • If the by-product from Route A can be sold or used, the low atom economy matters far less — little is truly wasted.

Conclusion (needed for Level 3) — any justified verdict scores: the better route depends on whether Route A's by-product is useful and on how slow Route B is. If the by-product can be sold, Route A is likely best, because its high yield and fast rate outweigh the lower atom economy. If the by-product is waste and sustainability is the priority, Route B is the better choice despite its low yield and slow rate. 6 marks

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