Percentage Yield & Atom Economy

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Key facts & methods

Percentage yield

In real reactions, the actual yield is usually less than the theoretical maximum.
Theoretical yield: the maximum mass calculated from the equation using moles.
Actual yield: the mass of product actually obtained in the experiment.
Formula: percentage yield = (actual yield ÷ theoretical yield) × 100
A 100% yield means all the theoretical product was obtained — rarely achieved.
Reasons for low yield: reversible reactions, side reactions, product lost during purification/transfer.

Atom economy

Atom economy: the proportion of the molar mass of reactants that ends up in the desired product.
Formula: atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all products) × 100
High atom economy = less waste, more sustainable, more efficient use of resources.
Addition reactions have 100% atom economy (all atoms go into one product).
Substitution and decomposition reactions have lower atom economy (some atoms end up in waste products).

Industrial relevance

Industries want high yield AND high atom economy — maximises profit, minimises waste.
Low atom economy means costly raw materials are wasted.
Green chemistry aims to design reactions with high atom economy and minimal hazardous by-products.
Catalysts improve efficiency but do not directly change atom economy (they're not used up).

Worked examples

Percentage yield: theoretical yield of CaCO₃ = 25 g. Actual = 20 g. % yield = (20/25) × 100 = 80%.
Atom economy: CaCO₃ → CaO + CO₂. Desired product = CaO (Mᵣ = 56). Total = 56 + 44 = 100. Atom economy = 56/100 = 56%.
Addition: C₂H₄ + H₂ → C₂H₆. Only one product. Atom economy = 30/30 = 100%.

Exam questions — 5 questions · 12 marks · AQA 8462 style

3 marksCalculate percentage yieldAQA 8462 P2 style 🔢 Calculator

A student reacts 50 g of calcium carbonate (CaCO₃) with excess acid. The theoretical yield of carbon dioxide is 22 g. The student collects 17.6 g of CO₂. Calculate the percentage yield. (2 marks)

Hint: Percentage yield = (actual/theoretical) × 100. The theoretical yield is given — you don't need to calculate it.

+30 XP

3 marksCalculate atom economyAQA 8462 P2 style 🔢 Calculator

In the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, the desired product is iron (Fe). Calculate the atom economy. (Mᵣ: Fe=56, C=12, O=16) (3 marks)

Hint: Sum the Mᵣ of ALL products in the equation. Then: atom economy = (Mᵣ of desired product/total Mᵣ of all products) × 100.

+30 XP

1 markAddition reaction atom economyAQA 8462 style

What is the atom economy for the addition reaction: C₂H₄ + Br₂ → C₂H₄Br₂?

A 50%

B 75%

C 100%

D Cannot be calculated without molar masses

3 marksExplain low yieldAQA 8462 style

A student obtained a 72% yield of copper sulfate crystals from a reaction of copper oxide and sulfuric acid. Give THREE reasons why the yield was not 100%. (3 marks)

Hint: Think practically: what happens during the procedure that loses product? (Filtration, transfer, incomplete reaction, reversal, impurities.)

+30 XP

1 markWhy maximise atom economyAQA 8462 style

Why do industrial chemists try to use reactions with high atom economy?

A High atom economy reactions are always faster

B To reduce waste and cost — less raw material is lost as unwanted by-products

C High atom economy reactions always have higher percentage yields

D High atom economy reactions do not require catalysts

Percentage Yield & Atom Economy

Percentage yield

Atom economy

Industrial relevance

Worked examples

Module complete! 🎉