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GCSE/Chemistry/AQA· Higher tier

C3.8Yield and atom economy (HT): calculating percentage yield, percentage atom economy and judging reaction efficiency

Notes

Percentage yield and atom economy (HT)

Real reactions rarely produce the maximum predicted product. Two metrics judge how efficient a reaction is: percentage yield (how much product was actually made) and atom economy (how much of the reactant atoms ended up in the desired product).

Percentage yield

% yield = (actual yield ÷ theoretical yield) × 100

  • Actual yield = the mass of product really obtained (measured in lab).
  • Theoretical yield = the maximum mass calculable from the balanced equation, assuming 100% conversion.

Why is yield always < 100%?

  • The reaction may be reversible and not go to completion.
  • Side reactions produce other products.
  • Loss during transfer (filtering, decanting, washing).
  • Reactants impure to start with.
  • Loss as gases or vapours that escape.

Worked exampleWorked example — yield

Theoretical yield of MgO from 4.8 g of Mg is 8.0 g (using mole ratios; Mg = 24, MgO = 40).

If the actual yield is 6.4 g:

% yield = (6.4 ÷ 8.0) × 100 = 80%

Atom economy

Atom economy = (M_r of desired product ÷ M_r of all products) × 100

Or, equivalently, since reactant mass = product mass:

Atom economy = (M_r of desired product ÷ sum of M_r of all reactants × balancing numbers) × 100

Atom economy measures the proportion of reactant atoms that end up in the wanted product, ignoring reaction completeness. High atom economy = less waste.

Worked exampleWorked example — atom economy

Hydrogen production by reacting methane with steam:

CH₄ + 2H₂O → CO₂ + 4H₂

Suppose H₂ is the desired product.

  • Reactants total M_r = 16 + 2(18) = 16 + 36 = 52
  • Desired product mass = 4 × 2 = 8
  • Atom economy = (8 ÷ 52) × 100 = 15.4%

So most of the atoms (≈85%) end up as CO₂ — a low atom-economy process.

Comparison

A reaction can have high yield but low atom economy (lots of product made, but lots of waste byproduct). Or high atom economy but low yield (little waste in principle, but reaction inefficient).

Industrial chemistry tries to maximise both — e.g. by finding uses for byproducts, or designing reactions that combine atoms more efficiently.

Why atom economy matters — sustainability

  • Less waste → lower environmental impact.
  • Less raw material → lower cost.
  • Cleaner industrial processes.

The Haber process (covered in C10.10) has 100% atom economy: N₂ + 3H₂ → 2NH₃

Every atom ends up in the desired product (ammonia).

Worked exampleWorked example — combined

Ethanol can be made by hydration of ethene: C₂H₄ + H₂O → C₂H₅OH

  • M_r reactants: 28 + 18 = 46; M_r product C₂H₅OH = 46
  • Atom economy = 46/46 × 100 = 100%

Or by fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

  • M_r reactants: 180; desired = 2 × 46 = 92
  • Atom economy = 92/180 × 100 = 51.1%

Common mistakes

  • Confusing yield and atom economy. Yield = how much you got; atom economy = how much of what you started with became wanted product.
  • Forgetting balancing numbers when summing reactant or product masses.
  • Mixing up actual and theoretical in the yield formula.
  • Using moles instead of M_r in atom economy — atom economy uses masses.

Links

Foundation for C9.4 (carbon footprint) and C10 (sustainability). Heavily examined in synthesis questions.

AI-generated · claude-opus-4-7 · v3-deep-chemistry

Practice questions

Try each before peeking at the worked solution.

  1. Question 12 marks

    % yield (H)

    (H1) A student calculates a theoretical yield of 12.5 g of CuSO₄ but obtains 9.0 g. Calculate the % yield.

    [Higher tier — 2 marks]

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  2. Question 23 marks

    Reasons low yield (H)

    (H2) State three reasons why the actual yield in a chemical reaction may be less than the theoretical yield.

    [Higher tier — 3 marks]

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    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  3. Question 33 marks

    Atom economy calc (H)

    (H3) Calculate the atom economy of the reaction:
    Fe₂O₃ + 3CO → 2Fe + 3CO₂
    Desired product: Fe. Use Fe = 56, C = 12, O = 16.

    [Higher tier — 3 marks]

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    AI-generated · claude-opus-4-7 · v3-deep-chemistry

  4. Question 43 marks

    Why atom economy matters (H)

    (H4) Explain why high atom economy is important in industry.

    [Higher tier — 3 marks]

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  5. Question 53 marks

    Compare reactions (H)

    (H5) Two routes to make ethanol: A) C₂H₄ + H₂O → C₂H₅OH; B) C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Which has higher atom economy and why?

    [Higher tier — 3 marks]

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  6. Question 62 marks

    Find theoretical (H)

    (H6) A student gets 4.5 g of CaO from limestone (CaCO₃ → CaO + CO₂). The percentage yield is 80%. Calculate the theoretical yield.

    [Higher tier — 2 marks]

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  7. Question 72 marks

    100% atom economy (H)

    (H7) Why does the Haber process (N₂ + 3H₂ → 2NH₃) have 100% atom economy?

    [Higher tier — 2 marks]

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Flashcards

C3.8 — Yield and atom economy

10-card SR deck on % yield and atom economy (HT).

10 cards · spaced repetition (SM-2)