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GCSE/Chemistry/AQA· Higher tier

C3.5Amounts of substance in equations (HT): using mole ratios to calculate masses of products and reactants

Notes

Using moles and balanced equations (HT)

Once you can convert between mass and moles, balanced equations let you predict the mass of product made from a known mass of reactant — or the mass of reactant needed to make a target product.

The four-step method

For any reaction question, follow this routine:

  1. Write the balanced equation.
  2. Convert known mass → moles (moles = mass ÷ M_r).
  3. Use the equation to find moles of the unknown (mole ratio).
  4. Convert moles → mass (mass = moles × M_r).

Worked exampleWorked example 1 — simple ratio

What mass of magnesium oxide is produced when 6 g of magnesium burns in oxygen?

Equation: 2Mg + O₂ → 2MgO

A_r: Mg = 24; M_r MgO = 40.

  1. moles Mg = 6 ÷ 24 = 0.25 mol.
  2. From the equation: 2 mol Mg → 2 mol MgO. Ratio 1:1, so moles MgO = 0.25 mol.
  3. mass MgO = 0.25 × 40 = 10 g.

Worked exampleWorked example 2 — non-1:1 ratio

What mass of carbon dioxide is produced when 5.6 g of iron(III) oxide is reduced by carbon monoxide?

Equation: Fe₂O₃ + 3CO → 2Fe + 3CO₂

A_r: Fe = 56; O = 16. M_r Fe₂O₃ = 160; M_r CO₂ = 44.

  1. moles Fe₂O₃ = 5.6 ÷ 160 = 0.035 mol.
  2. Mole ratio Fe₂O₃ : CO₂ = 1 : 3, so moles CO₂ = 0.035 × 3 = 0.105 mol.
  3. mass CO₂ = 0.105 × 44 = 4.62 g.

Worked exampleWorked example 3 — find a coefficient

When iron reacts with oxygen, 11.2 g iron forms 16.0 g iron oxide. Determine the formula of the oxide.

A_r Fe = 56; A_r O = 16.

  • moles Fe = 11.2 ÷ 56 = 0.2 mol.
  • mass O combined = 16.0 − 11.2 = 4.8 g.
  • moles O = 4.8 ÷ 16 = 0.3 mol.
  • ratio Fe : O = 0.2 : 0.3 = 2 : 3 → Fe₂O₃.

Worked exampleWorked example 4 — finding the balancing number

Sodium reacts with oxygen: Na + O₂ → Na₂O.

If 4.6 g of Na reacts completely:

  • moles Na = 4.6 ÷ 23 = 0.2 mol.

To form Na₂O, 2 Na are needed per O atom (so per ½ O₂). Na : O₂ ratio is 4 : 1.

So 4Na + O₂ → 2Na₂O is the balanced equation.

moles O₂ = 0.2 ÷ 4 = 0.05; moles Na₂O = 0.2 ÷ 2 = 0.1; mass Na₂O = 0.1 × 62 = 6.2 g.

Common mistakes

  • Skipping the balanced equation. Always write it first — coefficients are essential.
  • Using mass ratios instead of mole ratios. The equation gives mole ratios.
  • Forgetting to multiply by the coefficient. In 2Mg + O₂ → 2MgO, the Mg : MgO ratio is 1 : 1; in Fe₂O₃ + 3CO → 2Fe + 3CO₂, ratio Fe₂O₃ : Fe is 1 : 2.
  • Wrong M_r in step 4. Convert to the correct compound's M_r.

Links

Required for C3.6 (limiting reactants), C3.7 (concentration), C3.8 (yield/atom economy) and C3.9 (gas volumes).

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

Practice questions

Try each before peeking at the worked solution.

  1. Question 13 marks

    Mass of product (H)

    (H1) What mass of MgO is formed when 4.8 g of Mg burns in oxygen?
    2Mg + O₂ → 2MgO. Use Mg = 24, O = 16.

    [Higher tier — 3 marks]

    Ask AI about this

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

  2. Question 23 marks

    Mass of reactant (H)

    (H2) What mass of magnesium is needed to react with 9.8 g of sulfuric acid (H₂SO₄)?
    Mg + H₂SO₄ → MgSO₄ + H₂. Use Mg = 24, M_r(H₂SO₄) = 98.

    [Higher tier — 3 marks]

    Ask AI about this

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

  3. Question 33 marks

    Non-1:1 ratio (H)

    (H3) What mass of carbon dioxide is produced when 1.0 kg of CaCO₃ thermally decomposes?
    CaCO₃ → CaO + CO₂. M_r(CaCO₃) = 100; M_r(CO₂) = 44.

    [Higher tier — 3 marks]

    Ask AI about this

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

  4. Question 44 marks

    Find empirical from masses (H)

    (H4) When 5.6 g of iron reacts with sulfur, 8.8 g of iron sulfide is formed. Find the empirical formula. Use Fe = 56, S = 32.

    [Higher tier — 4 marks]

    Ask AI about this

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

  5. Question 53 marks

    Mass with 1:3 ratio (H)

    (H5) Calculate the mass of CO₂ produced when 16 g of Fe₂O₃ reacts with excess CO:
    Fe₂O₃ + 3CO → 2Fe + 3CO₂. M_r(Fe₂O₃) = 160; M_r(CO₂) = 44.

    [Higher tier — 3 marks]

    Ask AI about this

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

  6. Question 63 marks

    From the equation (H)

    (H6) Hydrogen is produced from natural gas: CH₄ + 2H₂O → CO₂ + 4H₂. Calculate the mass of H₂ from 8 g of CH₄. M_r(CH₄)=16; M_r(H₂)=2.

    [Higher tier — 3 marks]

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

  7. Question 73 marks

    Reverse calc (H)

    (H7) What mass of HCl is needed to react with 5.3 g of Na₂CO₃?
    Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. M_r(Na₂CO₃)=106; M_r(HCl)=36.5.

    [Higher tier — 3 marks]

    Ask AI about this

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

Flashcards

C3.5 — Mole ratios from equations

10-card SR deck on using mole ratios in calculations (HT).

10 cards · spaced repetition (SM-2)