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GCSE/Chemistry/AQA

C3.2Relative formula mass and percentage by mass: calculating Mr and using it for percentage composition

Notes

Relative formula mass (M_r) and percentage by mass

Atoms are tiny; chemists work with relative masses rather than actual masses in grams. M_r is one of the most-used calculations in GCSE chemistry.

Relative atomic mass (A_r)

The relative atomic mass of an element is the average mass of its atoms compared to ¹⁄₁₂ the mass of a carbon-12 atom. Found on the periodic table — usually the larger number above the symbol (e.g. carbon 12, oxygen 16).

Why it isn't always a whole number: it's a weighted average of the masses of the isotopes, accounting for their abundance.

Relative formula mass (M_r)

For a compound, add up the A_r of every atom in the formula.

Worked examples:

  • H₂O: 2(1) + 16 = 18
  • CO₂: 12 + 2(16) = 44
  • NaCl: 23 + 35.5 = 58.5
  • Ca(OH)₂: 40 + 2(16 + 1) = 40 + 34 = 74 (don't forget brackets multiply both atoms inside)
  • MgSO₄·7H₂O (epsom salt with water of crystallisation): 24 + 32 + 4(16) + 7(2 + 16) = 24 + 32 + 64 + 126 = 246

Conservation of mass and M_r

In a balanced equation, the sum of M_r of reactants (multiplied by their coefficients) equals the sum of M_r of products.

For 2H₂ + O₂ → 2H₂O:

  • Reactant total: 2(2) + 32 = 36
  • Product total: 2(18) = 36 ✓

Percentage by mass of an element in a compound

The standard formula:

% mass = (n × A_r of element) ÷ M_r of compound × 100

where n is the number of atoms of that element in the formula.

Worked example 1: % nitrogen in NH₄NO₃ (ammonium nitrate).

  • N atoms: 2 (one in NH₄, one in NO₃)
  • M_r: 2(14) + 4(1) + 3(16) = 28 + 4 + 48 = 80
  • % N = (2 × 14)/80 × 100 = 28/80 × 100 = 35%

Worked example 2: % oxygen in CaCO₃ (chalk).

  • M_r: 40 + 12 + 3(16) = 40 + 12 + 48 = 100
  • % O = (3 × 16)/100 × 100 = 48/100 × 100 = 48%

Empirical vs molecular formula

  • Empirical formula — the simplest whole-number ratio of atoms (e.g. CH₂O for glucose).
  • Molecular formula — the actual number of atoms in a molecule (e.g. C₆H₁₂O₆ for glucose).

To find an empirical formula from masses or %:

  1. Mass (or %) of each element ÷ A_r → moles.
  2. Divide all moles by the smallest.
  3. Round to the nearest whole-number ratio (multiply if necessary, e.g. 1, 1.5, 2 → 2, 3, 4).

Worked example: a compound contains 40% C, 6.7% H, 53.3% O.

  • C: 40/12 = 3.33; H: 6.7/1 = 6.7; O: 53.3/16 = 3.33
  • Divide by 3.33: C = 1, H = 2, O = 1
  • Empirical formula: CH₂O

Common mistakes

  • Forgetting brackets multiply. Ca(OH)₂ has 2 O and 2 H from the (OH) bracket.
  • Confusing A_r and M_r. A_r is per atom; M_r is per formula.
  • Wrong number for water of crystallisation. ·5H₂O means 5 water molecules joined to each formula unit — multiply through.
  • Empirical formula simplifying error. Always divide by the smallest mole value.

Links

Required for C3.4 onwards (moles use M_r). Percentage composition is standard in C3.7 (concentration), C3.8 (atom economy) and C8.1 (purity).

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

Practice questions

Try each before peeking at the worked solution.

  1. Question 14 marks

    Calculate Mr (F)

    (F1) Calculate the relative formula mass of (a) H₂SO₄, (b) NaOH.

    Use: H = 1, S = 32, O = 16, Na = 23.

    [Foundation — 4 marks]

    Ask AI about this

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

  2. Question 22 marks

    Mr with brackets (F)

    (F2) Calculate the M_r of Ca(NO₃)₂.

    Use: Ca = 40, N = 14, O = 16.

    [Foundation — 2 marks]

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

  3. Question 33 marks

    % by mass (F/H)

    (F/H3) Calculate the percentage by mass of nitrogen in ammonia, NH₃.

    Use: N = 14, H = 1.

    [Crossover — 3 marks]

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

  4. Question 43 marks

    % nitrogen in fertiliser (H)

    (H4) Calculate the percentage by mass of nitrogen in urea, CO(NH₂)₂.

    Use: C = 12, O = 16, N = 14, H = 1.

    [Higher tier — 3 marks]

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

  5. Question 53 marks

    Empirical formula (H)

    (H5) A compound contains 27.3% carbon and 72.7% oxygen by mass. Find its empirical formula.

    Use: C = 12, O = 16.

    [Higher tier — 3 marks]

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

  6. Question 62 marks

    Hydrate Mr (H)

    (H6) Calculate the M_r of CuSO₄·5H₂O.

    Use: Cu = 63.5, S = 32, O = 16, H = 1.

    [Higher tier — 2 marks]

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

  7. Question 73 marks

    Conservation check (H)

    (H7) Show that the equation 2H₂ + O₂ → 2H₂O obeys the law of conservation of mass.

    Use: H = 1, O = 16.

    [Higher tier — 3 marks]

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Flashcards

C3.2 — Mr and percentage by mass

10-card SR deck on relative formula mass and composition.

10 cards · spaced repetition (SM-2)