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K324 / 2027
Chemical Calculations

Lesson 6 of 6

Composition, formulae, yield and purity

Distinguish the composition of a compound from the success of a preparation.

O-Level 6092 (2026) / SEC G3 K324 (2027)

Pure only
  • I can calculate elemental mass percentages.Syllabus

    Syllabus K324, 4.2(d). Calculate elemental mass percentages

  • I can deduce empirical and molecular formulae.Syllabus

    Syllabus K324, 4.2(e). Deduce empirical and molecular formulae

  • I can calculate yield and purity.Syllabus

    Syllabus K324, 4.2(h). Calculate yield and purity

Composition, formulae, yield and purity

Distinguish the composition of a compound from the success of a preparation.

Percentage by mass of an element = (its total Ar contribution in the formula / Mr) x 100%. In CaCO3, Mr = 40 + 12 + 48 = 100, so oxygen contributes 48%. An empirical formula is the simplest whole-number atom ratio; a molecular formula gives actual atom counts in a molecule.

Worked exampleFrom percentage composition to molecular formula

A compound contains 40.0% C, 6.7% H and 53.3% O. Its Mr is 180.

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  1. Assume 100 g: C = 40.0/12 = 3.33 mol; H = 6.7/1 = 6.7 mol; O = 53.3/16 = 3.33 mol.
  2. Divide by the smallest: approximately 1:2:1, giving CH2O.
  3. Empirical formula mass = 30; multiplier = 180/30 = 6.
Answer

Molecular formula C6H12O6. If ratios are near halves, multiply every ratio by two instead of rounding one ratio alone.

Percentage yield = actual product/theoretical product x 100%. The theoretical yield comes from the limiting reactant. Percentage purity = mass of pure substance/mass of sample x 100%. A sample can be very pure even when the preparation gives a poor yield.

Worked exampleKeep the denominator meaningful

A reaction should produce 5.00 g of salt but gives 4.10 g. Separately, a 2.50 g limestone sample contains 2.00 g CaCO3.

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  1. Yield = 4.10/5.00 x 100 = 82.0%.
  2. Purity = 2.00/2.50 x 100 = 80.0%.
Answer

82.0% yield; 80.0% purity. Wet crystals can give an apparently excessive yield; impurities can change the measured sample mass.

Try it step by step

Work it out in steps: Moles of reactant, then Theoretical mass of product, then Percentage yield.

  1. Problem 1. Heating 5.00 g of calcium carbonate gives 2.24 g of calcium oxide. CaCO3(s) -> CaO(s) + CO2(g). Find the percentage yield to 3 significant figures. Use M(CaCO3) = 100 and M(CaO) = 56 g mol-1.
  2. Problem 2. 2.40 g of magnesium burns in excess oxygen and gives 3.60 g of magnesium oxide. 2Mg(s) + O2(g) -> 2MgO(s). Find the percentage yield to 3 significant figures. Use Mg = 24 and M(MgO) = 40 g mol-1.
  3. Problem 3. 4.80 g of magnesium reacts with excess dilute sulfuric acid. After the water is evaporated off, 21.0 g of dry anhydrous magnesium sulfate is obtained. Mg(s) + H2SO4(aq) -> MgSO4(aq) + H2(g). Give each answer to 3 significant figures. Use Mg = 24 and M(MgSO4) = 120 g mol-1.
Show the answer

Problem 1.

Moles of reactant: n(CaCO3) = 5.00 / 100 = 0.0500 mol.

Theoretical mass of product: The ratio is 1 CaCO3 to 1 CaO, so the theoretical mass of CaO = 0.0500 x 56 = 2.80 g.

Percentage yield: Percentage yield = actual / theoretical x 100 = 2.24 / 2.80 x 100 = 80.0%.

Problem 2.

Moles of reactant: n(Mg) = 2.40 / 24 = 0.100 mol.

Theoretical mass of product: The ratio is 2 Mg to 2 MgO, or 1:1, so the theoretical mass = 0.100 x 40 = 4.00 g.

Percentage yield: Percentage yield = 3.60 / 4.00 x 100 = 90.0%.

Problem 3.

Moles of reactant: n(Mg) = 4.80 / 24 = 0.200 mol.

Theoretical mass of product: The ratio is 1 Mg to 1 MgSO4, so the theoretical mass = 0.200 x 120 = 24.0 g.

Percentage yield: Percentage yield = 21.0 / 24.0 x 100 = 87.5%.

Worked exampleFind purity from the amount that reacts

A 2.80 g sample containing CaCO3 and inert impurities produces 600 cm3 CO2 with excess dilute acid at room temperature and pressure. Find its percentage purity. Use M(CaCO3) = 100 g mol-1; impurities do not produce gas.

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  1. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l). The carbonate:CO2 mole ratio is 1:1.
  2. 600 cm3 = 0.600 dm3; n(CO2) = 0.600/24 = 0.0250 mol.
  3. The sample contains 0.0250 x 100 = 2.50 g pure CaCO3.
  4. Percentage purity = 2.50/2.80 x 100 = 89.3%.
Answer

89.3% CaCO3 by mass. The denominator is the whole impure sample; use the gas to find the mass of the reactive pure component.

Check your understanding

Why can percentage composition alone not tell CH2O apart from C6H12O6?

  1. C6H12O6 has a higher percentage of carbon, but too close to measure.
  2. They have the same atom ratio, so their percentages by mass are identical.
  3. Percentages are reliable for ionic compounds, but not for molecules.
Show the answer

They have the same atom ratio, so their percentages by mass are identical.

C6H12O6 is six times CH2O, so the ratio of C : H : O is the same and so are the percentages. The relative molecular mass gives the multiplier for the molecular formula.