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Chemical Calculations

Topic 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
Pure only

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 example

From percentage composition to molecular formula

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

  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 example

Keep 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.

  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.

Worked example

Find 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.

  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 understandingWhy cannot percentage composition alone distinguish CH2O from C6H12O6?Think it through, then reveal the answer
They have the same simplest atom ratio and percentages. Relative molecular mass supplies the multiplier needed for the molecular formula.