Calculations using moles and percentage composition
Every mole sum passes through number of moles: from mass, particles or gas volume. Percentage composition is (mass of element ÷ molar mass) × 100.
Three ways to find the number of moles
| You are given | Formula | Quick example |
|---|---|---|
| Mass in grams | n = mass ÷ molar mass | 69 g of Na: 69 ÷ 23 = 3 mol |
| Number of particles | n = particles ÷ 6.023 × 10²³ | 6.023 × 10²² atoms: 0.1 mol |
| Volume of a gas at STP | n = volume ÷ 22.4 L | 44.8 L of N₂: 44.8 ÷ 22.4 = 2 mol |
Worked example: 88 g of carbon dioxide
For 88 g of CO₂ (C = 12, O = 16), find (i) the number of moles (ii) the number of molecules (iii) the volume at STP.
- Molar mass of CO₂ = 12 + 2(16) = 44 g mol⁻¹
- (i) n = mass ÷ molar mass = 88 g ÷ 44 g mol⁻¹ = 2 mol
- (ii) Molecules = n × 6.023 × 10²³ = 2 × 6.023 × 10²³ = 1.2046 × 10²⁴
- (iii) Volume = n × 22.4 L = 2 × 22.4 L = 44.8 L
2 mol, 1.2046 × 10²⁴ molecules and 44.8 L at STP.
Worked example: from atoms to grams
Find the mass of 3.0115 × 10²³ atoms of sodium (Na = 23).
- Formula: n = number of atoms ÷ 6.023 × 10²³
- n = 3.0115 × 10²³ ÷ 6.023 × 10²³ = 0.5 mol
- Mass = n × molar mass = 0.5 mol × 23 g mol⁻¹ = 11.5 g
Mass of sodium = 11.5 g.
Worked example: percentage composition of urea
Find the percentage of each element in urea, CO(NH₂)₂ (C = 12, O = 16, N = 14, H = 1).
- Molar mass = 12 + 16 + 2(14) + 4(1) = 12 + 16 + 28 + 4 = 60 g
- % C = (12 ÷ 60) × 100 = 20.00%
- % O = (16 ÷ 60) × 100 = 26.67%
- % N = (28 ÷ 60) × 100 = 46.67%
- % H = (4 ÷ 60) × 100 = 6.67%
- Check: 20.00 + 26.67 + 46.67 + 6.67 = 100.01, which is 100% after rounding
Urea: C 20.00%, O 26.67%, N 46.67%, H 6.67% (rounded to two decimal places).
A urea fertiliser bag says it has about 46% nitrogen. That number comes from percentage composition: 28 g of nitrogen in every 60 g of urea. Farmers use it to decide how much urea a paddy field needs.
2 marks: “Calculate the number of moles in 4.6 g of sodium (Na = 23).” n = mass ÷ molar mass = 4.6 ÷ 23 = 0.2 mol.
4 marks: “Calculate the percentage composition of sodium hydroxide, NaOH (Na = 23, O = 16, H = 1).” Molar mass = 23 + 16 + 1 = 40 g. % Na = (23 ÷ 40) × 100 = 57.5%. % O = (16 ÷ 40) × 100 = 40%. % H = (1 ÷ 40) × 100 = 2.5%. Total 100%.
7 marks: “(a) Find the mass of 0.5 mole of CO₂ and the number of molecules in it. (b) Find the percentage of nitrogen in the fertiliser ammonium nitrate, NH₄NO₃ (N = 14, H = 1, O = 16).” (a) Mass = 0.5 × 44 = 22 g; molecules = 0.5 × 6.023 × 10²³ = 3.0115 × 10²³. (b) Molar mass = 2(14) + 4(1) + 3(16) = 80; % N = (28 ÷ 80) × 100 = 35%.
Mass, particles or litres: first turn it into moles. Then go wherever the question wants.