Chemical Calculations rests on two foundations: correct formulae with correctly balanced equations, and the mole, which counts particles by weighing them. The mole links what you measure in the lab (mass, gas volume and solution concentration) to the number of particles that actually react.
Almost every calculation question in Paper 1 and Paper 2 uses the same three steps. You convert the given quantity into moles, apply the mole ratio from the balanced equation, and convert the moles into the quantity asked for. Several definitions in this topic are also tested directly as 1–2 mark recall questions.
What these notes cover
- •State symbols and chemical formulae
- •Balancing chemical equations
- •Ionic equations
- •Relative atomic and molecular mass
- •The mole and the Avogadro constant
- •Molar gas volume at r.t.p.
- •Concentration and dilution
- •Empirical and molecular formula
- •Reacting masses and titration calculations
- •Limiting reactant
- •Percentage yield and purity
- •Common mistakes and exam technique
Key definitions
- Relative atomic mass (Aᵣ)
- The average mass of one atom of the element relative to one-twelfth of the mass of one atom of carbon-12. It is a ratio, so it has no units.
- Relative molecular mass (Mᵣ)
- The sum of the relative atomic masses of all the atoms shown in the formula. For ionic compounds it is called relative formula mass. It has no units.
- Mole and Avogadro constant
- One mole of a substance contains 6 × 10²³ particles (atoms, molecules or ions). This number is the Avogadro constant.
- Empirical formula
- The simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in one molecule.
- Limiting reactant
- The reactant that runs out first. The amount of product depends only on the limiting reactant, and the other reactant is in excess.
Key formulas
Moles from mass
n = m ÷ Mᵣ
n is the number of moles, m is the mass in g and Mᵣ is the molar mass in g/mol. Use this for solids and liquids.
Moles of gas at r.t.p.
n = V ÷ 24
V is in dm³. One mole of any gas occupies 24 dm³ (24 000 cm³) at r.t.p. This applies to gases only.
Moles in solution
n = c × V
c is the concentration in mol/dm³ and V is the volume of solution in dm³. Divide a volume in cm³ by 1000 first.
Dilution
c₁V₁ = c₂V₂
c₁ and V₁ are the values before dilution, and c₂ and V₂ are the values after. The moles of solute stay the same. Both volumes only need to be in the same unit.
Percentage by mass
% by mass of element = (number of atoms of element × Aᵣ) ÷ Mᵣ of compound × 100%
The percentages of all the elements in a compound must add up to 100%.
Percentage yield
% yield = (actual mass of product obtained ÷ theoretical mass of product) × 100%
Calculate the theoretical mass from the limiting reactant.
Worked example
25.0 cm³ of sodium hydroxide solution is exactly neutralised by 20.0 cm³ of 0.100 mol/dm³ sulfuric acid. Calculate the concentration of the sodium hydroxide solution in mol/dm³.
- 1Write the balanced equation: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l).
- 2Find the moles of the known solution: n(H₂SO₄) = 0.100 × (20.0 ÷ 1000) = 2.00 × 10⁻³ mol.
- 3Apply the mole ratio H₂SO₄ : NaOH = 1 : 2, so n(NaOH) = 2.00 × 10⁻³ × 2 = 4.00 × 10⁻³ mol.
- 4Convert the volume of NaOH: 25.0 cm³ = 25.0 ÷ 1000 = 0.0250 dm³.
- 5Calculate the concentration: c(NaOH) = 4.00 × 10⁻³ ÷ 0.0250 = 0.160 mol/dm³.
Watch out: Sulfuric acid is diprotic, so each mole of acid neutralises two moles of NaOH. Assuming a 1 : 1 ratio gives 0.0800 mol/dm³, which is exactly half the correct answer.
Common mistakes
Putting cm³ straight into n = c × V
Wrong: For 25.0 cm³ of 0.100 mol/dm³ acid: n = 0.100 × 25.0 = 2.50 mol.
Right: Convert first: 25.0 cm³ = 0.0250 dm³, so n = 0.100 × 0.0250 = 0.00250 mol.
Ignoring brackets when finding Mᵣ
Wrong: Mᵣ of Ca(OH)₂ = 40 + 16 + 1 = 57.
Right: The 2 multiplies everything inside the bracket: Mᵣ of Ca(OH)₂ = 40 + 2 × (16 + 1) = 74.
Rounding an awkward ratio in an empirical formula
Wrong: Rounding a ratio of Fe : O = 1 : 1.5 down to 1 : 1.
Right: Multiply the whole ratio by 2 to give 2 : 3, so the empirical formula is Fe₂O₃.
Questions students ask
How do you find the limiting reactant in chemistry?
Convert each reactant to moles, then divide each by its coefficient in the balanced equation. The smallest result is the limiting reactant, and you calculate all products from it. It is not always the reactant with the smaller mass or fewer moles.
What is the molar gas volume at r.t.p.?
One mole of any gas occupies 24 dm³ (24 000 cm³) at room temperature and pressure, so n = V ÷ 24 with V in dm³. It applies to gases only. Combined Chemistry only requires gas calculations at r.t.p.
How do you write an ionic equation?
Write the full balanced equation with state symbols. Then split every (aq) compound into its ions and cross out the spectator ions that appear unchanged on both sides. Precipitates, water, gases and solid metals stay whole.
What is the difference between empirical and molecular formula?
The empirical formula is the simplest whole-number ratio of atoms, and the molecular formula is the actual number of atoms in one molecule. For example, ethene is C₂H₄ with empirical formula CH₂. To find the multiplier, use n = Mᵣ ÷ empirical formula mass.
In the full PDF
- •Tables of common cations and anions, with the criss-cross method for ionic formulae worked through
- •Worked ionic equations for precipitation, neutralisation and displacement reactions
- •Further worked examples on hydrated salts, gas volumes, both concentration units and dilution
- •Empirical and molecular formula examples from percentage composition and from reacting masses
- •Limiting reactant, percentage yield and percentage purity examples, plus all six common exam mistakes with fixes
- •The five-line exam routine for full marks, section summaries and a whole-topic summary