Formulae, Equations and the Mole: Revision Notes

Condensed for the final weeks. For worked explanations, use the Formulae, Equations and the Mole study guide, and test yourself with the practice questions.

Every equation you need

moles          n = mass / Mr
mass           mass = n x Mr
concentration  c = n / V          (V in dm3)
moles from conc n = c x V
gas volume     V = n x 24 dm3     (room temp and pressure)
particles      N = n x 6.02 x 10^23
% yield        (actual / theoretical) x 100
% purity       (mass of pure / total mass) x 100

Unit trap: 1000 cm³ = 1 dm³. Divide cm³ by 1000 before using c = n / V.

The universal method

Almost every stoichiometry question is the same four steps:

1. Balance the equation.
2. Convert the known substance to moles.
3. Use the mole ratio from the balanced equation.
4. Convert the unknown moles back to mass, volume or concentration.

Skipping step 1 invalidates step 3, which is where most lost marks originate — an unbalanced equation gives the wrong mole ratio, and every later step inherits that error.

Common ions to know by heart

Charge	Ions
1+	Na⁺, K⁺, Li⁺, Ag⁺, NH₄⁺, H⁺
2+	Mg²⁺, Ca²⁺, Zn²⁺, Cu²⁺, Fe²⁺, Ba²⁺
3+	Al³⁺, Fe³⁺
1−	Cl⁻, Br⁻, I⁻, OH⁻, NO₃⁻, HCO₃⁻
2−	O²⁻, S²⁻, SO₄²⁻, CO₃²⁻
3−	PO₄³⁻, N³⁻

Writing a formula: swap the charges, then simplify. Ca²⁺ and NO₃⁻ → Ca(NO₃)₂. Brackets are needed whenever more than one polyatomic ion is present.

When the charges don’t cancel neatly, use the lowest common multiple. Al³⁺ and O²⁻: the LCM of 3 and 2 is 6, so 2 Al³⁺ balances 3 O²⁻, giving Al₂O₃. The compound must always end up electrically neutral overall.

Empirical formula — the routine

1. Write mass (or %) of each element
2. Divide each by its Ar          -> moles
3. Divide all by the smallest      -> simplest ratio
4. Multiply up to whole numbers if needed

Molecular formula = empirical formula × n, where n = (molecular mass) / (empirical mass).

Percentage composition by mass

A different question type from percentage yield: what fraction of a compound’s total mass comes from one element?

% composition = (mass of that element in the formula / Mr of the compound) x 100

Worked example. Percentage of oxygen by mass in CaCO₃ (Mr = 100, three oxygen atoms, Ar(O) = 16).

% O = (3 x 16 / 100) x 100 = 48%

Set out the numerator (the element’s total contribution) and the denominator (the whole compound’s Mr) explicitly before dividing — combining them in the wrong order is the usual slip.

Limiting reagent

Convert both reactants to moles, divide each by its coefficient in the balanced equation, and the smaller result is limiting. All product calculations use the limiting reagent — never the one in excess.

What’s Core, what’s Extended

For IGCSE 0620, Core candidates need only state that concentration can be measured in g/dm³ or mol/dm³, and can calculate reacting masses in simple proportions without using moles at all. Everything else on this page — the mole itself, molar gas volume, concentration calculations, empirical formula from data, and percentage yield/composition/purity — is Extended content. O Level 5070 candidates require all of it, with no Core/Extended split.

Exam traps

• Balancing must never change a formula — only add coefficients in front.
• Concentration in mol/dm³ requires volume in dm³, not cm³.
• 24 dm³ per mole applies to gases at rtp only.
• Percentage yield can never exceed 100%; if it does, recheck the theoretical value.
• Ar is for atoms, Mr for molecules and formula units — use the right one.
• State symbols are often worth a mark: (s), (l), (g), (aq).
• Using the LCM method but forgetting to check the final formula is electrically neutral overall.
• Confusing percentage composition (one element’s share of a compound’s mass) with percentage yield (actual product vs theoretical) — they use completely different numerators and denominators.

Self-test

1. Calculate the mass of 0.25 mol of CaCO₃. (Ar: Ca 40, C 12, O 16)
2. What volume does 0.5 mol of CO₂ occupy at rtp?
3. 25.0 cm³ of 0.1 mol/dm³ HCl — how many moles?
4. A compound is 40% C, 6.7% H, 53.3% O. Find the empirical formula.
5. Why must the equation be balanced before using a mole ratio?
6. Deduce the formula of aluminium oxide from the charges Al³⁺ and O²⁻.
7. Find the percentage by mass of oxygen in CaCO₃ (Ar: Ca 40, C 12, O 16).

Answers: 1. Mr = 40+12+48 = 100; mass = 0.25 × 100 = 25 g. 2. 0.5 × 24 = 12 dm³. 3. V = 0.025 dm³; n = 0.1 × 0.025 = 0.0025 mol. 4. 40/12 = 3.33, 6.7/1 = 6.7, 53.3/16 = 3.33; divide by 3.33 → 1 : 2 : 1 → CH₂O. 5. The ratio of coefficients is the mole ratio; an unbalanced equation gives the wrong ratio and every subsequent step is wrong. 6. LCM of 3 and 2 is 6, so 2 Al³⁺ balances 3 O²⁻ → Al₂O₃. 7. Mr(CaCO₃) = 100, mass of O = 3 × 16 = 48; % O = (48 ÷ 100) × 100 = 48%.