ChemistryCore20 min read

Stoichiometry

Moles, equations and limiting reagents

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01

The mole is just a counting word

A dozen is 12. A mole is 6.022 × 10²³. That is all it is — a number chosen so that one mole of a substance weighs its relative formula mass in grams.

It exists because you cannot count atoms but you can weigh them. The mole is the bridge between the mass you measure on a balance and the particle counts a chemical equation talks about.

n = m / MN = n × 6.022×10²³n = moles · m = mass in g · M = molar mass in g/mol
02

Balanced equations are recipes

2H₂ + O₂ → 2H₂O does not say "two grams of hydrogen". It says two moles of hydrogen molecules react with one mole of oxygen molecules. The coefficients are a ratio of particles, never of mass.

Balancing enforces conservation of mass: atoms are rearranged, never created or destroyed. Count each element on both sides and they must match.

03

The limiting reagent

Reactants are almost never supplied in the exact ratio. Whichever runs out first stops the reaction dead — that is the limiting reagent, and it alone determines how much product you get. The rest is left over in excess.

The method: divide each reactant's moles by its coefficient. The smallest result is limiting. Then scale the product from that number.

Set H₂ = 5 and O₂ = 4. Dividing by coefficients gives 2.5 and 4 — hydrogen is limiting, you get 5 mol of water, and 1.5 mol of oxygen sits unused. Faded molecules are the leftovers.

04

Percentage yield

The calculation gives the theoretical yield — the perfect-world maximum. Real labs lose product to side reactions, incomplete reactions, and transfers between containers.

% yield = (actual / theoretical) × 100above 100% means your product is wet or impure, not that you beat chemistry

Worked example

Burn 4 g of hydrogen in excess oxygen. M(H₂) = 2 g/mol, so n = 4/2 = 2 mol H₂. The ratio 2H₂ : 2H₂O is 1:1, so you get 2 mol of water. M(H₂O) = 18 g/mol, giving 36 g theoretical. If you collect 30 g, the yield is 30/36 × 100 = 83%.

05

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound; the molecular formula is the actual number of each atom in a molecule. Ethene is C₂H₄ (molecular) but CH₂ (empirical). You find an empirical formula from masses or percentages by converting each element to moles and dividing by the smallest.

To get the molecular formula, work out the empirical formula mass and see how many times it divides into the true relative molecular mass.

moles of each element = mass ÷ Aᵣ → divide all by the smallestthe resulting ratio is the empirical formula
06

Moles in solutions and gases

For solutions, concentration links moles and volume: a 1 mol/dm³ solution has one mole of solute in every cubic decimetre (1000 cm³). For gases, one mole of any gas occupies the same volume at the same conditions — about 24 dm³ at room temperature and pressure. Together with n = m/M, these let you convert freely between mass, moles, solution volume and gas volume.

The number of particles in a mole is the Avogadro constant, 6.02 × 10²³ — so one mole of water contains 6.02 × 10²³ molecules.

c = n ÷ V(dm³)V(gas) = n × 24 dm³c = concentration (mol/dm³) · n = moles · V = volume · divide cm³ by 1000 for dm³
07

Atom economy (A Level)

A reaction can give a high percentage yield yet still waste most of its atoms as unwanted by-products. Atom economy measures how much of the reactant mass ends up in the desired product. A high atom economy means less waste — a central idea of green chemistry, and often more important industrially than yield alone.

atom economy = (Mr of desired product ÷ Σ Mr of all products) × 100%high atom economy = little waste; addition reactions are 100% atom-economical
08

A worked example

Worked example

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. (Ar: C = 12, H = 1, O = 16)

  1. Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O.Percentages become masses in a 100 g sample.
  2. Moles: C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33.moles = mass ÷ Ar for each element.
  3. Divide by the smallest (3.33): C = 1, H = 2, O = 1.This gives the simplest whole-number ratio.
  4. Empirical formula = CH₂O.The 1 : 2 : 1 ratio of C : H : O.

CH₂O (this is the empirical formula; the molecular formula could be C₂H₄O₂, C₆H₁₂O₆, etc., depending on the relative molecular mass).

Practice questions

6 questions · 24 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

4 · 10 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Calculate the number of moles in 8.0 g of methane, CH₄. (Ar: C = 12, H = 1)
Model answer

Mr of CH₄ = 12 + 4 = 16. Moles = mass ÷ Mr = 8.0 ÷ 16 = 0.50 mol.

Examiner tip. Always find the molar mass first, then moles = mass ÷ molar mass.

SQ2[2 marks]
Define the term relative molecular mass.
Model answer

The sum of the relative atomic masses of all the atoms shown in the molecular formula of a substance.

Examiner tip. For ionic compounds the equivalent term is relative formula mass — same calculation.

SQ3[3 marks]
Balance the equation: __ Al + __ O₂ → __ Al₂O₃.
Model answer

4 Al + 3 O₂ → 2 Al₂O₃.

Examiner tip. Balance the metal and oxygen by trial; using 2 Al₂O₃ gives 4 Al and 6 O, so 3 O₂.

SQ4[3 marks]
Calculate the concentration, in mol/dm³, of a solution made by dissolving 0.20 mol of sodium chloride in 500 cm³ of solution.
Model answer

V = 500 ÷ 1000 = 0.50 dm³. c = n ÷ V = 0.20 ÷ 0.50 = 0.40 mol/dm³.

Examiner tip. Convert cm³ to dm³ first (÷1000), then c = n ÷ V.

Long questions

2 · 14 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[7 marks]
Calcium carbonate decomposes: CaCO₃ → CaO + CO₂. (Ar: Ca = 40, C = 12, O = 16)
  1. Calculate the relative formula mass of CaCO₃. [1]
  2. Calculate the mass of CaO produced from 25 g of CaCO₃. [3]
  3. Calculate the volume of CO₂ produced at r.t.p. (1 mole of gas = 24 dm³). [3]
Mark scheme
  1. Mr(CaCO₃) = 40 + 12 + 48 = 100[1]
  2. Moles CaCO₃ = 25 ÷ 100 = 0.25 mol[1]
  3. Moles CaO = 0.25 (1:1 ratio); Mr(CaO) = 56[1]
  4. Mass CaO = 0.25 × 56 = 14 g[1]
  5. Moles CO₂ = 0.25 (1:1 ratio)[1]
  6. Volume = moles × 24 = 0.25 × 24[1]
  7. Volume CO₂ = 6.0 dm³[1]

Mr = 100; 14 g CaO; 6.0 dm³ CO₂.

Examiner tip. Convert to moles, use the equation's mole ratio, then convert back to mass (× Mr) or gas volume (× 24 dm³).

LQ2[7 marks]
Iron is extracted by reducing iron(III) oxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. (Ar: Fe = 56, O = 16, C = 12)
  1. Calculate the relative formula mass of Fe₂O₃. [1]
  2. Calculate the mass of iron produced from 320 tonnes of Fe₂O₃. [3]
  3. Calculate the atom economy of this reaction for making iron. [3]
Mark scheme
  1. Mr(Fe₂O₃) = 112 + 48 = 160[1]
  2. Moles Fe₂O₃ = 320 ÷ 160 = 2 (tonne-moles)[1]
  3. Moles Fe = 2 × 2 = 4; Ar(Fe) = 56[1]
  4. Mass Fe = 4 × 56 = 224 tonnes[1]
  5. Desired product mass in equation = 2 × 56 = 112; total product mass = 112 + 3 × 44 = 244[1]
  6. Atom economy = (112 ÷ 244) × 100[1]
  7. = 45.9% (≈ 46%)[1]

Mr = 160; 224 tonnes of iron; atom economy ≈ 46%.

Examiner tip. Atom economy uses the formula masses in the equation, not the actual masses.