Module A7 · CSEC Chemistry
The Mole Concept

Molar masses, Avogadro's constant, gas volumes, concentrations, empirical formulae, and stoichiometry — the calculation toolkit every chemist needs.

Section 1 — Relative Molecular & Formula Masses

Atoms are too tiny to weigh directly, so chemists compare masses to carbon-12. These ratios have no units.

Key Definitions Relative atomic mass (Aᵣ): Average mass of one atom compared to 1/12 mass of a C-12 atom. No units.
Relative molecular mass (Mᵣ): For covalent molecules — sum of all Aᵣ values.
Relative formula mass (Mᵣ): For ionic compounds — sum of all Aᵣ values in one formula unit.
Worked Examples — Calculating Mᵣ Mᵣ(H₂O) = (2×1) + 16 = 18 Mᵣ(H₂SO₄) = (2×1) + 32 + (4×16) = 98 Mᵣ(C₆H₁₂O₆) = (6×12) + (12×1) + (6×16) = 180 Mᵣ(MgCO₃) = 24 + 12 + (3×16) = 84 Mᵣ(Ca₃(PO₄)₂) = (3×40) + (2×31) + (8×16) = 310 Mᵣ((NH₄)₂CO₃) = (2×14)+(8×1)+12+(3×16) = 96

Section 2 — The Mole & Avogadro's Constant

A mole is chemistry's counting unit. Just as 'a dozen' = 12, a 'mole' = 6.0 × 10²³ particles. This number is called Avogadro's constant (Nₐ).

🧠 What Counts as a 'Particle'? For elements: atoms. For covalent compounds: molecules. For ionic compounds: formula units.
Examples: 40 g Ca = 1 mol = 6.0×10²³ Ca atoms · 28 g N₂ = 1 mol = 6.0×10²³ molecules · 180 g glucose = 1 mol
n = m / Mmoles = mass ÷ molar mass
N = n × Nₐparticles = moles × 6.0×10²³
n = V / Vₘmoles = volume ÷ molar volume

Section 3 — Mole-Mass Calculations

Molar mass (M) is the mass of 1 mole of a substance in g mol⁻¹. Its numerical value equals the relative mass.

Worked Examples Example 1: Moles in 10 g of CaCO₃? M(CaCO₃) = 40 + 12 + (3×16) = 100 g/mol n = 10/100 = 0.1 mol Example 2: Mass of 0.2 mol H₂SO₄? M(H₂SO₄) = (2×1) + 32 + (4×16) = 98 g/mol m = 0.2 × 98 = 19.6 g Example 3: Mass of 0.25 mol Na₂SO₄? M(Na₂SO₄) = (2×23) + 32 + (4×16) = 142 g/mol m = 0.25 × 142 = 35.5 g

Section 4 — Moles & Gas Volumes (Avogadro's Law)

Avogadro's Law Equal volumes of ALL gases at the same temperature and pressure contain the same number of molecules.

Molar Volume at stp (0°C, 101.3 kPa): Vₘ = 22.4 dm³ = 22 400 cm³
Molar Volume at rtp (25°C, 101.3 kPa): Vₘ = 24.0 dm³ = 24 000 cm³
Worked Examples — Gas Volumes Example 1: Volume of 0.25 mol N₂ at stp? V = 0.25 × 22.4 = 5.6 dm³ Example 2: Moles in 2.4 dm³ O₂ at rtp? n = 2.4/24.0 = 0.1 mol Example 3: Volume of 6.4 g O₂ at stp? n = 6.4/32 = 0.2 mol → V = 0.2 × 22.4 = 4.48 dm³ Example 4: Mass of 600 cm³ CO₂ at rtp? n = 600/24000 = 0.025 mol → m = 0.025 × 44 = 1.1 g
💡 Exam Tip — stp vs rtp Always check which conditions are given! Use 22.4 dm³ at stp (0°C) and 24.0 dm³ at rtp (25°C). This is the most common mistake on mole calculations in the exam.

Section 5 — Concentration of Solutions

Concentration tells us how much solute is dissolved in a given volume of solution.

c = m / V(dm³)Mass concentration — unit: g dm⁻³
c = n / V(dm³)Molar concentration — unit: mol dm⁻³
Worked Examples — Concentration Example 1: 10 g NaOH in 1 dm³. Find concentrations. Mass conc = 10.0 g dm⁻³ M(NaOH) = 40 → n = 10/40 = 0.25 mol Molar conc = 0.25 mol dm⁻³ Example 2: 6 g NaOH in 200 cm³. Find molar concentration. n = 6/40 = 0.15 mol in 200 cm³ In 1000 cm³: (0.15/200) × 1000 = 0.75 mol dm⁻³ Example 3: Mass of K₂CO₃ needed for 250 cm³ at 0.2 mol dm⁻³? Moles needed = (0.2/1000) × 250 = 0.05 mol M(K₂CO₃) = 138 g/mol → m = 0.05 × 138 = 6.9 g

Section 6 — Empirical & Molecular Formulae

Key Definitions Empirical formula: Simplest whole-number mole ratio of atoms. Always used for ionic compounds.
Molecular formula: Actual number of each atom in one molecule (covalent only).
To find molecular formula: n = Mᵣ(compound) ÷ Mᵣ(empirical), then multiply empirical by n.

Steps to Find Empirical Formula

  • 1Write the mass (g) or percentage (%) of each element.
  • 2Divide each mass by its molar mass to get moles.
  • 3Divide ALL mole values by the SMALLEST mole value.
  • 4Round to nearest whole number for the empirical formula. (If you get 1.5, multiply all by 2; if 1.33, multiply all by 3.)
Worked Examples Example 1: K=6.52g, Cr=4.34g, O=5.34g Moles: K=6.52/39=0.167 Cr=4.34/52=0.083 O=5.34/16=0.333 Ratio (/0.083): K=2 Cr=1 O=4 Empirical formula: K₂CrO₄ Example 2: 40%C, 6.7%H, 53.3%O (Mᵣ=180) Assume 100g: C=40g, H=6.7g, O=53.3g Moles: C=40/12=3.33 H=6.7/1=6.7 O=53.3/16=3.33 Ratio: C=1 H=2 O=1 → Empirical: CH₂O (Mᵣ=30) n = 180/30 = 6 → Molecular formula: C₆H₁₂O₆ (glucose!)
% Composition Formula % element = (mass of element in 1 mol ÷ molar mass) × 100

Example: % H in H₂O = (2/18) × 100 = 11.1%
% N in (NH₄)₂SO₄: M=132, N=2×14=28g → (28/132)×100 = 21.2%

Section 7 — Stoichiometry (Mole Ratios in Reactions)

The coefficients in a balanced equation tell you the mole ratio of reactants and products. Use this to link a known quantity to an unknown quantity.

The 4-Step Method (Works for ALL Stoichiometry)

  • 1Find moles of known substance: n = m/M, or n = V/Vₘ, or n = c × V/1000
  • 2Use the mole ratio from the balanced equation to find moles of unknown.
  • 3Convert moles of unknown to the required quantity (mass / volume / concentration).
  • 4State the answer with correct units.
Worked Examples — Stoichiometry Mass-mass: 12 g Mg burns in O₂. Mass of MgO produced? Equation: 2Mg + O₂ → 2MgO (ratio Mg:MgO = 1:1) n(Mg) = 12/24 = 0.5 mol → n(MgO) = 0.5 mol m(MgO) = 0.5 × 40 = 20 g Mass-volume: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) For 2.8 g C₂H₄: n(C₂H₄) = 2.8/28 = 0.1 mol Ratio C₂H₄:CO₂ = 1:2 → n(CO₂) = 0.2 mol m(CO₂) = 0.2 × 44 = 8.8 g V(CO₂) at stp = 0.2 × 22.4 = 4.48 dm³ Concentration: 50 cm³ NaOH (2.0 mol dm⁻³) + excess H₂SO₄ 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O (ratio 2:1:1) n(NaOH) = (2.0/1000) × 50 = 0.1 mol n(Na₂SO₄) = 0.1/2 = 0.05 mol m(Na₂SO₄) = 0.05 × 142 = 7.1 g
🌍 Real World — Fertiliser Manufacturing Ammonium nitrate (NH₄NO₃, Mᵣ=80) fertilisers are manufactured using stoichiometry calculations. Caribbean sugar and banana industries depend on precisely-formulated fertilisers — the calculations you just learned determine how much of each reactant to use at industrial scale.

Section 8 — Resources & Simulations

🔬 PhET — Reactants & Products ▶ Mole Concept Videos ▶ Empirical Formula Videos ▶ Stoichiometry Videos 📄 BBC Bitesize — Moles

Section 9 — CSEC Practice Questions

Question 1 — Molar Mass & Moles
Calculate: (a) the relative formula mass of Ca(HCO₃)₂   (b) the number of moles in 8.28 g of K₂CO₃   (c) the mass of 0.4 mol Zn(OH)₂
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a Ca(HCO₃)₂: Ca=40, H=1×2=2, C=12×2=24, O=16×6=96
Total = 40 + 2 + 24 + 96 = 162

b M(K₂CO₃) = (2×39) + 12 + (3×16) = 78 + 12 + 48 = 138 g/mol
n = 8.28/138 = 0.06 mol

c M(Zn(OH)₂) = 65 + (2×16) + (2×1) = 65 + 32 + 2 = 99 g/mol
m = 0.4 × 99 = 39.6 g

✅ (a) 162   (b) 0.06 mol   (c) 39.6 g
Question 2 — Gas Volumes
(a) What volume does 0.075 mol O₂ occupy at rtp?   (b) What mass is 1792 cm³ of NH₃ at stp?   (c) How many molecules are in a 4.8 dm³ sample of CO₂ at rtp?
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a At rtp, Vₘ = 24.0 dm³. V = 0.075 × 24.0 = 1.8 dm³

b At stp, Vₘ = 22.4 dm³ = 22400 cm³. n = 1792/22400 = 0.08 mol
M(NH₃) = 14 + 3 = 17 g/mol. m = 0.08 × 17 = 1.36 g

c At rtp: n = 4.8/24.0 = 0.2 mol
N = 0.2 × 6.0×10²³ = 1.2 × 10²³ molecules

✅ (a) 1.8 dm³   (b) 1.36 g   (c) 1.2 × 10²³ molecules
Question 3 — Empirical Formula
A compound contains 54.5% C, 9.1% H, and 36.4% O by mass. Its relative molecular mass is 88. Find: (a) the empirical formula, (b) the molecular formula.
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1 Assume 100 g: C = 54.5 g, H = 9.1 g, O = 36.4 g

2 Find moles: C = 54.5/12 = 4.54 · H = 9.1/1 = 9.1 · O = 36.4/16 = 2.275

3 Divide by smallest (2.275): C = 4.54/2.275 = 2 · H = 9.1/2.275 = 4 · O = 2.275/2.275 = 1

a Empirical formula: C₂H₄O   Mᵣ(empirical) = 24 + 4 + 16 = 44

b n = 88/44 = 2 → Molecular formula = 2 × C₂H₄O = C₄H₈O₂

✅ (a) Empirical: C₂H₄O   (b) Molecular: C₄H₈O₂
Question 4 — Stoichiometry (Mass-Mass)
When limestone (CaCO₃) is heated, it decomposes: CaCO₃(s) → CaO(s) + CO₂(g). What mass of CO₂ is released when 300 g of CaCO₃ is completely decomposed?
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1 M(CaCO₃) = 40 + 12 + (3×16) = 100 g/mol
n(CaCO₃) = 300/100 = 3.0 mol

2 Mole ratio from equation: CaCO₃ : CO₂ = 1 : 1
∴ n(CO₂) = 3.0 mol

3 M(CO₂) = 12 + (2×16) = 44 g/mol
m(CO₂) = 3.0 × 44 = 132 g

✅ 132 g of CO₂ is released.
Question 5 — Concentration & Stoichiometry
25.0 cm³ of 0.40 mol dm⁻³ HCl reacts completely with excess magnesium ribbon. 2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g). Calculate: (a) moles of HCl used, (b) moles of H₂ produced, (c) volume of H₂ at rtp.
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a n(HCl) = c × V/1000 = 0.40 × 25.0/1000 = 0.01 mol

b Mole ratio: 2HCl : 1H₂ = 2:1
n(H₂) = 0.01/2 = 0.005 mol

c V(H₂) at rtp = n × Vₘ = 0.005 × 24.0 = 0.12 dm³ (or 120 cm³)

✅ (a) 0.01 mol HCl   (b) 0.005 mol H₂   (c) 0.12 dm³ (120 cm³) at rtp
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