MODULE A7

The Mole Concept

Avogadro's number, molar mass, empirical formulas, concentration, stoichiometry — the language of quantitative chemistry! 🧮

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1. The Mole & Avogadro's Number

A mole is the chemist's counting unit — just like a "dozen" means 12, a "mole" means 6.02 × 10²³ particles (atoms, molecules, ions, etc.).

🔢 Avogadro's number (Lₐ) = 6.02 × 10²³ mol⁻¹
This is the number of particles in exactly 1 mole of any substance.
Real world: 1 mole of water (18 g) contains 6.02 × 10²³ molecules — that's more molecules than grains of sand on Earth!

The Three Mole Triangles

Triangle 1: Moles ↔ Mass

mass (g) moles Mᵣ n = m ÷ Mᵣ

Triangle 2: Moles ↔ Particles

No. of particles moles 6.02×10²³ n = N ÷ Lₐ

Triangle 3: Concentration

moles (n) c (mol/dm³) V (dm³) n = c × V

2. Molar Mass & Mole Calculations

The molar mass (Mᵣ) of a substance is its relative formula mass in grams per mole (g mol⁻¹). It is calculated by adding up the relative atomic masses (Aᵣ) of all atoms in the formula.

SubstanceFormulaCalculationMolar Mass (g/mol)
WaterH₂O2(1) + 1618
Carbon dioxideCO₂12 + 2(16)44
Sodium chlorideNaCl23 + 35.558.5
Calcium carbonateCaCO₃40 + 12 + 3(16)100
AmmoniaNH₃14 + 3(1)17
Sulfuric acidH₂SO₄2(1) + 32 + 4(16)98
Key formulas:
n = m / Mᵣ  →  moles = mass ÷ molar mass
m = n × Mᵣ  →  mass = moles × molar mass
Mᵣ = m / n  →  molar mass = mass ÷ moles

3. Molar Volume of Gases

One mole of any gas occupies the same volume under the same conditions of temperature and pressure (Avogadro's Law).

ConditionAbbreviationTemperaturePressureMolar Volume
Room temperature & pressurer.t.p.25°C (298 K)1 atm24 dm³/mol
Standard temperature & pressures.t.p.0°C (273 K)1 atm22.4 dm³/mol
V = n × 24 (at r.t.p., volume in dm³)
n = V / 24 (at r.t.p.)
Example: 0.5 mol CO₂ at r.t.p. = 0.5 × 24 = 12 dm³

4. Empirical & Molecular Formulas

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of atoms in one molecule (a multiple of the empirical formula).

Finding Empirical Formula from % Composition — 4 Steps

  1. Write the % (or mass) of each element
  2. Divide each by its Aᵣ → get moles ratio
  3. Divide all by the smallest value → get ratio
  4. Round to nearest whole number (×2 if needed)
📝 Worked Example: A compound is 40% C, 6.67% H, 53.33% O.
Step 2: C = 40/12 = 3.33 | H = 6.67/1 = 6.67 | O = 53.33/16 = 3.33
Step 3: Divide by 3.33 → C:1, H:2, O:1
Empirical formula: CH₂O (this is the formula for formaldehyde!)

Finding Molecular Formula from Empirical Formula

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

If empirical formula is CH₂O (Mᵣ = 30) and the actual Mᵣ = 180: n = 180/30 = 6 → molecular formula = C₆H₁₂O₆ (glucose!)

5. Concentration

The concentration of a solution is the amount of solute (in moles) per unit volume of solution.

c = n / V    concentration (mol/dm³) = moles ÷ volume (dm³)
n = c × V    moles = concentration × volume (dm³)
⚠️ Convert cm³ to dm³: divide by 1000. E.g. 250 cm³ = 0.25 dm³

Example: 0.5 mol NaCl dissolved in 250 cm³ (0.25 dm³) of solution:
c = 0.5 / 0.25 = 2 mol/dm³

6. Stoichiometry — 4-Step Method

Stoichiometry uses a balanced equation to calculate the mass (or moles) of products or reactants.

  1. Write the balanced equation
  2. Convert the given quantity to moles
  3. Use mole ratio from the equation to find moles of the wanted substance
  4. Convert moles back to the required unit (g, dm³, molecules…)
📝 Example: What mass of MgO is produced when 4.8 g of Mg burns? (2Mg + O₂ → 2MgO)
Step 2: n(Mg) = 4.8 / 24 = 0.2 mol
Step 3: Ratio Mg:MgO = 2:2 = 1:1 → n(MgO) = 0.2 mol
Step 4: m(MgO) = 0.2 × 40 = 8.0 g

⚡ Interactive Mole Calculators

Three calculators — enter any two known values and hit Calculate to find the third. All conversions shown step-by-step!

🔢 Calculator 1: Mass ↔ Moles ↔ Molar Mass

💧 Calculator 2: Concentration ↔ Moles ↔ Volume

💨 Calculator 3: Gas Volume ↔ Moles (at r.t.p., 1 mol = 24 dm³)

📈 Concentration vs Volume (c = n/V)

As volume increases at constant moles, concentration decreases hyperbolically. Use the slider to set the number of moles.

1.0 mol

🃏 Flashcards — The Mole Concept

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Answer
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❓ Quiz — The Mole Concept

🔢 Worked Example — Stoichiometry Problem

Problem: Calcium carbonate (CaCO₃) decomposes on heating: CaCO₃(s) → CaO(s) + CO₂(g). What mass of CaO is produced AND what volume of CO₂ (at r.t.p.) is released when 25 g of CaCO₃ decomposes? (Aᵣ: Ca=40, C=12, O=16)

Step 1: Calculate moles of CaCO₃

Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100 g/mol. Use n = m / Mᵣ. How many moles in 25 g?

Step 2: Moles of CaO produced

From the equation: CaCO₃ : CaO = 1:1. So moles of CaO = moles of CaCO₃. What is n(CaO)?

Step 3: Mass of CaO

Molar mass of CaO = 40 + 16 = 56 g/mol. Use m = n × Mᵣ. What is the mass of 0.25 mol CaO?

Step 4: Volume of CO₂ at r.t.p.

n(CO₂) = 0.25 mol (1:1 ratio). At r.t.p., 1 mol gas = 24 dm³. Use V = n × 24. What is the volume of CO₂?

🔗 Matching — Mole Concept Terms

Click a term, then its matching definition. Green = correct!

Term / Formula

Definition / Explanation

📝 CSEC-Style Questions

Q1. Calculate the number of moles in: (a) 8.8 g of CO₂, (b) 11.7 g of NaCl, (c) 5.4 g of Al. (Aᵣ: C=12, O=16, Na=23, Cl=35.5, Al=27) [6 marks]+
Mark Scheme

1 (a) Mᵣ(CO₂) = 12 + 2(16) = 44 g/mol ✓ | n = 8.8 / 44 = 0.2 mol ✓

2 (b) Mᵣ(NaCl) = 23 + 35.5 = 58.5 g/mol ✓ | n = 11.7 / 58.5 = 0.2 mol ✓

3 (c) Mᵣ(Al) = 27 g/mol | n = 5.4 / 27 = 0.2 mol ✓ ✓ (all three give the same number of moles!)

Q2. A compound contains 75% carbon and 25% hydrogen by mass. The molar mass of the compound is 16 g/mol. Find (a) the empirical formula and (b) the molecular formula. [5 marks]+
Mark Scheme

1 (a) Moles ratio: C = 75/12 = 6.25 | H = 25/1 = 25 ✓

2 Divide by smallest (6.25): C = 1, H = 4 ✓

3 Empirical formula: CH₄ ✓

4 (b) Mᵣ(CH₄) = 12 + 4 = 16 | n = 16/16 = 1 ✓

5 Molecular formula = CH₄ × 1 = CH₄ (methane) ✓

Q3. 25.0 cm³ of NaOH solution has a concentration of 0.100 mol/dm³. (a) Calculate the moles of NaOH. (b) This reacts with HCl: NaOH + HCl → NaCl + H₂O. Calculate the concentration of the HCl if 20.0 cm³ was required to neutralise the NaOH. [5 marks]+
Mark Scheme

1 (a) V(NaOH) = 25.0 cm³ = 0.025 dm³ ✓

2 n(NaOH) = c × V = 0.100 × 0.025 = 0.0025 mol ✓

3 (b) Mole ratio NaOH : HCl = 1:1 → n(HCl) = 0.0025 mol ✓

4 V(HCl) = 20.0 cm³ = 0.020 dm³ ✓

5 c(HCl) = n/V = 0.0025 / 0.020 = 0.125 mol/dm³ ✓

Q4. Iron reacts with hydrochloric acid: Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g). What volume of hydrogen gas (at r.t.p.) is produced when 5.6 g of iron reacts with excess HCl? (Aᵣ: Fe = 56) [4 marks]+
Mark Scheme

1 n(Fe) = 5.6 / 56 = 0.1 mol ✓

2 Ratio Fe : H₂ = 1:1 → n(H₂) = 0.1 mol ✓

3 V(H₂) = n × 24 = 0.1 × 24 ✓

4 V(H₂) = 2.4 dm³ ✓

⭐ Key Formulas

Mass ↔ Moles

n = m / Mᵣ m = n × Mᵣ Mᵣ = m / n

Particles

Lₐ = 6.02 × 10²³ mol⁻¹ n = N / Lₐ N = n × Lₐ

Concentration

c = n / V (V in dm³) n = c × V 250 cm³ = 0.25 dm³

Gas Volume

V = n × 24 (r.t.p., dm³) n = V / 24 s.t.p.: 1 mol = 22.4 dm³

Empirical Formula

÷ Aᵣ → ÷ smallest → round n = Mᵣ(mol) / Mᵣ(emp)

Stoichiometry

1. Balance equation 2. → moles 3. Use ratio 4. → required unit

📚 Resources

📖 ChemGuide: Moles ▶ YouTube: Mole Concept ▶ YouTube: Empirical Formula 🔬 PhET: Molarity ▶ YouTube: Stoichiometry
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