The Mole & The Avogadro Constant

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Notes

The Mole & Avogadro Constant

The **mole** (mol) is the SI unit for amount of substance.
One mole contains **6.02 × 10²³ particles** (atoms, molecules, or ions) – the **Avogadro constant**.
One mole of sodium (Na) contains 6.02 × 10²³ Na atoms; one mole of H₂ contains 6.02 × 10²³ H₂ molecules.
**Molar mass** of an element = relative atomic mass in grams; for a compound = relative formula mass in grams.
At **RTP** (20 °C, 1 atm), one mole of any gas occupies **24 dm³** (or 24 000 cm³) – the **molar gas volume**.
Gas volume (dm³) = moles × 24; moles = volume (dm³) ÷ 24.

Linking Moles, Mass & Mr

**Moles = mass (g) ÷ Mᵣ** (or Aᵣ for elements).
**Mass = moles × Mᵣ**.
Example: 0.250 mol Zn (Aᵣ=65) has mass = 0.250 × 65 = 16.25 g.
Example: 2.64 g sucrose (Mᵣ=342) gives moles = 2.64 ÷ 342 = 7.72 × 10⁻³ mol.
Use the formula triangle: cover the quantity you want to find.

Reacting Masses & Limiting Reactants

Use balanced equation to find **molar ratio** between reactants and products.
Convert given mass to moles, then use ratio to find moles of required substance, then convert back to mass.
**Limiting reactant** is the one used up first; it determines the amount of product.
To identify limiting reactant: calculate moles of each reactant, compare with the molar ratio from the equation.
Example: 2Mg + O₂ → 2MgO; 6.0 g Mg (0.25 mol) gives 0.25 mol MgO (mass = 10 g).

Calculating Concentration

**Concentration** in g/dm³ = mass of solute (g) ÷ volume (dm³).
**Concentration** in mol/dm³ = moles of solute ÷ volume (dm³).
To convert g/dm³ to mol/dm³, divide by molar mass; to convert mol/dm³ to g/dm³, multiply by molar mass.
Always convert cm³ to dm³ by dividing by 1000.
Example: 10 g NaOH in 2 dm³ gives 5 g/dm³.

Titration Calculations

Use the balanced equation to find the **mole ratio** between acid and base.
Calculate moles of known solution: moles = concentration × volume (dm³).
Use ratio to find moles of unknown, then concentration = moles ÷ volume (dm³).
Example: 25.0 cm³ HCl titrated with 12.1 cm³ 0.100 mol/dm³ NaOH (1:1 ratio) gives [HCl] = 0.0484 mol/dm³.

Empirical & Molecular Formula

**Empirical formula** is the simplest whole‑number ratio of atoms in a compound.
To find empirical formula: divide each element's mass (or % mass) by its Aᵣ, then divide by the smallest result to get a ratio; multiply to get whole numbers if needed.
**Molecular formula** = (empirical formula)ₙ, where n = Mᵣ(compound) ÷ Mᵣ(empirical formula).
Example: empirical formula CH₂O (Mᵣ=30); if Mᵣ=60, n=2, so molecular formula = C₂H₄O₂.
**Water of crystallisation**: use similar method to find x in hydrated salt formula (e.g., CuSO₄·xH₂O).

Percentage Yield & Purity

**Percentage yield** = (actual yield ÷ theoretical yield) × 100%.
Actual yield is the mass obtained; theoretical yield is calculated from the balanced equation.
**Percentage purity** = (mass of pure substance ÷ total mass of sample) × 100%.
**Percentage composition by mass** = (total mass of element in compound ÷ Mᵣ of compound) × 100%.
Example: Fe₂O₃ (Mᵣ=160) contains 112 g Fe → %Fe = (112÷160)×100 = 70%.

Bohr model of sodium atom (11 protons, 12 neutrons, electron configuration 2.8.1). One mole of sodium contains 6.02×10²³ such atoms.

Particle arrangement in solid, liquid, and gas. At RTP, one mole of any gas occupies 24 dm³.

Practice questions

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1.Which row correctly describes the mole and the value of Avogadro's constant?
Easy
- Aone mole of a substance is equal to the substance's relative atomic or molecular mass in grams; one mole of a substance contains 6.02 × 10²³ atoms, molecules or formula units
- Bone mole of a substance is equal to the substance's atomic number in grams; one mole of a substance contains 6.02 × 10²³ atoms, molecules or formula units
- Cone mole of a substance is equal to the substance's relative atomic or molecular mass in grams; one mole of a substance contains 12.04 × 10²³ atoms, molecules or formula units
- Done mole of a substance is equal to the substance's atomic number in grams; one mole of a substance contains 12.04 × 10²³ atoms, molecules or formula units
2.What is the mass of one mole of carbon dioxide (CO₂)? (Ar: C = 12, O = 16)
Easy
- A28 g
- B44 g
- C32 g
- D12 g
3.How many molecules are there in one mole of hydrogen gas (H₂)?
Easy
- A6.02 × 10²³ molecules
- B1.204 × 10²⁴ molecules
- C3.01 × 10²³ molecules
- D12.04 × 10²³ molecules
4.The complete combustion of methane produces carbon dioxide and steam. CH₄(g) + 2O₂(g) → 2H₂O(g) + CO₂(g). Which statements about the reaction are correct? 1. The empirical formula of methane is CH₄ 2. The number of atoms in 1 mole of methane is 4 × Avogadro's constant 3. 1 mole of methane produces 72 dm³ of gaseous products at r.t.p. 4. 1 mole of methane occupies a volume of 12 dm³ at r.t.p.
Medium
- A1, 2 and 3
- B1 and 2
- C1 and 3
- D2 and 4
5.Magnesium carbonate and hydrochloric acid react: MgCO₃(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) + CO₂(g). What is the volume of CO₂ produced when 21 g of magnesium carbonate (Mr = 84) reacts completely with excess hydrochloric acid? (One mole of gas occupies 24.0 dm³ at r.t.p.)
Medium
- A4 dm³
- B8 dm³
- C6 dm³
- D2 dm³
6.Calcium carbonate undergoes thermal decomposition: CaCO₃ → CaO + CO₂. What mass of calcium oxide (Mr = 56) is formed when 60 g of calcium carbonate (Mr = 100) is completely decomposed?
Medium
- A28 g
- B18.5 g
- C60 g
- D33.6 g
7.Substance X was analysed and found to contain 40.00% carbon, 6.67% hydrogen and 53.33% oxygen by mass. What is the empirical formula of X?
Medium
- AC₄H₈O₄
- BC₆H₁₂O₈
- CC₂H₄O₂
- DCH₂O
8.What is the concentration in mol dm⁻³ of a solution of sodium hydroxide that contains 16 g of NaOH in 200 cm³ of distilled water? (Mr NaOH = 40)
Hard
- A2 mol dm⁻³
- B2.5 mol cm⁻³
- C0.2 mol dm⁻³
- D1 mol cm⁻³

Slides

The Mole & Avogadro's Constant

The mole (mol) is the SI unit for amount of substance.
One mole contains 6.02 × 10²³ particles (Avogadro's constant).
Particles can be atoms, molecules, ions, or formula units.

How many particles are in one mole of any substance?

6.02 × 10²³
12.04 × 10²³
6.02 × 10²²
1.00 × 10²³

Molar Mass

The mass of one mole of a substance is its molar mass.
For an element, molar mass = relative atomic mass in grams.
For a compound, molar mass = relative molecular/formula mass in grams.
Example: 1 mole of carbon has mass 12 g.

One mole of water (H₂O) has a mass of ____ g.

Moles, Mass & Mr

Number of moles = mass (g) ÷ molar mass (g/mol).
Mass = moles × molar mass.
Molar mass = mass ÷ moles.
Use the formula triangle to rearrange.

What is the mass of 0.250 moles of zinc? (Ar Zn = 65)

16.25 g
65 g
0.250 g
260 g

Mole & Volume of Gas

At RTP (20°C, 1 atm), one mole of any gas occupies 24 dm³ (24,000 cm³).
This is the molar gas volume.
Volume (dm³) = moles × 24; moles = volume (dm³) ÷ 24.

3 moles of hydrogen gas occupy ____ dm³ at RTP.

Reacting Masses

Use balanced equations to find mole ratios.
Convert given mass to moles, then use ratio to find moles of other substance.
Convert moles back to mass using molar mass.

2Mg + O₂ → 2MgO. What mass of MgO is formed from 6.0 g Mg? (Ar Mg=24, O=16)

10 g
20 g
5 g
12 g

Limiting Reactants

The limiting reactant is used up first and determines the amount of product.
The other reactant is in excess.
To find the limiting reactant: calculate moles of each, compare with the balanced equation ratio.

Which statements about limiting reactants are true?

The limiting reactant is the one with the smallest mass.
The reaction stops when the limiting reactant is used up.
The amount of product depends on the limiting reactant.

Concentration

Concentration in g/dm³ = mass of solute (g) ÷ volume (dm³).
Concentration in mol/dm³ = moles of solute ÷ volume (dm³).
To convert: g/dm³ ÷ molar mass = mol/dm³.
Volume in cm³ ÷ 1000 = volume in dm³.

80 g of NaOH (Mr=40) dissolved in 500 cm³ water gives a concentration of ____ mol/dm³.

Titration Calculations

Use the balanced equation to find the mole ratio.
Calculate moles of known solution (concentration × volume in dm³).
Use ratio to find moles of unknown.
Concentration of unknown = moles ÷ volume (dm³).

25.0 cm³ of 0.100 mol/dm³ NaOH neutralises 20.0 cm³ of HCl. What is [HCl]? (1:1 ratio)

0.125 mol/dm³
0.100 mol/dm³
0.080 mol/dm³
0.200 mol/dm³

Empirical & Molecular Formula

Empirical formula is the simplest whole number ratio of atoms.
To find it: divide each element's mass (or % by mass) by its Ar, then divide by the smallest result.
Molecular formula = (empirical formula) × n, where n = Mr(compound) / Mr(empirical).

A compound contains 40% C, 6.67% H, 53.33% O. Its empirical formula is ____.

Percentage Yield & Purity

Percentage yield = (actual yield / theoretical yield) × 100%.
Percentage purity = (mass of pure substance / total mass) × 100%.
Theoretical yield is calculated from the balanced equation.

A reaction gives 1.6 g of product; theoretical yield is 2.0 g. What is the percentage yield?

80%
125%
20%
60%

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Notes

The Mole & Avogadro Constant

Linking Moles, Mass & Mr

Reacting Masses & Limiting Reactants

Calculating Concentration

Titration Calculations

Empirical & Molecular Formula

Percentage Yield & Purity

Bohr model of sodium atom (11 protons, 12 neutrons, electron configuration 2.8.1). One mole of sodium contains 6.02×10²³ such atoms.

Particle arrangement in solid, liquid, and gas. At RTP, one mole of any gas occupies 24 dm³.

Practice questions

Unlock all 40 questions, slides & more

Past papers

Particle arrangement in solid, liquid, and gas. At RTP, one mole of any gas occupies 24 dm³.