The Mole and the Avogadro ConstantCambridge IGCSE Chemistry: Revision notes

The mole (mol) is the SI unit of amount of substance. One mole of any substance contains exactly 6.02 × 10²³ particles (atoms, molecules, ions, or electrons). This number is called the Avogadro constant (Nₐ).

The mole provides a bridge between the atomic scale and the macroscopic scale we can measure in the laboratory:

• 1 mole of carbon-12 atoms has a mass of exactly 12 g
• 1 mole of any element has a mass (in grams) equal to its relative atomic mass (Aᵣ)
• 1 mole of any compound has a mass (in grams) equal to its relative molecular mass (Mᵣ)

This means the molar mass of a substance is numerically equal to its relative mass, but with units of g/mol.

Use the fundamental relationship:

amount of substance (mol) = mass (g) / molar mass (g/mol)

Or rearranged:

• mass (g) = amount (mol) × molar mass (g/mol)
• molar mass (g/mol) = mass (g) / amount (mol)

To find the molar mass of a compound:

1. Write the chemical formula
2. Look up the relative atomic masses (usually on the periodic table)
3. Multiply each by the number of atoms of that element
4. Add all values together

Example: Calculate molar mass of calcium carbonate, CaCO₃

• Ca: 40 × 1 = 40
• C: 12 × 1 = 12
• O: 16 × 3 = 48
• Total = 100 g/mol

Once you know the amount in moles, calculate the number of particles using:

number of particles = amount (mol) × Avogadro constant (6.02 × 10²³)

This works for any type of particle: atoms, molecules, ions, or electrons.

Step-by-step process:

1. Calculate amount in moles using mass and molar mass (if not already given)
2. Multiply by 6.02 × 10²³
3. Check your answer is in standard form if very large

Example: How many atoms of oxygen are in 4.0 g of oxygen gas (O₂)?

• Molar mass of O₂ = 16 × 2 = 32 g/mol
• Amount = 4.0 g ÷ 32 g/mol = 0.125 mol
• Number of O₂ molecules = 0.125 × 6.02 × 10²³ = 7.53 × 10²²
• Number of O atoms = 7.53 × 10²² × 2 = 1.51 × 10²³ (because each O₂ has 2 O atoms)

At room temperature and pressure (r.t.p.), defined as 25°C and 100 kPa, 1 mole of any gas occupies 24 dm³ (or 24 litres).

This relationship allows you to calculate:

• Volume of a gas from moles: volume (dm³) = amount (mol) × 24 dm³/mol
• Amount of gas from volume: amount (mol) = volume (dm³) / 24 dm³/mol

Why this matters: You don't need to know the identity of the gas — the volume depends only on the number of moles.

Example: What volume does 2.0 mol of nitrogen gas occupy at r.t.p.?

• Volume = 2.0 mol × 24 dm³/mol = 48 dm³

Example: What amount of oxygen is in 60 dm³ at r.t.p.?

• Amount = 60 dm³ ÷ 24 dm³/mol = 2.5 mol

Concentration measures the amount of solute dissolved in a solution:

Converting between the two:

• Molar concentration × molar mass = mass concentration
• Mass concentration ÷ molar mass = molar concentration

Example: A solution contains 58.5 g of NaCl in 1 dm³ of solution. What is the molar concentration?

• Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
• Mass concentration = 58.5 g/dm³
• Molar concentration = 58.5 g/dm³ ÷ 58.5 g/mol = 1.0 mol/dm³

Rearranged formulas:

• amount (mol) = concentration (mol/dm³) × volume (dm³)
• mass (g) = concentration (g/dm³) × volume (dm³)
• volume (dm³) = amount (mol) / concentration (mol/dm³)

Stoichiometry uses balanced equations to calculate the quantities of reactants and products.

Step-by-step method for reacting masses:

1. Write the balanced equation
2. Find molar masses of relevant substances
3. Calculate moles from given mass
4. Use mole ratio from equation to find moles of required substance
5. Convert back to mass if needed

For gases at r.t.p.:

• Replace step 5 with volume using 24 dm³/mol

Limiting reactant: The reactant that runs out first and determines how much product forms. To find it:

1. Calculate moles of each reactant
2. Divide by the stoichiometric coefficient in the equation
3. The substance with the smallest value is limiting

Example: In 2Mg + O₂ → 2MgO, if 2.4 g Mg (Aᵣ = 24) reacts with excess O₂, calculate mass of MgO (Mᵣ = 40).

• Moles of Mg = 2.4 g ÷ 24 g/mol = 0.1 mol
• Mole ratio: 2 Mg : 2 MgO = 1 : 1
• Moles of MgO = 0.1 mol
• Mass of MgO = 0.1 mol × 40 g/mol = 4.0 g

The empirical formula shows the simplest whole-number ratio of atoms. The molecular formula shows the actual number of atoms.

Finding empirical formula from composition data:

1. Convert masses or percentages to moles (divide by Aᵣ)
2. Divide all values by the smallest number of moles
3. Multiply by whole numbers if needed to get whole-number ratios
4. Write the empirical formula

Example: A compound contains 40% C, 6.7% H, 53.3% O by mass. Find empirical formula (assume 100 g).

• C: 40 g ÷ 12 g/mol = 3.33 mol
• H: 6.7 g ÷ 1 g/mol = 6.7 mol
• O: 53.3 g ÷ 16 g/mol = 3.33 mol
• Divide by smallest (3.33): C=1, H≈2, O=1
• Empirical formula: CH₂O

Finding molecular formula:

• Calculate molar mass of empirical formula
• Use: n = molar mass (from data) / molar mass (of empirical formula)
• Multiply empirical formula by n

If empirical formula is CH₂O (molar mass 30) and actual molar mass is 60:

• n = 60 ÷ 30 = 2
• Molecular formula = (CH₂O)₂ = C₂H₄O₂

Percentage yield compares actual to theoretical amount:

% yield = (actual amount / theoretical amount) × 100%

Theoretical amount is calculated using stoichiometry assuming all reactants react perfectly.

Percentage composition by mass shows the mass fraction of each element:

% composition = (mass of element / total mass of compound) × 100%

Or using molar mass: % composition = (Aᵣ × number of atoms / Mᵣ of compound) × 100%

Example: In CaCO₃ (molar mass 100), percentage of calcium:

• % Ca = (40 / 100) × 100% = 40%

Percentage purity indicates how pure a sample is:

% purity = (mass of pure substance / total mass of sample) × 100%

Example: If 10 g of impure NaCl contains 9.2 g pure NaCl:

• % purity = (9.2 / 10) × 100% = 92%

Titration is an analytical method using a standard solution (known concentration) to find the concentration of an unknown solution.

Standard procedure:

1. Record the initial volume of titrant in the burette
2. Add indicator to the solution in the flask
3. Add titrant until the end point is reached (indicator changes colour)
4. Record the final volume and calculate volume used
5. Repeat until two consecutive results agree within 0.1 cm³
6. Calculate the mean volume of the repeated concordant results

Calculating concentration or moles:

• From titration data you get the volume of titrant used and its concentration
• Calculate moles of titrant = concentration (mol/dm³) × volume (dm³)
• Use the mole ratio from the balanced equation
• Calculate moles of unknown solution
• If needed, find concentration = moles / volume of unknown

Example: 25.0 cm³ of NaOH (0.1 mol/dm³) neutralises 20.0 cm³ of HCl. Find concentration of HCl.

• Moles of NaOH = 0.1 mol/dm³ × 0.025 dm³ = 0.0025 mol
• Equation: HCl + NaOH → NaCl + H₂O (1:1 ratio)
• Moles of HCl = 0.0025 mol
• Concentration of HCl = 0.0025 mol ÷ 0.020 dm³ = 0.125 mol/dm³