The Mole Concept Explained

The mole is the central concept in stoichiometry and underpins every calculation question on Papers 3 and 4. Many students struggle not because the maths is hard, but because the concept feels abstract. This page builds the idea from the ground up.

Why the mole exists

Atoms are far too small to count individually. A single drop of water contains roughly 1,000,000,000,000,000,000,000 molecules. Chemists needed a practical counting unit, just as eggs come in dozens.

The mole is that unit. One mole of any substance contains exactly 6.02 x 10^23 particles. This number is the Avogadro constant (L or NA).

The value 6.02 x 10^23 was chosen so that one mole of carbon-12 atoms has a mass of exactly 12 g, making the relative atomic mass numerically equal to the mass of one mole in grams.

The key relationship: mass, moles, and Mr

The relative formula mass (Mr) of a substance tells you the mass of one mole in grams.

Formula: n = m / Mr

Where:

• n = number of moles (mol)
• m = mass (g)
• Mr = relative formula mass

Substance	Formula	Mr	Mass of 1 mol
Carbon	C	12	12 g
Water	H2O	18	18 g
Sodium chloride	NaCl	58.5	58.5 g
Calcium carbonate	CaCO3	100	100 g
Sulfuric acid	H2SO4	98	98 g

Worked example

How many moles in 5.6 g of iron (Ar = 56)?

n = m / Mr = 5.6 / 56 = 0.1 mol

How many moles in 9 g of water (Mr = 18)?

n = m / Mr = 9 / 18 = 0.5 mol

Moles of gases: the molar gas volume

At room temperature and pressure (r.t.p., approximately 25 C and 1 atm), one mole of any gas occupies 24 dm3 (24,000 cm3). This applies to all gases equally because gas particles are so spread apart that their size does not matter.

Formula: n = V / 24 (volume in dm3)

Or: n = V / 24,000 (volume in cm3)

Gas	Formula	Volume of 1 mol at r.t.p.
Hydrogen	H2	24 dm3
Oxygen	O2	24 dm3
Carbon dioxide	CO2	24 dm3
Chlorine	Cl2	24 dm3

Worked example

What volume does 0.25 mol of CO2 occupy at r.t.p.?

V = n x 24 = 0.25 x 24 = 6 dm3

What is the mass of 480 cm3 of oxygen at r.t.p.? (Mr of O2 = 32)

n = 480 / 24,000 = 0.02 mol m = n x Mr = 0.02 x 32 = 0.64 g

Moles in solution: concentration

Concentration links moles to the volume of a solution.

Formula: n = c x V (where c is in mol/dm3 and V is in dm3)

Or equivalently: c = n / V

To convert cm3 to dm3, divide by 1000.

Worked example

What is the concentration of a solution containing 0.5 mol NaOH in 250 cm3?

V = 250 / 1000 = 0.25 dm3 c = n / V = 0.5 / 0.25 = 2 mol/dm3

How many moles of HCl in 50 cm3 of 0.1 mol/dm3 solution?

V = 50 / 1000 = 0.05 dm3 n = c x V = 0.1 x 0.05 = 0.005 mol

Concentration in g/dm3 can be converted using: concentration (g/dm3) = concentration (mol/dm3) x Mr

See concentration of solutions for full worked examples including titration calculations.

Using moles in equations

The coefficients in a balanced equation give the mole ratio. This is the foundation of all reacting mass calculations.

For Mg + 2HCl -> MgCl2 + H2:

• 1 mol Mg reacts with 2 mol HCl
• 1 mol Mg produces 1 mol MgCl2 and 1 mol H2

Step-by-step reacting mass method

1. Write the balanced equation
2. Calculate moles of the substance you know (from mass, volume, or concentration)
3. Use the mole ratio from the equation to find moles of the substance you want
4. Convert back to mass, volume, or concentration as required

Worked example

What mass of magnesium is needed to react with 100 cm3 of 1 mol/dm3 HCl?

Mg + 2HCl -> MgCl2 + H2

1. Moles of HCl = c x V = 1 x (100/1000) = 0.1 mol
2. From equation: 2 mol HCl reacts with 1 mol Mg, so moles of Mg = 0.1 / 2 = 0.05 mol
3. Mass of Mg = n x Ar = 0.05 x 24 = 1.2 g

Summary of all mole formulae

To find	Formula	Units
Moles from mass	n = m / Mr	g and g/mol
Moles of gas	n = V / 24	V in dm3
Moles of gas	n = V / 24,000	V in cm3
Moles in solution	n = c x V	c in mol/dm3, V in dm3
Number of particles	N = n x 6.02 x 10^23

These five formulae cover every mole calculation in the 0620 syllabus. Link them through the mole as the central quantity, and every stoichiometry problem reduces to three steps: convert to moles, use the ratio, convert out of moles.

Common exam mistakes

1. Forgetting to convert cm3 to dm3: Divide by 1000 before using in concentration formulae.
2. Using Ar instead of Mr: For molecules like O2, the molar mass is Mr = 2 x 16 = 32, not Ar = 16.
3. Ignoring the mole ratio: Not all ratios are 1:1. Always check the balanced equation.
4. Mixing up g/dm3 and mol/dm3: These are different units of concentration. Convert using Mr.

Exam-style worked questions

Calculate the volume of hydrogen gas produced at r.t.p. when 0.48 g of magnesium reacts with excess hydrochloric acid. [Ar: Mg = 24, molar gas volume = 24 dm3] [3 marks]

Mg + 2HCl -> MgCl2 + H2

• Moles of Mg = 0.48 / 24 = 0.02 mol [1]
• From equation, moles of H2 = moles of Mg = 0.02 mol [1]
• Volume of H2 = 0.02 x 24 = 0.48 dm3 (or 480 cm3) [1]

25.0 cm3 of 0.10 mol/dm3 sodium hydroxide is exactly neutralised by 20.0 cm3 of hydrochloric acid. Calculate the concentration of the hydrochloric acid. [3 marks]

NaOH + HCl -> NaCl + H2O

• Moles of NaOH = 0.10 x (25.0/1000) = 0.0025 mol [1]
• From equation, moles of HCl = moles of NaOH = 0.0025 mol [1]
• Concentration of HCl = 0.0025 / (20.0/1000) = 0.125 mol/dm3 [1]