The Mole and the Avogadro Constant

The mole links measurable laboratory quantities to numbers of particles. In Cambridge IGCSE Chemistry 0620, the mole and Avogadro constant are Supplement content, so this page is for Extended candidates.

Essential definitions (Supplement)

A mole is an amount of substance containing 6.02 × 10²³ particles.

The Avogadro constant is 6.02 × 10²³ particles per mole.

The word particles must be interpreted correctly:

• atoms for monatomic elements such as copper
• molecules for substances such as oxygen, O2
• ions where an ionic particle is specified
• formula units where the context uses an ionic compound as a whole

One mole of O2 contains 6.02 × 10²³ oxygen molecules but twice as many oxygen atoms.

Mass and moles

Use:

number of moles, n = mass, m ÷ molar mass, M

or

n = m / M

The numerical molar mass in g/mol is obtained from Ar or Mr.

Example:

Mr of CO2 = 12 + (2 × 16) = 44
Moles in 8.8 g of CO2 = 8.8 ÷ 44 = 0.20 mol

Rearrangements:

• m = n × M
• M = m ÷ n

Write the formula, substitute values and include the unit. This keeps method marks visible.

Gas volume at room temperature and pressure

At r.t.p., one mole of any gas occupies 24 dm³, or 24,000 cm³.

Use:

n = gas volume in dm³ ÷ 24

or

n = gas volume in cm³ ÷ 24,000

Example:

600 cm³ of ammonia at r.t.p.:

n = 600 ÷ 24,000 = 0.0250 mol

Do not divide a value in cm³ by 24. Either convert cm³ to dm³ first or use 24,000.

Particles and moles

Use:

number of particles = moles × 6.02 × 10²³

and

moles = number of particles ÷ 6.02 × 10²³

Example:

Particles in 0.20 mol of CO2:

0.20 × 6.02 × 10²³ = 1.204 × 10²³ molecules

If the question asks for atoms of oxygen rather than CO2 molecules, multiply the number of molecules by two.

Conversion map

Starting quantity	Conversion	Result
Mass	divide by molar mass	Moles
Gas volume at r.t.p.	divide by 24 dm³ or 24,000 cm³	Moles
Number of particles	divide by 6.02 × 10²³	Moles
Moles	multiply by molar mass	Mass
Moles	multiply by 24 dm³ or 24,000 cm³	Gas volume at r.t.p.
Moles	multiply by 6.02 × 10²³	Number of particles

Many multi-step questions become manageable when every route passes through moles.

Worked examination question

A sample of ammonia, NH3, occupies 600 cm³ at r.t.p.

(a) Calculate the number of moles of ammonia. [2]

• n = 600 ÷ 24,000 [1]
• n = 0.0250 mol [1]

(b) Calculate the mass of ammonia. [Ar: N = 14, H = 1] [2]

• Mr of NH3 = 17 [1]
• mass = 0.0250 × 17 = 0.425 g [1]

(c) Calculate the number of ammonia molecules. [1]

• 0.0250 × 6.02 × 10²³ = 1.51 × 10²² molecules [1]

The same mole value is reused in each part. Clear working also allows error carried forward where the mark scheme permits it.

Common mistakes

1. Treating the topic as Core. These calculations are Supplement for the current syllabus cycle.
2. Using 24 with a volume in cm³. Use 24,000 or convert to dm³.
3. Using molar gas volume for liquids or solids. The 24 dm³ relation is for gases at r.t.p.
4. Choosing the wrong particle. State atoms, molecules or ions as required.
5. Using Ar instead of Mr for a compound. Add all atoms in the formula.
6. Losing powers of ten. Write 6.02 × 10²³ clearly.
7. Hiding the method. A bare final number can lose method marks if the answer is wrong.

Exam-ready working

A strong calculation is laid out vertically:

Mr(CO2) = 44
n = m / M
n = 8.8 / 44
n = 0.20 mol

This is safer than compressing every operation into one line.

Test yourself

Q1. Calculate the number of moles in 6.4 g of sulfur dioxide, SO2. [Ar: S = 32, O = 16] [2]

Show answer

Mr = 64 [1]; n = 6.4 ÷ 64 = 0.10 mol [1].

Q2. Calculate the volume in cm³ occupied by 0.015 mol of hydrogen at r.t.p. [2]

Show answer

V = 0.015 × 24,000 = 360 cm³.

Q3. A gas sample has mass 2.2 g and amount 0.050 mol. Calculate its relative molecular mass. [2]

Show answer

M = m ÷ n = 2.2 ÷ 0.050 = 44.

Next, apply the method to reacting-mass calculations and the mole calculations technique guide.