The Mole and the Avogadro Constant

One mole of water sits in a tablespoon; counting its molecules at one per second would take 19 quadrillion years. That absurd number, 6.02 × 10²³ (the Avogadro constant), is the bridge between masses you can weigh and particles you cannot, and it is Supplement material that anchors every Extended calculation. Paper 4 gives 4-6 marks to candidates who can move freely between moles, grams, gas volumes and particle counts; this page sets up those conversions, and reacting-mass calculations puts them to work.

Definitions that earn marks (Supplement)

A mole is the amount of substance that contains 6.02 × 10²³ particles. The Avogadro constant is that number of particles per mole. State which particles when it matters: one mole of oxygen molecules (O2) contains 6.02 × 10²³ molecules but 1.204 × 10²⁴ atoms, a favourite multiple-choice distractor on Paper 2.

Molar mass is the mass of one mole of a substance: numerically equal to the Ar or Mr, with units g/mol. So one mole of carbon weighs 12 g, one mole of CO2 weighs 44 g. This is why Ar and Mr had to come first.

The conversion triangle (Supplement)

Conversion	Formula	Watch for
Mass ↔ moles	n = m ÷ M (mass in g, M = molar mass)	Use the Mr of the right substance
Gas volume ↔ moles (r.t.p.)	n = V ÷ 24 (V in dm³) or V ÷ 24,000 (cm³)	Gases only; convert cm³ first
Particles ↔ moles	particles = n × 6.02 × 10²³	Specify atoms vs molecules

Worked conversions, one each way. Mass to moles: 8.8 g of CO2 → n = 8.8 ÷ 44 = 0.20 mol. Moles to gas volume: 0.20 mol of any gas at r.t.p. → V = 0.20 × 24 = 4.8 dm³. Moles to particles: 0.20 × 6.02 × 10²³ = 1.204 × 10²³ molecules.

The molar gas volume deserves emphasis: at room temperature and pressure, one mole of any gas occupies 24 dm³. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, which is why the constant does not care whether the gas is H2 (2 g per 24 dm³) or Cl2 (71 g per 24 dm³).

Rearranging is where students wobble. From n = m/M: m = n × M and M = m ÷ n. The last form answers “calculate the Mr of gas X given 0.25 mol weighs 11 g” → M = 11 ÷ 0.25 = 44. Our mole calculations technique guide drills all three arrangements until they are reflexes.

Worked exam question

At r.t.p., a sample of ammonia gas, NH3, has a volume of 600 cm³. (a) Calculate the number of moles of ammonia in the sample. (2) (b) Calculate the mass of the sample. [Ar: N 14, H 1] (1) (c) Calculate the number of ammonia molecules in the sample. (1)

Model answer: (a) 600 cm³ = 0.600 dm³ (1); n = 0.600 ÷ 24 = 0.025 mol (1). (b) Mr of NH3 = 17, so mass = 0.025 × 17 = 0.425 g (1). (c) 0.025 × 6.02 × 10²³ = 1.5 × 10²² molecules (1).

Mark-by-mark: (a) awards the unit conversion separately from the division. Candidates who divide 600 by 24 get 25 mol, an answer so large it should trigger a sanity check. (b) is one mark for n × M with the correct Mr of 17. (c) multiplies the same n by the Avogadro constant. Error carried forward protects (b) and (c) if (a) slipped, provided working is visible.

The mistakes that cost marks

1. cm³ fed into n = V ÷ 24. Either convert to dm³ first or use 24,000, and write the conversion as its own line, because it is frequently its own mark.
2. Using 24 dm³ for liquids or solids. Molar gas volume applies to gases only; for everything else, use mass and Mr.
3. Avogadro number applied to the wrong particle: moles of O2 × 6.02 × 10²³ gives molecules; double it for atoms. Read which the question wants.
4. Quoting “6.02 × 10²³” with no per-mole context, or writing 6.02 × 1023 by losing the power. Superscripts matter; on paper, write the index clearly.

How examiners want it phrased

Typical student wording	Accepted mark-scheme wording
”A mole is like a dozen but bigger"	"The amount of substance containing 6.02 × 10²³ particles"
"24 is the magic gas number"	"One mole of any gas occupies 24 dm³ at room temperature and pressure"
"Divide by the big number on the Periodic Table"	"n = mass ÷ molar mass = 8.8 g ÷ 44 g/mol = 0.20 mol"
"Loads of molecules"	"0.025 mol × 6.02 × 10²³ = 1.5 × 10²² molecules”

The Malaysia note

The 24 dm³ value catches Malaysian students with SPM-trained older siblings or tutors, because SPM works at 22.4 dm³ (s.t.p.) and the habit transfers invisibly, producing a wrong constant that looks like a method error. 0620 uses 24 at r.t.p., printed nowhere on the paper, so it must be memorised. Most of our students fix mole-conversion fluency in 2-3 sessions, and the free 1-hour trial usually reveals within twenty minutes whether the gap is the constant, the conversions or the rearranging.

Test yourself

Three conversions, no peeking at the triangle above. The answers stay hidden until you click.

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

Show answer

• Mr of SO2 = 32 + 2(16) = 64 [1] • n = 6.4 ÷ 64 = 0.10 mol [1]

Q2 (2 marks). Calculate the volume, in cm³, occupied by 0.015 mol of hydrogen gas at r.t.p.

Show answer

• V = 0.015 × 24 = 0.36 dm³ [1] • = 0.36 × 1000 = 360 cm³ [1]

Q3 (2 marks). 0.050 mol of gas Y has a mass of 2.2 g. Calculate the relative molecular mass of Y and suggest its identity.

Show answer

• M = m ÷ n = 2.2 ÷ 0.050 = 44 [1] • Carbon dioxide, CO2 (Mr = 44) [1]