BORING EDUCATION — LIVE EXPERIMENTProduct is live. Experiments are running. Feedback is being collected. Improvements are underway.Welcome to Boring Education. We're glad you're here while we're figuring out how to make learning better.
BORINGEDUCATION
← All notes

Mole Concept and Avogadro's Number: Worked Examples for 1st Year Chemistry

28 September 2026

A mole is the amount of a substance that contains 6.022 × 10²³ particles (atoms, molecules, ions or formula units). This number is Avogadro's number, Nₐ. One mole of any substance has a mass equal to its relative atomic, molecular or formula mass in grams, and one mole of any gas occupies 22.414 dm³ at STP.

Think of a kiryana store that sells eggs by the dozen: nobody counts eggs one by one, they count dozens. Chemists do the same with atoms, except their "dozen" is 6.022 × 10²³. That counting unit is the mole.

What is Avogadro's number?

Avogadro's number is Nₐ = 6.022 × 10²³ mol⁻¹. It is the number of particles in one mole:

  • 1 mole of carbon atoms = 6.022 × 10²³ atoms
  • 1 mole of water molecules = 6.022 × 10²³ molecules
  • 1 mole of NaCl = 6.022 × 10²³ formula units

Always name the particle. "One mole of oxygen" is ambiguous: one mole of O atoms weighs 16 g, but one mole of O₂ molecules weighs 32 g.

Molar mass: from grams to moles

The molar mass is the mass of one mole, in g mol⁻¹. It has the same number as the relative mass:

Substance Relative mass Molar mass
C 12 12 g mol⁻¹
H₂O 2(1) + 16 = 18 18 g mol⁻¹
CO₂ 12 + 2(16) = 44 44 g mol⁻¹
NaCl 23 + 35.5 = 58.5 58.5 g mol⁻¹

The three formulas you need:

  • moles = mass ÷ molar mass
  • particles = moles × 6.022 × 10²³
  • volume of a gas at STP = moles × 22.414 dm³

Worked example 1: molecules in 9 g of water

  1. Molar mass of H₂O = 18 g mol⁻¹.
  2. Moles = 9 ÷ 18 = 0.5 mol.
  3. Molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules.

Now a favourite follow-up: how many atoms? Each H₂O molecule has 3 atoms (2 H + 1 O), so atoms = 3 × 3.011 × 10²³ = 9.033 × 10²³ atoms.

Worked example 2: volume of 11 g of CO₂ at STP

  1. Molar mass of CO₂ = 44 g mol⁻¹.
  2. Moles = 11 ÷ 44 = 0.25 mol.
  3. Volume = 0.25 × 22.414 = 5.60 dm³ at STP.

Worked example 3: mass of 3.011 × 10²³ glucose molecules

  1. Moles = 3.011 × 10²³ ÷ 6.022 × 10²³ = 0.5 mol.
  2. Molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16) = 180 g mol⁻¹.
  3. Mass = 0.5 × 180 = 90 g.

Notice the pattern: every question is "convert to moles first, then convert to what is asked". Moles are the bridge between grams, particles and litres.

Common mistakes in the exam

  • Not naming the particle. Write "0.5 mol of H₂O molecules", not just "0.5 mol".
  • Using 22.414 dm³ for a solid or liquid. Molar volume works only for gases, and only at STP (0 °C and 1 atm).
  • Forgetting atoms per molecule. Molecules and atoms are different counts; multiply by the number of atoms in the formula.
  • Mixing up relative mass and molar mass. Relative mass has no unit; molar mass is in g mol⁻¹.

Quick revision

  • 1 mole = 6.022 × 10²³ particles.
  • moles = mass ÷ molar mass.
  • 1 mole of any gas = 22.414 dm³ at STP.
  • Always go through moles.

Want to practise with an interactive mole converter, board-style model answers and MCQs? Open the free lesson 1.5 The Mole.

Quick answers

What is a mole in chemistry?

A mole is the amount of a substance containing 6.022 × 10²³ particles, which may be atoms, molecules, ions or formula units. Its mass in grams equals the substance's relative atomic, molecular or formula mass.

What is the value of Avogadro's number?

Avogadro's number is 6.022 × 10²³ per mole. It is the number of particles in one mole of any substance.

How many molecules are in 9 g of water?

9 g ÷ 18 g mol⁻¹ = 0.5 mol, and 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules of water.

What volume does one mole of a gas occupy at STP?

One mole of any gas occupies 22.414 dm³ at STP, which is 0 °C and 1 atmosphere pressure. This molar volume applies only to gases.

How do you convert grams to moles?

Divide the mass in grams by the molar mass in g mol⁻¹. For example, 11 g of CO₂ ÷ 44 g mol⁻¹ = 0.25 mol.