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IGCSE · Cambridge (CIE) · 0620

Moles, molar mass and particle counts

Connect a measured mass with an amount of substance and a specified particle count.

Stoichiometry · 3.3

Extended only (Supplement) · Chemistry 0620, 2026–2028. Core reacting-mass proportions are taught separately.

Before you begin

  • Calculate relative atomic, molecular or formula mass.
  • Use powers of ten and rearrange a simple equation.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What is water's molar mass when its relative molecular mass is 18?

What you will learn

  • Use the mole and Avogadro constant to count specified particles.
  • Rearrange n=m/M to calculate amount, mass or molar mass.
Mass mgramsAmount nmoles÷ M (g/mol)× M (g/mol)Specified particle countName the particlesCount: n × 6.02×10²³Moles: N ÷ 6.02×10²³
Calculation routes meet at amount n in moles. Divide mass in grams by molar mass in grams per mole to find n. Multiply n by molar mass to return to mass. Multiply n by 6.02 times ten to the power 23 per mole to count specified particles; divide a particle count by that constant to return to moles. Name the particles: molecules, atoms, ions or formula units.

Specify the particles you are counting

A mole, symbol mol, is the unit for amount of substance. For this syllabus, use the Avogadro constant 6.02×10²³ per mole: one mole of specified particles contains 6.02×10²³ of those particles. Name the entities, such as atoms, molecules, ions or formula units. One mole of O₂ molecules contains two moles of oxygen atoms because each molecule contains two atoms.

Do not call a mole a fixed mass. Different substances have different masses per mole. Molar mass M has units g/mol; using supplied relative masses, its numerical value matches Aᵣ for an element's atoms or Mᵣ for the specified molecular or formula unit. Relative mass itself has no unit. For example, water has Mᵣ=18 and molar mass 18 g/mol using H=1 and O=16.

Use the mass route and the particle route

For mass m in grams and molar mass M in g/mol, the amount n in mol is m/M. Rearranging gives m=nM and M=m/n. Keep the units visible so that a kilogram sample is converted to grams before using a g/mol value. An unknown molar mass can help determine an unknown relative atomic or molecular mass when the particle identity is stated.

Particle count N equals n times the Avogadro constant. To go backwards, divide N by that constant. If the question asks for all atoms in a molecular sample, first find its molecule count, then multiply by the number of atoms per molecule. Round the final answer sensibly after the complete calculation, rather than rounding each intermediate amount.

n = m/M; m = nM; M = m/n; N = n × 6.02×10²³

Pause and explain

How many moles of oxygen atoms are in 0.40 mol of O₂ molecules?

Put the idea to work

Worked example

Find the amount, molecule count and total atom count in 9.0 g H₂O using M=18 g/mol and Avogadro constant 6.02×10²³/mol.

Show the worked solution
  1. Amount is n=9.0/18=0.50 mol of water molecules.
  2. Molecule count is 0.50×6.02×10²³=3.01×10²³.
  3. Each molecule has three atoms, so total atom count is 3×3.01×10²³=9.03×10²³ before final rounding.

Answer 0.50 mol; 3.01×10²³ molecules; 9.03×10²³ atoms (unrounded counts).

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Calculate the amount and molecule count in 22 g CO₂ using C=12 and O=16.

Give me a hint

Find molar mass, then divide mass by it before counting molecules.

Compare my reasoning
  1. Molar mass is 12+2×16=44 g/mol.
  2. n=22/44=0.50 mol.
  3. N=0.50×6.02×10²³=3.01×10²³ CO₂ molecules.

0.50 mol and 3.01×10²³ CO₂ molecules.

Look for these in your work

  • The formula gives the molar mass.
  • The final count names molecules, not all atoms.
2 · Independent

An element's atoms have amount 0.20 mol and mass 5.4 g. Find molar mass, relative atomic mass and atom count.

Give me a hint

Use M=m/n; then separate the units of M and Aᵣ.

Compare my reasoning
  1. M=5.4/0.20=27 g/mol.
  2. The corresponding relative atomic mass is 27 with no unit.
  3. Atom count is 0.20×6.02×10²³=1.204×10²³ before rounding.

27 g/mol; Aᵣ=27; 1.204×10²³ atoms before rounding.

Look for these in your work

  • Relative mass and molar mass have distinct units.
  • Atom count uses the stated amount.
3 · Transfer

A sample contains 3.01×10²³ O₂ molecules. Calculate its amount of molecules, mass using O=16, and number of oxygen atoms.

Give me a hint

Divide by Avogadro's constant; O₂ has molar mass 32 g/mol and two atoms per molecule.

Compare my reasoning
  1. n=3.01×10²³/(6.02×10²³)=0.50 mol O₂.
  2. Mass is 0.50×32=16 g.
  3. There are 2×3.01×10²³=6.02×10²³ oxygen atoms.

0.50 mol O₂; 16 g; 6.02×10²³ oxygen atoms.

Look for these in your work

  • The reverse particle calculation is explicit.
  • The particle identity determines both molar mass and the atom multiplier.

Common mistakes

  • Using the same mass for one mole of every substance.
  • Answering an atom-count question with a molecule count.
  • Putting a gram unit on Aᵣ or Mᵣ instead of distinguishing molar mass.
Recall without your notes

What can you explain now?

A sample contains 0.25 mol of a substance and has mass 11.0 g. Find its molar mass.

Compare with the explanation

44 g/mol

Rearrange n=m/M to M=m/n: 11.0/0.25=44 g/mol.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Explain the difference between relative mass and molar mass. Convert a mass into a molecule count, then into an atom count, naming the particles at every step.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Chemistry

Course scope and assessment

143 sequenced lessons cover all twelve syllabus areas with 429 guided, independent and application problems. 49 of the 49 syllabus sections have linked teaching and practice. This measures availability of mapped content, not full programme completion or guaranteed exam readiness. Three earlier starter lessons remain available. Extend your formula notebook as new reagents and ions appear; the starter notebook is not a universal formula reference. Paper-based method exercises support laboratory learning but do not demonstrate practical competence or complete either practical-assessment route. Written practice is self-checked.

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