Notes › Chemistry › IGCSE 0620 › The mole and stoichiometry

Cambridge IGCSE Chemistry 0620: The Mole and Stoichiometry Notes

The mole is the unit chemists use to count particles. One mole of any substance contains 6.02 × 1023 particles, and its mass in grams equals its relative formula mass (Mr). These notes cover the mole and stoichiometry calculations in the Cambridge IGCSE Chemistry (0620) syllabus, with worked examples, an interactive mole calculator, practice questions and exam tips.

Cambridge 0620StoichiometryCalculations
Use the mole calculatorBook a 1-to-1 lesson

Written by Ms. Humaira Asim, a science teacher with 15+ years of teaching experience. Written for the Cambridge IGCSE Chemistry 0620 syllabus (2026–2028).

Core or Extended? Some stoichiometry calculations (for example limiting reactants and titration calculations) are Extended-tier content. Check your own syllabus to see which parts apply to you.

Relative atomic mass (Ar) and relative formula mass (Mr)

The relative atomic mass (Ar) of an element compares the average mass of its atoms with 1/12 of the mass of a carbon-12 atom. You look it up in the Periodic Table. The relative formula mass (Mr) is the sum of the Ar values of all the atoms in the formula. Neither has units.

Mr = sum of all Ar in the formulaUse the Ar values given in your exam Periodic Table, for example Cl = 35.5 and Cu = 63.5.

Worked example 1: Mr of calcium hydroxide, Ca(OH)2

Use Ar: Ca = 40, O = 16, H = 1.

  1. Ca(OH)2 has 1 Ca, 2 O and 2 H.
  2. Mr = 40 + (2 × 16) + (2 × 1) = 40 + 32 + 2

Mr = 74

Watch the brackets. In Ca(OH)2 the small 2 multiplies everything inside the brackets, so there are two O atoms and two H atoms.

The mole and the Avogadro constant

A mole is the amount of substance that contains 6.02 × 1023 particles. This number is the Avogadro constant. The mass of one mole of a substance, in grams, is numerically equal to its Mr. So 1 mol of water (Mr = 18) has a mass of 18 g and contains 6.02 × 1023 water molecules.

moles (mol) = mass (g) ÷ MrRearranged: mass = moles × Mr and Mr = mass ÷ moles

Worked example 2: moles from mass

How many moles are in 5.3 g of sodium carbonate, Na2CO3? (Ar: Na = 23, C = 12, O = 16)

  1. Mr = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106
  2. moles = 5.3 ÷ 106

moles = 0.050 mol

Worked example 3: mass from moles

What is the mass of 0.25 mol of sulfuric acid, H2SO4? (Ar: H = 1, S = 32, O = 16)

  1. Mr = (2 × 1) + 32 + (4 × 16) = 98
  2. mass = 0.25 × 98

mass = 24.5 g

Reacting masses from balanced equations

The numbers in front of formulae in a balanced equation tell you the ratio of moles. Use this method for any reacting-mass question:

  1. Write the balanced equation.
  2. Convert the mass of the substance you know into moles (mass ÷ Mr).
  3. Use the equation ratio to find the moles of the substance you want.
  4. Convert those moles into mass (moles × Mr).

Worked example 4: mass of product

Mg + 2HCl → MgCl2 + H2. What mass of magnesium chloride forms from 6.0 g of magnesium? (Ar: Mg = 24, Cl = 35.5)

  1. moles of Mg = 6.0 ÷ 24 = 0.25 mol
  2. Ratio Mg : MgCl2 = 1 : 1, so moles of MgCl2 = 0.25 mol
  3. Mr of MgCl2 = 24 + (2 × 35.5) = 95
  4. mass = 0.25 × 95

mass = 23.75 g (about 24 g)

Gas volumes at RTP

At room temperature and pressure (RTP), one mole of any gas occupies 24 dm3 (24 000 cm3). Equal volumes of different gases, measured at the same temperature and pressure, contain the same number of molecules.

gas volume (dm3) = moles × 24Divide by 24 to go back to moles. 1 dm3 = 1000 cm3.

Worked example 5: volume of gas

CaCO3 + 2HCl → CaCl2 + H2O + CO2. What volume of carbon dioxide at RTP forms from 10 g of calcium carbonate? (Ar: Ca = 40, C = 12, O = 16)

  1. Mr of CaCO3 = 40 + 12 + 48 = 100
  2. moles of CaCO3 = 10 ÷ 100 = 0.10 mol
  3. Ratio CaCO3 : CO2 = 1 : 1, so moles of CO2 = 0.10 mol
  4. volume = 0.10 × 24

volume = 2.4 dm3

Concentration of solutions

Concentration tells you how much solute is dissolved in 1 dm3 of solution. It can be given in g/dm3 or mol/dm3. Always convert volumes in cm3 to dm3 by dividing by 1000.

concentration (mol/dm3) = moles ÷ volume (dm3)concentration (g/dm3) = mass ÷ volume (dm3)  |  moles = concentration × volume (dm3)

Worked example 6: concentration

4.0 g of sodium hydroxide, NaOH, is dissolved to make 250 cm3 of solution. Find the concentration in mol/dm3 and g/dm3. (Ar: Na = 23, O = 16, H = 1)

  1. Mr of NaOH = 40, so moles = 4.0 ÷ 40 = 0.10 mol
  2. volume = 250 ÷ 1000 = 0.250 dm3
  3. concentration = 0.10 ÷ 0.250 = 0.40 mol/dm3
  4. in g/dm3: 4.0 ÷ 0.250 = 16 g/dm3

0.40 mol/dm3 (16 g/dm3)

Limiting reactants

The limiting reactant is the one that is used up first, so it decides how much product forms. The other reactant is in excess. To find it, work out the moles of each reactant, then compare them using the equation ratio.

Worked example 7: which reactant is limiting?

2.4 g of magnesium reacts with 100 cm3 of 1.0 mol/dm3 hydrochloric acid. Mg + 2HCl → MgCl2 + H2. What volume of hydrogen forms at RTP?

  1. moles of Mg = 2.4 ÷ 24 = 0.10 mol
  2. moles of HCl = 1.0 × 0.100 = 0.10 mol
  3. 0.10 mol of Mg needs 0.20 mol of HCl, but only 0.10 mol is available, so HCl is limiting
  4. Ratio HCl : H2 = 2 : 1, so moles of H2 = 0.05 mol
  5. volume = 0.05 × 24

volume of H2 = 1.2 dm3

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in one molecule.

  1. Write the mass (or percentage) of each element.
  2. Divide each by its Ar to get moles.
  3. Divide all the moles by the smallest number.
  4. Round to a simple whole-number ratio (multiply if needed to remove halves or thirds).
molecular formula = empirical formula × nn = Mr of the compound ÷ mass of the empirical formula

Worked example 8: empirical and molecular formula

A hydrocarbon contains 2.4 g of carbon and 0.6 g of hydrogen. Its Mr is 30.

  1. C: 2.4 ÷ 12 = 0.2 mol   H: 0.6 ÷ 1 = 0.6 mol
  2. Divide by 0.2: C = 1, H = 3, so the empirical formula is CH3
  3. Empirical formula mass = 12 + 3 = 15, so n = 30 ÷ 15 = 2

Molecular formula = C2H6

Percentage yield and percentage purity

percentage yield = (actual yield ÷ theoretical yield) × 100percentage purity = (mass of pure substance ÷ mass of impure sample) × 100

The theoretical yield is the maximum mass of product the equation predicts. The actual yield is what you really collect, which is usually less because of incomplete reactions, side reactions and losses during separation.

Worked example 9: percentage yield

A reaction should produce 15.0 g of product, but only 12.0 g is collected.

  1. percentage yield = (12.0 ÷ 15.0) × 100

percentage yield = 80%

Titration calculations

In a titration you find the unknown concentration of one solution using a solution of known concentration. Use moles = concentration × volume (in dm3), then the equation ratio.

Worked example 10: finding a concentration

25.0 cm3 of sodium hydroxide solution is neutralised by 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid. NaOH + HCl → NaCl + H2O. Find the concentration of the NaOH.

  1. moles of HCl = 0.100 × (20.0 ÷ 1000) = 0.00200 mol
  2. Ratio HCl : NaOH = 1 : 1, so moles of NaOH = 0.00200 mol
  3. concentration = 0.00200 ÷ (25.0 ÷ 1000)

concentration = 0.0800 mol/dm3

Common mistakes in mole calculations

  • Ignoring the equation ratio. If the equation has 2HCl, the moles of HCl are twice the moles of Mg that react.
  • Using cm3 instead of dm3. Divide cm3 by 1000 before using concentration.
  • Using the wrong Mr. Gases like H2, O2 and Cl2 are diatomic, so their Mr values are 2, 32 and 71.
  • Rounding too early. Keep extra digits in your calculator and round only the final answer, usually to 2 or 3 significant figures.
  • Missing the brackets in a formula. In Ca(OH)2 the 2 applies to both O and H.
  • Leaving out units. Write g, mol, dm3 or mol/dm3 in your answer.
Exam tip: Write the formula first, show every step, and check your answer is sensible. Examiners award marks for method even when the final answer is wrong.

Interactive mole calculator

Type a formula such as Ca(OH)2 or CuSO4.5H2O and enter either the mass or the moles. The other value updates as you type. It uses the Periodic Table Ar values taught at IGCSE.

Want more tools? Try our other science calculators.

Moles, mass and Mr

Mr = ?

Concentration

Gas volume at RTP

1 mol of any gas = 24 dm³ at RTP.

Practice questions with answers

Try each question first, then open the answer to check your working. All questions use these values: Ar H = 1, C = 12, O = 16, Mg = 24, Ca = 40, Cu = 63.5.

Q1. Calculate the Mr of Ca(OH)2.

40 + (2 × 16) + (2 × 1) = 74

Q2. How many moles are in 14.8 g of Ca(OH)2?

14.8 ÷ 74 = 0.20 mol

Q3. 2Cu + O2 → 2CuO. What mass of CuO forms from 0.50 mol of copper?

Ratio Cu : CuO = 1 : 1, so 0.50 mol of CuO. Mr = 63.5 + 16 = 79.5. Mass = 0.50 × 79.5 = 39.75 g (about 39.8 g)

Q4. What volume of hydrogen at RTP forms when 1.2 g of magnesium reacts with excess acid? Mg + 2HCl → MgCl2 + H2

Moles of Mg = 1.2 ÷ 24 = 0.050 mol. Ratio Mg : H2 = 1 : 1. Volume = 0.050 × 24 = 1.2 dm3

Q5. An 8.0 g sample of impure limestone contains 6.4 g of calcium carbonate. What is the percentage purity?

(6.4 ÷ 8.0) × 100 = 80%

Q6. (Extended) A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mr is 180. Find its molecular formula.

C: 40.0 ÷ 12 = 3.33   H: 6.7 ÷ 1 = 6.7   O: 53.3 ÷ 16 = 3.33. Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH2O (mass 30). n = 180 ÷ 30 = 6. Molecular formula = C6H12O6

Q7. (Extended) 2.4 g of Mg is added to 100 cm3 of 1.0 mol/dm3 HCl. Which reactant is in excess?

Moles of Mg = 0.10 mol. Moles of HCl = 0.10 mol, but 0.10 mol of Mg needs 0.20 mol of HCl. So magnesium is in excess and HCl is limiting.

Frequently asked questions

What is a mole in IGCSE Chemistry?

A mole is the amount of substance that contains 6.02 × 1023 particles (the Avogadro constant). Its mass in grams equals the relative formula mass.

How do you calculate moles from mass?

Divide the mass in grams by the relative formula mass (Mr): moles = mass ÷ Mr.

What is the volume of one mole of gas at RTP?

24 dm3 (24 000 cm3) at room temperature and pressure.

What is the difference between an empirical and a molecular formula?

The empirical formula is the simplest whole-number ratio of atoms. The molecular formula is the actual number of atoms in one molecule, for example CH2O and C6H12O6.

How do you find the limiting reactant?

Work out the moles of each reactant, divide each by its coefficient in the balanced equation, and the smaller value is the limiting reactant.

Stuck on mole calculations?

Get personal help with Ms. Asim in a 1-to-1 online lesson, scheduled around your time zone.

Book an appointmentMore Chemistry notes

This is an independent study resource. It is not affiliated with or endorsed by Cambridge University Press & Assessment. Always check the latest official syllabus for your exam series.

Scroll to Top