Quantitative chemistry · GCSE Chemistry
Moles
GCSE Chemistry revision on moles: n = m / Mᵣ, Avogadro’s constant, mole ratios from balanced equations, and a full worked mass-to-mass calculation.
Moles = mass in grams ÷ Mᵣ. One mole contains 6.02 × 10²³ particles. The big numbers in the equation are the mole ratio. Convert, ratio, convert.
The important bits
What you need to know
- 1
The mole is the amount of substance that contains the Avogadro number of particles, 6.02 × 10²³. One mole of carbon-12 atoms has a mass of 12 g.
- 2
Moles n = mass m (g) / relative formula mass Mᵣ. Rearrange: mass = n × Mᵣ. Mᵣ is the sum of Aᵣ values in the formula, with no units.
- 3
Mᵣ examples: H₂O = 18, CO₂ = 44, CaCO₃ = 100, NaOH = 40, HCl = 36.5, H₂SO₄ = 98, Mg = 24, O₂ = 32. Use the Aᵣ values given in the paper.
- 4
The balanced equation gives the mole ratio. 2H₂ + O₂ → 2H₂O means 2 mol hydrogen react with 1 mol oxygen to make 2 mol water.
- 5
Mass-to-mass method: write the equation, find Mᵣ of both substances, change the given mass into moles, use the ratio, change the unknown moles back into mass.
- 6
On Higher tier, moles of a gas at room temperature and pressure: volume (dm³) = n × 24, because 1 mol of any gas occupies 24 dm³ at RTP (check your specification).
- 7
Limiting reactant: the reactant that is used up first. Calculate moles of each, compare with the ratio, and use the smaller (limiting) amount to find the product.
- 8
This is Higher tier on Combined Science. Foundation still needs Mᵣ and conservation of mass. Show every step: a naked number without moles rarely scores full marks.
Quotations worth analysing
Short evidence. Real method.
“n = m / Mᵣ”
Mass must be in grams, not kilograms. Mᵣ comes from the formula and the periodic table. This one line starts almost every calculation question.
“CaCO₃ → CaO + CO₂”
One mole of calcium carbonate (100 g) makes one mole of carbon dioxide (44 g) and one mole of calcium oxide (56 g). 100 = 56 + 44: mass is conserved.
“N_A = 6.02 × 10²³ mol⁻¹”
Number of particles = n × 6.02 × 10²³. 0.50 mol of H₂O contains 3.01 × 10²³ molecules, and 9.03 × 10²³ atoms (three atoms per molecule).
Go deeper
Every calculation is convert, ratio, convert
Underline the two substances in the question. Write the balanced equation if it is not given. Find both Mᵣ values. Change the known mass into moles. Look at the big numbers: if the ratio is 2:1, double or halve as needed. Change the new moles into the mass you were asked for. If you skip the mole step and scale masses with Mᵣ values that do not match the ratio, you will be wrong whenever the equation is not 1:1. Check the order of magnitude: 4 g of hydrogen (Mᵣ = 2, so 2 mol) cannot make 4 g of water. Two moles of H₂ make two moles of H₂O (Mᵣ 18), so 36 g, because oxygen joined as well.
Go deeper
A worked mass-to-mass calculation, written as the exam wants it
Calculate the mass of carbon dioxide from 10.0 g of calcium carbonate: CaCO₃ → CaO + CO₂. Mᵣ(CaCO₃) = 40 + 12 + 48 = 100. Mᵣ(CO₂) = 12 + 32 = 44. Moles of CaCO₃ = 10.0 / 100 = 0.100 mol. Ratio 1:1, so moles of CO₂ = 0.100 mol. Mass of CO₂ = 0.100 × 44 = 4.40 g. Five short lines: Mᵣ, moles, ratio, moles, mass. If the equation were 2CaCO₃ something, you would multiply or divide at the ratio line. Significant figures: 10.0 g has three, so 4.40 g is appropriate. Units on every line stop you mixing g with mol.
Go deeper
Limiting reactants are two mole calculations, then a comparison
Suppose 6.0 g of magnesium (Mᵣ 24) reacts with 8.0 g of oxygen (Mᵣ 32): 2Mg + O₂ → 2MgO. Moles Mg = 6.0/24 = 0.25 mol. Moles O₂ = 8.0/32 = 0.25 mol. The ratio needs 2 mol Mg per 1 mol O₂, so 0.25 mol O₂ would need 0.50 mol Mg. You only have 0.25 mol Mg, so magnesium is limiting and oxygen is in excess. Product from 0.25 mol Mg: ratio 2 Mg : 2 MgO, so 0.25 mol MgO. Mᵣ(MgO) = 40, mass = 10.0 g. Using the oxygen as if it all reacted would invent extra product that never forms. Always name the limiting reactant in words.
See the idea in action
What mass of water is made when 8.0 g of hydrogen burns in excess oxygen? 2H₂ + O₂ → 2H₂O. Mᵣ(H₂) = 2.0, so moles of H₂ = 8.0 / 2.0 = 4.0 mol. Ratio 2H₂ : 2H₂O is 1:1, so moles of H₂O = 4.0 mol. Mᵣ(H₂O) = 18, mass = 4.0 × 18 = 72 g. Check: oxygen used = 4.0 mol H₂ needs 2.0 mol O₂ (ratio 2:1), mass of O₂ = 2.0 × 32 = 64 g. 8.0 + 64 = 72 g of water. Mass conserved. If you had only 16 g of oxygen, oxygen would be limiting (0.50 mol O₂ makes 1.0 mol water = 18 g) and leftover hydrogen would remain.
Exam technique
Turn knowledge into marks
Write Mᵣ, moles, ratio, moles, mass as five labelled lines. Keep mass in grams. For limiting questions, calculate moles of both reactants before you touch the product. Show the Avogadro step only if the question asks for number of particles.
Common mistakes
Do not give these marks away
- 01
Using mass numbers without dividing by Mᵣ, or leaving mass in kilograms.
- 02
Ignoring the mole ratio, especially 2:1 equations, or inventing product from the excess reactant.
- 03
Calculating Mᵣ of H₂ as 1, or of O₂ as 16, forgetting they are diatomic molecules.
How many moles are in 11 g of CO₂? (Mᵣ of CO₂ = 44)
A0.25 mol
B0.50 mol
C2.0 mol
D11 mol
Show the answer
0.25 mol. n = m / Mᵣ = 11 / 44 = 0.25 mol. 44 g would be 1.0 mol; 11 g is a quarter of that.
Quick questions
If this is the bit you searched
How do you calculate moles from mass GCSE Chemistry?
Moles = mass in grams ÷ relative formula mass (n = m / Mᵣ). Rearrange to mass = moles × Mᵣ when you need a mass.
What is the Avogadro constant?
6.02 × 10²³, the number of particles in one mole of a substance. Number of particles = moles × 6.02 × 10²³.
How do you use a balanced equation in a mole calculation?
Convert the given mass to moles, then use the big numbers in the equation as the mole ratio, then convert the new moles to mass (or volume) of the unknown.
What is a limiting reactant?
The reactant that is used up first. Extra of the other reactant cannot make more product. Work in moles and compare with the equation ratio.