Pearson Edexcel International GCSE Chemistry 4CH1: Chemical formulae, equations and calculations – Study Guide

This study guide teaches sub-topic 1(e), Chemical formulae, equations and calculations, from Topic 1 Principles of chemistry of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). It covers specification points 1.25 to 1.36. The qualification is untiered. Points 1.34C and 1.35C are printed in bold with a C reference, so they are Paper 2 only; everything else can appear on Paper 1 and Paper 2. A calculator may be used on both papers.

Use it with the course hub and the printable 4CH1 checklist. Condense it with the chemical formulae, equations and calculations revision notes, then test yourself with the formulae, equations and calculations practice questions. For all of Topic 1, see the Principles of chemistry study guide, revision notes and practice set. Find your weak spots with the free 4CH1 all-topics diagnostic.

What this sub-topic covers

Point	What you must be able to do	Paper
1.25	Write word and balanced equations with state symbols, including for unfamiliar reactions	1 and 2
1.26	Calculate relative formula masses (M_r) from relative atomic masses (A_r)	1 and 2
1.27	Know that the mole (mol) is the unit for amount of substance	1 and 2
1.28	Carry out calculations linking amount of substance, A_r and M_r	1 and 2
1.29	Calculate reacting masses from experimental data and equations	1 and 2
1.30	Calculate percentage yield	1 and 2
1.31	Explain how formulae are found by experiment: metal oxides, water, hydrated salts	1 and 2
1.32	Know the terms empirical formula and molecular formula	1 and 2
1.33	Calculate empirical and molecular formulae from experimental data	1 and 2
1.34C	Calculate with amount of substance, volume and concentration (mol/dm³)	Paper 2 only
1.35C	Calculate with gas volumes and the molar volume, 24 dm³ (24 000 cm³) at rtp	Paper 2 only
1.36	Practical: formula of a metal oxide by combustion (magnesium oxide) or reduction (copper(II) oxide)	1 and 2

1.25 Word equations and balanced equations

A balanced equation has the same number of atoms of each element on both sides. Include state symbols unless told otherwise:

• (s) solid, including a precipitate
• (l) liquid, including water itself and molten substances
• (g) gas
• (aq) aqueous: dissolved in water

Method for balancing

1. Write correct formulae. Never change a formula to make it balance.
2. Balance one element at a time with numbers in front of formulae, leaving elements on their own (H₂, O₂) until last.
3. Recount every element, then add state symbols.

Worked example 1. Aluminium reacts with dilute hydrochloric acid to give aluminium chloride solution and hydrogen gas.

word:      aluminium + hydrochloric acid → aluminium chloride + hydrogen
formulae:  Al + HCl → AlCl₃ + H₂
Cl, H:     3HCl balances Cl; its 3 H need 1.5 H₂
fix:       double: 2Al + 6HCl → 2AlCl₃ + 3H₂
check:     Al 2 = 2; H 6 = 6; Cl 6 = 6
final:     2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)

Worked example 2 (unfamiliar reaction). You are told that colourless lead(II) nitrate solution and potassium iodide solution react to form a yellow precipitate of lead(II) iodide, PbI₂, and potassium nitrate solution. “Precipitate” tells you PbI₂ is (s).

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

Count NO₃ as one unit: two on each side.

1.26 Relative formula mass

The relative formula mass, M_r, is the sum of the relative atomic masses of all the atoms in the formula. For a substance made of molecules it is also called the relative molecular mass. M_r has no units.

A number after a bracket multiplies everything inside; in a hydrated salt the dot means “add”.

Al₂(SO₄)₃:   (2 × 27) + 3 × (32 + 4 × 16) = 54 + 288 = 342
CuSO₄·5H₂O:  63.5 + 32 + (4 × 16) + 5 × 18 = 249.5

(A_r: H = 1, O = 16, Al = 27, S = 32, Cu = 63.5)

1.27 and 1.28 The mole

The mole (mol) is the unit for amount of substance. One mole has a mass in grams equal to the A_r or M_r, so 1 mol of water has a mass of 18 g.

amount (mol) = mass (g) ÷ M_r
mass (g)     = amount (mol) × M_r
M_r          = mass (g) ÷ amount (mol)

Worked example 3.

• Amount in 9.0 g of water: 9.0 ÷ 18 = 0.50 mol.
• Mass of 0.125 mol of aluminium sulfate: 0.125 × 342 = 42.75 = 42.8 g (3 s.f.).
• 0.0400 mol of a white solid has a mass of 4.24 g. M_r = 4.24 ÷ 0.0400 = 106, which matches sodium carbonate, Na₂CO₃ (2 × 23 + 12 + 3 × 16 = 106).

1.29 Reacting masses

The balanced equation gives the mole ratio. Moles, not masses, react in that ratio.

Method

1. Convert the mass you know into moles.
2. Use the equation’s ratio to find moles of the substance you want.
3. Convert those moles back into a mass.

Worked example 4. Iron(III) oxide is reduced by carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the maximum mass of iron from 40.0 g of iron(III) oxide. (A_r: O = 16, Fe = 56)

M_r(Fe₂O₃) = 2 × 56 + 3 × 16 = 160
n(Fe₂O₃)   = 40.0 ÷ 160 = 0.250 mol
ratio Fe₂O₃ : Fe = 1 : 2, so n(Fe) = 0.500 mol
mass of Fe = 0.500 × 56 = 28.0 g

From experimental data. 5.60 g of iron is heated in chlorine and forms 16.25 g of iron chloride. (A_r: Cl = 35.5, Fe = 56)

mass of chlorine reacted = 16.25 − 5.60 = 10.65 g
n(Fe)  = 5.60 ÷ 56 = 0.100 mol
n(Cl₂) = 10.65 ÷ 71 = 0.150 mol
ratio Fe : Cl₂ = 0.100 : 0.150 = 2 : 3
equation: 2Fe + 3Cl₂ → 2FeCl₃

1.30 Percentage yield

Theoretical yield: the maximum mass calculated from the equation. Actual yield: the mass obtained.

percentage yield = (actual yield ÷ theoretical yield) × 100

Worked example 5. In worked example 4, a furnace sample gives 25.2 g of iron instead of 28.0 g. Percentage yield = (25.2 ÷ 28.0) × 100 = 90.0%.

A value above 100% means an error, often a product that is still wet.

1.31 and 1.36 Finding formulae by experiment

Every method: find the mass of each element (or of salt and water), convert to moles, find the simplest whole-number ratio.

Metal oxide by combustion (magnesium oxide)

1. Weigh a crucible and lid, add clean magnesium ribbon and weigh again.
2. Heat strongly, lifting the lid now and then to let air in without letting magnesium oxide smoke escape.
3. Cool, weigh, and repeat heating until the mass is constant (reaction complete).
4. Mass of oxygen = mass of oxide − mass of magnesium.

Metal oxide by reduction (copper(II) oxide)

1. Weigh a dry reduction tube, add copper(II) oxide and weigh again.
2. Pass hydrogen or methane over the oxide, burning off the excess gas at the far end.
3. Heat until the black solid turns pink-brown; keep the gas flowing while the copper cools so it is not reoxidised.
4. Weigh, and heat to constant mass. Mass of oxygen = mass lost.

Worked example 6. 4.64 g of an iron oxide is reduced by hydrogen, leaving 3.36 g of iron. (A_r: O = 16, Fe = 56)

mass of O = 4.64 − 3.36 = 1.28 g
n(Fe) = 3.36 ÷ 56 = 0.0600     n(O) = 1.28 ÷ 16 = 0.0800
divide by smallest: Fe 1, O 1.33
multiply by 3:      Fe 3, O 4      → Fe₃O₄

A ratio like 1 : 1.33 or 1 : 1.5 is not an error: multiply up.

Water

Reduce copper(II) oxide with hydrogen and absorb the water formed in a weighed drying tube. Mass lost by the oxide = mass of oxygen; mass of water − oxygen = mass of hydrogen. If the oxide loses 1.60 g and 1.80 g of water is collected, hydrogen = 0.20 g. n(H) = 0.20 ÷ 1 = 0.20; n(O) = 1.60 ÷ 16 = 0.10. Ratio 2 : 1, so H₂O.

Salts containing water of crystallisation

Weigh the hydrated salt, heat to drive off the water of crystallisation, and heat to constant mass.

Worked example 7. 3.66 g of hydrated barium chloride, BaCl₂·xH₂O, leaves 3.12 g of anhydrous BaCl₂. (A_r: H = 1, O = 16, Cl = 35.5, Ba = 137)

mass of water = 3.66 − 3.12 = 0.54 g
n(BaCl₂) = 3.12 ÷ 208 = 0.0150 mol
n(H₂O)   = 0.54 ÷ 18 = 0.0300 mol
x = 0.0300 ÷ 0.0150 = 2      → BaCl₂·2H₂O

1.32 and 1.33 Empirical and molecular formulae

• Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula: the actual number of atoms of each element in one molecule.

To go from empirical to molecular, divide M_r by the empirical formula mass and multiply every subscript by the answer.

Worked example 8. An oxide of phosphorus contains 43.7% phosphorus and 56.3% oxygen by mass. Its M_r is 284. (A_r: O = 16, P = 31)

P: 43.7 ÷ 31 = 1.41      O: 56.3 ÷ 16 = 3.52
divide by smallest: P 1, O 2.50
multiply by 2: P 2, O 5          → empirical formula P₂O₅
empirical formula mass = 2 × 31 + 5 × 16 = 142
284 ÷ 142 = 2                    → molecular formula P₄O₁₀

Treat percentages as masses in 100 g, and keep three significant figures in the mole values.

1.34C Concentration of solutions (Paper 2 only)

Concentration in mol/dm³ is the amount of solute in 1 dm³ of solution. 1 dm³ = 1000 cm³.

concentration (mol/dm³) = amount (mol) ÷ volume (dm³)
volume (dm³) = volume (cm³) ÷ 1000

Worked example 9. 18.75 cm³ of 0.200 mol/dm³ sulfuric acid neutralises 25.0 cm³ of sodium hydroxide solution. Find the concentration of the sodium hydroxide.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
n(H₂SO₄) = 0.200 × 18.75 ÷ 1000 = 0.00375 mol
n(NaOH)  = 2 × 0.00375 = 0.00750 mol
c(NaOH)  = 0.00750 ÷ 0.0250 = 0.300 mol/dm³

To make 500 cm³ of 0.100 mol/dm³ sodium hydroxide you need 0.100 × 0.500 = 0.0500 mol, which is 0.0500 × 40 = 2.00 g.

1.35C Gas volumes (Paper 2 only)

One mole of any gas occupies 24 dm³ (24 000 cm³) at room temperature and pressure (rtp).

volume (dm³)  = amount (mol) × 24
volume (cm³)  = amount (mol) × 24 000

Worked example 10. Calculate the volume of carbon dioxide at rtp when 7.50 g of calcium carbonate is heated until it fully decomposes: CaCO₃(s) → CaO(s) + CO₂(g). (A_r: C = 12, O = 16, Ca = 40)

M_r(CaCO₃) = 100; n = 7.50 ÷ 100 = 0.0750 mol
n(CO₂) = 0.0750 mol (1 : 1)
volume = 0.0750 × 24 = 1.80 dm³

For gases at the same conditions, mole ratio = volume ratio. In CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), 50 cm³ of methane needs 100 cm³ of oxygen and gives 50 cm³ of carbon dioxide.

Common errors

• Changing a formula (writing H₂O₂ for water) to make an equation balance.
• Writing water or a precipitate as (aq).
• Putting the coefficient into the M_r: 2H₂O still has M_r 18; the 2 belongs in the ratio step.
• Dividing a percentage by the atomic number instead of A_r.
• Rounding 1.33 to 1, giving FeO instead of Fe₃O₄.
• Using cm³ directly in a mol/dm³ calculation (Paper 2).
• Using 24 when the answer is wanted in cm³ (Paper 2).

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1) specification, Issue 3, September 2024, published by Pearson Education Limited – Topic 1 Principles of chemistry, sub-topic (e) Chemical formulae, equations and calculations, points 1.25 to 1.36.