Pearson Edexcel International GCSE Chemistry 4CH1: Chemical formulae, equations and calculations – Revision Notes

These revision notes condense sub-topic 1(e), Chemical formulae, equations and calculations (points 1.25 to 1.36), from Topic 1 Principles of chemistry of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024). The qualification is untiered. Points 1.34C and 1.35C are Paper 2 only; the rest can be tested on both papers, and a calculator is allowed on both. For full explanations and worked examples, read the chemical formulae, equations and calculations study guide first.

Links: Edexcel IGCSE Chemistry course hub, printable 4CH1 checklist, formulae, equations and calculations practice questions, Principles of chemistry revision notes for the rest of Topic 1, and the free 4CH1 all-topics diagnostic.

Key definitions

• Relative formula mass (M_r): the sum of the relative atomic masses (A_r) of all the atoms in a formula. For a molecular substance it is the relative molecular mass. No units.
• Mole (mol): the unit for amount of substance (1.27). One mole has a mass in grams equal to the A_r or M_r.
• Empirical formula: the simplest whole-number ratio of atoms of each element in a compound (1.32).
• Molecular formula: the actual number of atoms of each element in one molecule (1.32).
• Water of crystallisation: water that is part of the crystal structure of a salt, shown after a dot, as in CuSO₄·5H₂O.
• Theoretical yield: the maximum mass of product calculated from the equation. Actual yield: the mass you obtain.
• Concentration (Paper 2 only): amount of solute per dm³ of solution, in mol/dm³.
• Molar volume (Paper 2 only): the volume of one mole of any gas, 24 dm³ or 24 000 cm³ at room temperature and pressure (rtp).

State symbols (1.25)

Symbol	Meaning	Typical example
(s)	solid, including a precipitate	metals, insoluble salts, carbonates heated as solids
(l)	liquid	water, molten compounds, bromine
(g)	gas	H₂, O₂, CO₂, NH₃
(aq)	dissolved in water	acids, alkalis, soluble salts in solution

Formula table

Quantity	Formula	Units	Paper
amount	n = mass ÷ M_r	mol	1 and 2
mass	mass = n × M_r	g	1 and 2
percentage yield	(actual ÷ theoretical) × 100	%	1 and 2
empirical to molecular	factor = M_r ÷ empirical formula mass	none	1 and 2
concentration	c = n ÷ V	mol/dm³ (V in dm³)	Paper 2 only
amount in solution	n = c × V	mol	Paper 2 only
volume conversion	dm³ = cm³ ÷ 1000	dm³	Paper 2 only
gas volume	V = n × 24	dm³	Paper 2 only
gas volume	V = n × 24 000	cm³	Paper 2 only

Method in steps

Balancing an equation (1.25)

1. Write every formula correctly. Never change a subscript.
2. Balance one element at a time using numbers in front of formulae.
3. Leave free elements (O₂, H₂, Cl₂) until last; double everything to clear a ½.
4. Recount every element, then add state symbols.

For an unfamiliar reaction, take the formulae and states from the question. “Precipitate” means (s); “solution” means (aq); “fizzing” or “gas given off” means (g).

Reacting mass (1.29)

1. Mass → moles of the known substance (÷ M_r).
2. Moles × ratio from the balanced equation.
3. Moles → mass of the wanted substance (× M_r).

Worked reminder. Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq). 3.31 g of lead(II) nitrate (M_r 331) is 0.0100 mol, so 0.0100 mol of PbI₂ (M_r 461) forms: 4.61 g. (A_r: N = 14, O = 16, I = 127, Pb = 207)

If the question gives masses of reactant and product, find moles of each and use them to deduce the ratio, and so the equation.

Empirical formula (1.33)

1. Write the mass (or %) of each element.
2. Divide each by its A_r.
3. Divide every answer by the smallest.
4. If you get .5, ×2; if .33 or .67, ×3.
5. Molecular formula: M_r ÷ empirical formula mass, then multiply all subscripts.

Water of crystallisation (1.31)

1. Mass of water = hydrated mass − anhydrous mass.
2. Moles of anhydrous salt and moles of water.
3. x = moles of water ÷ moles of salt.

Concentration and titration (1.34C, Paper 2 only)

1. Convert every volume to dm³ (÷ 1000).
2. n = c × V for the solution you know fully.
3. Use the equation’s ratio.
4. c = n ÷ V for the unknown.

Gas volume (1.35C, Paper 2 only)

1. Moles of the known substance.
2. Ratio to moles of gas.
3. × 24 for dm³ or × 24 000 for cm³.

Worked reminder. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g). 2.30 g of sodium (A_r 23) is 0.100 mol, giving 0.0500 mol of hydrogen: 0.0500 × 24 000 = 1200 cm³ (1.20 dm³) at rtp.

For reacting gases only, volume ratio = mole ratio, so no conversion to moles is needed.

Practical 1.36: formula of a metal oxide

	Combustion (magnesium oxide)	Reduction (copper(II) oxide)
What happens	metal gains oxygen	oxide loses oxygen
Apparatus	crucible and lid, pipe-clay triangle, Bunsen	reduction tube, gas supply (hydrogen or methane), Bunsen
Mass of oxygen	final mass − mass of metal	mass of oxide − mass of metal left
Key precaution	lift lid now and then for air; keep it on to stop MgO smoke escaping	keep gas flowing while copper cools so it is not reoxidised; burn excess gas
Completion check	heat, cool, weigh to constant mass	heat, cool, weigh to constant mass
Error if not complete	too little oxygen measured	too little oxygen measured

The formula of water can be found the same way: reduce copper(II) oxide with hydrogen and absorb the water formed in a weighed drying tube. Oxygen = mass lost by the oxide; hydrogen = mass of water − mass of oxygen.

Must-know distinctions

• A_r vs M_r: A_r is for one element’s atoms; M_r is for a whole formula.
• Coefficient vs subscript: 3H₂O means three molecules, each with M_r 18. The 3 goes in the ratio step, not the M_r.
• Empirical vs molecular: C₂H₆ is molecular; CH₃ is its empirical formula. Ionic compounds are given by empirical formulae.
• Theoretical vs actual yield: actual goes on top. Percentage yield cannot exceed 100%.
• (l) vs (aq): water is (l). A substance dissolved in water is (aq).
• cm³ vs dm³ (Paper 2): divide cm³ by 1000 before using mol/dm³.
• 24 vs 24 000 (Paper 2): match the molar volume to the units the answer asks for.

Quick self-test

1. Balance: C₂H₆ + O₂ → CO₂ + H₂O.
2. Calculate the M_r of calcium hydroxide, Ca(OH)₂. (A_r: H = 1, O = 16, Ca = 40)
3. Calculate the M_r of magnesium nitrate, Mg(NO₃)₂. (A_r: N = 14, O = 16, Mg = 24)
4. How many moles are in 4.4 g of carbon dioxide? (A_r: C = 12, O = 16)
5. What is the mass of 0.30 mol of sodium hydroxide, NaOH? (A_r: H = 1, O = 16, Na = 23)
6. Zn + S → ZnS. Calculate the mass of zinc sulfide formed from 13.0 g of zinc. (A_r: S = 32, Zn = 65)
7. In question 6, 16.5 g of zinc sulfide is obtained. Calculate the percentage yield.
8. A hydrocarbon contains 2.4 g of carbon and 0.6 g of hydrogen. Its M_r is 30. Find its empirical and molecular formulae. (A_r: H = 1, C = 12)
9. 6.44 g of hydrated sodium sulfate, Na₂SO₄·xH₂O, leaves 2.84 g of anhydrous salt on heating to constant mass. Find x. (A_r: H = 1, O = 16, Na = 23, S = 32)
10. (Paper 2 only) How many moles of HCl are in 20.0 cm³ of 0.150 mol/dm³ hydrochloric acid?
11. (Paper 2 only) What volume, in cm³, does 0.0250 mol of hydrogen occupy at rtp?
12. (Paper 2 only) 0.0400 mol of solute is dissolved to make 250 cm³ of solution. Calculate the concentration.

Answers

1. 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O (C 4 = 4; H 12 = 12; O 14 = 8 + 6).
2. 40 + 2 × (16 + 1) = 74.
3. 24 + 2 × (14 + 3 × 16) = 148.
4. 4.4 ÷ 44 = 0.10 mol.
5. 0.30 × 40 = 12.0 g.
6. n(Zn) = 13.0 ÷ 65 = 0.200 mol; ratio 1 : 1; M_r(ZnS) = 97; mass = 0.200 × 97 = 19.4 g.
7. (16.5 ÷ 19.4) × 100 = 85.1%.
8. C: 2.4 ÷ 12 = 0.20; H: 0.6 ÷ 1 = 0.60; ratio 1 : 3, empirical formula CH₃; 30 ÷ 15 = 2, molecular formula C₂H₆.
9. Water = 3.60 g = 0.200 mol; Na₂SO₄ (M_r 142) = 0.0200 mol; x = 10.
10. 0.150 × 20.0 ÷ 1000 = 0.00300 mol.
11. 0.0250 × 24 000 = 600 cm³.
12. 0.0400 ÷ 0.250 = 0.160 mol/dm³.

Where marks are usually lost

• Balancing by changing a formula (for example writing H₂O₂ for water) instead of adding coefficients.
• Leaving out state symbols when the question asks for them, or writing (aq) for water or a precipitate.
• Forgetting the bracket multiplier, so Mg(NO₃)₂ is worked out as 24 + 14 + 6 × 16.
• Using the mass ratio instead of the mole ratio in a reacting-mass question.
• Rounding 1.33 or 1.5 to a whole number in an empirical formula instead of multiplying up.
• Dividing the masses of water and salt (not their moles) when finding x in a hydrated salt.
• Giving a percentage yield above 100% by putting the theoretical yield on top.
• In a practical write-up, not mentioning heating to constant mass, or letting copper cool in air after reduction.
• (Paper 2) Using a volume in cm³ in c = n ÷ V without dividing by 1000.
• (Paper 2) Multiplying by 24 when cm³ is asked for, or by 24 000 when dm³ is asked for.

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.