Overview

What you'll learn

Write formulae. Build the formula of a compound from the ions or atoms in it by balancing charges — cation first, brackets around a polyatomic ion when you need more than one.

Construct and balance equations. Adjust coefficients so every element balances, add state symbols, and strip out spectator ions to write ionic equations.

Use the mole. Convert between mass, gas volume, concentration and number of particles — everything routes through moles.

Solve stoichiometry problems. Handle reacting masses, the limiting reactant, percentage yield, percentage purity and empirical formulae with one repeatable method.

Tutor's Insight

"The maths that runs through everything."
Both subtopics here are non-negotiable. Acids, electrolysis, energetics — almost every structured question in Paper 1 and Paper 2 leans on these calculations somewhere. Remember 24 dm³ is the volume of any gas at r.t.p. There are no shortcuts: the marks go to people who've practised the method until it's automatic. Drill the four steps now, every time, and you stop losing easy marks later.

4.1 Formulae & Equation Writing

4.1 Formulae & Equation Writing

What this subtopic asks of you
  • Write the formulae of compounds from the ions or atoms in them.
  • Construct and balance chemical equations — including ionic equations — with state symbols.

4.1 Formulae & Equation Writing

Build a formula by balancing the charges

The Rule
Total + charge = total − charge
An ionic compound's formula is the simplest ratio of ions where + and − charges cancel exactly. Write the cation first, then the anion. Put brackets around a polyatomic ion when you need more than one of it.
Worked — Calcium Nitrate
Ca²⁺ and NO₃⁻
One Ca²⁺ carries 2+. Each NO₃⁻ carries 1−, so you need two of them to balance it.

Formula: Ca(NO₃)₂ — brackets, because there's more than one nitrate ion.
Quick method — swap the charge numbers and drop them in as subscripts, then simplify.

4.1 Formulae & Equation Writing

The ions worth knowing by heart

Type Charge Examples
Group 1 cations 1+ Na⁺, K⁺
Group 2 cations 2+ Mg²⁺, Ca²⁺
Other cations varies Al³⁺, Zn²⁺, Ag⁺, H⁺
Transition metal cations Roman numeral Fe²⁺, Fe³⁺, Cu²⁺
Group 16 anions 2− O²⁻, S²⁻
Group 17 anions 1− Cl⁻, Br⁻, I⁻
Hydroxide / Nitrate 1− OH⁻, NO₃⁻
Carbonate / Sulfate 2− CO₃²⁻, SO₄²⁻
Ammonium 1+ NH₄⁺
For the simple ions, the group number points straight to the charge. Polyatomic ions (OH⁻, NO₃⁻, CO₃²⁻, SO₄²⁻, NH₄⁺) act as one unit — learn them as a block.

Free Notes · O-Level Pure Chemistry

Read the full chapter

Balancing and ionic equations, the mole concept, molar mass and volume, the four-step method, limiting reactant, percentage yield and purity, empirical formulae and 4 worked exam questions.

4.1 Formulae & Equation Writing

Balance the atoms, then strip the spectators

Step 01
Balancing an equation
Every element must have the same number of atoms on both sides. Adjust the coefficients in front of formulae — never change a formula itself. Add state symbols: (s), (l), (g), (aq).
Step 02
Writing an ionic equation
Split aqueous acids and aqueous ionic compounds into their ions. Leave solids, liquids and gases whole. Then cancel the spectator ions — the ones unchanged on both sides.
Exam Habit

Sanity check — Count every atom on each side before you move on. One missed atom costs the mark.

4.2 The Mole Concept & Stoichiometry

4.2 The Mole Concept & Stoichiometry

What this subtopic asks of you
  • Use the mole to convert between mass, gas volume, concentration and number of particles.
  • Solve reacting-mass and volume problems — including limiting reactants, % yield, % purity and empirical formulae.

4.2 The Mole Concept & Stoichiometry

Mass, measured relative to carbon-12

Relative Atomic Mass · Aᵣ
The average mass of an atom
Compared to 1/12 the mass of a carbon-12 atom. It's an average because most elements exist as a mix of isotopes.
Relative Molecular Mass · Mᵣ
Add up the atoms
Add up the Aᵣ values of every atom in the formula. For an ionic compound it's the relative formula mass.
% by mass of an element = (number of atoms × Aᵣ of the element) ÷ Mᵣ of the compound × 100%.

4.2 The Mole Concept & Stoichiometry

One mole is just a counted amount

The Number
6.02 × 10²³ particles
One mole of anything contains 6.02 × 10²³ particles — the Avogadro constant.
Molar Mass
Grams per mole
Mass of one mole, in grams. Numerically equal to the Aᵣ or Mᵣ. Units: g/mol.
Molar Volume
24 dm³ for any gas
One mole of any gas occupies 24 dm³ at room temperature and pressure (r.t.p.).
Why a 'mole'? Atoms are too small to count directly. The mole lets us weigh out a known number of them.

4.2 The Mole Concept & Stoichiometry

Everything routes through moles

MASS → MOLES → VOLUME · CONCENTRATION
The Map
Get to moles, then convert out

Mass ↔ moles: ÷ Mᵣ to reach moles; × Mᵣ to go back to grams.


Gas volume ↔ moles: ÷ 24 dm³ to reach moles; × 24 dm³ to go back to a volume at r.t.p.


Solution ↔ moles: moles = concentration × volume.


Get to moles, use the mole ratio from the equation, then convert out. To go the other way, reverse the operation.

Key Formula · Moles
number of moles =
mass (in grams) Mᵣ
Also: moles of gas = volume ÷ 24 dm³ (at r.t.p.)  ·  moles in solution = concentration × volume.

4.2 The Mole Concept & Stoichiometry

Four steps, every time

Step 01
Balanced equation
Write it out and balance it before anything else.
Step 02
Moles of the known
From its mass, gas volume or concentration.
Step 03
Use the mole ratio
Scale, using the ratio from the equation, to what you want.
Step 04
Convert to the answer
Back to a mass, a volume or a concentration.
It never changes — the same four steps solve reacting-mass, gas-volume and titration problems alike.

4.2 The Mole Concept & Stoichiometry

The limiting reactant sets the limit

What it means
One reactant runs out first
When reactants aren't mixed in the exact equation ratio, one runs out first. That one — the limiting reactant — decides how much product forms. The rest is in excess.
The Analogy
Car bodies and wheels
One car body + four wheels → one car. With 10 bodies but only 12 wheels, you build just 3 cars — the wheels run out first. The wheels are the limiting reactant; the car bodies are in excess.
To find it — work out the moles of each reactant, then compare against the mole ratio in the equation.

4.2 The Mole Concept & Stoichiometry

Two ratios that measure how well it went

Percentage Yield
actual ÷ theoretical × 100%
Theoretical yield = the maximum the balanced equation predicts. Actual yield = what you really obtain. Reactions rarely give 100%.
Percentage Purity
pure mass ÷ sample mass × 100%
A sample often has impurities. From the amount of product formed, you back-calculate the mass of pure substance in it.
Exam Habit

Don't mix them up — yield is about the reaction's efficiency. Purity is about the sample you started with.

4.2 The Mole Concept & Stoichiometry

From masses to a formula

Step 01
Empirical formula
The simplest whole-number ratio of atoms. Divide each element's mass (or %) by its Aᵣ, then simplify the ratio.
Step 02
Molecular formula
The actual number of atoms in one molecule. It is (empirical formula)ₙ — find n by comparing the empirical formula mass with the given Mᵣ.
Worked example

Empirical formula CH₂, Mᵣ = 42 → (CH₂)ₙ with 14n = 42, so n = 3 → molecular formula C₃H₆.

Practice

Exam-style questions

Question 01
The Question
Structured

Reacting masses

48 g of magnesium reacts completely with excess hydrochloric acid.


What volume of hydrogen gas is produced at r.t.p.?


[ Aᵣ: Mg = 24 ]

Worked Answer
48 dm³ of hydrogen
  1. Mg + 2HCl → MgCl₂ + H₂
  2. moles of Mg = 48 ÷ 24 = 2 mol
  3. mole ratio Mg : H₂ = 1 : 1, so 2 mol of H₂
  4. volume of H₂ = 2 × 24 = 48 dm³
Question 02
The Question
MCQ

Number of moles in a mass

How many moles of calcium carbonate, CaCO₃, are present in 25 g of the compound?

[ Aᵣ: Ca = 40, C = 12, O = 16 ]

  • A   0.25 mol
  • B   0.40 mol
  • C   2.50 mol
  • D   4.00 mol
Worked Answer
A — 0.25 mol
  1. Mᵣ of CaCO₃ = 40 + 12 + (3 × 16) = 100.
  2. moles = mass ÷ Mᵣ = 25 ÷ 100 = 0.25 mol.
  3. Answer: A.
Question 03
The Question
Structured

Limiting reactant

Hydrogen reacts with oxygen to form water:

2H₂ + O₂ → 2H₂O

6 mol of H₂ is mixed with 2 mol of O₂. Which reactant is limiting, and how many moles of water form?

Worked Answer
O₂ limiting → 4 mol H₂O
  1. The equation needs H₂ : O₂ in a 2 : 1 ratio.
  2. 2 mol O₂ would need 4 mol H₂, but 6 mol H₂ is present — so H₂ is in excess and O₂ runs out first. O₂ is the limiting reactant.
  3. Work from the limiting reactant: 2 mol O₂ → 4 mol H₂O (ratio O₂ : H₂O = 1 : 2).
  4. Answer: 4 mol of water (2 mol of H₂ left over in excess).
Question 04
The Question
Structured

Empirical formula from percentages

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mᵣ is 60.

Find its empirical formula and molecular formula.

[ Aᵣ: C = 12, H = 1, O = 16 ]

Worked Answer
CH₂O → C₂H₄O₂
  1. Divide each % by its Aᵣ: C = 40.0 ÷ 12 = 3.33; H = 6.7 ÷ 1 = 6.7; O = 53.3 ÷ 16 = 3.33.
  2. Divide by the smallest (3.33): C = 1, H = 2, O = 1 → empirical formula CH₂O.
  3. Empirical formula mass = 12 + 2 + 16 = 30. n = 60 ÷ 30 = 2.
  4. Molecular formula = (CH₂O)₂ = C₂H₄O₂.

Frequently Asked Questions

Chemical Calculations — FAQ

What is Chemical Calculations in the O-Level Chemistry syllabus (6092)?
Chemical Calculations (Topic 4 of Syllabus 6092) has two subtopics: 4.1 Formulae & Equation Writing — writing formulae of compounds from their ions, and constructing and balancing chemical and ionic equations with state symbols; and 4.2 The Mole Concept & Stoichiometry — using the mole to convert between mass, gas volume, concentration and number of particles, and solving reacting-mass problems including limiting reactants, % yield, % purity and empirical formulae. Almost every structured question in Paper 1 and Paper 2 leans on these calculations.
How do you calculate the number of moles from a mass?
Number of moles = mass (in grams) ÷ Mᵣ (the relative molecular or formula mass). For example, 48 g of magnesium (Aᵣ = 24) is 48 ÷ 24 = 2 mol. To go the other way, multiply: mass = moles × Mᵣ. Everything routes through moles — get to moles first, apply the mole ratio from the balanced equation, then convert out.
What is the mole and the Avogadro constant?
One mole of any substance contains 6.02 × 10²³ particles — the Avogadro constant. Atoms are far too small to count directly, so the mole lets us weigh out a known number of them. The molar mass (g/mol) is numerically equal to the Aᵣ or Mᵣ, and one mole of any gas occupies 24 dm³ at room temperature and pressure (r.t.p.).
How do you write the formula of an ionic compound?
Balance the charges so the total positive charge equals the total negative charge, since a compound is neutral. Write the cation first, then the anion. The quick method is to swap the charge numbers and drop them in as subscripts, then simplify. Put brackets around a polyatomic ion when you need more than one of it — e.g. calcium nitrate is Ca(NO₃)₂.
What is the limiting reactant and how do you find it?
When reactants are not mixed in the exact ratio of the equation, one runs out first — that one is the limiting reactant, and it decides how much product forms. The rest is in excess. To find it, work out the moles of each reactant and compare them against the mole ratio in the balanced equation; the reactant that provides fewer moles relative to the ratio is limiting.
What is the difference between empirical and molecular formula?
The empirical formula is the simplest whole-number ratio of atoms — divide each element's mass (or %) by its Aᵣ, then simplify. The molecular formula is the actual number of atoms in one molecule; it is (empirical formula)ₙ, where n is found by comparing the empirical formula mass with the given Mᵣ. For example, empirical formula CH₂ with Mᵣ = 42 gives 14n = 42, so n = 3 and the molecular formula is C₃H₆.

O-Level Pure Chemistry  ·  Syllabus 6092  ·  Topic 4 of 12  ·  © 2026 Overmugged. For personal study use only.