The big idea: A chemical equation is a recipe for a reaction: the reactants are written on the left, an arrow points to the products on the right.
Atoms are never created or destroyed in a reaction — they are only rearranged. So a correct equation must be balanced: the same number of each kind of atom appears on both sides.
Stoichiometry is the study of these whole-number ratios — how much of each substance reacts and is made.
Balancing keeps the count of every kind of atom the same on both sides — the atoms are only rearranged, never created or destroyed.
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Two words to know: - Stoichiometry — the whole-number ratios in which substances react and form, read straight from a balanced equation. - Mole ratio (stoichiometric ratio) — the ratio of the big numbers (coefficients) in front of the formulas. It tells you how many moles of one substance react with, or make, another.
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To balance an equation you change only the coefficients — the big numbers in front of each formula. You must never change a formula itself (a subscript), because that would describe a different substance.
How to balance
- Write the correct formulas of every reactant and product (do not touch these).
- Count each kind of atom on the left and on the right.
- Adjust the coefficients to make every atom count match — balance one element at a time.
- Leave oxygen and hydrogen until last (they often appear in more than one formula).
- Check the coefficients are the smallest whole numbers (divide through if they share a factor).
State symbols are written after each formula to show its physical state in the reaction:
| State symbol | Means | Example |
|---|---|---|
| (s) | solid | CaCO3(s) |
| (l) | pure liquid | H2O(l) |
| (g) | gas | CO2(g) |
| (aq) | aqueous — dissolved in water | HCl(aq), NaCl(aq) |
Worked example — balancing the combustion of propane
Balance the equation for the complete combustion of propane:
C3H8 + O2 → CO2 + H2O
Solution
- Carbon first: there are 3 C on the left, so put 3 in front of CO2.
- Hydrogen next: 8 H on the left means 4 H2O (4 × 2 = 8 H).
- Oxygen last: the right now has (3 × 2) + (4 × 1) = 10 O atoms, so we need 5 O2 on the left.
- Check: C 3 = 3, H 8 = 8, O 10 = 10. Balanced, smallest whole numbers.
Final answer
C3H8 + 5 O2 → 3 CO2 + 4 H2O.
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Once an equation is balanced, the coefficients are a mole ratio. They tell you the proportions in which the substances react and are formed — this is the link you use for every 'how much?' calculation in this unit.
What the coefficients tell you: For N2 + 3 H2 → 2 NH3:
- 1 mol of N2 reacts with 3 mol of H2. - This makes 2 mol of NH3. - The mole ratio N2 : H2 : NH3 is 1 : 3 : 2 — scale it up or down for any amount.
Worked example — amount from a mole ratio
For the reaction N2 + 3 H2 → 2 NH3, what amount (in mol) of ammonia, NH3, is produced from 0.60 mol of hydrogen, H2 (with nitrogen in excess)?
Solution
- Read the ratio first — from the coefficients, H2 : NH3 is 3 : 2.
- Multiply the amount of H2 by the ratio:
- Work it out — keep the unit:
Final answer
0.40 mol of NH3.
How this is tested: Balanced equations are asked for across the whole exam.
- Paper 1A (MCQ): determine the coefficient of one species when an equation is balanced with the smallest whole numbers (combustion is a favourite). - Paper 2: write / state the balanced equation, with state symbols for a described reaction, then use the mole ratio in a later part. - Paper 1B: deduce the coefficients for an unfamiliar reaction from the formulas given.
The marks that get dropped: a missing or wrong state symbol, and changing a formula instead of a coefficient to 'balance' it.
Score it cleanly: (1) Only change the big numbers (coefficients), never a subscript. (2) Reduce to the smallest whole numbers. (3) Add the state symbol to every formula when asked — (aq) means dissolved in water.
IB-style question — neutralising an acid (a)
Dilute sulfuric acid reacts with solid calcium carbonate to form aqueous calcium sulfate, water and carbon dioxide gas. (a) Write the balanced equation, including state symbols. [2]
How to score the marks
- Mark 1 — correct formulas and balancing. Each carbonate gives one CO2 and one water, and the ratio CaCO3 : H2SO4 is 1 : 1:
- Mark 2 — correct state symbols. Solid carbonate (s), aqueous acid and salt (aq), liquid water (l), gas CO2 (g):
Final answer
CaCO3(s) + H2SO4(aq) → CaSO4(aq) + H2O(l) + CO2(g).
IB-style question — using the mole ratio (b)
(b) Calculate the amount, in mol, of carbon dioxide produced when 0.250 mol of calcium carbonate reacts completely.
Solution
- Read the ratio — from the balanced equation, CaCO3 : CO2 is 1 : 1.
- So the amount of CO2 equals the amount of CaCO3:
Final answer
0.250 mol of CO2.