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.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
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.
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 N_{2} + 3 H_{2} → 2 NH_{3}:
- 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.