IB Chemistry SLTopic 5 — Quantifying Chemical ChangePaper 1 & 2Core skill~11 min read
Balancing Chemical Equations
A chemical reaction rearranges atoms; it never creates or destroys them. So whatever atoms go into a reaction must come out of it, and a balanced equation is simply that fact written down honestly.
📚 What you need to know
Atoms are conserved: the number of each type must be identical on both sides.
Balance by putting coefficients in front of formulae. Never change a formula to make it fit.
Treat polyatomic ions (SO42–, NO3–, CO32–) as single blocks when they survive the reaction intact.
Seven elements are diatomic as elements: H2, N2, O2, F2, Cl2, Br2, I2.
For combustion, balance carbon, then hydrogen, then oxygen last.
State symbols are (s), (l), (g) and (aq) — add them when the question uses them.
Why equations have to balance
Mass is conserved in a chemical reaction because atoms are conserved. Nothing vanishes and nothing appears; bonds break and re-form, and the same collection of atoms ends up arranged differently. An unbalanced equation is therefore not a slightly wrong equation — it describes something impossible.
The coefficient 2 in front of H2O means two whole water molecules. Writing H2O2 instead would balance the arithmetic and describe an entirely different substance.
This is the rule that decides most marks in the topic: coefficients go in front, subscripts are untouchable. The subscript is part of the substance’s identity. Change it and you have quietly swapped water for hydrogen peroxide.
The formulae come first
You cannot balance an equation until every formula in it is correct, so that is genuinely step one. Two things trip people up here.
Only as the uncombined element. Hydrogen in water is written H2O, not H2O2.
The second is polyatomic ions. Groups such as SO42–, NO3–, CO32– and OH– usually pass through a reaction unchanged, so counting them as single units saves a great deal of work. If sulfate appears on both sides, count “sulfates”, not sulfurs and oxygens separately.
Only treat an ion as a block if it really does survive intact. In a reaction where carbonate is destroyed — CaCO3 decomposing to CaO and CO2, for instance — you have to go back to counting individual atoms.
The method
🧩 Balancing, step by step
Write the correct formulae for every reactant and product. Do not touch these again.
Tally the atoms on each side, treating intact polyatomic ions as single units.
Balance one element at a time, starting with the one that appears in the fewest formulae.
Leave until last any element that appears on its own as an element, such as O2 — it can absorb whatever is left over.
Re-check every element, then add state symbols.
Step 4 is the one worth internalising. An element on its own has no other job in the equation, so its coefficient can be adjusted freely at the end without disturbing anything you have already balanced. Fix it early and you will only have to fix it again.
WORKED EXAMPLE
Write a balanced equation for aluminium reacting with hydrochloric acid to give aluminium chloride and hydrogen.
Step 1 — formulaeAl³⁺ with Cl− gives AlCl₃, and hydrogen is diatomic.Al + HCl → AlCl₃ + H₂Step 2 — chlorine first, it is in the fewest places3 Cl on the right needs 3 HCl on the left.Al + 3HCl → AlCl₃ + H₂Step 3 — hydrogen3 H cannot make a whole number of H₂, so double everything.2Al + 6HCl → 2AlCl₃ + 3H₂2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)Check: Al 2 = 2, H 6 = 6, Cl 6 = 6.
WORKED EXAMPLE
Balance: Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
Step 1 — spot the blocksPhosphate survives intact, so count PO₄ units, not P and O separately.Step 2 — calcium and phosphate3 Ca on the right → 3Ca(OH)₂2 PO₄ on the right → 2H₃PO₄Step 3 — hydrogen decides the waterLeft now has 6 H from the hydroxides and 6 H from the acid = 12 H.12 H → 6H₂O3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂OOxygen check: left 6 + 8 = 14; right 8 + 6 = 14. Treating phosphate as a block turned a nightmare into three lines.
WORKED EXAMPLE
Balance the complete combustion of propene, C3H6.
Step 1 — carbon, then hydrogen3 C → 3CO₂ 6 H → 3H₂OStep 2 — oxygen last(3 × 2) + 3 = 9 O atoms → 4½O₂C₃H₆ + 4½O₂ → 3CO₂ + 3H₂OStep 3 — double for whole numbers2C₃H₆ + 9O₂ → 6CO₂ + 6H₂OBoth versions are correct. Keep the half if the question wants one mole of fuel, as enthalpy of combustion questions do.
💡 Exam tip
Check every element at the end, not just the one you were working on. It takes ten seconds.
Start with the element in the fewest formulae and finish with any uncombined element.
Half coefficients are allowed unless the question says otherwise. Double through if it does.
Write state symbols whenever the question includes them — they are frequently worth a mark of their own.
If an equation refuses to balance, check the formulae first. It is almost always a wrong formula, not bad arithmetic.
⚠️ Common mix-up
Changing a subscript to balance. That changes the substance, not the amount.
Forgetting an element is diatomic and writing O or H instead of O2 or H2.
Splitting a polyatomic ion that passed through unchanged, and drowning in oxygen atoms.
Balancing oxygen too early in a combustion equation, then having to redo it.
Ignoring brackets: Ca(OH)2 contains two oxygens and two hydrogens, and 3Ca(OH)2 contains six of each.
Up next: Reacting Masses — because once an equation is balanced, those coefficients become the exchange rate that lets you turn a mass of one substance into a mass of another.
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