IB Chemistry HLTopic 5 — How Much? Quantifying Chemical ChangePaper 1 & 2Core skill~9 min read
Balancing Chemical Equations
Every calculation in this whole topic sits on top of a balanced equation. Get the balancing numbers wrong and your moles, masses, volumes and yields are all wrong too — even if the rest of your maths is perfect. So it is worth ten minutes to make this automatic.
📚 What you need to know
Atoms are never created or destroyed in a reaction, so each element must have the same number of atoms on both sides.
You balance by putting numbers in front of formulae — never by changing a formula.
A number in front multiplies everything in that formula.
Treat polyatomic ions like SO42− and NO3− as single units when they survive the reaction.
For combustion, balance carbon first, then hydrogen, then oxygen last.
Seven elements are diatomic: H2, N2, O2, F2, Cl2, Br2, I2.
Add state symbols (s), (l), (g), (aq) — they are often worth a mark on their own.
Why balancing is not just a rule
Burn 10 g of magnesium and you get more than 10 g of magnesium oxide. Nothing has been created — oxygen from the air joined in. Weigh everything, including the gases, and the mass before equals the mass after, every time.
That is what a balanced equation records: the same atoms, rearranged. So when you count 6 chlorine atoms on the left, there must be 6 on the right. No exceptions, no rounding.
The tempting move is to change FeCl3 into FeCl2 so the chlorines match. That is not balancing — it invents a different compound.
If you find yourself editing a subscript, stop. Subscripts are part of the compound’s identity, fixed by its bonding. The only dial you are allowed to turn is the big number out front.
The method
🧩 How to balance anything
Write the correct formulae for everything, reactants on the left, products on the right.
Count each element on both sides. Write the counts down — do not do it in your head.
Balance one element at a time. Start with the element that appears in the fewest places.
Leave elements that appear on their own until last — usually O2 or H2, because they are easy to adjust without upsetting anything else.
Recount everything at the end. Every element, both sides.
Add state symbols. (s), (l), (g) or (aq) for each species.
The odd-number trick. Stuck with an odd number on one side and an even one on the other? Double everything. If you need 1½O2, multiply the entire equation by 2. Fractions are allowed in your rough working but the final answer should use whole numbers.
Combustion: always in the same order
Burning a hydrocarbon looks messy because oxygen ends up in two different products. Fix that by dealing with oxygen last, once the carbons and hydrogens are locked in.
Count the oxygens on the product side once carbon and hydrogen are fixed: 3 × 2 from CO2 plus 4 × 1 from H2O gives 10, so you need 5O2.
State symbols and the diatomic seven
Symbol
Means
Typical example
(s)
Solid
CaCO3(s), Mg(s), any metal or precipitate
(l)
Pure liquid
H2O(l), Br2(l)
(g)
Gas
CO2(g), H2(g), O2(g)
(aq)
Dissolved in water
HCl(aq), NaOH(aq), CuSO4(aq)
The seven diatomic elements have to be written as pairs when they are free: H2, N2, O2, F2, Cl2, Br2, I2. Write “O” instead of “O2” and your balancing will look right while being chemically wrong.
Water is the sneaky one. In combustion at room temperature it is H2O(l), but in a hot engine or a gas-volume question it is H2O(g). That choice changes the total gas volume, so read the question.
Worked examples
WORKED EXAMPLE
Balance: Fe + Cl2 → FeCl3
Step 1: count what you have
Fe: 1 and 1. Cl: 2 on the left, 3 on the right.
Step 2: find a number both sides can reach
The lowest common multiple of 2 and 3 is 6, so aim for 6 chlorines.
Step 3: put the numbers in3Cl₂ gives 6 Cl; 2FeCl₃ gives 6 ClStep 4: fix the iron, then recount
2FeCl3 needs 2 Fe, so write 2Fe. Fe: 2 and 2. Cl: 6 and 6.
2Fe(s) + 3Cl₂(g) → 2FeCl₃(s)Balance the awkward element first, then let the easy one follow. Iron only appears in one place on each side, so it is the easy one.
WORKED EXAMPLE
Write a balanced equation, with state symbols, for aluminium reacting with hydrochloric acid to give aluminium chloride solution and hydrogen gas.
Step 1: formulae first
Al + HCl → AlCl3 + H2Step 2: chlorine3 Cl in AlCl₃, so 3HCl on the leftStep 3: hydrogen
3HCl gives 3 H, but H2 comes in pairs. Double everything: 2Al + 6HCl → 2AlCl₃ + 3H₂Step 4: recount and add states
Al 2 and 2, H 6 and 6, Cl 6 and 6.
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)The odd hydrogen count is the clue to double up. Aluminium chloride is (aq) because it is dissolved, and the hydrogen bubbles off as (g).
WORKED EXAMPLE
Balance: Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O
Step 1: treat PO₄ as one block
Do not split it into P and O. There are 2 PO₄ blocks on the right, so you need 2H3PO4.
Step 2: calcium3 Ca on the right, so 3Ca(OH)₂Step 3: hydrogen and oxygen fall out as water
H on the left: (3 × 2) + (2 × 3) = 12, so 6H₂OStep 4: check the oxygens outside the blocks
Left has 6 O from the hydroxides; right has 6 O in 6H2O. Balanced.
3Ca(OH)₂ + 2H₃PO₄ → Ca₃(PO₄)₂ + 6H₂OKeeping the phosphate together turns a nightmare into three quick steps. Split it and you will be juggling seven oxygens per side.
💡 Exam tip
Write the atom count under each formula. It is quick, and it stops you convincing yourself an equation balances when it does not.
Balance the element that appears in the fewest compounds first. Usually the metal, and usually it takes one step.
Save free elements for last. O2, H2 and Cl2 can be adjusted without touching anything else.
Never leave a fraction in a final answer unless the question specifically allows it. Double through instead.
Add state symbols every time. They cost you three seconds and they are frequently a separate mark.
Check the total charge too for ionic equations — both sides must match.
⚠ Common mix-up
Changing a subscript to make it fit. H2O is not H2O2, and CO2 is not CO3.
Forgetting a coefficient multiplies everything. 2H2SO4 contains 4 H, 2 S and 8 O.
Writing O instead of O2 for oxygen gas. Same for the other diatomics.
Splitting up polyatomic ions that pass through the reaction unchanged. It makes the counting far harder than it needs to be.
Balancing the oxygens first in a combustion and then having to redo them twice.
Skipping the final recount. Most balancing errors are caught by simply counting again.
A balanced equation is a recipe written in moles. Next you will turn those moles into grams, which is where the real exam marks live. Up next: Reacting Masses.
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