Balancing Chemical Equations

What is it?

A chemical equation describes a reaction using formulas. Reactants go on the left of the arrow and products on the right:

\ce{2H2 + O2 -> 2H2O}

An equation is balanced when every element has the same number of atoms on both sides. You balance an equation by placing whole-number coefficients (the big numbers in front of formulas).

Key idea

In a chemical reaction, atoms are rearranged, never created or destroyed (the law of conservation of mass). A balanced equation shows this: the atoms you start with are exactly the atoms you end with.

Why does it matter?

It obeys a law of nature. Mass is conserved in every chemical reaction; an unbalanced equation describes something impossible.
It’s the starting point for calculations. The coefficients give the mole ratios used in stoichiometry. An unbalanced equation gives wrong answers.
It shows the recipe. $\ce{2H2 + O2 -> 2H2O}$ tells you hydrogen and oxygen react in a 2 : 1 ratio by molecules (and by moles).

How does it work?

1. Coefficients, not subscripts

A subscript is part of the formula. It defines which substance it is: $\ce{H2O}$ is water, $\ce{H2O2}$ is hydrogen peroxide.
A coefficient says how many units of that substance: $\ce{2H2O}$ means two water molecules, containing 4 H and 2 O atoms in total.

Never change subscripts to balance an equation. That changes the substances, not the amounts.

2. Counting atoms

Multiply the coefficient by each subscript. Brackets multiply everything inside:

$\ce{3CO2}$ : 3 C and 3 × 2 = 6 O
$\ce{Ca(NO3)2}$ : 1 Ca, 2 N and 2 × 3 = 6 O
$\ce{2Al2(SO4)3}$ : 4 Al, 6 S and 2 × 3 × 4 = 24 O

3. The step-by-step method

Write the correct formulas of all reactants and products.
Count the atoms of each element on each side.
Balance one element at a time by changing coefficients. Start with an element that appears in only one substance on each side, usually a metal. Leave hydrogen and oxygen until last, and leave any element on its own (such as $\ce{O2}$ or $\ce{Fe}$ ) to the very end.
Re-count every element. Changing one coefficient can unbalance another element.
Check the coefficients are the smallest whole numbers. If they all share a factor, divide by it.

4. Useful tricks

Polyatomic ions: if an ion such as $\ce{SO4^2-}$ or $\ce{NO3-}$ appears unchanged on both sides, balance it as one unit instead of counting S and O separately.
Odd–even trick: if one side has an odd number of an atom and the other side must be even (as with $\ce{O2}$ ), double everything else first.
Fractions: you can use a fraction such as $\tfrac{7}{2}$ temporarily, then multiply the whole equation to clear it.

Think of it like this

Balancing an equation is like checking a recipe for building toy models from Lego bricks. You can make more copies of a model (change a coefficient), but you can’t change the design of a model (a subscript). When you’re done, every brick you started with must be used, with none left over and none missing.

More precisely

Coefficients count particles (molecules, atoms or formula units) and therefore also moles. They do not give the ratio of masses: in $\ce{2H2 + O2 -> 2H2O}$ , about 4 g of hydrogen reacts with 32 g of oxygen, not in a 2 : 1 ratio by mass. Equations can also include state symbols: (s) solid, (l) liquid, (g) gas and (aq) dissolved in water, e.g. $\ce{2H2(g) + O2(g) -> 2H2O(l)}$ . Ionic equations must also balance charge.

Visualise it

Two particle diagrams. Top: H2 + O2 → H2O, not balanced. One hydrogen molecule and one oxygen molecule on the left make one water molecule on the right, so hydrogen has 2 atoms on each side but oxygen has 2 on the left and only 1 on the right. Bottom: 2H2 + O2 → 2H2O, balanced. Two hydrogen molecules and one oxygen molecule make two water molecules: 4 hydrogen atoms and 2 oxygen atoms on each side. — Adding coefficients makes the atom counts match on both sides.

Worked example

Worked example: Burning propane

Question: Balance $\ce{C3H8 + O2 -> CO2 + H2O}$ .

C: 3 on the left, so put 3 in front of $\ce{CO2}$ : $\ce{C3H8 + O2 -> 3CO2 + H2O}$
H: 8 on the left, so put 4 in front of $\ce{H2O}$ (4 × 2 = 8): $\ce{C3H8 + O2 -> 3CO2 + 4H2O}$
O (last, because $\ce{O2}$ is an element on its own): right side has 3 × 2 + 4 × 1 = 10 O, so put 5 in front of $\ce{O2}$ .
Balanced: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$ . Check: C 3 = 3, H 8 = 8, O 10 = 10 ✓

Worked example: A polyatomic ion

Question: Balance $\ce{Pb(NO3)2 + KI -> PbI2 + KNO3}$ .

Pb is already balanced (1 = 1).
Treat $\ce{NO3}$ as one unit: 2 on the left, 1 on the right, so put 2 in front of $\ce{KNO3}$ .
Now K is 2 on the right, so put 2 in front of $\ce{KI}$ . That also balances I (2 = 2).
Balanced: $\ce{Pb(NO3)2 + 2KI -> PbI2 + 2KNO3}$ ✓

Worked example: Using the odd–even trick

Question: Balance $\ce{Fe + O2 -> Fe2O3}$ .

O: 2 on the left but 3 on the right. Odd and even don’t match, so make the right side even: put 2 in front of $\ce{Fe2O3}$ (6 O).
O: 6 on the right, so put 3 in front of $\ce{O2}$ .
Fe: 2 × 2 = 4 on the right, so put 4 in front of $\ce{Fe}$ .
Balanced: $\ce{4Fe + 3O2 -> 2Fe2O3}$ . Check: Fe 4 = 4, O 6 = 6 ✓

Common mistake

Common mistake: Changing a subscript

To “balance” $\ce{H2 + O2 -> H2O}$ , writing $\ce{H2 + O2 -> H2O2}$ makes the atoms match, but it turns the product into hydrogen peroxide. Change coefficients only: $\ce{2H2 + O2 -> 2H2O}$ .

Common mistake: Forgetting to multiply through brackets

$\ce{3Ca(OH)2}$ contains 3 Ca, 6 O and 6 H, not 3 O and 3 H.

Common mistake: Stopping before re-counting everything

After each change, re-count all the elements. Fixing one element often unbalances another.

Common mistake: Leaving coefficients that can be simplified

$\ce{4H2 + 2O2 -> 4H2O}$ has equal atom counts, but it should be simplified to $\ce{2H2 + O2 -> 2H2O}$ . Use the smallest whole numbers.

Notation note

A coefficient of 1 is not written: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$ has a coefficient of 1 in front of $\ce{CH4}$ and $\ce{CO2}$ .
Some courses write ”=” or ”⟶” instead of ”→” in equations.

Remember this

Balanced = same number of atoms of each element on both sides (conservation of mass).
Change coefficients, never subscripts.
Balance metals and other elements first, then H, then O; leave free elements (e.g. $\ce{O2}$ ) to last.
Treat unchanged polyatomic ions as one unit; use the smallest whole-number coefficients.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/balancing-equations/