Balancing chemical equations: a step-by-step guide with lab examples

A chemical equation is a short way of writing what happens in a reaction: what goes in, what comes out, and in what proportions. A balanced equation is one where nothing appears or disappears along the way. This guide explains why equations must balance, gives a step-by-step method, and works through examples from the lab, including how a peptide bond forms and breaks.

Terms are defined in the glossary.

Balancing in four steps: (1) write the correct formula for every substance; (2) count the atoms of each element on each side; (3) change coefficients, never subscripts, leaving lone elements such as O₂ until last; (4) recount and reduce to the smallest whole numbers. The full method, with worked examples, is below.

Why equations must balance

Atoms aren't created or destroyed in a chemical reaction. They only rearrange. So every atom on the left of the arrow must turn up on the right. This is the law of conservation of mass: the total mass of the products equals the total mass of the reactants.

An equation that doesn't balance describes something impossible. It also gives the wrong proportions, which matters whenever you use it to work out amounts. See moles, Avogadro's number and percentage yield.

The parts of an equation

Take the burning of methane:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

How to balance an equation, step by step

1. Write the correct formula for every substance. Get these right first. They decide everything else.
2. Count the atoms of each element on each side. A small table helps.
3. Balance one element at a time by changing coefficients. Never change a subscript. Changing H₂O to H₂O₂ would turn water into hydrogen peroxide, a different substance.
4. Leave elements that appear on their own, such as O₂, until last. Their coefficient doesn't disturb any other element.
5. Recount everything. Every element should now match.
6. Use the smallest whole numbers. If every coefficient can be divided by the same number, divide.

Worked example: methane

Start with the unbalanced skeleton: CH₄ + O₂ → CO₂ + H₂O.

Carbon already balances. Hydrogen doesn't: four on the left, two on the right. Put a 2 in front of H₂O. The right side now has 4 hydrogens and 2 + 2 = 4 oxygens. Finally, put a 2 in front of O₂ to give 4 oxygens on the left:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Worked example: propane

C₃H₈ + O₂ → CO₂ + H₂O

1. Carbon: three on the left, so put 3 in front of CO₂.
2. Hydrogen: eight on the left, so put 4 in front of H₂O.
3. Oxygen: the right now has 3 × 2 + 4 = 10 oxygens, so put 5 in front of O₂.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Example: forming a peptide bond

When two amino acids join, the carboxylic acid group of one reacts with the amino group of the other. A molecule of water is lost. Glycine and alanine joining to make the dipeptide glycylalanine is:

C₂H₅NO₂ + C₃H₇NO₂ → C₅H₁₀N₂O₃ + H₂O

The masses balance too. Glycine (75.07 g/mol) plus alanine (89.09 g/mol) is 164.16 g/mol. Glycylalanine (146.14 g/mol) plus water (18.02 g/mol) is also 164.16 g/mol. That lost water is why each amino acid in a chain is called a residue. See what is a peptide.

In practice, peptides aren't made by simply mixing amino acids. The acid group has to be activated first, and other reactive groups protected. See solid-phase peptide synthesis. The balanced equation shows the overall change, not the route.

Example: hydrolysis of a peptide bond

Hydrolysis is the reverse. Water breaks the peptide bond and the two amino acids are released:

C₅H₁₀N₂O₃ + H₂O → C₂H₅NO₂ + C₃H₇NO₂

The atom counts are the same table read from right to left. A peptide or polypeptide of n residues has n − 1 peptide bonds, so breaking it completely into amino acids takes n − 1 molecules of water.

Example: neutralisation

An acid and a base react to give a salt and water. With hydrochloric acid and sodium hydroxide:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

That balances as written. Sulfuric acid has two acidic hydrogens, so it needs two units of sodium hydroxide:

H₂SO₄(aq) + 2 NaOH(aq) → Na₂SO₄(aq) + 2 H₂O(l)

Acetic acid, the solvent in our acetic acid guide, reacts one to one:

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

See buffers and pH for what happens when a weak acid is only partly neutralised.

Ionic equations

Many substances in water exist as separate ions. In the hydrochloric acid example, the sodium and chloride ions are present before and after the reaction without changing. Ions that take no part are called spectator ions. Leaving them out gives the net ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)

An ionic equation must balance in two ways: in atoms, and in electric charge. Here the left has a total charge of +1 − 1 = 0, and so does the water on the right. Every strong acid neutralised by a strong base shares this same net ionic equation.

Example: the Karl Fischer reaction

The reaction behind Karl Fischer titration, which measures water, is a good test of balancing. Written in its simplest classic form, iodine and sulfur dioxide react with water:

I₂ + SO₂ + 2 H₂O → 2 HI + H₂SO₄

The working version in a titrator also includes an alcohol (ROH) and a base (written RN). The Karl Fischer guide gives it in two steps:

ROH + SO₂ + RN → (RNH)SO₃R

(RNH)SO₃R + I₂ + H₂O + 2 RN → (RNH)SO₄R + 2 (RNH)I

Treat R and RN as single units and count them like atoms: both steps balance. In the second step, one iodine molecule is used for each water molecule. That fixed one-to-one ratio is what lets the instrument calculate the water content from the iodine used.

Common mistakes

• Changing a subscript to make the numbers work. That changes the substance.
• Forgetting water in reactions that make or use it, such as forming or breaking a peptide bond.
• Not recounting after the last change. A coefficient that fixes one element often unbalances another.
• Leaving fractions or common factors. Write 2 H₂ + O₂ → 2 H₂O, not H₂ + ½ O₂ → H₂O, unless a fraction is needed for a reason.
• Ignoring charge in ionic equations.

Practice questions

Balance each equation. The answers are in the block below.

1. H₂ + O₂ → H₂O
2. N₂ + H₂ → NH₃
3. Mg + HCl → MgCl₂ + H₂
4. Na + H₂O → NaOH + H₂
5. Al + O₂ → Al₂O₃
6. Ca(OH)₂ + HCl → CaCl₂ + H₂O
7. C₂H₅OH + O₂ → CO₂ + H₂O (burning ethanol)
8. C₆H₁₁N₃O₄ + H₂O → C₂H₅NO₂ (hydrolysis of triglycine, Gly-Gly-Gly, into glycine)

Show the answers

1. 2 H₂ + O₂ → 2 H₂O
2. N₂ + 3 H₂ → 2 NH₃
3. Mg + 2 HCl → MgCl₂ + H₂
4. 2 Na + 2 H₂O → 2 NaOH + H₂
5. 4 Al + 3 O₂ → 2 Al₂O₃
6. Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O
7. C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
8. C₆H₁₁N₃O₄ + 2 H₂O → 3 C₂H₅NO₂. Three residues means two peptide bonds, so two waters.

Related guides

For turning a balanced equation into amounts, see moles, Avogadro's number and percentage yield. For how formulas are written, see empirical and molecular formulas. For more on acids, bases and neutralisation, see acids and bases. For how a reaction's speed is measured, see rates of reaction.