How to Balance Chemical Equations

Conservation of Mass — Why You Can't Change What's There

When iron rusts, iron atoms don't disappear and oxygen atoms don't appear from nowhere. Every atom that exists before a chemical reaction exists after it — just rearranged into different molecules. This is the Law of Conservation of Mass, and it's why chemical equations must be balanced.

An unbalanced equation like Fe + O₂ → Fe₂O₃ is chemically meaningful (iron reacts with oxygen to form iron oxide) but mathematically wrong. The left side has 1 iron atom and 2 oxygen atoms; the right side has 2 iron atoms and 3 oxygen atoms. Atoms have appeared from nowhere. The balanced version — 4Fe + 3O₂ → 2Fe₂O₃ — has 4 iron and 6 oxygen on each side.

Here's the systematic method for balancing equations, with three worked examples.

The Method: Inspection and Coefficient Adjustment

Key rule: You can only add coefficients (numbers in front of formulas) — never change the subscripts within a formula. Changing Fe₂O₃ to FeO₃ would make it a completely different substance.

Step 1: Write the unbalanced equation and count atoms on each side. Step 2: Identify the element that appears in the fewest compounds and balance it first. Step 3: Balance remaining elements, working left to right. Step 4: Balance hydrogen and oxygen last (they're usually the most affected by other changes). Step 5: Verify all atoms balance and that coefficients are the smallest whole numbers possible.

Example 1: Iron Rusting — Fe + O₂ → Fe₂O₃

Step 1: Count atoms in the unbalanced equation

• Left side: 1 Fe, 2 O
• Right side: 2 Fe, 3 O
• Neither side balances.

Step 2: Balance iron (Fe) first The product side has 2 Fe per molecule. Put coefficient 2 in front of Fe: 2Fe + O₂ → Fe₂O₃

• Left: 2 Fe, 2 O
• Right: 2 Fe, 3 O
• Iron is balanced; oxygen still isn't.

Step 3: Balance oxygen Left side has 2 O (from O₂), right side has 3 O. Need to find a coefficient for O₂ and Fe₂O₃ that makes oxygen equal on both sides. The LCM of 2 and 3 is 6.

If we use 3/2 O₂ on the left: 2Fe + 3/2 O₂ → Fe₂O₃ gives 3 O on each side — balanced, but coefficients must be whole numbers.

Multiply everything by 2: 4Fe + 3O₂ → 2Fe₂O₃

Verify:

• Left: 4 Fe, 6 O
• Right: 4 Fe (2 molecules × 2 Fe each), 6 O (2 molecules × 3 O each)
• Balanced.

Example 2: Combustion — CH₄ + O₂ → CO₂ + H₂O

Methane combustion is a common exam question because it introduces balancing hydrogen and oxygen last.

Step 1: Unbalanced equation CH₄ + O₂ → CO₂ + H₂O

• Left: 1 C, 4 H, 2 O
• Right: 1 C, 2 H, 3 O
• Not balanced.

Step 2: Carbon is simplest — balance first Both sides have 1 C. Carbon is already balanced.

Step 3: Balance hydrogen Left has 4 H, right has 2 H. Put coefficient 2 in front of H₂O: CH₄ + O₂ → CO₂ + 2H₂O

• Left: 1 C, 4 H, 2 O
• Right: 1 C, 4 H, 4 O
• Hydrogen balanced; oxygen not.

Step 4: Balance oxygen Right side now has 4 O (2 from CO₂ + 2×1 from H₂O). Left side has 2 O. Put coefficient 2 in front of O₂: CH₄ + 2O₂ → CO₂ + 2H₂O

Verify:

• Left: 1 C, 4 H, 4 O
• Right: 1 C, 4 H, 4 O
• Balanced.

This is the balanced combustion equation for methane (natural gas). One mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water vapor.

Example 3: Harder Case — Al + H₂SO₄ → Al₂(SO₄)₃ + H₂

Aluminum reacting with sulfuric acid. This is harder because of the polyatomic sulfate ion (SO₄²⁻).

Tip for polyatomic ions: Treat them as units when they appear unchanged on both sides. Here, SO₄²⁻ appears in H₂SO₄ on the left and Al₂(SO₄)₃ on the right — same ion, intact on both sides. Balance it as a unit.

Step 1: Count

• Left: 1 Al, 2 H, 1 SO₄
• Right: 2 Al, 2 H, 3 SO₄
• Nothing balanced.

Step 2: Balance aluminum Right side has 2 Al. Put coefficient 2 in front of Al: 2Al + H₂SO₄ → Al₂(SO₄)₃ + H₂

• Left: 2 Al, 2 H, 1 SO₄
• Right: 2 Al, 2 H, 3 SO₄

Step 3: Balance sulfate Right has 3 SO₄; left has 1. Put coefficient 3 in front of H₂SO₄: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + H₂

• Left: 2 Al, 6 H, 3 SO₄
• Right: 2 Al, 2 H, 3 SO₄
• Sulfate balanced; hydrogen not.

Step 4: Balance hydrogen Left has 6 H (from 3H₂SO₄), right has 2 H (from H₂). Put coefficient 3 in front of H₂: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Verify:

• Left: 2 Al, 6 H, 3 SO₄ (= 12 O in SO₄ + 12 S)
• Right: 2 Al, 6 H, 3 SO₄ (= 12 O in SO₄ + 12 S)
• Balanced.

Balancing Redox Equations: The Half-Reaction Method

For more complex reactions involving electron transfer (redox reactions), simple inspection fails. The half-reaction method is more systematic:

1. Identify what's oxidized (loses electrons) and what's reduced (gains electrons)
2. Write separate half-reactions for oxidation and reduction
3. Balance each half-reaction for atoms (using H₂O for O and H⁺ for H in acidic solution)
4. Balance each half-reaction for charge by adding electrons
5. Multiply half-reactions so electrons cancel, then add them

Example: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic solution)

Oxidation half-reaction: Fe²⁺ → Fe³⁺ + e⁻

Reduction half-reaction: MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Multiply oxidation by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻

Add the half-reactions: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Verify charge: Left = (-1) + (5×2) + (8×1) = 17+; Right = 2+ + 15+ = 17+. Balanced.

Common Mistakes

Changing subscripts instead of coefficients: Writing H₂O as HO to "balance" hydrogen is wrong — HO is a completely different molecule (hydroxyl radical). Only coefficients can be changed.

Forgetting polyatomic ions can be treated as units: If SO₄²⁻ appears intact on both sides, balance it as one unit (4 oxygens together), not as individual oxygen atoms.

Not checking all atoms after every step: After each coefficient adjustment, recount every element. A change to balance one element often throws another off balance.

Not simplifying to lowest whole numbers: If you end up with 2CH₄ + 4O₂ → 2CO₂ + 4H₂O, divide everything by 2 to get CH₄ + 2O₂ → CO₂ + 2H₂O.