Balancing chemical equations is one of the first major skills in chemistry — and one of the most satisfying once it clicks. The core principle is simple:
Atoms are neither created nor destroyed in a chemical reaction. The number of each type of atom must be equal on both sides of the equation.
This is the Law of Conservation of Mass.
Why Do We Balance Equations?
An unbalanced equation like:
H₂ + O₂ → H₂O
isn't wrong — it just says hydrogen and oxygen react to form water. But it doesn't tell us how much of each. The balanced version:
2H₂ + O₂ → 2H₂O
tells us: 2 molecules of hydrogen gas react with 1 molecule of oxygen to produce 2 molecules of water.
Step-by-Step: The Inspection Method
This works for most equations you'll encounter in high school chemistry.
Step 1: Write the Unbalanced Equation
Start by writing out reactants and products with correct formulas. Don't change any subscripts — you can only add coefficients in front.
Example: Iron reacts with oxygen to form iron(III) oxide.
Fe + O₂ → Fe₂O₃
Step 2: Count Atoms on Each Side
| Element | Left (Reactants) | Right (Products) |
|---|---|---|
| Fe | 1 | 2 |
| O | 2 | 3 |
Neither is balanced.
Step 3: Balance One Element at a Time
Start with elements that appear in the fewest compounds. Here, balance Fe first.
Put a 4 in front of Fe on the left (we'll see why in a moment):
4Fe + 3O₂ → 2Fe₂O₃
Now re-count:
| Element | Left | Right |
|---|---|---|
| Fe | 4 | 4 ✓ |
| O | 6 | 6 ✓ |
Balanced!
Worked Example 2: Combustion Reaction
Balance the combustion of propane (C₃H₈):
C₃H₈ + O₂ → CO₂ + H₂O
Step 1: Balance carbon. There are 3 C on the left, so put 3 in front of CO₂: C₃H₈ + O₂ → 3CO₂ + H₂O
Step 2: Balance hydrogen. There are 8 H on the left, H₂O has 2 H each, so put 4 in front of H₂O: C₃H₈ + O₂ → 3CO₂ + 4H₂O
Step 3: Balance oxygen. Right side: 3×2 + 4×1 = 10 O atoms. Left side needs 10/2 = 5 O₂: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Final check:
| Element | Left | Right |
|---|---|---|
| C | 3 | 3 ✓ |
| H | 8 | 8 ✓ |
| O | 10 | 10 ✓ |
Balanced: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O ✓
The Half-Reaction Method (for Redox Equations)
For oxidation-reduction (redox) reactions in acidic or basic solution, the inspection method gets complicated. Use half-reactions instead.
Example: Balance in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Step 1 — Separate into half-reactions:
- Oxidation: Fe²⁺ → Fe³⁺
- Reduction: MnO₄⁻ → Mn²⁺
Step 2 — Balance atoms (not O, not H yet): Both are already balanced for main atoms.
Step 3 — Balance O using H₂O: MnO₄⁻ → Mn²⁺ + 4H₂O
Step 4 — Balance H using H⁺ (acidic solution): 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
Step 5 — Balance charge using electrons:
- Oxidation: Fe²⁺ → Fe³⁺ + e⁻
- Reduction: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
Step 6 — Equal electrons in both half-reactions: Multiply oxidation ×5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
Step 7 — Add and cancel: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ ✓
Common Mistakes
Changing subscripts — Never change subscripts (e.g., H₂O to H₃O). Only add coefficients in front of whole formulas.
Forgetting polyatomic ions — If a polyatomic ion like SO₄²⁻ appears unchanged on both sides, balance it as a unit.
Skipping the check — Always recount every atom after you think it's balanced.
Leaving fractions — If you end up with fractional coefficients (like ½), multiply everything through to get whole numbers.
Tips for Success
Save oxygen and hydrogen for last — they appear in many compounds, so balance them after everything else.
Use trial and error confidently — it's expected. Start with a guess, adjust, and check.
Odd-even trick — if you have an odd count on one side and even on the other, double the odd side's coefficient.
For a complete reference, check our Chemistry Cheatsheets or use our Stoichiometry Calculator to check your balanced equations instantly.