Chemical equation balancer
Enter an equation with species separated by +, and the two sides joined by =, -> or → — coefficients are computed algebraically (exact, no trial-and-error).
Try an example
Common balancing methods
Different equations call for different methods — open each one below to see the steps and a fully worked example:
| Method | Best for |
|---|---|
| Trial and error (odd–even) | Simple equations with few species |
| Valence (criss-cross) | Oxides, salts, basic inorganic reactions |
| Oxidation number (electron balance) | Redox reactions |
| Ion–electron (half-reaction) | Redox in aqueous solution |
| Algebraic | Complex equations — the method this tool uses (see “Step-by-step solution”) |
| Element conservation | Quick numeric problem solving without the full equation |
| Electron conservation | Redox calculation problems |
Trial and error (odd–even trick)
The familiar lower-secondary approach — effective for simple equations with few species.
- Give the most complex formula (usually on the right) a coefficient of 1.
- Balance the elements one by one, starting with the element that appears in the fewest species; a common order is metal → non-metal → H → O.
- Odd–even trick: if an element has an odd atom count on one side but is always even on the other, double the coefficient of the odd-count species and continue.
- If a fractional coefficient appears, multiply the whole equation by its denominator. Finally recount every element on both sides.
Example: Al + O₂ → Al₂O₃
Al + O₂ → Al₂O₃ — O is 3 (odd) on the right but always even on the left ⇒ double Al₂O₃.
Al + O₂ → 2Al₂O₃ — the right side now has 4 Al and 6 O ⇒ put 4 before Al and 3 before O₂.
4Al + 3O₂ → 2Al₂O₃ ✓ (Al: 4 = 4, O: 6 = 6)
Oxidation number method (electron balance)
The standard grade-10 method for redox reactions: total electrons lost must equal total electrons gained.
- Assign oxidation numbers to find the reducing agent (oxidation number increases) and the oxidising agent (oxidation number decreases).
- Write the oxidation process and the reduction process, showing the electrons lost / gained.
- Find the least common multiple of the electron counts so that electrons lost = electrons gained, giving the coefficients of the reducing and oxidising agents.
- Place those coefficients into the equation, then balance the rest in the order metal → acid radical → H, and check with O.
Example: Cu + HNO₃ (dilute) → Cu(NO₃)₂ + NO + H₂O
Oxidation numbers: Cu⁰ → Cu⁺² (reducing agent); N⁺⁵ (in HNO₃) → N⁺² (in NO) (oxidising agent).
Oxidation: Cu⁰ → Cu⁺² + 2e (×3)
Reduction: N⁺⁵ + 3e → N⁺² (×2)
LCM(2, 3) = 6 ⇒ put 3 before Cu and Cu(NO₃)₂, and 2 before NO.
N in the salt Cu(NO₃)₂ keeps its oxidation number ⇒ HNO₃ = 3×2 (salt) + 2 (reduced) = 8; H gives 4H₂O.
3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO↑ + 4H₂O ✓ (O check: 24 = 18 + 2 + 4)
Ion–electron method (half-reactions)
For redox reactions in aqueous solution with an acidic or basic medium.
- Split strong electrolytes into ions and write two half-reactions: one oxidation, one reduction.
- Balance each half-reaction: main element first → add H₂O to balance O → add H⁺ to balance H (acidic medium; use OH⁻ in basic medium) → add electrons to balance charge.
- Multiply each half so the electron counts match, add them together, and cancel any H₂O or H⁺ appearing on both sides.
- If needed, recombine the ions into a molecular equation.
Example: KMnO₄ + FeSO₄ + H₂SO₄
Reduction half: MnO₄⁻ + 8H⁺ + 5e → Mn²⁺ + 4H₂O (×2)
Oxidation half: Fe²⁺ → Fe³⁺ + 1e (×10)
Adding both halves: 2MnO₄⁻ + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 10Fe³⁺ + 8H₂O
Molecular form: 2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → 2MnSO₄ + 5Fe₂(SO₄)₃ + K₂SO₄ + 8H₂O ✓
How to enter equations
- Separate species with “+”; join the two sides with “=”, “->” or “→”.
- Element symbols are case-sensitive: CO (carbon monoxide) differs from Co (cobalt).
- Use parentheses for groups: Ca(OH)2, Fe2(SO4)3.
- Write hydrates with a dot: CuSO4.5H2O.
- No need to type coefficients — any you type (2H2O) are ignored and recomputed.
How does it work?
Each element gives one atom-conservation equation between the two sides. The tool builds a linear system from all elements, solves it with Gaussian elimination over exact rational numbers, then scales to the smallest positive integer coefficients. This balances even complex redox reactions without guessing.