Chemical equation balancer — smallest whole-number coefficients, with an atom check

How to use the chemical equation balancer

1. In Unbalanced equation, type the reactants and products, separated by +.
2. Separate the two sides with →, -> or =.
3. For ions, type charges as Fe^3+ or Fe³⁺, Na+ or Cl- for single charges, and e- for an electron. State symbols such as (aq) can stay.
4. Press Balance. Read the Balanced equation (with subscripts and in plain text), the Coefficients, and the Atom check table.

What it does and when to use it

Balancing by inspection works for simple reactions, but redox reactions and equations with five or more species can take many attempts. The balancer gives the answer and a table you can use to check your own working: if your coefficients differ, the atom counts show where.

Once balanced, the coefficients give the mole ratio. Find the mass of each species with the molar mass calculator, look up elements in the periodic table, and test yourself on reactions with the flashcard maker.

How it works

Each species gets an unknown coefficient. For every element, the atoms on the left must equal the atoms on the right, which gives one equation per element; for ions, a further equation makes the charges equal. The tool solves this system exactly with whole-number fractions (no rounding), using Gauss–Jordan elimination to find its null space. It then scales the solution to the smallest whole numbers.

If the only solution is all zeros, the equation cannot be balanced. If there are two or more independent solutions, the reaction is a mix of separate reactions. A species that would need a zero or negative coefficient, or an element found on one side only, is reported. It all runs in your browser.

Worked examples

Aluminium burning. Al + O2 -> Al2O3 balances to 4 Al + 3 O2 → 2 Al2O3. The atom check shows 4 Al and 6 O on each side.

Copper and dilute nitric acid. Cu + HNO3 → Cu(NO3)2 + NO + H2O gives coefficients 3, 8, 3, 2, 4. The check confirms 3 Cu, 8 H, 8 N and 24 O on each side; only 2 of the 8 nitrogen atoms end up in NO.

Permanganate and iron(II) in acid. MnO4- + Fe^2+ + H+ → Mn^2+ + Fe^3+ + H2O becomes MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O. The extra charge row shows +17 on both sides.

Impossible. H2O -> H2O2 is rejected: no positive coefficients can make the oxygen match.

Limits and tips

• Write correct formulas; the tool cannot spot a wrong product, only an unbalanced one.
• Hydrates such as CuSO4·5H2O work, but organic shorthand such as Ph or Me does not. Write the full formula.
• In half-equations, include e- so the charges can balance.