How to Balance a Chemical Equation

Balancing equations is conservation of mass made literal — the same atoms have to come out the other side of the arrow that went in, and your job is to find the integer coefficients that make the bookkeeping match. The process is mechanical once you know the order to attack things in, but the failure mode that blows up most early problem sets is changing a subscript instead of a coefficient. NaCl is sodium chloride. NaCl₂ would be a different compound that doesn’t exist in any meaningful sense. Coefficients only — they’re the dial you’re allowed to turn.

Pick the rarest element first, leave H and O for last

The reason to start with the element that appears in the fewest formulas is simple: every coefficient you set affects every other element in that formula, so locking down a coefficient that touches just one or two compounds limits the cascading damage. Hydrogen and oxygen show up in almost everything, so balancing them first means you’ll undo your work three times.

Take the combustion of methane: CH₄ + O₂ → CO₂ + H₂O. Carbon shows up once on each side and is already balanced — leave it. Hydrogen: four on the left in CH₄, two on the right per water molecule. Drop a 2 in front of H₂O. Now CH₄ + O₂ → CO₂ + 2 H₂O. Oxygen on the right is 2 (from CO₂) + 2 (from 2 H₂O) = 4. Left side is 2. Coefficient of 2 in front of O₂ closes it. Final: CH₄ + 2 O₂ → CO₂ + 2 H₂O. Took three adjustments, no fractions, done.

When the LCM is awkward

Rusting is the textbook trap: Fe + O₂ → Fe₂O₃. Iron goes 1 → 2, so put a 2 on the left iron. Oxygen goes 2 → 3, and the lowest common multiple of 2 and 3 is 6 — so 3 O₂ on the left and 2 Fe₂O₃ on the right gives you 6 oxygens both ways. But now iron rebalances: 2 Fe₂O₃ has 4 iron atoms, so the Fe coefficient on the left jumps from 2 to 4. Final: 4 Fe + 3 O₂ → 2 Fe₂O₃. The thing to notice — and this catches people repeatedly — is that fixing oxygen forced you back to fix iron again. Always recount everything after a coefficient change.

Neutralization keeps the polyatomic ion intact

H₂SO₄ + NaOH → Na₂SO₄ + H₂O. Treat sulfate (SO₄) as one unit — it appears intact on both sides, so don’t break it apart and count S and O separately. Sodium needs a 2 in front of NaOH. That gives 4 hydrogens on the left (2 from H₂SO₄ + 2 from 2 NaOH), so a 2 in front of H₂O closes it. Oxygen check: 4 + 2 = 6 on the left, 4 + 2 = 6 on the right. Balanced: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O.

Photosynthesis — the satisfying one

CO₂ + H₂O → C₆H₁₂O₆ + O₂. Carbon: 6 in glucose, so 6 CO₂. Hydrogen: 12 in glucose, so 6 H₂O. Now oxygen: 6×2 + 6×1 = 18 on the left, and 6 (in glucose) + however many O₂ on the right. You need 12 more, so 6 O₂. Final: 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂. The symmetry of the 6-6-1-6 pattern is part of why this equation gets memorized — it’s also genuinely how the reaction stoichiometry works at the net level.

Where this goes wrong

The single most common error is changing subscripts. If you find yourself “balancing” H₂O by writing H₂O₂, stop — peroxide is a different molecule. The second most common is forgetting to recheck after a coefficient change, like the iron case above. Third: leaving a fraction in the answer. If you end up with ½ O₂, multiply every coefficient by 2 to clear it. And once you’re done, check whether all coefficients share a common factor — if you wrote 2 H₂ + 2 Cl₂ → 4 HCl, divide through to get H₂ + Cl₂ → 2 HCl.

For combustion, the reliable order is carbon → hydrogen → oxygen. For redox reactions, inspection breaks down quickly — switch to the half-reaction method instead. For anything with polyatomic ions surviving intact across the arrow, treat them as single units. The Stoichiometry Calculator will verify your balanced equation and let you explore mole ratios between species once it’s right.