Balancing Chemical Equations

Every atom on the reactant side must appear on the product side. Combustion of methane CH₄: CH₄ + 2 O₂ → CO₂ + 2 H₂O balances one C, four H, and four O. Unbalanced equations give wrong mole ratios and wrong gram predictions. Molar masses enter only after balancing — coefficients multiply moles, not atomic masses directly.

Start with the most complex molecule (often an organic or polyatomic compound). Balance metals first, then nonmetals, then hydrogen and oxygen last. For aluminum sulfate Al₂(SO₄)₃ reactions, treat SO₄ as a unit if it appears unchanged on both sides. Neutralization of NaOH with H₂SO₄: 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O — note water H₂O as a product, not a reactant.

Strong electrolytes in aqueous solution may be written as ions. Ag⁺(aq) + Cl⁻(aq) → AgCl(s) is the net ionic precipitation from AgNO₃ and NaCl. Spectator ions (Na⁺, NO₃⁻) cancel. Net ionic equations clarify the actual 1:1 mole ratio for precipitate mass. Molar mass of AgCl still uses the full formula unit.

Thermal decomposition: 2 NaHCO₃ → Na₂CO₃ + H₂O + CO₂ — baking soda releases gas moles predictable from mass. Ammonia synthesis N₂ + 3 H₂ → 2 NH₃ sets industrial mole ratios. Contact process steps leading to sulfuric acid H₂SO₄ require balanced sulfur oxidation states. Practice until coefficients become quick checks before any multi-step calculation.

Balancing by trial and error is a genuinely reliable technique once practiced, not a haphazard guessing game. Take propane combustion: C₃H₈ + O₂ → CO₂ + H₂O. First balance carbon: propane has 3 carbons, so place a 3 in front of CO₂: C₃H₈ + O₂ → 3 CO₂ + H₂O. Next balance hydrogen: propane has 8 hydrogens, so place a 4 in front of H₂O (since each water molecule has 2 hydrogens, 4 × 2 = 8): C₃H₈ + O₂ → 3 CO₂ + 4 H₂O.

Finally balance oxygen, which usually needs to be done last since it appears in multiple places. The right side now has (3 × 2) + (4 × 1) = 6 + 4 = 10 oxygen atoms, so place a 5 in front of O₂ on the left (5 × 2 = 10): C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Checking every element confirms the equation is now fully balanced: 3 C, 8 H, and 10 O on both sides.

It's tempting, especially under time pressure, to jump straight to molar mass and stoichiometry calculations using an equation that "looks about right" without formally verifying it's balanced. This is a risky habit, because an unbalanced equation's coefficients do not represent true mole ratios — any downstream mass or mole calculation built on unbalanced coefficients will be confidently wrong, often without any obvious red flag in the arithmetic itself.

A reliable personal rule: never begin a molar-mass or stoichiometry calculation on any equation you have not explicitly verified is balanced, by counting every atom of every element on both sides one final time. This verification step takes only a few seconds but prevents an entire category of otherwise hard-to-catch errors.