Stoichiometry: moles, mass and balanced equations

Stoichiometry: moles, mass and balanced equations

Stoichiometry uses a balanced chemical equation as a ratio of particles and therefore of moles. Never use a coefficient ratio until the equation is balanced. A reliable calculation path is: write the equation, convert the known quantity to moles, apply the coefficient ratio, then convert to the requested mass, volume or concentration. Use n = m/M for mass-to-mole conversion, where M is molar mass in g mol⁻¹. Include units at each step; they expose incorrect conversions. A mole is a counting amount, not a mass: one mole of different substances contains the same number of particles but can have different masses.

Work it through

For 2H₂ + O₂ → 2H₂O, the coefficient ratio H₂:H₂O is 2:2, or 1:1. If 4.0 mol H₂ reacts completely with enough oxygen, 4.0 mol H₂O can form. If the question instead gives 9.0 g water, calculate n(H₂O) = 9.0/18.0 = 0.50 mol before using the ratio.

Mastery target

Solve a multi-step mole calculation with a balanced equation and defend the answer using units and a reasonableness check.