Stoichiometry
The idea
Stoichiometry turns a balanced equation into quantitative predictions: given an amount of one substance, it tells you how much of any other substance the reaction consumes or produces. It works because the coefficients you learned to balance are mole ratios — exact exchange rates between substances — and because the mole, which you can reach from any mass, is the common currency.
Every mass-to-mass problem travels the same three-leg highway: convert the given mass to moles using its molar mass, hop across the equation using the coefficient ratio, then convert the new moles back to mass with the other substance's molar mass. Writing each leg with units and canceling them is your built-in error detector — if the units do not cancel to what you want, the setup is wrong.
The trap to avoid is applying coefficients directly to grams. A ratio of 1 : 3 in the equation means 1 mole to 3 moles, never 1 gram to 3 grams, because different substances pack different masses into each mole. Skipping the mole conversion is the single most common stoichiometry error.
Worked example
Propane burns by C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. What mass of CO₂ is produced when 11.0 g of propane burns completely? Use C = 12.0, H = 1.0, O = 16.0 g/mol.
- Confirm the equation is balanced before trusting its ratios: C is 3 = 3, H is 8 = 8, and O is 10 on the left versus 3 × 2 + 4 × 1 = 10 on the right. Safe to proceed.
- Leg one — mass to moles of the given: M(C₃H₈) = 3 × 12.0 + 8 × 1.0 = 44.0 g/mol, so 11.0 g ÷ 44.0 g/mol = 0.250 mol propane.
- Leg two — cross the equation: the coefficients say 3 mol CO₂ form per 1 mol C₃H₈, so 0.250 mol × 3 = 0.750 mol CO₂.
- Leg three — moles back to mass: M(CO₂) = 12.0 + 2 × 16.0 = 44.0 g/mol, so 0.750 mol × 44.0 g/mol = 33.0 g of CO₂.
- Sanity-check: propane and CO₂ happen to share the molar mass 44.0 g/mol, so the 3:1 mole ratio shows up directly as a 3:1 mass ratio here — 33.0 g really is three times 11.0 g, a coincidence worth noticing rather than expecting.
Answer. Burning 11.0 g of propane produces 33.0 g of CO₂.
Check your understanding
- Why must every stoichiometry path pass through moles rather than jumping from grams to grams?
- How would the calculation change if only a limited amount of oxygen were available instead of excess?
- How would you extend the same three-leg method to find the mass of O₂ consumed by the 11.0 g of propane?
- What goes wrong, numerically and conceptually, if someone reads the coefficients as gram ratios?
Build the foundations first
Stoichiometry builds on these concepts. If any feel shaky, start there.