Equations and Stoichiometry
What to know
Stoichiometry is mole-ratio arithmetic. The core skill: balance the equation, convert given quantities to moles, apply the mole ratio from coefficients, and convert back to the requested unit (grams, liters of gas, particles). Know molar mass, molar volume of an ideal gas at STP (22.4 L/mol), and Avogadro's number. Identify limiting reactant by comparing available moles to required ratios; the excess reactant remains. Calculate theoretical yield and percent yield = (actual/theoretical) × 100.
For solutions, molarity M = mol/L drives dilution (M1V1 = M2V2) and titration calculations. Percent composition and empirical/molecular formula problems use mass ratios and molar masses.
Traps
- Using grams directly in a mole ratio instead of converting to moles.
- Picking the reactant with the smaller mass rather than the one that runs out first.
- Forgetting to balance before reading coefficients.
- Mixing up actual and theoretical yield in percent-yield problems.
Worked example
How many grams of water form from 4.0 g H2 + excess O2? Reaction: 2H2 + O2 → 2H2O. Moles H2 = 4.0/2.02 ≈ 1.98 mol. Ratio H2:H2O is 2:2 = 1:1, so 1.98 mol H2O. Mass = 1.98 × 18.0 ≈ 35.6 g. Rule of thumb: grams → moles → ratio → moles → grams, never skip a step.