Chemical Reactions, Stoichiometry and Quantitative Calculations
Stoichiometry and Quantitative Calculations tests the mole as a currency: converting mass, gas volume and concentration into moles, identifying the limiting reagent, and carrying units through to the answer.
Core relations
- n = m/M; n = V/22.4 for a gas at STP (0 °C, 101.3 kPa); n = c × V for a solution, with V in litres.
- Empirical formula from percentage composition; molecular formula = empirical formula × n, where n = M_molecular/M_empirical.
- Percentage yield = actual/theoretical × 100; percentage purity compares the mass of pure substance with the sample mass.
- Balance the equation before applying mole ratios, and convert mL to L before multiplying by molarity.
Worked example
10.0 g of CaCO₃ (M = 100 g/mol) is added to 100 mL of 1.50 M HCl. What volume of CO₂ at STP is produced? n(CaCO₃) = 10.0/100 = 0.100 mol; n(HCl) = 1.50 × 0.100 = 0.150 mol. The equation CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ needs 2 mol HCl per mol CaCO₃, so 0.100 mol CaCO₃ would require 0.200 mol HCl. Only 0.150 mol is present, so HCl is limiting. n(CO₂) = 0.150/2 = 0.075 mol; V = 0.075 × 22.4 = 1.68 L.
Traps
- Calling the reagent with the larger mass the excess one without applying the mole ratio.
- Using 22.4 L/mol at room temperature, where the molar volume is about 24.0 L/mol.
- Rounding intermediate mole values too early and losing the last significant figure.