Basic Concepts of Chemistry and Chemical Calculations
Unit 1 tests your ability to convert between mass, moles, particle number and gas volume, and to carry those conversions through stoichiometric problems.
Core rules
- One mole contains 6.022 × 10²³ particles (Avogadro constant); molar mass is expressed in g mol⁻¹.
- At 273 K and 1 atm, one mole of any gas occupies 22.4 L.
- Percentage composition gives the empirical formula; molecular formula = n × empirical formula, with n = molar mass ÷ empirical formula mass.
- The laws of conservation of mass, definite proportions, multiple proportions and Avogadro's law underpin every calculation.
- For a limiting reagent, divide the moles of each reactant by its stoichiometric coefficient; the smallest quotient identifies the limiting reagent.
Worked example
For 2H₂ + O₂ → 2H₂O, mix 3.0 g H₂ (1.5 mol) with 32.0 g O₂ (1.0 mol). The required ratio is 2:1, so 1.0 mol O₂ needs 2.0 mol H₂; only 1.5 mol is present, so H₂ is limiting. Water formed = 1.5 mol × 18 g mol⁻¹ = 27.0 g. Oxygen used = 0.75 mol, leaving 0.25 mol = 8.0 g.
Traps
- Do not use 22.4 L mol⁻¹ for liquids or solids.
- H₂, O₂, N₂ and Cl₂ are diatomic; 1 mol of H₂ contains 2 mol of H atoms.
- Round only at the end; premature rounding changes the last significant figure.
- The limiting reagent is not necessarily the reactant with the smaller mass.
- Keep units consistent and report answers to the correct number of significant figures.