Revise: Quantities of Substances: Chemical Calculations

Mr = sum of every atom's Ar in the formula.

Mr(H₂O) = (2×1) + 16 = 18.

1 mole = 6.02 × 10²³ particles (Avogadro's constant).

2 mol water = 1.204 × 10²⁴ molecules.

n = m/M

36 g water ÷ 18 g/mol = 2 mol.

c = n/V, with V in dm³.

0.5 mol in 2 dm³ = 0.25 mol/dm³.

% yield = (actual/theoretical) × 100

15 g actual, 20 g theoretical → 75%.

Ar = one element's relative mass; Mr = sum for the whole formula.

Mr(CO₂) = 12 + (2×16) = 44.

The mole is a counting unit: 1 mol = 6.02 × 10²³ particles.

Same particle count, different mass, for any 1 mol sample.

n = m/M — rearrange to m = n × M for mass.

88 g CO₂ ÷ 44 g/mol = 2 mol.

Empirical formula = simplest whole-number atom ratio.

C₆H₁₂O₆ → CH₂O (divide by 6).

Balance with coefficients, never subscripts — atoms are conserved.

2H₂ + O₂ → 2H₂O.

Balanced-equation coefficients are mole ratios.

2H₂ + O₂ → 2H₂O: 5 mol H₂ → 5 mol H₂O.

The limiting reactant runs out first and caps the product.

4 mol H₂ needs 2 mol O₂; only 1 mol O₂ available → O₂ limits.

c = n/V (mol/dm³) — watch cm³ vs. dm³.

1 mol in 4 dm³ = 0.25 mol/dm³.

Actual product vs. theoretical maximum, as a percentage.

40 g actual / 50 g theoretical = 80%.

Desired product mass vs. total reactant mass — different from % yield.

36 desired mass / 90 total mass = 40%.