Key ideas
Quantitative organic chemistry combines general formulas for the families with molar-mass and combustion calculations. A formula tells you how many hydrogens a family has, and a balanced equation tells you how much oxygen it needs.
- Each pi bond or ring removes two hydrogens: alkanes CₙH₂ₙ₊₂, alkenes and cycloalkanes CₙH₂ₙ, alkynes CₙH₂ₙ₋₂, and saturated alcohols CₙH₂ₙ₊₂O.
- Molar mass is the sum of the atomic masses, and the percentage of an element is its total mass in the formula divided by the molar mass.
- Complete combustion of CₓH_y needs x + y ÷ 4 moles of O₂ and gives x moles of CO₂ and y ÷ 2 moles of H₂O.
- Check a balanced equation by counting each element on both sides, and multiply by 2 if a fractional coefficient appears.
Key equations
alkane CₙH₂ₙ₊₂alkene CₙH₂ₙalkyne CₙH₂ₙ₋₂CₓH_y + (x + y ÷ 4) O₂ → x CO₂ + (y ÷ 2) H₂O% C = 12.01 x ÷ M × 100
Worked example
Butane, C₄H₁₀: M = 4 × 12.01 + 10 × 1.008 = 58.12 g/mol, and the carbon content is 48.04 ÷ 58.12 = 82.7 %. Combustion needs 4 + 10 ÷ 4 = 6.5 mol O₂ per mole, giving 4 CO₂ and 5 H₂O. Doubled to avoid a fraction: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O.
Common mistakes
- Forgetting that oxygen in the molecule lowers the amount of O₂ needed for combustion.
- Using the number of atoms rather than their atomic masses when calculating a percentage by mass.
- Leaving a half-integer coefficient in the final balanced equation.
Practice
Test yourself with practice questions on this topic: Formulas and stoichiometry