Find the empirical formula and the molecular formula of an organic compound from the data given below: C = 75.92%, H = 6.32% and N = 17.76% The vapour density of the compound is 39.5. [C = 12, H = 1, N = 14]
Topic: Mole concept, gas volumes and stoichiometry
Answer
Given: C = 75.92 %, H = 6.32 %, N = 17.76 %, vapour density = 39.5.
[C = 12, H = 1, N = 14]
Empirical formula:
| Carbon | Hydrogen | Nitrogen | |
|---|---|---|---|
| Percentage | 75.92 | 6.32 | 17.76 |
| ÷ atomic mass | 75.92 ÷ 12 = 6.33 | 6.32 ÷ 1 = 6.32 | 17.76 ÷ 14 = 1.27 |
| ÷ smallest (1.27) | 4.98 ≈ 5 | 4.98 ≈ 5 | 1 |
Empirical formula = C₅H₅N Empirical formula mass = (5 × 12) + (5 × 1) + 14 = 60 + 5 + 14 = 79
Molecular formula: Molecular mass = 2 × vapour density = 2 × 39.5 = 79 n = molecular mass ÷ empirical formula mass = 79 ÷ 79 = 1
Molecular formula = (C₅H₅N)₁ = C₅H₅N
Empirical formula C₅H₅N and molecular formula C₅H₅N (this is pyridine).
A four-mark question in two halves: the empirical formula from the percentages, then the molecular formula from the vapour density. Show both halves — the second is worth marks even when n turns out to be 1.
Do not be thrown when the two formulae come out the same. n = 1 is a perfectly good answer and simply means the molecule IS the empirical unit. Students who expect them to differ often go back and "correct" a right answer.
The other slip is forgetting to double the vapour density. Using 39.5 as the molecular mass gives n = 0.5, which is impossible — n is always a whole number, so a fraction is your signal that the factor of 2 is missing.