A compound gave the following data : C = 57.82%, O = 38.58% and the rest hydrogen. It's vapour density is 83. Find it's empirical and molecular formula. [C = 12, O = 16, H = 1]

C : O : H = 2 : 1 : $\dfrac{3}{2}$ = 4 : 2 : 3

Simplest ratio of whole numbers = 4 : 2 : 3

Hence, empirical formula is C₄O₂H₃ or C₄H₃O₂

Empirical formula weight = (4 x 12) + (3 x 1) + (2 x 16) = 48 + 3 + 32 = 83

Vapour density (V.D.) = 83

Molecular weight = 2 x V.D. = 2 x 83 = 166

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{166}{83} = 2$

Molecular formula = n[E.F.] = 2[C₄H₃O₂] = C₈H₆O₄

∴ Molecular formula = C₈H₆O₄