• # question_answer An organic compound has an empirical formula $C{{H}_{2}}O$, its vapour density is 45. The molecular formula of the compounds is  [DCE  2004] A) $C{{H}_{2}}O$ B) ${{C}_{2}}{{H}_{5}}O$ C) ${{C}_{2}}{{H}_{2}}O$ D) ${{C}_{3}}{{H}_{6}}{{O}_{3}}$

Mol. wt = 2 $\times$ Vap. Density $=2\times 45=90$ Empirical formula weight $=12+2+16=30$ $\therefore \,\,\,n=\frac{\text{mol}\text{. wt}\text{.}}{\text{empirical formula wt}\text{.}}$ $=\frac{90}{30}=3$ $\therefore$  Molecular formula of the compounds $={{(C{{H}_{2}}O)}_{3}}$$={{C}_{3}}{{H}_{6}}{{O}_{3}}$