• # question_answer Empirical formula of a compound is ${{C}_{2}}{{H}_{5}}O$ and its molecular weight is 90. Molecular formula of the compound is [NCERT 1971] A) ${{C}_{2}}{{H}_{5}}O$ B) ${{C}_{3}}{{H}_{6}}{{O}_{3}}$ C) ${{C}_{4}}{{H}_{10}}{{O}_{2}}$ D) ${{C}_{5}}{{H}_{14}}O$

Empirical formula mass =${{C}_{2}}{{H}_{5}}O$ = 24+ 5 +16= 45. $n=\frac{\text{Mol}\text{. mass}}{\text{Emp}\text{. mass}}=\frac{90}{45}=2$ Mol. formula = ${{({{C}_{2}}{{H}_{5}}O)}_{2}}={{C}_{4}}{{H}_{10}}{{O}_{2}}$.