$60 g$ of a compound on analysis gave $C=24 g$, $H=4 g$ and $O=32 g$. Its empirical formula is

mass of oxygen $=32 g=2$ atoms of $O^{\text{'}}$ 

mass of carbon $=24 g=2$ atoms of ' $C$ ' 

mass of Hydrogen $=4 g=4$ atoms of ' $H$ '

The molecular Formula of the compound is $C_2{H}_4{O}_2$. 

By simplifying it further, i.e, dividing the coefficient of elements by 2, we get the empirical formula as ${CH}_{2}O$.

Hence, option (B) is correct.