A compound $\left(60 g\right)$ on analysis gave $C=24 g,H=4 g$ and $O=32 g$. Its empirical formula is

The mass of the given compound is $60 \mathrm{g}$.

The molar mass of carbon is $12 \mathrm{g}$, the molar mass of hydrogen is $1 \mathrm{g}$ and that of oxygen is $16 \mathrm{g}$.

The mass of carbon, hydrogen and oxygen given is $24 \mathrm{g},4 \mathrm{g}$ and $32 \mathrm{g}$.

So, the no. of atoms of carbon $=24/12=2$

The no. of atoms of hydrogen $=4/1=4$

And the no. of atoms of oxygen $=32/16=2$

So, the simple ratio of carbon: hydrogen: oxygen $=2:4:2=1:2:1$

Hence, the compound is ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2$ and the empirical formula is ${\mathrm{CH}}_2\mathrm{O}$.