60 gm of an organic compound containing C, H, and O atoms on complete combustion gave 88 gm ${\mathrm{CO}}_{2 }$and $36 \mathrm{gm}$ ${\mathrm{H}}_2\mathrm{O}$. The empirical formula of the organic compound is :

Given, 

Amount of organic compound = 60g

Amount of ${\mathrm{CO}}_2$ produced = 88g

Amount of ${\mathrm{H}}_2\mathrm{O}$ produced = 36g

• 44g of ${\mathrm{CO}}_2$ has 12g carbon

Therefore, 88g ${\mathrm{CO}}_2$ has $\frac{12\times 88}{44}=24g C$

$\% \mathrm{C}=\frac{24}{60}\times 100=44\%$

• 18 of ${\mathrm{H}}_2\mathrm{O}$ has 2g hydrogen

Therefore, 36g ${\mathrm{H}}_2\mathrm{O}$ has $\frac{2\times 36}{18}=4g \mathrm{H}$

$\% \mathrm{H}=\frac{4}{60}\times 100=6.66\%$

• $\% \mathrm{O}=100- (40+6.66)= 53.33\%$

Therefore C: H: O :: $\frac{40}{12}:\frac{6.66}{1}:\frac{53.33}{16}=3.33:6.66:3.33$

Therefore $\mathrm{C}:\mathrm{H}: \mathrm{O}::1:2:1$

Thus, the empirical formula of the organic compound is ${\mathrm{CH}}_2\mathrm{O}$.