Sol. $$ \begin{aligned} & \text { Moles of } \mathrm{CO}_2=\frac{0.22}{44}=\frac{1}{200} \\ & \therefore \text { Moles of carbon } \\ & =\left(\text { Moles of } \mathrm{CO}_2\right) \times 1=\frac{1}{200} \\ & \therefore \text { wt. of } \mathrm{C}=\frac{1}{200} \times 12=0.06 \\ & \% \text { of } \mathrm{C}=\frac{0.06}{\mathrm{~W}} \times 100=24 \\ & \begin{aligned} (\mathrm{W}=\text { Wt. of Organic Compound }) \\ \mathrm{W}=0.25 \end{aligned} \\ & \text { Moles of } \mathrm{H}_2 \mathrm{O}=\frac{0.126}{18}=0.007 \\ & \therefore \text { Moles of atom }=2 \times 0.007=0.014 \\ & \% \text { of Hydrogen }=\frac{0.014 \times 1}{\mathrm{~W}} \times 100 \\ & \qquad=\frac{0.014 \times 1}{0.25} \times 100=5.6 \\ & =56 \times 10^{-1} \end{aligned} $$