Given, mass of ${\mathrm{CO}}_2=0.535$
Mass of compound $=0.765\mathrm{g}$
Now, percentage of carbon
$$\begin{array}{rl}& =\frac{\text{ weight }}{\text{ molecular wt. }}\times \frac{\text{ mass of }{\mathrm{CO}}_2}{\text{ mass of compound }}\times 100\\ & =\frac{12}{44}\times \frac{0.535}{0.765}\times 100\\ \mathrm{C}& =19.07\mathrm{\%}\\ & =19\mathrm{\%}\end{array}$$

Similarly, mass of ${\mathrm{H}}_2\mathrm{O}=0.13\mathrm{g}$
mass of compound $=0.765\mathrm{g}$
Now, percentage of hydrogen $=\frac{2}{18}\times \frac{0.138}{0.765}\times 100$
$\mathrm{\%}\mathrm{H}=2.00\mathrm{\%}$
$\therefore$ Ratio of $\mathrm{C}$ and $\mathrm{H}$ is $19:2$
Hence, the ratio of percentage of $\mathrm{C}$ and $\mathrm{H}$ is $19:2$.