#### Answer

15.585%

#### Work Step by Step

Mass of H in the compound = $22\times1.008g$ = 22.176 g
Molar mass of $C_{10}H_{22}$ = $(10\times12.011g)+(22\times1.008g)$ = 142.29 g
Mass % of H = (Mass of H in $C_{10}H_{22}$$\times$100)/(Molar mass of $C_{10}H_{22}$) = $\frac{22.176g\times100}{142.29g}$ = 15.585%