Percentage composition of hydrogen in benzene ( $C_{6} H_{6}$ ) is :

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Solution

Atomic mass of hydrogen = 1 u
Atomic mass of carbon = 12 u

Molecular mass of benzene $(C_{6} H_{6})$ is $= (12 \times 6) + (6 \times 1) = 78 u$

Mass percentage of an element in a compound = $\frac{Mass of an element}{Mass of compound} \times 100$

Mass % of hydrogen in benzene = $\frac{6}{78}$ $\times 100$ $= 7.7 %$

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