Correct Answer - Option 3 : 24 g of carbon and 1g of hydrogen

__Calculation__:

Let the hydrocarbon is C_{x}H_{y}

\({{\rm{C}}_{\rm{x}}}{{\rm{H}}_{\rm{y}}} + {{\rm{O}}_2} \to {\rm{xC}}{{\rm{O}}_2} + \frac{{\rm{y}}}{2}{\rm{\;}}{{\rm{H}}_2}{\rm{O}}\)

Mole of water H_{2}O = 18.01528 g/mol

Mole of carbon dioxide CO_{2} = 44.01 g/mol

Weight of hydrogen (H) = 1.00784 u

Weight of carbon (C) = 12.0107 u

Weight of carbon = \(\frac{{88}}{{44}} \times 12 = 24\;g\)

Weight of Hydrogen = \(\frac{9}{{18}} \times 1 = 0.5\;g \approx 1\)

Therefore the unknown

**hydrocarbons contains 24 g of carbon and 1 g of hydrogen**.