Combustion analysis of a hydrocarbon produced 33.01 g co2 and 13.52 g h2o. calculate the empirical formula of the hydrocarbon.

Answer is: the empirical formula of the hydrocarbon is CH₂.
Chemical reaction: CₓHₐ + O₂ → xC + a/2H₂O.
m(CO₂) = 33.01 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 33.01 g ÷ 44.01 g/mol.
n(CO₂) = n(C) = 0.75 mol.
m(H₂O) = 13.52 g.
n(H₂O) = 13.52 g ÷ 18 g/mol.
n(H₂O) = 0.75 mol.
n(H) = 2 · n(H₂O) = 1.5 mol.
n(C) : n(H) = 0.75 mol : 1.5 mol /0.75 mol.
n(C) : n(H) = 1 : 2.

Ibelieve your answer is true 

The answer is b because a,c,and d are opinons