Answer
The empirical formula of the hydrocarbon is: $$CH_2$$
Work Step by Step
1. Calculate the number of moles of each compound:
$$M \space (CO_2) = 12.0 \space g + 2 \times 16.0 \space g = 44.0 \space g/mol$$ $$M \space (H_2O) = 1.01 \space g \times 2 + 16.0 \space g = 18.02 \space g/mol$$
$$33.01 \space g \space (CO_2) \times \frac{1 \space mol \space (CO_2)}{44.0 \space g \space (CO_2)} = 0.750 \space moles$$ $$13.51 \space g \space (H_2O) \times \frac{1 \space mol \space (H_2O)}{18.02 \space g \space (H_2O)} = 0.750 \space moles$$
2. Find the number of moles of $C$ and $H$:
$$n \space (C) = 1 \times n(CO_2) = 0.750 \space moles$$ $$n(H) = 2 \times n(H_2O) = 1.50 \space moles$$
3. Thus, the formula of this hydrocarbon is proportional to: $C_{0.750}H_{1.50}$. We just have to multiply the subscripts by a number to get whole numbers.
$$C_{0.750}H_{1.50} \times 4 = C_3H_6$$
Both 3 and 6 are divisible by 3, so: $$C_3H_6 \div 3 = CH_2$$
