A hydrocarbon is completely combusted, producing 0.66 g of $CO_2$ and 0.36 g of $H_2O$. We need to find the empirical formula of the hydrocarbon, given that the atomic mass of C = 12, H = 1, and O = 16. The possible answers are: (1) $C_6H_4$ (2) $C_3H_4$ (3) $C_3H_{10}$ (4) $C_3H_8$ (5) $C_4H_{12}$

1. Problem Description

A hydrocarbon is completely combusted, producing 0.66 g of $CO_2$ and 0.36 g of $H_2O$. We need to find the empirical formula of the hydrocarbon, given that the atomic mass of C = 12, H = 1, and O =

1

6. The possible answers are:

(1) $C_6H_4$

(2) $C_3H_4$

(3) $C_3H_{10}$

(4) $C_3H_8$

(5) $C_4H_{12}$

2. Solution Steps

First, we need to determine the number of moles of carbon and hydrogen in the products.

Moles of $CO_2 = \frac{0.66 \ g}{44 \ g/mol} = 0.015 \ mol$

Since each mole of $CO_2$ contains 1 mole of carbon, moles of $C = 0.015 \ mol$.

Moles of $H_2O = \frac{0.36 \ g}{18 \ g/mol} = 0.02 \ mol$

Since each mole of $H_2O$ contains 2 moles of hydrogen, moles of $H = 2 \times 0.02 \ mol = 0.04 \ mol$.

Now, we can find the ratio of moles of carbon to moles of hydrogen:

$\frac{Moles \ of \ C}{Moles \ of \ H} = \frac{0.015}{0.04} = \frac{15}{40} = \frac{3}{8}$

This means the empirical formula is $C_3H_8$.

3. Final Answer

(4) $C_3H_8$