A gaseous hydrocarbon gives upon combustion, 0.72 g of water and 3.08 g of `CO_(2)`. The empirical formula of the hydrocarbon is

Options

A. `C_(3)H_(4)`
B. `C_(6)H_(5)`
C. `C_(7)H_(8)`
D. `C_(2)H_(4)`

Answer

To find the empirical formula of the gaseous hydrocarbon that produces 0.72 g of water and 3.08 g of carbon dioxide upon combustion, we can follow these steps:

### Step 1: Calculate the moles of water and carbon dioxide produced.

1. **Moles of Water (H₂O)**:
- Molar mass of water (H₂O) = 18 g/mol
- Moles of water = mass of water / molar mass of water
\[ \text{Moles of H₂O} = \frac{0.72 \, \text{g}}{18 \, \text{g/mol}} = 0.04 \, \text{mol} \]

2. **Moles of Carbon Dioxide (CO₂)**:
- Molar mass of carbon dioxide (CO₂) = 44 g/mol
- Moles of carbon dioxide = mass of CO₂ / molar mass of CO₂
\[ \text{Moles of CO₂} = \frac{3.08 \, \text{g}}{44 \, \text{g/mol}} \approx 0.07 \, \text{mol} \]

### Step 2: Calculate the moles of carbon and hydrogen.

1. **Moles of Hydrogen**:
- Each mole of water contains 2 moles of hydrogen.
\[ \text{Moles of H} = 2 \times \text{Moles of H₂O} = 2 \times 0.04 \, \text{mol} = 0.08 \, \text{mol} \]

2. **Moles of Carbon**:
- Each mole of carbon dioxide contains 1 mole of carbon.
\[ \text{Moles of C} = \text{Moles of CO₂} = 0.07 \, \text{mol} \]

### Step 3: Determine the simplest ratio of moles of carbon to moles of hydrogen.

1. **Ratio of C to H**:
- We have 0.07 moles of carbon and 0.08 moles of hydrogen.
- To find the simplest ratio, divide both by the smallest number of moles (0.07):
\[ \text{Ratio of C} = \frac{0.07}{0.07} = 1 \]
\[ \text{Ratio of H} = \frac{0.08}{0.07} \approx 1.14 \]

2. **Convert to whole numbers**:
- To convert the ratio to whole numbers, we can multiply both by a factor that will give us whole numbers. Here, multiplying by 7 gives:
\[ C: 1 \times 7 = 7 \]
\[ H: 1.14 \times 7 \approx 8 \]

### Step 4: Write the empirical formula.

The empirical formula of the hydrocarbon is \( C_7H_8 \).

### Final Answer:
The empirical formula of the hydrocarbon is \( C_7H_8 \).

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