A gaseous hydrocarbon on complete combustion gives 3.38 g of `CO_(2)` and 0.690 g of `H_(2)O` and no other product. The empirical formula of the hydrocarbon is

Options

A. `CH`
B. `CH_(2)`
C. `CH_(3)`
D. The data is not complete

Explanation

To find the empirical formula of the hydrocarbon from the given data, we will follow these steps:

### Step 1: Calculate the moles of CO₂ produced
Given the mass of CO₂ produced is 3.38 g. The molar mass of CO₂ is 44 g/mol.

\[ \text{Moles of CO₂} = \frac{\text{mass of CO₂}}{\text{molar mass of CO₂}} = \frac{3.38 \, \text{g}}{44 \, \text{g/mol}} = 0.0768 \, \text{moles} \]

### Step 2: Calculate the moles of H₂O produced
Given the mass of H₂O produced is 0.690 g. The molar mass of H₂O is 18 g/mol.

\[ \text{Moles of H₂O} = \frac{\text{mass of H₂O}}{\text{molar mass of H₂O}} = \frac{0.690 \, \text{g}}{18 \, \text{g/mol}} = 0.0767 \, \text{moles} \]

### Step 3: Determine the moles of carbon and hydrogen
From the combustion reaction:
- Each mole of CO₂ produced corresponds to 1 mole of carbon (C).
- Each mole of H₂O produced corresponds to 2 moles of hydrogen (H).

Thus, the moles of carbon (C) is equal to the moles of CO₂:
\[ \text{Moles of C} = 0.0768 \, \text{moles} \]

The moles of hydrogen (H) is twice the moles of H₂O:
\[ \text{Moles of H} = 2 \times \text{Moles of H₂O} = 2 \times 0.0767 \, \text{moles} = 0.1534 \, \text{moles} \]

### Step 4: Find the simplest mole ratio
Now we have:
- Moles of C = 0.0768
- Moles of H = 0.1534

To find the simplest ratio, we divide both by the smallest number of moles (which is 0.0768):

\[ \text{Ratio of C} = \frac{0.0768}{0.0768} = 1 \]
\[ \text{Ratio of H} = \frac{0.1534}{0.0768} \approx 2 \]

### Step 5: Write the empirical formula
The simplest whole number ratio of C to H is 1:2. Therefore, the empirical formula of the hydrocarbon is:

\[ \text{Empirical formula} = \text{C}_1\text{H}_2 \text{ or } \text{CH}_2 \]

### Summary
The empirical formula of the hydrocarbon is CH₂.

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