The empirical formula of a gaseous compound `CH_(2)`. The density of the compound is 1.25 gm/lit at S.T.P. The molecular formula of the compound is 'X'

Options

A. `C_(2)H_(4)`
B. `C_(3)H_(6)`
C. `C_(6)H_(12)`
D. `C_(4)H_(8)`

Explanation

To find the molecular formula of the compound with the empirical formula \( CH_2 \) and a density of \( 1.25 \, \text{g/L} \) at STP, we can follow these steps:

### Step 1: Calculate the molar mass of the compound using density
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies \( 22.4 \, \text{L} \). We can use the density to find the mass of 1 mole of the gas.

\[ \text{Molar mass} = \text{Density} \times \text{Volume at STP} \]

Given:
- Density = \( 1.25 \, \text{g/L} \)
- Volume at STP = \( 22.4 \, \text{L} \)

Calculating the molar mass:

\[ \text{Molar mass} = 1.25 \, \text{g/L} \times 22.4 \, \text{L} = 28 \, \text{g/mol} \]

### Step 2: Calculate the molar mass of the empirical formula
The empirical formula is \( CH_2 \). Now, we calculate its molar mass:

- Atomic mass of Carbon (C) = \( 12 \, \text{g/mol} \)
- Atomic mass of Hydrogen (H) = \( 1 \, \text{g/mol} \)

Calculating the molar mass of \( CH_2 \):

\[ \text{Molar mass of } CH_2 = 12 + (2 \times 1) = 12 + 2 = 14 \, \text{g/mol} \]

### Step 3: Determine the value of \( n \)
The molecular formula is related to the empirical formula by the equation:

\[ \text{Molecular formula} = (\text{Empirical formula}) \times n \]

Where \( n \) is the ratio of the molar mass of the compound to the molar mass of the empirical formula:

\[ n = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}} = \frac{28 \, \text{g/mol}}{14 \, \text{g/mol}} = 2 \]

### Step 4: Write the molecular formula
Now that we have \( n = 2 \), we can find the molecular formula:

\[ \text{Molecular formula} = (CH_2) \times 2 = C_2H_4 \]

Thus, the molecular formula of the compound is \( C_2H_4 \).

### Final Answer:
The molecular formula of the compound is \( C_2H_4 \).

---