The empirical formula of a compound is `CH_2O` and its vapour density is 30. The molecular formula of the compound is:

Options

A. `C_3H_6O_3`
B. `C_2H_4O_2`
C. `CH_2O`
D. `C_2H_4O`

Explanation

To find the molecular formula of the compound with the empirical formula `CH₂O` and a vapor density of 30, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Empirical Formula**:
The empirical formula of the compound is given as `CH₂O`.

2. **Calculate the Molecular Weight**:
The molecular weight can be calculated using the formula:
\[ \text{Molecular Weight} = 2 \times \text{Vapor Density} \]
Given that the vapor density is 30:
\[ \text{Molecular Weight} = 2 \times 30 = 60 \text{ g/mol} \]

3. **Calculate the Empirical Formula Weight**:
The empirical formula weight (EFW) for `CH₂O` can be calculated as follows:
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol (2 H = 2 g/mol)
- Oxygen (O) = 16 g/mol
\[ \text{EFW} = 12 + 2 + 16 = 30 \text{ g/mol} \]

4. **Determine the Value of n**:
The value of \( n \) can be calculated using the formula:
\[ n = \frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}} \]
Substituting the values we found:
\[ n = \frac{60}{30} = 2 \]

5. **Calculate the Molecular Formula**:
The molecular formula can be determined by multiplying the empirical formula by \( n \):
\[ \text{Molecular Formula} = \text{Empirical Formula} \times n = CH₂O \times 2 \]
This gives:
\[ \text{Molecular Formula} = C_{2}H_{4}O_{2} \]

6. **Final Answer**:
The molecular formula of the compound is \( C_{2}H_{4}O_{2} \).