The molecular formula of a gas with vapour density 15 and empirical formula CH3 is :

Answer

To find the molecular formula of a gas with a given vapor density and empirical formula, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Information:**
- Vapour Density (VD) = 15
- Empirical Formula = CH₃

2. **Calculate the Molecular Weight:**
- The molecular weight (MW) can be calculated using the formula:
\[ \text{Molecular Weight} = 2 \times \text{Vapor Density} \]
- Substituting the given vapor density:
\[ \text{Molecular Weight} = 2 \times 15 = 30 \]

3. **Calculate the Empirical Formula Weight:**
- The empirical formula is CH₃. To find its weight, we add the atomic weights of carbon (C) and hydrogen (H):
- Atomic weight of C = 12 g/mol
- Atomic weight of H = 1 g/mol
- Therefore, the empirical formula weight (EFW) is:
\[ \text{Empirical Formula Weight} = 12 + (3 \times 1) = 12 + 3 = 15 \text{ g/mol} \]

4. **Calculate 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 have:
\[ n = \frac{30}{15} = 2 \]

5. **Determine the Molecular Formula:**
- The molecular formula can be found by multiplying the empirical formula by n:
\[ \text{Molecular Formula} = n \times \text{Empirical Formula} \]
- Thus:
\[ \text{Molecular Formula} = 2 \times \text{CH}_3 = \text{C}_2\text{H}_6 \]

### Final Answer:
The molecular formula of the gas is **C₂H₆**.