A compound having empirical formula `(C_3H_4O)n` has vapour density 84. The molecular formula of this compound is

Options

A. `C_3H_4O`
B. `C_6H_3O_2`
C. `C_6H_12O_3`
D. `C_9H_12O_3`

Answer

To find the molecular formula of the compound with the empirical formula \( (C_3H_4O)_n \) and a vapor density of 84, we can follow these steps:

### Step 1: Understand the relationship between vapor density and molecular mass
Vapor density (VD) is related to molecular mass (M) by the formula:
\[ \text{Vapor Density} = \frac{\text{Molecular Mass}}{2} \]
Given that the vapor density is 84, we can express this as:
\[ 84 = \frac{M}{2} \]

### Step 2: Calculate the molecular mass
To find the molecular mass, we rearrange the equation:
\[ M = 84 \times 2 = 168 \]

### Step 3: Calculate the molar mass of the empirical formula
The empirical formula is \( C_3H_4O \). We need to calculate its molar mass:
- Carbon (C): 12 g/mol, so for 3 carbon atoms: \( 3 \times 12 = 36 \) g/mol
- Hydrogen (H): 1 g/mol, so for 4 hydrogen atoms: \( 4 \times 1 = 4 \) g/mol
- Oxygen (O): 16 g/mol, so for 1 oxygen atom: \( 1 \times 16 = 16 \) g/mol

Adding these together gives:
\[ \text{Molar mass of } C_3H_4O = 36 + 4 + 16 = 56 \text{ g/mol} \]

### Step 4: Determine the value of n
The molecular mass is related to the empirical formula mass by the equation:
\[ M = n \times \text{(molar mass of empirical formula)} \]
Substituting the values we have:
\[ 168 = n \times 56 \]
To find \( n \), we rearrange the equation:
\[ n = \frac{168}{56} = 3 \]

### Step 5: Write the molecular formula
Now that we have \( n = 3 \), we can write the molecular formula by multiplying the subscripts in the empirical formula by \( n \):
\[ \text{Molecular formula} = C_{3 \times 3}H_{4 \times 3}O_{1 \times 3} = C_9H_{12}O_3 \]

### Final Answer
The molecular formula of the compound is \( C_9H_{12}O_3 \).

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