A compound with empirical formula `C_2H_5O` has a vapour density of 45. The molecular formula of the compound will be

Options

A. `C_(2)H_(5)O`
B. `C_(2)H_(2)O_(4)`
C. `C_(5)H_(14)O_(4)`
D. `C_(4)H_(10)O_(2)`

Explanation

To determine the molecular formula of the compound with the empirical formula \( C_2H_5O \) and a vapor density of 45, we can follow these steps:

### Step 1: 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 45:
\[ \text{Molecular Weight} = 2 \times 45 = 90 \]

### Step 2: Calculate the Empirical Formula Weight
Next, we need to calculate the weight of the empirical formula \( C_2H_5O \):
- Carbon (C): 2 atoms × 12 g/mol = 24 g/mol
- Hydrogen (H): 5 atoms × 1 g/mol = 5 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

Now, add these together:
\[ \text{Empirical Formula Weight} = 24 + 5 + 16 = 45 \text{ g/mol} \]

### Step 3: Determine the Number of Empirical Units in the Molecular Formula
Now, we can find the number of empirical units (n) in the molecular formula using:
\[ n = \frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}} \]
Substituting the values we calculated:
\[ n = \frac{90}{45} = 2 \]

### Step 4: Calculate the Molecular Formula
The molecular formula can be obtained by multiplying the empirical formula by n:
\[ \text{Molecular Formula} = n \times \text{Empirical Formula} = 2 \times C_2H_5O \]
This gives:
\[ \text{Molecular Formula} = C_{2 \times 2}H_{5 \times 2}O_{1 \times 2} = C_4H_{10}O_2 \]

### Final Answer
Thus, the molecular formula of the compound is:
\[ \text{Molecular Formula} = C_4H_{10}O_2 \]

---