An organic compound has an empirical formula `CH_2 O`. Its vapour density is 45. The molecular formula of the compound is

Options

A. `CH_2 O`
B. `C_2 H_5 O`
C. `C_3 H_6 O_3`
D. `C_2 H_2`

Explanation

To find the molecular formula of the organic compound with an empirical formula of CH₂O 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 CH₂O. The molecular weights of the individual elements are:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol (and there are 2 H in CH₂O, so \(2 \times 1 = 2\))
- Oxygen (O): 16 g/mol

Now, we can calculate the empirical formula weight:
\[ \text{Empirical Weight} = 12 + 2 + 16 = 30 \]

### Step 3: Determine the Number of Empirical Units (n)
To find the number of empirical units (n), we use the formula:
\[ n = \frac{\text{Molecular Weight}}{\text{Empirical Weight}} \]

Substituting the values we have:
\[ n = \frac{90}{30} = 3 \]

### Step 4: Calculate the Molecular Formula
The molecular formula can be determined by multiplying the empirical formula by n:
\[ \text{Molecular Formula} = n \times \text{Empirical Formula} \]

Thus:
\[ \text{Molecular Formula} = 3 \times \text{CH}_2\text{O} = \text{C}_3\text{H}_6\text{O}_3 \]

### Conclusion
The molecular formula of the compound is **C₃H₆O₃**.

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