The empirical formula of an organic compound is `CH_(2)O`. Its vapour density is 45. The molecular formula of the compound is

Options

A. `CH_(2)O`
B. `C_(2)H_(4)O_(2)`
C. `C_(3)H_(6)O_(3)`
D. `C_(6)H_(12)O_(6)`

Explanation

To determine 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: Understand the relationship between vapor density and molar mass
Vapor density (VD) is related to molar mass (M) by the formula:
\[ \text{Vapor Density} = \frac{M}{2} \]
From this, we can rearrange to find the molar mass:
\[ M = \text{Vapor Density} \times 2 \]

### Step 2: Calculate the molar mass using the given vapor density
Given that the vapor density is 45, we can calculate the molar mass:
\[ M = 45 \times 2 = 90 \text{ g/mol} \]

### Step 3: Calculate the molar mass of the empirical formula
Next, we need to calculate the molar mass of the empirical formula `CH₂O`:
- Carbon (C) has a molar mass of 12 g/mol
- Hydrogen (H) has a molar mass of 1 g/mol, and there are 2 hydrogen atoms: \(1 \times 2 = 2\) g/mol
- Oxygen (O) has a molar mass of 16 g/mol

Now, we can sum these values:
\[ \text{Molar mass of CH}_2\text{O} = 12 + 2 + 16 = 30 \text{ g/mol} \]

### Step 4: Determine the value of n
The molecular formula can be expressed in terms of the empirical formula as:
\[ \text{Molecular formula} = (CH_2O)_n \]
To find n, we use the formula:
\[ n = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}} = \frac{90}{30} = 3 \]

### Step 5: Write the molecular formula
Now that we have n, we can write the molecular formula:
\[ \text{Molecular formula} = (CH_2O)_3 = C_3H_6O_3 \]

### Final Answer
The molecular formula of the compound is \(C_3H_6O_3\).

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