A compound on analysis gave the following percentage composition by weight: hydrogen = 9.09, oxygen = 36.36 carbon = 54.55
Its `VD` is 44. Find the molecular formula of the compound.

Explanation

To find the molecular formula of the compound based on the given percentage composition and vapor density, follow these steps:

### Step 1: Calculate the number of moles of each element
We start by converting the percentage composition of each element into moles.

- **Carbon (C)**:
\[ \text{Moles of C} = \frac{\text{mass of C}}{\text{molar mass of C}} = \frac{54.55 \, \text{g}}{12 \, \text{g/mol}} = 4.54 \, \text{moles} \]

- **Hydrogen (H)**:
\[ \text{Moles of H} = \frac{\text{mass of H}}{\text{molar mass of H}} = \frac{9.09 \, \text{g}}{1 \, \text{g/mol}} = 9.09 \, \text{moles} \]

- **Oxygen (O)**:
\[ \text{Moles of O} = \frac{\text{mass of O}}{\text{molar mass of O}} = \frac{36.36 \, \text{g}}{16 \, \text{g/mol}} = 2.27 \, \text{moles} \]

### Step 2: Determine the simplest mole ratio
Next, we need to find the simplest ratio of moles for each element by dividing each by the smallest number of moles calculated.

- **Carbon**:
\[ \frac{4.54}{2.27} = 2 \]

- **Hydrogen**:
\[ \frac{9.09}{2.27} = 4 \]

- **Oxygen**:
\[ \frac{2.27}{2.27} = 1 \]

### Step 3: Write the empirical formula
From the simplest mole ratios, we can write the empirical formula:
\[ \text{Empirical formula} = C_2H_4O \]

### Step 4: Calculate the empirical formula weight
Now, we calculate the empirical formula weight:
\[ \text{Empirical formula weight} = (2 \times 12) + (4 \times 1) + (1 \times 16) = 24 + 4 + 16 = 44 \, \text{g/mol} \]

### Step 5: Use vapor density to find molecular weight
Given that the vapor density (VD) is 44, we can find the molecular weight:
\[ \text{Molecular weight} = 2 \times \text{Vapor Density} = 2 \times 44 = 88 \, \text{g/mol} \]

### Step 6: Calculate the n factor
Now, we calculate the n factor:
\[ n = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} = \frac{88}{44} = 2 \]

### Step 7: Determine the molecular formula
Finally, we can find the molecular formula:
\[ \text{Molecular formula} = n \times \text{Empirical formula} = 2 \times C_2H_4O = C_4H_8O_2 \]

### Conclusion
The molecular formula of the compound is:
\[ \text{Molecular formula} = C_4H_8O_2 \]

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