In a compound Carbond `=52.2%` , Hydrogen `=13%`, Oxygen `=34.8%` are present vapour density of the compound is 46. Calculate molecular formula of the compound ?

Options

A. `C_(2)H_(8)O_(2)`
B. `C_(4)H_(8)O_(2)`
C. `C_(4)H_(10)O`
D. `C_(4)H_(12)O_(2)`

Explanation

To calculate the molecular formula of the compound with the given percentages of carbon, hydrogen, and oxygen, we will follow these steps:

### Step 1: Determine the Moles of Each Element
We will convert the percentage composition of each element into moles by dividing by their respective atomic masses.

1. **Carbon (C)**:
- Percentage = 52.2%
- Atomic mass of Carbon = 12 g/mol
- Moles of Carbon = \( \frac{52.2}{12} \approx 4.35 \)

2. **Hydrogen (H)**:
- Percentage = 13%
- Atomic mass of Hydrogen = 1 g/mol
- Moles of Hydrogen = \( \frac{13}{1} = 13 \)

3. **Oxygen (O)**:
- Percentage = 34.8%
- Atomic mass of Oxygen = 16 g/mol
- Moles of Oxygen = \( \frac{34.8}{16} \approx 2.18 \)

### Step 2: Find the Simplest Ratio
Next, we will find the simplest ratio of the moles calculated above.

- Moles of Carbon = 4.35
- Moles of Hydrogen = 13
- Moles of Oxygen = 2.18

To find the simplest ratio, we divide each value by the smallest number of moles (which is approximately 2.18):

1. **Carbon**:
- \( \frac{4.35}{2.18} \approx 2 \)

2. **Hydrogen**:
- \( \frac{13}{2.18} \approx 6 \)

3. **Oxygen**:
- \( \frac{2.18}{2.18} = 1 \)

Thus, the simplest ratio is approximately:
- C: 2
- H: 6
- O: 1

### Step 3: Write the Empirical Formula
From the simplest ratio, the empirical formula of the compound is:
- \( C_2H_6O \)

### Step 4: Calculate the Empirical Mass
Now, we will calculate the empirical mass of the compound:
- Mass of Carbon = \( 2 \times 12 = 24 \)
- Mass of Hydrogen = \( 6 \times 1 = 6 \)
- Mass of Oxygen = \( 1 \times 16 = 16 \)

Total empirical mass = \( 24 + 6 + 16 = 46 \) g/mol

### Step 5: Calculate the Molecular Mass
Given that the vapor density of the compound is 46, we can find the molecular mass:
- Molecular mass = \( 2 \times \text{Vapor Density} = 2 \times 46 = 92 \) g/mol

### Step 6: Determine the Number of Empirical Units in the Molecular Formula
To find how many empirical units are in the molecular formula, we divide the molecular mass by the empirical mass:
- \( n = \frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{92}{46} = 2 \)

### Step 7: Calculate the Molecular Formula
Finally, we multiply the subscripts in the empirical formula by \( n \):
- Molecular formula = \( C_{2 \times 2}H_{6 \times 2}O_{1 \times 2} = C_4H_{12}O_2 \)

### Conclusion
The molecular formula of the compound is:
- **C₄H₁₂O₂**