An organic compound contains `49.3%` carbon `6.84%` hydrogen and its vapour density is 73 . Molecular formula of the compund is

Options

A. `C_(3)H_(5)O_(2)`
B. `C_(6)H_(10)O_(4)`
C. `C_(3)H_(10)O_(2)`
D. `C_(4)H_(10)O_(2)`

Explanation

To find the molecular formula of the organic compound given the percentage composition and vapor density, we will follow these steps:

### Step 1: Determine the percentage of oxygen
Given:
- Carbon (C) = 49.3%
- Hydrogen (H) = 6.84%

To find the percentage of oxygen (O):
\[ \text{Percentage of O} = 100\% - (\text{Percentage of C} + \text{Percentage of H}) \]
\[ \text{Percentage of O} = 100\% - (49.3\% + 6.84\%) = 100\% - 56.14\% = 43.86\% \]

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound:
- Mass of C = 49.3 g
- Mass of H = 6.84 g
- Mass of O = 43.86 g

### Step 3: Calculate the number of moles of each element
Using the molar masses:
- Molar mass of C = 12 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of O = 16 g/mol

Calculating the number of moles:
\[ \text{Moles of C} = \frac{49.3 \text{ g}}{12 \text{ g/mol}} \approx 4.11 \text{ moles} \]
\[ \text{Moles of H} = \frac{6.84 \text{ g}}{1 \text{ g/mol}} = 6.84 \text{ moles} \]
\[ \text{Moles of O} = \frac{43.86 \text{ g}}{16 \text{ g/mol}} \approx 2.74 \text{ moles} \]

### Step 4: Determine the simplest mole ratio
To find the simplest ratio, divide each mole value by the smallest number of moles (2.74):
\[ \text{Ratio of C} = \frac{4.11}{2.74} \approx 1.50 \]
\[ \text{Ratio of H} = \frac{6.84}{2.74} \approx 2.50 \]
\[ \text{Ratio of O} = \frac{2.74}{2.74} = 1 \]

### Step 5: Convert to whole numbers
To convert the ratios to whole numbers, multiply each by 2:
\[ \text{C} = 1.50 \times 2 = 3 \]
\[ \text{H} = 2.50 \times 2 = 5 \]
\[ \text{O} = 1 \times 2 = 2 \]

Thus, the empirical formula is \( C_3H_5O_2 \).

### Step 6: Calculate the empirical formula mass
\[ \text{Empirical formula mass} = (3 \times 12) + (5 \times 1) + (2 \times 16) = 36 + 5 + 32 = 73 \text{ g/mol} \]

### Step 7: Calculate the molecular formula
Using the vapor density to find the molecular mass:
\[ \text{Molecular mass} = 2 \times \text{Vapor Density} = 2 \times 73 = 146 \text{ g/mol} \]

### Step 8: Determine the factor \( n \)
\[ n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{146}{73} = 2 \]

### Step 9: Write the molecular formula
Multiply the subscripts in the empirical formula by \( n \):
\[ \text{Molecular formula} = C_{3 \times 2}H_{5 \times 2}O_{2 \times 2} = C_6H_{10}O_4 \]

### Final Answer
The molecular formula of the compound is \( C_6H_{10}O_4 \).

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