An organic compound containing carbon, hydrogen and oxygen gave the following percentage composition :
C = 40.68%, H = 5.08%
The vapour density of the compound is 59. Calculate the molecular formula of the compound.

Explanation

To calculate the molecular formula of the organic compound containing carbon, hydrogen, and oxygen based on the given percentage composition and vapor density, we can follow these steps:

### Step 1: Determine the mass of each element
Assume the total mass of the compound is 100 g. This makes it easier to convert percentages to grams.

- Mass of Carbon (C) = 40.68 g
- Mass of Hydrogen (H) = 5.08 g
- Mass of Oxygen (O) = 100 g - (40.68 g + 5.08 g) = 100 g - 45.76 g = 54.24 g

### Step 2: Calculate the number of moles of each element
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- Moles of Carbon (C):
\[ \text{Moles of C} = \frac{40.68 \text{ g}}{12 \text{ g/mol}} = 3.39 \text{ moles} \]

- Moles of Hydrogen (H):
\[ \text{Moles of H} = \frac{5.08 \text{ g}}{1 \text{ g/mol}} = 5.08 \text{ moles} \]

- Moles of Oxygen (O):
\[ \text{Moles of O} = \frac{54.24 \text{ g}}{16 \text{ g/mol}} = 3.39 \text{ moles} \]

### Step 3: Find the simplest mole ratio
Now, we divide the number of moles of each element by the smallest number of moles calculated:

- Ratio of C:
\[ \frac{3.39}{3.39} = 1 \]

- Ratio of H:
\[ \frac{5.08}{3.39} \approx 1.5 \]

- Ratio of O:
\[ \frac{3.39}{3.39} = 1 \]

### Step 4: Convert to whole numbers
To convert the ratios to whole numbers, we can multiply all ratios by 2 (since 1.5 is not a whole number):

- C: \(1 \times 2 = 2\)
- H: \(1.5 \times 2 = 3\)
- O: \(1 \times 2 = 2\)

Thus, the empirical formula is \(C_2H_3O_2\).

### Step 5: Calculate the empirical formula mass
Now we calculate the molar mass of the empirical formula:
\[ \text{Empirical mass} = (2 \times 12) + (3 \times 1) + (2 \times 16) = 24 + 3 + 32 = 59 \text{ g/mol} \]

### Step 6: Calculate the molecular mass
The molecular mass can be calculated using the vapor density:
\[ \text{Molecular mass} = 2 \times \text{Vapor density} = 2 \times 59 = 118 \text{ g/mol} \]

### Step 7: Determine the molecular formula
Now, we find the ratio of the molecular mass to the empirical mass:
\[ \text{Ratio} = \frac{\text{Molecular mass}}{\text{Empirical mass}} = \frac{118}{59} = 2 \]

Thus, the molecular formula is obtained by multiplying the subscripts in the empirical formula by this ratio:
\[ \text{Molecular formula} = C_{2 \times 2}H_{3 \times 2}O_{2 \times 2} = C_4H_6O_4 \]

### Final Answer:
The molecular formula of the compound is \(C_4H_6O_4\).

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