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 can follow these steps:

### Step 1: Assume a Total Mass
Assume the total mass of the compound is 100 grams. This makes it easier to convert percentages into grams.

- Carbon (C) = 52.2 grams
- Hydrogen (H) = 13 grams
- Oxygen (O) = 34.8 grams

### Step 2: Calculate the Number of Moles
Use the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- For Carbon:
\[ \text{Moles of C} = \frac{52.2 \text{ g}}{12 \text{ g/mol}} = 4.35 \text{ moles} \]

- For Hydrogen:
\[ \text{Moles of H} = \frac{13 \text{ g}}{1 \text{ g/mol}} = 13 \text{ moles} \]

- For Oxygen:
\[ \text{Moles of O} = \frac{34.8 \text{ g}}{16 \text{ g/mol}} = 2.175 \text{ moles} \]

### Step 3: Divide by the Smallest Number of Moles
Identify the smallest number of moles calculated, which is for Oxygen (2.175 moles), and divide all mole values by this number to find the simplest ratio.

- For Carbon:
\[ \frac{4.35}{2.175} = 2 \]

- For Hydrogen:
\[ \frac{13}{2.175} \approx 6 \]

- For Oxygen:
\[ \frac{2.175}{2.175} = 1 \]

### Step 4: Write the Empirical Formula
From the ratios obtained, we can write the empirical formula:
\[ \text{Empirical formula} = C_2H_6O_1 \text{ or } C_2H_6O \]

### Step 5: Calculate the Empirical Formula Mass
Calculate the molar mass of the empirical formula \(C_2H_6O\):
\[ \text{Molar mass} = (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46 \text{ g/mol} \]

### Step 6: Use Vapour Density to Find Molecular Weight
Given that the vapour density of the compound is 46, we can find the molecular weight using the formula:
\[ \text{Vapour Density} = \frac{\text{Molecular Weight}}{2} \]
Thus,
\[ \text{Molecular Weight} = 46 \times 2 = 92 \text{ g/mol} \]

### Step 7: Determine the Value of n
Now, we can find \(n\) using the formula:
\[ n = \frac{\text{Molecular Weight}}{\text{Empirical Formula Mass}} = \frac{92}{46} = 2 \]

### Step 8: Calculate the Molecular Formula
To find the molecular formula, multiply the subscripts in the empirical formula by \(n\):
\[ \text{Molecular Formula} = C_{2 \times 2}H_{6 \times 2}O_{1 \times 2} = C_4H_{12}O_2 \]

### Final Answer
The molecular formula of the compound is:
\[ \text{Molecular Formula} = C_4H_{12}O_2 \]

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