A hydrocarbon contain `80%` C . The vapour density of compound in 30 . Molecular formula of compound is :-

Options

A. (a)`CH_(3)`
B. (b)`C_(2)H_(6)`
C. (c)`C_(4)H_(12)`
D. (d)All of these

Answer

To determine the molecular formula of the hydrocarbon that contains 80% carbon and has a vapor density of 30, we can follow these steps:

### Step 1: Calculate the Molecular Mass
The molecular mass (M) of a compound can be calculated using the formula:
\[ M = 2 \times \text{Vapor Density} \]

Given that the vapor density is 30:
\[ M = 2 \times 30 = 60 \]

### Step 2: Determine the Mass of Carbon and Hydrogen
We know that the hydrocarbon contains 80% carbon. To find the mass of carbon in the molecular mass:
\[ \text{Mass of Carbon} = 80\% \text{ of } 60 = \frac{80}{100} \times 60 = 48 \text{ g} \]

The remaining mass will be the mass of hydrogen:
\[ \text{Mass of Hydrogen} = \text{Molecular Mass} - \text{Mass of Carbon} = 60 - 48 = 12 \text{ g} \]

### Step 3: Calculate the Number of Moles of Carbon and Hydrogen
Using the atomic masses (Carbon = 12 g/mol, Hydrogen = 1 g/mol):
- Moles of Carbon:
\[ \text{Moles of Carbon} = \frac{\text{Mass of Carbon}}{\text{Molar Mass of Carbon}} = \frac{48}{12} = 4 \text{ moles} \]

- Moles of Hydrogen:
\[ \text{Moles of Hydrogen} = \frac{\text{Mass of Hydrogen}}{\text{Molar Mass of Hydrogen}} = \frac{12}{1} = 12 \text{ moles} \]

### Step 4: Determine the Simplest Ratio
The simplest ratio of moles of Carbon to Hydrogen is:
\[ \text{C:H} = 4:12 \]

To simplify this ratio, divide both numbers by the greatest common divisor (4):
\[ \text{C:H} = 1:3 \]

### Step 5: Write the Empirical Formula
From the simplest ratio, the empirical formula of the hydrocarbon is:
\[ \text{CH}_3 \]

### Step 6: Determine the Molecular Formula
Since the empirical formula mass of CH₃ is:
\[ \text{Molar Mass of CH}_3 = 12 + 3 \times 1 = 15 \text{ g/mol} \]

Now, we can find the number of empirical units in the molecular formula:
\[ \text{Number of units} = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{60}{15} = 4 \]

Thus, the molecular formula is:
\[ \text{C}_{4}\text{H}_{12} \]

### Final Answer
The molecular formula of the compound is:
\[ \text{C}_{4}\text{H}_{12} \]

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