A hydrocarbon contains 80% carbon. The weight of `dm^(3)` gas at NTP is 1.35. The molecular formula of the compound is

Options

A. `CH_(4)`
B. `C_(6)H_(6)`
C. `C_(2)H_(4)`
D. `C_(2)H_(6)`

Explanation

To find the molecular formula of the hydrocarbon given that it contains 80% carbon by mass and the weight of the gas at NTP is 1.35 g/dm³, we can follow these steps:

### Step 1: Calculate the Molar Mass of the Hydrocarbon
At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 dm³. Given that the weight of the gas is 1.35 g/dm³, we can calculate the molar mass (M) of the hydrocarbon using the formula:

\[ \text{Molar Mass} = \text{Density} \times \text{Volume at NTP} \]

Substituting the values:

\[ M = 1.35 \, \text{g/dm}^3 \times 22.4 \, \text{dm}^3 = 30.24 \, \text{g/mol} \]

### Step 2: Determine the Mass of Carbon and Hydrogen
Since the hydrocarbon contains 80% carbon, we can find the mass of carbon in 30.24 g of the hydrocarbon:

\[ \text{Mass of Carbon} = 0.80 \times 30.24 \, \text{g} = 24.192 \, \text{g} \]

The remaining mass will be hydrogen:

\[ \text{Mass of Hydrogen} = 30.24 \, \text{g} - 24.192 \, \text{g} = 6.048 \, \text{g} \]

### Step 3: Calculate the Number of Moles of Carbon and Hydrogen
Now, we can calculate the number of moles of carbon and hydrogen using their molar masses (C = 12 g/mol, H = 1 g/mol):

\[ \text{Moles of Carbon} = \frac{24.192 \, \text{g}}{12 \, \text{g/mol}} = 2.016 \, \text{mol} \]

\[ \text{Moles of Hydrogen} = \frac{6.048 \, \text{g}}{1 \, \text{g/mol}} = 6.048 \, \text{mol} \]

### Step 4: Find the Simplest Ratio
Next, we need to find the simplest whole number ratio of moles of carbon to moles of hydrogen:

\[ \text{Ratio of C:H} = \frac{2.016}{2.016} : \frac{6.048}{2.016} \approx 1 : 3 \]

### Step 5: Write the Molecular Formula
From the simplest ratio, we can deduce the molecular formula of the hydrocarbon:

\[ \text{Molecular Formula} = C_1H_3 = CH_4 \]

### Final Answer
The molecular formula of the hydrocarbon is **CH₄**.

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