A hydrocabron coantaing `80%` carbon by mass, then the hydrocarbon is :-

Options

A. `CH_(4)`
B. `C_(2)H_(4)`
C. `C_(2)H_(2)`
D. `C_(3)H_(9)`

Answer

To determine the hydrocarbon that contains 80% carbon by mass, we can follow these steps:

### Step 1: Define the Hydrocarbon
Let the hydrocarbon be represented as \( C_xH_y \), where \( x \) is the number of carbon atoms and \( y \) is the number of hydrogen atoms.

### Step 2: Set Up the Mass Percentage Equation
The mass percentage of carbon in the hydrocarbon can be expressed as:
\[ \text{Mass percentage of carbon} = \frac{\text{Mass of carbon}}{\text{Total mass of hydrocarbon}} \times 100 \]
Given that the hydrocarbon contains 80% carbon by mass, we can write:
\[ 80 = \frac{12x}{12x + y} \times 100 \]
where the mass of carbon is \( 12x \) (since the atomic mass of carbon is 12) and the mass of hydrogen is \( y \) (since the atomic mass of hydrogen is 1).

### Step 3: Simplify the Equation
Rearranging the equation gives:
\[ 80 = \frac{12x}{12x + y} \times 100 \]
Dividing both sides by 100:
\[ 0.8 = \frac{12x}{12x + y} \]
Cross-multiplying yields:
\[ 0.8(12x + y) = 12x \]
Expanding this gives:
\[ 9.6x + 0.8y = 12x \]

### Step 4: Rearranging the Equation
Rearranging the equation to isolate \( y \):
\[ 0.8y = 12x - 9.6x \]
This simplifies to:
\[ 0.8y = 2.4x \]
Dividing both sides by 0.8:
\[ y = 3x \]

### Step 5: Determine the Ratio of Carbon to Hydrogen
The ratio of \( y \) to \( x \) is:
\[ \frac{y}{x} = 3 \]
This means for every 1 carbon atom, there are 3 hydrogen atoms. Therefore, the ratio of \( C \) to \( H \) is \( 1:3 \).

### Step 6: Identify Possible Hydrocarbons
From the ratio \( C:H = 1:3 \), we can identify possible hydrocarbons:
- \( C_1H_3 \) (Propene)
- \( C_2H_6 \) (Ethane)
- \( C_3H_9 \) (Propane)
- \( C_4H_{12} \) (Butane)

### Conclusion
Thus, the hydrocarbon that contains 80% carbon by mass can be represented in the form of \( C_xH_y \) where \( y = 3x \). The possible hydrocarbons include \( C_1H_3 \), \( C_2H_6 \), \( C_3H_9 \), and \( C_4H_{12} \).