10 ml gaseous hydrocarbon was burnt completely in 80 ml of `O_(2)` at NTP. The remaining gas occupied 70 ml at NTP. This volume became 50 ml on treatment with KOH solution. The empirical formula of the hydrocarbon is :

Options

A. `C_(3)H_(6)`
B. `C_(2)H_(4)`
C. `CH_(4)`
D. `C_(6)H_(6)`

Answer

To find the empirical formula of the hydrocarbon, we will follow these steps:

### Step 1: Write the general reaction for the combustion of the hydrocarbon.
Let the hydrocarbon be represented as \( C_xH_y \). The combustion reaction can be written as:
\[ C_xH_y + \left( \frac{x + y}{4} \right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]

### Step 2: Determine the volumes of gases involved.
From the problem, we know:
- Volume of hydrocarbon burnt = 10 ml
- Volume of \( O_2 \) used = 80 ml
- Remaining gas after combustion = 70 ml
- Volume after treatment with KOH = 50 ml

### Step 3: Calculate the volume of \( CO_2 \) produced.
The volume of gas remaining after combustion (70 ml) consists of unused \( O_2 \) and \( CO_2 \). When treated with KOH, which absorbs \( CO_2 \), the volume decreases to 50 ml. Therefore, the volume of \( CO_2 \) absorbed by KOH is:
\[ 70 \, \text{ml} - 50 \, \text{ml} = 20 \, \text{ml} \]

### Step 4: Relate the volume of \( CO_2 \) to the hydrocarbon.
Since the volume of \( CO_2 \) produced is equal to \( 10x \) (from the stoichiometry of the reaction), we can set up the equation:
\[ 10x = 20 \implies x = 2 \]

### Step 5: Calculate the volume of unused \( O_2 \).
The total volume of \( O_2 \) used can be calculated as follows:
\[ \text{Unused } O_2 = 70 \, \text{ml} - 20 \, \text{ml} = 50 \, \text{ml} \]
Thus, the volume of \( O_2 \) used is:
\[ 80 \, \text{ml} - 50 \, \text{ml} = 30 \, \text{ml} \]

### Step 6: Relate the volume of \( O_2 \) used to the hydrocarbon.
From the stoichiometry, the volume of \( O_2 \) used is given by:
\[ \text{Volume of } O_2 = 10 \left( \frac{x + y}{4} \right) \]
Setting this equal to the volume used:
\[ 10 \left( \frac{2 + y}{4} \right) = 30 \]
Solving for \( y \):
\[ \frac{2 + y}{4} = 3 \implies 2 + y = 12 \implies y = 10 \]

### Step 7: Write the empirical formula.
Now we have \( x = 2 \) and \( y = 10 \). Therefore, the empirical formula of the hydrocarbon is:
\[ C_2H_4 \]

### Final Answer:
The empirical formula of the hydrocarbon is \( C_2H_4 \).

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