`10mL` of gaseous hydrocarbon on combustion gives `40ml` of `CO_(2)(g)` and `50mL` of `H_(2)O` (vapour) The hydrocarbon is .

Options

A. `C_(4)H_(5)`
B. `C_(8)H_(10)`
C. `C_(4)H_(8)`
D. `C_(4)H_(10)`

Explanation

To determine the formula of the hydrocarbon from the given combustion data, we can follow these steps:

### Step 1: Understand the combustion reaction
When a hydrocarbon combusts, it reacts with oxygen to produce carbon dioxide (CO₂) and water (H₂O). The general combustion reaction can be represented as:
\[ \text{C}_x\text{H}_y + O_2 \rightarrow CO_2 + H_2O \]

### Step 2: Analyze the given data
We are given:
- Volume of hydrocarbon = 10 mL
- Volume of CO₂ produced = 40 mL
- Volume of H₂O produced = 50 mL

### Step 3: Determine the moles of carbon and hydrogen
From the volumes of CO₂ and H₂O produced, we can find the moles of carbon and hydrogen in the hydrocarbon.

1. **Carbon from CO₂**:
- Each mole of CO₂ contains 1 mole of carbon.
- Therefore, 40 mL of CO₂ corresponds to 40 mL of carbon.
- The ratio of hydrocarbon to CO₂ is:
\[ \text{Hydrocarbon} : CO_2 = 10 \text{ mL} : 40 \text{ mL} = 1 : 4 \]
This means that 1 mole of the hydrocarbon produces 4 moles of carbon.

2. **Hydrogen from H₂O**:
- Each mole of H₂O contains 2 moles of hydrogen.
- Therefore, 50 mL of H₂O corresponds to 50 mL of water, which contains:
\[ \text{Hydrogen} = 50 \text{ mL} \times 2 = 100 \text{ mL of H atoms} \]
The ratio of hydrocarbon to H₂O is:
\[ \text{Hydrocarbon} : H_2O = 10 \text{ mL} : 50 \text{ mL} = 1 : 5 \]
This means that 1 mole of the hydrocarbon produces 5 moles of hydrogen.

### Step 4: Write the empirical formula
From the above ratios:
- The hydrocarbon has 4 moles of carbon (C) and 10 moles of hydrogen (H).
- Therefore, the empirical formula can be represented as:
\[ C_4H_{10} \]

### Step 5: Conclusion
The molecular formula of the hydrocarbon is:
\[ \text{C}_4\text{H}_{10} \]