Correct Answer - Option 2 : C

_{4} H

_{6}
**Concept:**

Let, the hydrocarbon be C_{x}H_{y}

The general equation for hydrocarbon combustion be

\({C_x}{H_y} + \left( {x\, + \frac{y}{4}} \right){O_2}{\rm{\;}} \to xC{O_2}{\rm{\;}} + {\rm{\;}}\frac{y}{2}{H_2}0\)

Here the carbon has x number of moles and hydrogen has y number of moles in LHS.

So, x is placed before carbon dioxide, and \(\frac{y}{2}\) is placed before water and \(\left( {x\, + \frac{y}{4}} \right)\) by adding the RHS.

For 1 ml of hydrocarbon, the general equation will be written as,

\(1\;ml\; + \;\left( {x\, + \frac{y}{4}} \right)ml{\rm{\;}} \to x\;ml + {\rm{\;}}\frac{y}{2}{\rm{\;ml}}\)

Similarly, for 10 ml of hydrocarbon the general equation will be written as,

\(10\;ml\; + \;10\left( {x\, + \frac{y}{4}} \right)ml{\rm{\;}} \to 10x\;ml + 10.{\rm{\;}}\frac{y}{2}{\rm{\;ml}}\)

**Calculation:**

From the question, 10 mL of a hydrocarbon releases 40 mL of CO_{2}

⇒ 10x = 40

⇒ x = 4

Similarly, from the question, oxygen produced is 55 ml, thus from the general equation given above,

\(10\left( {x + \frac{y}{4}} \right) = 55\)

\(\Rightarrow 10\left( {4 + \frac{y}{4}} \right) = 55\)

\(4 + \frac{y}{4} = \frac{{55}}{{10}}\)

\(\frac{y}{4} = 5.5 - 4\)

y = 1.5 × 4 = 6

∴ The required formula for Hydrocarbon is C

_{4} H

_{6}