Let the organic compound be $C_xH_y$. The combustion reaction can be written as:

$C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$

Given:

Volume of organic compound = 80 mL

Volume of $O_2$ = 264 mL

Total volume of gaseous mixture = 224 mL

Volume of gas remaining after passing through KOH = 64 mL (This is $O_2$ remaining)

Volume of $CO_2$ absorbed by KOH = 224 - 64 = 160 mL

From the stoichiometry of the reaction:

80 mL of $C_xH_y$ produces 160 mL of $CO_2$

Therefore, $x = \frac{160}{80} = 2$

Volume of $O_2$ used = 264 - 64 = 200 mL

From the stoichiometry:

$x + \frac{y}{4} = \frac{200}{80} = 2.5$

Since $x = 2$:

$2 + \frac{y}{4} = 2.5$

$\frac{y}{4} = 0.5$

$y = 2$

Therefore, the organic compound is $C_2H_2$.