Question

5.6 g of an organic compound on burning with excess of oxygen gave 17.6 g of $C{O}_2$ and $7.2g{H}_{2}O$. The organic compound is:

• A. $C_6{H}_6$
• B. $C_4{H}_8$
• C. $C_3{H}_8$
• D. $C{H}_{3}COOH$
• E. $C{H}_{3}CHO$

Answer 1

The correct option is B $C_4{H}_8$

C${}_x{H}_y$ +$O_2$ $\to$ $xC{O}_2$ +$\frac{y}{2}{H}_{2}O$

17.6g of $C{O}_2$=0.4moles

7.2g of $H_{2}O$=0.4 moles

From reaction $x=0.4$ and $\frac{y}{2}=0.4$

$\therefore \frac{x}{y}=\frac{1}{2}$

From given options , the organic compound is C${}_4{H}_8$ is in ratio 1:2.

Hence, option $B$ is correct.