Question

An organic compound contains $C,H$ and $O$. If $C$(%) $:$ $H$(%) = $6:1$, what is the simplest formula of the compound, given that one mole of the compound contains half as much oxygen as would be required to burn all the $C$ and $H$ atoms in it to $C{O}_2$ and $H_{2}O$?

• A. $C{H}_{2}O$
• B. $C_2{H}_4{O}_3$
• C. $C_3{H}_{6}O$
• D. $C_3{H}_6{O}_2$

Answer 1

Answer: (B)

$C_2{H}_4{O}_3$

Molar ratio of $C:H=\frac{6}{12}:\frac{1}{1}=2:1$
Let the formula of compound be $C_x{H}_{2x}{O}_y$.
$C_x{H}_{2x}{O}_y+O_2\to xC{O}_2+x{H}_{2}O$
From the question, $\frac{3x}{2}=y$
$\therefore x:y=2:3$
Hence, the formula is $C_2{H}_4{O}_3$.