Question

$64g$ of an organic compound contains $24g$ of carbon, $8g$ of hydrogen and the rest oxygen. The empirical formula of the compound is

• A. $C{H}_{2}O$
• B. $C_2{H}_{4}O$
• C. $C{H}_{4}O$
• D. $C_2{H}_8{O}_2$

Answer 1

Answer: (C)

Weight of organic compound $=64g$
Weight of carbon $=24g$
Weight of liydrogen $=8g$
Weight of oxygen $=64-[24+8]g$
$=32g$ Percentage of carbon $=\frac{24}{64}\times 100$
$\therefore$ Percentage of carbon $=37.5$
Percentage of hydrogen $=\frac{8}{64}\times 100$
$\therefore$ Percentage of hydrogen $=12.5$
Percentage of oxygen $=\frac{32}{64}\times 100$
$\therefore$ Percentage of oxygen $=50$

$\%$ of element	$C$	$H$	$O$
	$37.5$	$12.5$	$50$
Mole	$\frac{37.5}{12}$	$\frac{12.5}{1}$	$\frac{50}{16}$
Relative number of atoms	$3.125$	$12.5$	$3.125$
Simplest atomic ratio	$\frac{3.125}{3.125}$	$\frac{12.5}{3.125}$	$\frac{3.125}{3.125}$
	$=1$	$=4$	$=1$