Question

A compound contains $54.55\%$ carbon, $9.09\%$ hydrogen and $36.36\%$ oxygen . The empirical formula of this compound is

• A. $C_3{H}_{5}O$
• B. $C_4{H}_8{O}_2$
• C. $C_2{H}_4{O}_2$
• D. $C_2{H}_{4}O$

Answer 1

Answer: (D)

Element	$\%$	At. weight	$\%$ Atomic weight	Simplest molar ratio
$C$	$54.55$	$12$	$\frac{54.55}{12}=4.54$	$\frac{4.54}{2.275}=2$
$H$	$9.09$	$1$	$\frac{9.09}{1}=9.09$	$\frac{9.09}{2.27}=4$
$O$	$36.36$	$16$	$\frac{36.36}{16}=2.27$	$\frac{2.27}{2.27}=1$

$\therefore$ Empirical formula is $C_2{H}_{4}O$.