An organic compound with C=40 % and H=6.7 % will have the empirical formula

Question

An organic compound with C=40 % and H=6.7 % will have the empirical formula (A)  CH2  (B)  CH2O  (C) C 3 H 6 O 3 (D) C 2 H 4 O 2

Tardigrade Admin · Accepted Answer

Elements % No.of moles Simple Ratio C =40 % 40 / 12=3.33 3.33 / 3.33=1 H =6.7 % 6.7 / 1=6.7 6.7 / 3.33=2 O =53.3 % 53.3 / 16=3.33 3.33 / 3.33=1 Thus, emipirical formula = CH 2 O

Elements $%$	No.of moles	Simple Ratio
$C = 40%$	$40/12 = 3.33$	$3.33/3.33 = 1$
$H = 6.7%$	$6.7/1 = 6.7$	$6.7/3.33 = 2$
$O = 53.3%$	$53.3/16 = 3.33$	$3.33/3.33 = 1$

Q. An organic compound with $C = 40%$ and $H = 6.7%$ will have the empirical formula

$C H_{2}$

$C H_{2} O$

$C_{3} H_{6} O_{3}$

$C_{2} H_{4} O_{2}$

Elements $%$ No.of moles Simple Ratio

$C = 40%$ $40/12 = 3.33$ $3.33/3.33 = 1$

$H = 6.7%$ $6.7/1 = 6.7$ $6.7/3.33 = 2$

$O = 53.3%$ $53.3/16 = 3.33$ $3.33/3.33 = 1$

Thus, emipirical formula $= C H_{2} O$

Q. An organic compound with C=40% and H=6.7% will have the empirical formula

CH2​

CH2​O

C3​H6​O3​

C2​H4​O2​