60 gm of a compound on analysis produce 24 gm carbon, 4 gm hydrogen and 32 gm oxygen. The empirical formula of the compound is:

Question

60 gm of a compound on analysis produce 24 gm carbon, 4 gm hydrogen and 32 gm oxygen. The empirical formula of the compound is: (A) CH2O2 (B) CH2O (C) H2 (D) CH4

Tardigrade Admin · Accepted Answer

Element Mass Percentage Mass Ratio Simplest Ratio C 24 40 40/12= 3.33 3.33/ 3.33=1 H 4 6.66 6.66/1= 6.66 6.66/3.33=2 O 32 53.33 53.33/16=3.33 3.33/3.3 =1 Hence empirical formula = CH 2 O

Element	Mass	Percentage	Mass Ratio	Simplest Ratio
$C$	$24$	$40$	$40/12 = 3.33$	$3.33/3.33 = 1$
$H$	$4$	$6.66$	$6.66/1 = 6.66$	$6.66/3.33 = 2$
$O$	$32$	$53.33$	$53.33/16 = 3.33$	$3.33/3.3 = 1$

Q. $60 g m$ of a compound on analysis produce $24 g m$ carbon, $4 g m$ hydrogen and $32 g m$ oxygen. The empirical formula of the compound is:

$C H_{2} O_{2}$

$C H_{2} O$

$H_{2}$

$C H_{4}$

Element Mass Percentage Mass Ratio Simplest Ratio

$C$ $24$ $40$ $40/12 = 3.33$ $3.33/3.33 = 1$

$H$ $4$ $6.66$ $6.66/1 = 6.66$ $6.66/3.33 = 2$

$O$ $32$ $53.33$ $53.33/16 = 3.33$ $3.33/3.3 = 1$

Hence empirical formula $= C H_{2} O$

Q. 60gm of a compound on analysis produce 24gm carbon, 4gm hydrogen and 32gm oxygen. The empirical formula of the compound is:

CH2​O2​

CH2​O

H2​

CH4​