A compound (60 g ) on analysis produce carbon, hydrogen and oxygen 24 g, 4 g and 32 g respectively. The empirical formula is :

Question

A compound (60 g ) on analysis produce carbon, hydrogen and oxygen 24 g, 4 g and 32 g respectively. The empirical formula is : (A) C2H2O2 (B) C2H4O2 (C)  CH2O (D) C2H4O6

Tardigrade Admin · Accepted Answer

∵ 60 g of compound contains 24 g of carbon ∴ % of carbon =(24 × 100/60)=40 % Similarly, percentage of hydrogen =(4 × 100/60)=6.66 % ∴ percentage of oxygen =100-(40+6.66) =53.34 % Element t Percentage Atomic c mass Relative number of atoms Ratio C 40 % 12 40/12=3.33 3.33/3.3=3=1 H 6.66 % 1 6.66/1=6.66 6.66/3.33=2 O 53.34 % 16 53.34/16=3.33 3.33/3.33=1 Therefore, empirical formula = CH 2 O

Element $t$	Percentage	Atomic $c$ mass	Relative number of atoms	Ratio
$C$	$40%$	$12$	$40/12 = 3.33$	$3.33/3.3 = 3 = 1$
$H$	$6.66%$	$1$	$6.66/1 = 6.66$	$6.66/3.33 = 2$
$O$	$53.34%$	$16$	$53.34/16 = 3.33$	$3.33/3.33 = 1$

Q. A compound $(60 g)$ on analysis produce carbon, hydrogen and oxygen $24 g, 4 g$ and $32 g$ respectively. The empirical formula is :

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

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

$C H_{2} O$

$C_{2} H_{4} O_{6}$

Q. A compound (60g) on analysis produce carbon, hydrogen and oxygen 24g,4g and 32g respectively. The empirical formula is :

C2​H2​O2​

C2​H4​O2​

CH2​O

C2​H4​O6​

Q. A compound $(60 g)$ on analysis produce carbon, hydrogen and oxygen $24 g, 4 g$ and $32 g$ respectively. The empirical formula is :

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

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

$C H_{2} O$

$C_{2} H_{4} O_{6}$