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

ManipalManipal 2002Some Basic Concepts of Chemistry

$CH_{2}O_{2}$

25%

$CH_{2}O$

60%

$H_{2}$

$CH_{4}$

13%

Solution:

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
$C$	$24$	$40$	$40/12= 3.33$	$3.33/ 3.33=1$
$H$	$4$	$6.66$	$6.66/1= 6.66$	$6.66/3.33=2$
$O$	$32$	$53.33$	$53.33/16=3.33$	$3.33/3.3 =1$