one organic compound has c=40%, o=53.34% and h=6.60% then it's empirical formula?
Here the percentages given as :
C = 40%
as we know, atomic weight of carbon is 12 so, divide 40 by 12 ,weget:
40/12= 3.3
H= 6.60%
Same as the above method ,the atomic mass of hydrogen is 1 so, divide 6.60 by 1
=6.60/1=6.6.
O = 53.53%
th e atomic weight of oxygen is 16
Therefore ,divide 53.35 by 16
53.35/16= 3.3
So here the ratio of compound is 1:2:1
So, the empirical formula is CH2O.
thanks for read.
Hey!
C is 40% ( given.) 40/12=3.3 ( 12 is Atomic weight of carbon)
H is 6.60% (given) 6.6/1=6.6 (1 is Atomic weight of hydrogen)
O is 53.35% (given) 53.34/16=3.3 ( 16 is Atomic weight of oxygen)
the ratios of these compound is 1:2:1
Therefore the empirical formula of compound is CH2O
Hope it helped to clear your delimma.
Thank you.
