An organic compound containing only C, H and K. A 4.24 MG sample of this compound is completely burnt it gives 8.45 mg of Co2 and 3.46 MG H2O. Determine the empirical formula of organic compound .
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i) Determination of percentage of each element
%C = 12 × wt. of CO2 × 100/44 × wt. of org. compound
= 12 × 8.45 × 100/ 44 × 4.24
= 54.35%
%H = 2/18 × wt. of H2O × 100/ wt. of org. compound = 2 × 3.46 × 100/ 18×4.24
= 9.07%
%O = 100 -(54.35+9.07)36.58%
ii)Determination of atomic ratio
C= 54.35/12=4.53
H=9.07/1=9.07
O=36.58/16=2.28
iii)Dividing by smallest quotient and converting to whole number.
C=4.53/2.28=1.98~2
H=9.07/2.28=3.98~4
O=2.28/2.28=1
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