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An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be :
A
$CHO$
B
$CH_4O$
C
$CH_3O$
D
$CH_2O$
Detailed Solution
Percentage of oxygen $= 100 - (38.71 + 9.67) = 51.62$
Atomic mass of C = 12, H = 1 and O = 16.
C: mole ratio $= \dfrac{38.71}{12} = 3.22$; simple ratio $= \dfrac{3.22}{3.22} = 1$
H: mole ratio $= \dfrac{9.67}{1} = 9.67$; simple ratio $= \dfrac{9.67}{3.22} = 3$
O: mole ratio $= \dfrac{51.62}{16} = 3.22$; simple ratio $= \dfrac{3.22}{3.22} = 1$
C : H : O = 1 : 3 : 1
Thus the empirical formula of the compound is $CH_3O$.
Atomic mass of C = 12, H = 1 and O = 16.
C: mole ratio $= \dfrac{38.71}{12} = 3.22$; simple ratio $= \dfrac{3.22}{3.22} = 1$
H: mole ratio $= \dfrac{9.67}{1} = 9.67$; simple ratio $= \dfrac{9.67}{3.22} = 3$
O: mole ratio $= \dfrac{51.62}{16} = 3.22$; simple ratio $= \dfrac{3.22}{3.22} = 1$
C : H : O = 1 : 3 : 1
Thus the empirical formula of the compound is $CH_3O$.