An alcohol contains 64.81% C, 13.60% H, and 21.59% O. The molecular
mass of this compound is 74 g/mole. Determine the Empirical formula
and Molecular formula.
Molecular Mass of H = 1 g/mole
Molecular Mass of O = 16 g/mole
Molecular Mass of C = 12 g/mole
Molecular Mass of Compound = 74 g/mole
% of C present in Compound = 64.81%
Mass of C in the compound = (64.81/100) * 74 = 48.02g
So no of C atoms = 48.02/12 = 4
% of H present in Compound = 13.60%
Mass of H in the compound = (13.60/100) * 74 = 10g
So no of H atoms = 10/1 = 10
% of O present in Compound = 21.59%
Mass of O in the compound = (21.59/100) * 74 = 15.99g
So no of O atoms = 15.99/16 = 1
So Empirical Formula of Compound = (C4H10O)n
and Molecular Formula = C4H10O
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