%C = 40
%H = 6.67
%O = 53.33
Further explanation
Given
Glucose C₆H₁₂O₆
Required
The percent composition
Solution
MW Gulucose = 180 g/mol
Ar C = 12 g/mol
Ar H = 1 g/mol
Ar O = 16 g/mol
%C = 6.12/180 x 100% = 40%
%H = 12.1/180 x 100% = 6.67%
%O = 6.16/180 x 100% = 53.33%
