Find the percentage composition of glucose whose formula is C6H12O6
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C = 40 %
H = 6.66 %
O = 53.33 %
Explanation:
Given
glucose formula = C6H12O6
percentage composition = ??
Solution
As we know that
C = 12 a. m. u.
H = 1 a. m. u.
O = 16 a. m. u.
molar mass of glucose = 180 g/mol
% composition = molar mass of atom/ molar mass of compound X 100
% C = 72/180 X 100
% C = 40
% H = 12/180 X 100
% H = 6.66
% O = 96/180 X 100
% O = 53.33
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