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Calculate percentage composition of glucose (C6H12O6). 

(Atomic mass: C = 12, H = 1, O = 16)

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Molecular mass of glucose (C6H12O6) = 6 × 12 + 1 × 12 + 6 × 16 = 72 + 12 + 96 = 180 g 

%of carbon(C) in glucose = 72 /180 × 100 = 40 

% of hydrogen (H) in glucose = 12/ 180 × 100 = 6.66 

% of oxygen(O) in glucose = 96/ 180 × 100 = 53.33.

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