Question

calculate percentage composition of glucose (C6,H12,O6) (c=12, H=1,O=1​

Answer 1

Explanation:

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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Answer 2

Answer:

Atomic mass: c=12,H=1,o=10)

Explanation:

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

% of carbon (c) in glucose=72/180multiply 100=40.

%of oxygen(o)in glucose=96/180multiply100=53.33

%of hydrogen (h) in glucose=12/180multiply100=6.66