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0.2 g of an organic compound on complete combustion produces 0.44 g of CO2, then the percentage of carbon in it is :- (1) 50% (2) 60% (3) 70% (4) 80%

0.2 g of an organic compound on complete combustion produces 0.44 g of CO2, then the percentage of carbon in it is :-
(1) 50% (2) 60% (3) 70% (4) 80%

Grade:

3 Answers

Ravleen Kaur
askIITians Faculty 1452 Points
3 years ago
Hello student,

Let the organic compond be CxHy.
On combustion
CxHy + y/2 O2– xCO2+ y/2 H2O
Given 0.44 g ofCO2 is formed, It means 0.44/44 = 0.01 mol of CO2.
that organic compound contains 0.01 mol of Carbon.
Mass of 0.01 mol of C = 0.01 X 12 =0.12 g
Percentage of carbon = 0.12/0.2 x 100 = 60 %
Hence option (2) is correct answer.

Regards,
IHE
Phanindra
226 Points
3 years ago
%C = 12/44 × [(wt. of CO2 ) / (wt. of org. comp.) ] × 100
Hope you understand this formula...
Also there are formulas to calculate %N, %S , % P ,%Cl , %Br , %I...
SR Roy
128 Points
3 years ago
is correct answer. (2)Hence option60 %OGiven 0.44 g ofCO2 is formed, It means 0.44/44 = 0.01 mol of CO2.that organic compound contains 0.01 mol of Carbon.Mass of 0.01 mol of C = 0.01 X 12 =0.12 gPercentage of carbon = 0.12/0.2 x 100 = 2+ y/2 H2– xCO2.On combustionCxHy + y/2 OyHxLet the organic compond be C

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