• # question_answer 0.24 g of an organic compound gave 0.22 g $C{{O}_{2}}$ on complete combustion. If it contains 1.66 % hydrogen, then the percentage of C and O will be [MP PET 1986] A) 12.5 and 36.6 B) 25 and 75 C) 25 and 36.6 D) 25 and 80

% of C = $\frac{12}{44}\times \frac{\text{Mass of }C{{O}_{2}}}{\text{Mass of substance}}\times 100$ = $\frac{12\times 0.22}{44\times 0.24}\times 100$= 25; C = 25, H = 1.66 Total = 26.6 = 100 - 26.6 = 73.4.