Question

If $0.765$g of an organic compound on combustion gives $0.535$g $C{O}_2$ and $0.138$g of $H_{2}O$, then the ratio of the percentage of carbon and hydrogen is:

• A. 19 : 2
• B. 18 : 11
• C. 20 : 17
• D. 1 : 7

Answer 1

Answer: (A)

19 : 2

% $C=\frac{12}{44}\times \frac{mofc{o}_2}{mofcompd}\times 100$
$=\frac{12}{44}\times \frac{0.535}{0.765}\times 100$
%$C=19.07$%
%$H=\frac{2}{18}\times \frac{mof{H}_{2}o}{mofcompd}\times 100$
$=\frac{2}{18}\times \frac{0.138}{0.765}\times 100$
%$H=2.00$C:H = 19:2$