0.22 g of organic compound C(x)H(y)O whi...

0.22 g of organic compound `C_(x)H_(y)O` which occupied 112 mL at NTP and on combustion gave 0.44 g `CO_(2)`. If the percentage of oxygen is 36.45%, then the ratio of x to y in the compound is:

`1:1`

`1:2`

`1:3`

`1:4`

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The correct Answer is:

Molar mass of compound `=("wRT")/("PV")=(0.22xx0.0821xx273xx1000)/(1xx112)=44`
`0.44g CO_(2)` will contain carbon `=(12)/(44)xx0.44g`.
This is the mass of carbon present in 0.22 g of the compound.
% of C in the compound `=(12)/(44)xx0.44xx(100)/(0.22)=54.54%`
% of H = 100 - 54.54 - 36.45 = 9.01%
Then, the ratio of number of atoms of .C. and .H. = `(54.54)/(12):(9.01)/(1)=1:2`

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0.22 g of organic compound `C_(x)H_(y)O` which occupied 112 mL at NTP and on combustion gave 0.44 g `CO_(2)`. If the percentage of oxygen is 36.45%, then the ratio of x to y in the compound is:

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