1.

0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g `CO_(2)` and 0.18 g `H_(2)O`. If one mole of compound weighs 60, then molecular formula of the compound isA. `C_(3)H_(8)O`B. `C_(2)H_(4)O_(2)`C. `CH_(2)O`D. `C_(4)H_(6)O`

Answer» Correct Answer - B
`%C = (12)/(44) xx (0.44)/(0.30) xx 100 = 40`
`%H = (2)/(18) xx (0.8)/(0.30) xx 100 = 6.66`
`%O = 100 - (40.0 + 6.66) = 53.34`
Now `C:H:O = (40)/(12) : (6.66)/(1.0) : (53.34)/(16)`
= 3.33: 6.66 : 3.33
`:. E.F. = CH_(2)O`
E.F. wt. = 12+2+16 = 30
But Mol. wt. = 60
`:. M.F. = CH_(2)O xx (60)/(30) = C_(2)H_(4)O_(2)`


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