1.

0.30g of an organic compound containing C, Hand Oxygen on combustion yields 0.44g CO_(2) and 0.18g H_(2)O. If one mole of compound weighs 60, then molecular formula of the compound is

Answer»

`C_(3)H_(8)O`
`C_(2)H_(4)O_(2)`
`CH_(2)O`
`C_(4)H_(6)O`

Solution :PERCENTAGE of `C=(12)/(44)XX(0.44)/(0.30)xx100=40%`
Percentage of `=(2)/(18)xx(0.18)/(0.30)xx100=6.6%`
percentate of `O=100-(40+6.6)=53.4%`

Hence, EMPIRICAL formula `=CH_(2)O`
`n=("Molecular mass")/("Empirical formula mass")=(60)/(30)=2`
`rArr` Molecular formula of COMPOUND `=(CH_(2)O)_(2)`
`=C_(2)H_(4)O_(2)`


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