1.

Butyric acid contains only C, H and O. A 4.24 mg sample of butyric acid is completely burned. It gives 8.45 mg of CO_(2) and 3.46 mg og H_(2)O. The molecular mass of butyric acid was determined by experiment to be 88 amu. What is molecular formula?

Answer»


Solution :1 mole `CO_(2)` contains 1 g atom of C
`"i.e.,44 g "CO_(2)" contains C= 12 g"`
`THEREFORE 8.45 MG CO_(2)" will contain C"=(12)/(44)xx8.45mg`
This is present in 4.24 mg of the compound.
`therefore %" of C in the compound "=(12)/(44)xx(8.45)/(4.24)xx100=54.4%`
`"or use the formula directly"%" of C"=(12)/(44)xx("Mass of CO"_(2))/("Mass of compound")xx100`
`"Similarly, "%" of H"=(2)/(18)xx("Mass of "H_(2)O)/("Mass of compound")xx100=(2)/(18)xx(3.46)/(4.24)xx100=9.1%`
`therefore""%" or O"=100-(54.4+9.1)=36.5%`
CALCULATE E.F. Its comes out to be `C_(2)H_(4)O`.
E.F. mass = 44 u, Mol mass = 88 u.
Hence, n = Mol. mass/E.F mass = 2 `therefore"Mol. formula"=2xxE.F.=C_(4)H_(8)O_(2)`


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