1.

An organic compound on analysis gave the following results : C = 54.5% , O = 36.4 %, H = 9.1% . The Empirical formula of the compound is

Answer»

`C_(2)H_(4)`
`C_(3)H_(4)O`
`C_(3)H_(4)O`
`C_(4)H_(8)O`

SOLUTION :`{:("ELEMENT"," No. of moles"," Simple RATIO"),(C=54.5,54.5//12=4.54,=2(2.2)),(H=9.1,9.1//1=9.1,=4(4.5)),(O=36.4,36.4//16=2.27,=1(1)):}`
Hence `C_(2)H_(4)O`.


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