1.

An organic compound on analysis gaveC=48 g,H=8g and N=56 g. Volume of 1.0 g of the compound was foud to be 200ml at NTP. Molecular formula of the compound is

Answer»

`C_(4)H_(8)N_(4)`
`C_(2)H_(4)N_(2)`
`C_(12)H_(24)N_(12)`
`C_(16)H_(32)N_(16)`

Solution :`{:("Element",%," NO. of moles"," Simple ratio"),(C,48,48//12=4,1),(H,8,8//1=8,2),(N,56,56//14=4,1):}`
Empirical FORMULA ` = CH_(2)N`
Empirical formula MASS `= 28`
Now, `200ml` of compound `= 1 gm`
`22400 ml` of compound `1/200 XX 22400 = 112`
`n '(" Mol.mass")/(" Emp.formula mass") = 112/28 = 4`
Therefore, molecularformula `= (CH_(2)N)_(4) = C_(4)H_(8)N_(4)`.


Discussion

No Comment Found

Related InterviewSolutions