`A` compound containing `Ca,C,N` and `S` was subjected to quantitative analysis and formula mass determination. `A 0.25 g` of this compound was mixed with `Na_(2)CO_(3)` to convert all `Ca` into `0.16 " "g " "CaCO_(3^(.) A 0.115 g` sample of compound was carried through a series of reaction until all its `S` was changed into `SO_(4)^(-2)` and precipitated as `0.344 g` of `BaSO_(4). A 0.712 g` sample was processed to liberate all of its `N` as `NH_(3)` and `0.155 g NH_(3)` was obtained. The formula mass was found to be `156`. Determine the empirical and molecular formula of the compound :

Answer

`CaC_(2)N_(2)S_(2),CaC_(2)N_(2)S_(2)`

Explanation

Mass `%` of `Ca`
`=(0.16)/(100)xx40xx(100)/(0.25)=25.6`
Mass `%` of `S`
`= (0.344)/(233)xx(32xx100)/(0.115)=41`
Mass `%` of `N`
`=(0.155)/(17)xx(14xx100)/(0.712)=17.9`
implies Mass `%` of `C =15.48`
Now :
`{:("Elements",Ca,S,N,C,),("Mass"%,25.6,41,17.9,15.48,),("Mol ratio",0.64,1.28,1.28,1.29,),("Simple ratio",1,2,2,2,):}`
Empirical formula `=CaC_(2)N_(2)S_(2)`
Empirical formula weight `=156`
Hence, molecular formula `=CaC_(2)N_(2)S_(2)`