`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 :

Explanation

Moles of `CaCO_(3)=0.16/100=` Moles of Ca
Wt of `Ca=0.16/100 times 40`
Mass % of Ca `=0.16/100 times 40 times 100/0.25 times 25.6`
Similarly Mass % of `S=0.344/233 times (32 times 100)/0.115=41`
Similarly Mass % of N `=0.155/17 times 14/0.712 times 100=17.9`
`rArr` 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)`,
Molecular formula wt = 156, `n times 156=156 rArr n=1`
Hence, molecular formula `=CaC_(2)N_(2)S_(2)`