Correct Answer - `CaC_(2)N_(2)S_(2),CaC_(2)N_(2)S_(2)`
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)`