1, Staring material 0.25g compound
Whole Ca of the compound gives 0.16g CaCO_3
Molar mass of CaCO_3->100g"/"mol
So 0.25g compound contains 0.16xx40/100g Ca
Hence %of Ca->0.16xx40/100xx100/0.25=25.60
2, Staring material 0.115g compound
Whole S of the compound gives 0.344g BaSO_4
Molar mass of BaSO_4->233g"/"mol
So 0.115g compound contains 0.344xx32/233g Sulphur
Hence %of S->0.344xx32/233xx100/0.115=40.87
3, Staring material 0.712g compound
Whole N of the compound gives 0.155g NH_3
Molar mass of NH_3->17g"/"mol
So 0.712g compound contains 0.115xx14/17g Nitrogen
Hence %of N->0.115xx14/17xx100/0.712=17.98
So %of C=(100-25.60+40.87+17.98)=15.55
The ratio of number atoms of
Ca:C:N:S=25.6/40:15.55/12:17.98/14:40.87/32
=>Ca:C:N:S=0.64:1.29:1.28:1.28
=>Ca:C:N:S=1:2:2:2
So empirical formula Ca(CNS)_2
The empirical formula mass is 40+2(12+14+32)=156g"/"mol
The given molar mass being same , the molecular formula is also same.
