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.
