Writing The Formula for Cerium(IV) Sulfide



Writing The Formula for Cerium(IV) Sulfide

Consider the binary compound Cerium(IV) sulfide. We look on an ion table or common cation and anion list. The Cerium(IV) and sulfide are respectively Ce+4 and S-2. The beginning formula is (Ce+4)x(S-2)y. The charges multiplied by the subscripts must result in zero:  x(+4) + y(-2)=0. There are many pairs of subscripts that will allow the result to be zero: 6:12, 4:8, 2:4 and 1:2. We must use the lowest ratio. That is 1:2. The charges add to zero: 1(+4) +2(-2)=0. That will give us (Ce+4)1(S-2)2. The final equation is CeS2. The positive and negative ion charges are not left in the final formula. A subscript of “1” is assumed to be there with out it being written. And the parenthesis are left in the final formula only if they enclose a polyatomic ion with a subscript greater than “1" , as in the next example.


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