**How To Write Formulas**

An inorganic chemical formula represents an empirical combination of elements determined by experiment. The inorganic chemical formula can also be viewed as being constructed from commonly identified cations and anions. The cations have a positive charge and the anions have a negative charge. The resulting compound must have the smallest whole number ratio of each ion that allows the ion charges to cancel to zero.

One way to view the process is to start with the idea of a generic chemical formula:(Cation)_{x}(Anion)_{y}. A cation and an anion are selected form a list based on a given compound name. The cation has a specific positive charge and the anion a specific negative charge. The subscripts, x and y, multiply the ion charges so that they cancel to zero: x(+) + y(-)=0. The lowest ratio that allows the charges to cancel to zero is considerred to be that formula's subscripts.

For source of ions for writing or naming formulas select either **Download Ion Table pdf** or **Download Ion Table rtf**.

**Formula Example One: 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 Ce^{+4} and S^{-2}. The beginning generic 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 CeS_{2}. 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.

**Formula Example Two: Ammonium Phosphate**

Find the formula for ammonium phosphate. The ions are NH_{4}^{+1}and PO_{4}^{-3}. The generic equation is (NH_{4}^{+1})_{x}(PO_{4}^{-3})_{y}._{ }Again there are many pairs of subscripts that will allow the result to be zero: 12:4, 9:3, 6:2 and 3:1. The lowest ratio is 3:1. The charges add to zero: 3(+1) +1(-3)=0. Which gives the generic formula (NH_{4}^{+1})_{3}(PO_{4}^{-3})_{1}. The NH_{4}^{+1} is polyatomic with the subscript of 3 and therefor the parenthesis must be retained. The SO_{4}^{-2 }is polyatomic and requires the use of parenthesis, if the subscript is bigger than “1". The final equation is (NH_{4})_{3}PO_{4}.

**Formula Example Three: Aluminum Sulfate**

Find the formula for aluminum sulfate. Aluminum is Al^{+3}. Sulfate is SO_{4}^{-2}. The generic equation is (Al^{+3})_{x}(SO_{4}^{-2})_{y}._{ }Again there are many pairs of subscripts that will allow the result to be zero: 8:12, 6:9, 4:6 and 2:3. The lowest ratio is 2:3.The charges add to zero: 2(+3 +3(-2)=0. That will give us (Al^{+3})_{2}(SO_{4}^{-2})_{3}. The Al^{+3} is not polyatomic and therefor never requires the use of parenthesis. The SO_{4}^{-2 }is polyatomic and requires the use of parenthesis, if the subscript is bigger than “1". The final equation is Al_{2}(SO_{4})_{3}.