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Writing Formulas from Names of Ionic Compounds
Dr. MJ Patterson

To write the formula of an ionic compound correctly, there is one guiding principle:  the total charge on the compound must be zero.

The following procedure will help you achieve the goal of electrical neutrality.
 

  1. Write down the symbols for the ions with charges.
  2. Add the charges.
  3. If the total charge is zero, you are done.
  4. If the total charge is not zero, take the magnitude (absolute value) of the charge on the cation, and write it as the subscript on the anion.  Likewise, take the magnitude of the charge on the anion, and write it as the subscript on the cation.  Verify that the total charge is zero.

 

Example 1:
Write the formulas for sodium chloride and barium sulfate.

Solution 1:
The sodium cation is Na+, and the chloride anion is Cl-, so you would write Na+Cl-.  The total charge is (+1) + (-1) = 1-1 = 0.  So, the formula is NaCl.

The barium cation is Ba2+, and the sulfate ion is SO42-, so you would write Ba2+SO42-.  The total charge is (+2) + (-2) = 2-2 = 0.  So, the formula is BaSO4.

Example 2:
Write the formula for barium chloride.

Solution 2:
The barium cation is Ba2+, and the chloride ion is Cl-, so you would write Ba2+Cl-.  The total charge is (+2) + (-1) = 2-1 = 1.  As written, the compound is not neutral.  So, we want to take the charges and place them as subscripts on the opposite ion.
Ba2+Cl-=>Ba2+1Cl-2
So , the formula would be written BaCl2.  (The subscript 1 is understood and omitted.)

To double check that the formula is correct, we should verify that it is electrically neutral.  The formula indicates that there is one barium ion with a +2 charge.  This means that the total positive charge is 1(+2) = +2.  There are two chloride ions with a -1 charge.  The total negative charge is 2(-1) = -2.  The net charge is (+2) + (-2) = 2-2 = 0.  Therefore, the formula is written correctly.

Example 3:
Write the formulas for iron(II) phosphate and iron(III) phosphate.

Solution 3:
Iron(II) means iron with a +2 charge, while iron(III) means iron with a +3 charge.  Phosphate is PO43-.

iron(II) phosphate = Fe2+ PO43- =>  (Fe2+)3(PO43-)2 =>  Fe3(PO4)2

iron(III) phosphate = Fe3+ PO43- =>  FePO4

When more than one polyatomic ion is needed, parentheses are used around the entire ion.
 

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