**Writing Formulas for Ionic Compounds**

Before you can write the chemical formula for an ionic compound, you must be able to recognize the ions in a compound. This includes cations, monoatomic ions and polyatomic ions. The inability to recognize ions is the number one source for failure in writing chemical formulas. Your instructor will tell you which ions are important and in order to successfully write the formulas, you **must** memorize them.

Cations from Groups IA, IIA, IIIA (except for T1) will form only one ion. If a cation is capable of forming more than one ion, it will have a Roman numeral in parenthesis (stock number) with it. Monoatomic ions have a charge equal to their group number. Also, it is common practice to write the cation (+) first when writing chemical formulas.

Let's work through some examples.

**Write the chemical formula for magnesium oxide.**

Let's break it down by ions. Magnesium is a Group IIA ion so it will have a +2 charge. Oxide is one of the ions you should have memorized and you know it has a -2 charge. Also, oxygen is a Group VIA element which carries a -2 charge. We want the compound to have a net zero charge so if we do a little math, we find that plus 2 minus 2 equals zero. This means we need one of each to balance the compound (A 1:1 ratio). Placing the cation first, we have our formula for magnesium oxide, MgO.

**Compounds where the charges don't come out even**

Write the formula for aluminum chloride.

Again, let's break it down by ions. Aluminum is a Group IIIA element and carries a +3 charge. Chloride is a Group VIIA element and carries a -1 charge. If we do the math, we see that +3 -1 does not equal zero. It equals +2. We need two more negative ions to zero out the compound. Add two more chloride ions and add them up. We now have -3 for the chloride ions. Back to our math, +3 -3 does equal zero. So we need one aluminum ion and 3 chloride ions to zero out the compound. As before, place the cation first in the formula and we have our formula for aluminum chloride, AlCl_{3}.

**Polyatomic anions**

Most, not all, polyatomic anions end with the suffix -ite or -ate. Make sure you memorize the polyatomic ions so you can recognize them.

Write the formula for sodium sulfate.

As before, we'll examine the ions. Sodium is a Group IA element and carries a +1 charge. Sulfate is a polyatomic ion (SO_{4}^{2-}) which carries a -2 charge. If we do our math, we'll see that the two do not zero out. + 1 - 2 equals -1. To make it zero out, we need to add another +1 charge (Sodium ion). When we add another sodium ion, we end up with a balanced formula of Na_{2}SO_{4}.

We use the subscript of 2, Na_{2}SO_{4}, rather than 2NaSO_{4} because a leading number, (in this case the 2), indicates that there are two of the entire element, which is incorrect. Our formula, Na_{2}SO_{4} is correctly written.

**Compounds involving Stock numbers**

Write the formula for chromium(III) nitrate.

Break it down into ions. Chromium(III) carries a charge of +3. Nitrate is a polyatomic ion that you should be able to recognize. The nitrate ion, NO_{3}^{-}, carries a -1 charge. Time for the math. + 3 -1 equals +2. This does not zero out and we need to add 2 of the -1 ions to zero it out. The notation is slightly different when you are indicating more than one polyatomic ion. We want to indicate that there are more than one of the group NO_{3} so we enclose the entire group in parenthesis and use a subscript number to indicate how many of them we are using. Our balanced formula for chromium(III) nitrate is Cr(NO_{3})_{3}.

**Let's try a complicated one. Write the formula for iron(III) carbonate.**

Break it down into ions. Iron (III) carries a +3 charge. We know this because of the Roman numeral III. You should recognize the carbonate ion as a polyatomic ion (CO_{3}^{2-}) and know that it carries a -2 charge. Do the math, + 3 -2 equals +1 which does not zero out. We need another -1 charge but neither of our ions carries a -1 charge. So, what we need to do is find a common multiple of the two charges. We have a 3 and a 2. The least common multiple will be 6.

To make the iron(III) have a ^{+}6 charge, we need two of them (3 x 2 = 6). To make the carbonate ion have a ^{-}6 charge, we need 6 of them (^{-}1 x 6 = ^{-}6). We can now write our formula, Fe_{2}(CO_{3})_{3}. Don't forget the parenthesis around the polyatomic ion.

**Writing Formulas for Binary Molecular Compounds**

These are very easy. You'll need to memorize these prefixes in order to write the formulas.

1 - mono

2 - di

3 - tri

4 - tetra

5 - penta

6 - hexa

7 - hepta

8 - octa

9 - nona

10 - deca

11 - hendeca

12 - dodeca

The prefix in the compound tells you how many elements there are. The "o" or "a" at the end of a prefix is generally omitted when the word following the prefix begins with another vowel.

Carbon dioxide - Because there is no prefix in front of the "carbon", it is understood that there is only one of them. The di- prefix in front of the "oxide" tells us there are two of them. Our formula is CO_{2}.

Disulfur tetrafluoride - The di- prefix in front of sulfur tells us there are two of them. The tetra- prefix in front of the fluoride indicates that there are four of them, so our formula is S_{2}F_{4}.

Here's a few more examples:

**di**nitrogen **mono**xide - N_{2}O

nitrogen **mono**xide - NO

nitrogen **di**oxide - NO_{2}

**di**nitrogen **tri**oxide - N_{2}O_{3}

**di**nitrogen **tetra**oxide - N_{2}O_{4}

**di**nitrogen **pent**oxide - N_{2}O_{5}