Lesson 7: Ionic Compound Formulas & Names
It is possible to figure out or predict the formulas of ionic compounds just from knowing which elements are involved by considering the charges on the ions.
In the following sections are several examples of how to determine formulas and names of ionic compounds. (They are also shown in example 9 of your workbook.) Each example tries to model the step-by-step thought process you can use to figure out ionic formulas and names. In sequence, each presents an additional factor or level of complexity. It will be handy to have a periodic table with you as you go through this. Note as you go through these examples that the metal or positive ion is always written first in the formulas. Study these examples (here or in your workbook) as needed. Then try your hand at predicting formulas and names for ionic compounds in the "Practice" section (also Ex. 10, 11, and 13).
First Example | Second Example | Third Example
Fourth Example | Fifth Example | Practice
First Example
Sodium and Chlorine
Sodium is in the first column of the periodic table, so it loses one electron and becomes a sodium ion with a +1 charge. Chlorine is a nonmetal in the next to the last column. Because it is a nonmetal, it will tend to gain electrons, and it will gain as many electrons as it needs to fill up its outermost energy level. There's room for only one more electron so it will take on a -1 charge and is called a chloride ion. So we have a +1 charge for sodium and a -1 charge for chloride. The charges are opposite but equal; therefore, there is a 1:1 ratio of sodium ions to chloride ions and NaCl is the formula of the compound. The name of this ionic compound is sodium chloride because it is made of sodium ions and chloride ions. |
|
|||||||||||||||
Second Example
Barium and Chlorine
|
Barium is in Group II-A and so it will have a +2 charge when it forms an ion. The ion is called a barium ion. Chlorine, again has a -1 charge on its chloride ion. With a +2 charge and a -1 charge, we will need two of the chloride ions to match the amount of charge on one barium ion. The formula for this compound is BaCl2. Note that the charge ratio is two-to-one, so the ion (or atom) ratio is one-to-two. The name of this ionic compound is barium chloride because it is made of barium ions and chloride ions. |
|
|||||||||||||||
Third Example
Iron and Chlorine
|
Iron is a transition metal. With transition metals you cannot count on knowing the ionic charge just from its position in the periodic table. For most transition metals, you will have to look up what charges actually exist for that element. You can look these up in your textbook or you can look back at the list in your workbook. However, you should memorize the charges for copper and iron. Iron has an ion with a +2 charge called iron(II) ion or ferrous ion, and one with a +3 charge called iron(III) ion or ferric ion. The chloride ion, of course, has a -1 charge. Because there are two types of cations, two different compounds can be formed. When the Fe2+ combines with chloride, you will get FeCl2. FeCl2 is named iron(II) chloride because it is made of iron(II) ions and chloride ions. It can also be called ferrous chloride. If the iron +3 ion combines with chloride, then that will give you FeCl3 because three chloride ions are needed to balance the charge on a +3 ion. FeCl3 is named iron(III) chloride because it is made of iron(III) ions and chloride ions. It can also be called ferric chloride. |
|
Note that in in these compounds the charge ratios can be reversed to get the atom ratios.
Fourth Example
Magnesium and Oxygen
Magnesium is a Group II-A element; therefore, it will have a +2 charge on the magnesium ion. Oxygen is in Group VI-A. It has six electrons in its outer shell. It has room for two more so when it forms an ion, it has a -2 charge and it is called an oxide ion. Here again, we have equal charges on the two ions: a +2 on the magnesium and a -2 charge on the oxygen. So they will combine in a 1:1 ratio and the formula for that compound is MgO. It is called magnesium oxide because it is made of magnesium ions and oxide ions. |
|
|||||||||||||||
Fifth Example
Aluminum and Oxygen
Aluminum is in Group III-A. It is a metal. Metals lose electrons. It has three electrons in the outermost energy level; therefore, it loses three electrons and forms a +3 ion called an aluminum ion. Oxygen, of course, takes on a -2 charge to form an oxide ion. The charge ratio is 3:2 therefore the atom ratio will be 2:3. Since two aluminum ions each with +3 charge will balance the charge on three oxide ions with -2 charge, the formula will be Al2O3. The name is aluminum oxide because it is made from aluminum ions and oxide ions. |
|
|||||||||||||||
Practice
Now you should do the practice in Ex. 10 and 11. When working through Ex. 11, you will use the same process for determining the formulas for ionic compounds but this time you'll be working with polyatomic ions. Don't be thrown off just because the ions look more complicated; the ratios for the formulas are still determined by the charges on the ions, nothing else. Here are two examples to help you.
Lithium and Phosphate Lithium is a Group IA metal, so its ion will have a +1 charge. Phosphate is a polyatomic ion with a -3 charge. The compound needs three lithium ions to balance the -3 charge of the phosphate, so the formula is Li3PO4. Notice that we didn't do anything with the "4" subscript on the phosphate. That is part of the ion formula and doesn't have anything to do with determining the charge ratios. The name of the compound is lithium phosphate. |
|
|||||||||||||||
Magnesium and Acetate Magnesium is a Group IIA metal, so its ion will have a +2 charge. Acetate is a polyatomic ion with a -1 charge. The compound needs two acetate ions to balance the +2 charge on the magnesium. Therefore, the formula is Mg(C2H3O2)2. We need parentheses around the acetate ion to show that we need two of that whole polyatomic ion - without changing the formula of the acetate. The name of the compound is magnesium acetate. |
|
|||||||||||||||
Please notice that when we wrote the formula for lithium phosphate, we did not use parentheses. We did not need more than one phosphate ion, so we did not use them. Only use parentheses when you need to use a subscript for your polyatomic ion in the formula, like we did with the magnesium acetate.
If you are having trouble naming ionic compounds, try answering the questions in Ex.12 (this is misnumbered in the workbook as "Ex. 9" on p. 7). If you need help in answering those questions, talk to a lab instructor or your lecture (or online) instructor.
You will also need to be able to come up with a formula for a compound given the name; try your hand at Ex. 13 (misnumbered as "Ex. 10") on p. 8 of your workbook. The answers to exercises 10, 11, and 13 are given below.
Answers
Ex. 10:
Formula |
Name | |
a |
MgBr2 |
magnesium bromide |
b |
K2S |
potassium sulfide |
c |
AlCl3 |
aluminum chloride |
d |
Cu2S CuS |
copper(I) sulfide, cuprous sulfide copper(II) sulfide, cupric sulfide |
e |
ZnF2 |
zinc fluoride |
f |
CoO Co2O3 |
cobalt(II) oxide, cobaltous oxide cobalt(III) oxide, cobaltic oxide |
g |
Mg3N2 |
magnesium nitride |
h |
CaI2 |
calcium iodide |
Ex. 11:
Formula |
Name |
|
a |
Na2CO3 |
sodium carbonate |
b |
Ca3(PO4)2 |
calcium phosphate |
c |
Al(OH)3 |
aluminum hydroxide |
d |
KClO3 |
potassium chlorate |
e |
FeSO4 |
iron(II) sulfate, ferrous sulfate |
f |
Fe(NO3)3 |
iron(III) nitrate, ferric nitrate |
g |
Cu(C2H3O2)2 |
copper(II) acetate, cupric acetate |
h |
NH4NO3 |
ammonium nitrate |
Ex. 13:
| a | NH4OH |
| b | BaCl2 |
| c | Co2(CO3)3 |
| d | Fe(ClO3)2 |