Chemistry Foundations

4.4.5 - Working Out the Formula of Any Ionic Compound

# Working Out the Formula of Any Ionic Compound

Working out the formula of an ionic compound based on its name is something that many students find difficult. However, once you learn the steps it is actually fairly straightforward.

We will start by going through the steps and then we will work through some examples.

## Step 1: Name the cation and anion

As we learnt on the page on ionic compounds, every ionic compound's name is simply the name of its cation followed by the name of its anion. Therefore, if we are given the name of an ionic compound, the name of the cation is simply the first word and the name of the anion is the second word.

For example, in the ionic compound potassium oxide, the cation is potassium and the anion is oxide:

Potassium oxide

Cation: Potassium

Anion: Oxide

## Step 2: Work out the formulae of the cation and anion

The next step is to work out the chemical formulae of the cation and anion.

How we work out the chemical formula of an ion depends on what type of ion it is.

If it is a polyatomic ion, then we need to have its formula memorised. For example, if it is a sulfate ion, then we need to know from memory that sulfate's formula is SO42-. If you have not yet memorised the formulae of all the common polyatomic ions, go back and do that now.

If it is a monatomic ion that is not a transition element, then we need to work out its relative charge based on its group number. If you can't remember how to do that, then go back and refresh your memory now. We combine this with the element's chemical symbol to get the ion's formula.

For example, to work out the formula of an oxide ion, we start by identifying that oxygen is in group 6 and therefore the ion has a charge of -2. We can then write down the formula of the ion: O2-.

If it is an ion of a transition element, then the relative charge will be included in the name as roman numerals in brackets. For example, an iron (III) ion has a relative charge of +3 (transition element ions are always positively charged). We then combine this with the element's chemical symbol to get the ion's formula: Fe3+.

Going back to our potassium oxide example, we can now add in the formulae of the two ions. The are both monoatomic ions that are not transition elements, so we work out their relative charges using their elements' group numbers. Potassium is in group 1, so a potassium ion has a relative charge of +1. Oxygen is in group 6, so an oxide ion has a relative charge of -2:

Potassium oxide

Cation: Potassium - K+

Anion: Oxide - O2-

## Step 3: Work out the ratio of cations to anions and write down the compound's formula

Next we need to work out the ratio of cations to anions needed for the ionic compound to be neutral.

In some cases it may be obvious what the correct ratio is. For example, in our potassium oxide example, the cation is K+ with a relative charge of +1 and the anion is O2- with a relative charge of -2. Therefore in order for the charges to balance, we need to have twice as many potassium ions as oxide ions. Once we realise this, we can write down potassium oxide's formula:

K2O

In other cases, the ratio may be less obvious. For example, iron (III) sulfide is made up of iron (III) ions - Fe3+ - and sulfide ions - S2-.

What ratio of Fe3+ to S2- is needed in order for the compound to be neutral? The answer may not be so obvious. If we can't work it out intuitively, there is a step by step process we can follow.

We need to start by finding the lowest common multiple of the charges (ignoring the + and - signs). In this particular case, we need to find the lowest common multiple of 3 and 2, which is 6. Then we figure out of many of each ion we need to get a charge equal to that number.

We need two Fe3+ ions to get a charge of +6. And we need three S2- ions to get a charge of -6. Therefore we need a ratio of 2:3 of Fe3+ to S2- in order for the charges to cancel. We can now write down the formula of iron (III) sulfide:

Fe2S3

We will now work through some examples.

## Worked example 1: Calcium fluoride

Question: Write down the chemical formula of calcium fluoride

Step 1 - Name the cation and anion

The cation's name is simply the first word of the ionic compound's name and the anion's name is the second word of the ionic compound's name:

Calcium fluoride

Cation: Calcium

Anion: Fluoride

Step 2 - Work out the formulae of the cation and anion

Both of the ions are monatomic and they are not transition elements. Therefore we use their group numbers.

Calcium is in group 2, so a calcium ion has a charge of +2, making its formula Ca2+.

Fluorine is in group 7, so a fluoride ion has a charge of -1, making its formula F-.

Calcium fluoride

Cation: Calcium - Ca2+

Anion: Fluoride - F-

Step 3 - Work out the ratio of cations to anions and write down the compound's formula

In order for the ionic compound to be neutral, we need twice as many fluoride ions as calcium ions.

Therefore, the formula is:

CaF2

## Worked example 2: Copper (I) carbonate

Question: Write down the chemical formula of copper (I) carbonate

Step 1 - Name the cation and anion

Copper (I) carbonate

Cation: Copper (I)

Anion: Carbonate

Step 2 - Work out the formulae of the cation and anion

Copper (I) is a transition element ion. The roman numeral in brackets shows us that its charge is +1, making the ion's formula Cu+.

Carbonate is a polyatomic ion and we need to have its formula memorised. It is CO32-.

Copper (I) carbonate

Cation: Copper (I) - Cu+

Anion: Carbonate - CO32-

Step 3 - Work out the ratio of cations to anions and write down the compound's formula

Cu+ has a relative charge of +1. CO32- has a relative charge of -2. Therefore we need twice as many copper (I) ions as carbonate ions in order for the ionic compound to be neutral.

Therefore, copper (I) carbonate's formula is:

Cu2CO3

## Worked example 3: Magnesium phosphate

Step 1 - Name the cation and anion

Magnesium phosphate

Cation: Magnesium

Anion: Phosphate

Step 2 - Work out the formulae of the cation and anion

The magnesium ion is a monatomic non-transition element ion. Based on its group number, we can work out that its formula is Mg2+.

Phosphate is a polyatomic ion. Its formula is PO43-.

Magnesium phosphate

Cation: Magnesium - Mg2+

Anion: Phosphate - PO43-

Step 3 - Work out the ratio of cations to anions and write down the compound's formula

The magnesium ion has a relative charge of +2 and the phosphate ion has a relative charge of -3.

The lowest common multiple of 2 and 3 is 6. Three magnesium ions gives a charge of +6 and two phosphate ions gives a charge of -6. Therefore, in order for the charges of the cations and anions to cancel, we need a ratio of 3:2.

Therefore, magnesium phosphate's formula is:

Mg3(PO4)2

## Flashcards

Flashcards help you memorise information quickly. Copy each question onto its own flashcard and then write the answer on the other side. Testing yourself on these regularly will enable you to learn much more quickly than just reading and making notes.

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How do you work out the formula of an ionic compound from its name?