Chemguide: Support for CIE A level Chemistry

These statements deal with the meanings of molecular formulae and empirical formulae and how you can calculate them. I am taking them together even if it is out of syllabus order, because one follows logically from the other.

Before you go on, you should find and read the statements in your copy of the syllabus.

Explaining what a molecular formula is

This can obviously only apply to substances which consist of molecules - substances with a fixed number of atoms joined together by covalent bonds. The term "molecular formula" can't be applied to ionic substances like sodium chloride.

The molecular formula of the hydrocarbon ethane is C₂H₆, for example, and the molecular formula of hydrogen peroxide is H₂O₂.

The molecular formula gives a precise count of the numbers of each sort of atom which make up the molecule.

Explaining what an empirical formula is

By contrast, the empirical formula tells you the simplest ratio of the various atoms present in a substance. For example, in ethane (above) the ratio of the number of carbon to hydrogen atoms is 1:3. The empirical formula is CH₃. For hydrogen peroxide, where the simplest ratio is 1:1, the empirical formula is HO.

The empirical formulae for most molecular substances are virtually never used, unless they happen to be the same as the molecular formula - as in H₂O, for example. So what is the point of them?

The word "empirical" means "derived from observation or experiment" - so an empirical formula is one which you can find by doing experiments. In other words, you can calculate results from an experiment which will tell you that the empirical formula of a particular hydrocarbon (not ethane) is CH₂, for example.

That in itself isn't very helpful. The hydrocarbon could be C₂H₄, C₃H₆, C₄H₈, and so on and so on - anything with a carbon to hydrogen ratio of 1:2. To find out the correct molecular formula from the empirical formula, you would have to do another calculation. You will find out how to do that in further down the page as well.

The empirical formula is just a stage on the way to finding out the molecular formula of something.

The empirical formula and ionic compounds

For ionic compounds, like sodium chloride, the formula quoted is almost always the empirical formula. In an ionic compound, there are no fixed numbers of ions - it depends on how big the crystal is. So the formula of sodium chloride is simply given as NaCl, showing the 1:1 ratio. The formula of sodium oxide is Na₂O, showing a 2:1 ratio.

In a real crystal of sodium chloride or sodium oxide, there will be some huge variable number of positive and negative ions. The formula we write just tells us what the ratio is.

Calculating empirical formulae from masses or percentages

If you have a formula like, say, CH₄, you can read this as saying that 1 mole of carbon atoms are combined with 4 moles of hydrogen atoms.

In a different example, if you could work out that phosphorus and oxygen atoms combined together in the ratio of 2 moles of phosphorus atoms to 3 moles of oxygen atoms, then you would know that the empirical formula was P₂O₃.

You don't, of course know anything about the molecular formula. All you know is that the ratio is 2:3. The molecular formula could equally well be P₄O₆ or P₆O₉ or whatever.

You can find mole ratios from data involving either the masses or percentages of the combining atoms.

Finding empirical formulae from mass data

Suppose you found that 0.46 g of sodium formed 0.78 g of sodium sulfide. That means that 0.46 g of sodium combines with (0.78 - 0.46) g = 0.32 g of sulfur.

It is clearest if you set your answer out as a simple table:

That would tell you that the empirical formula was Na₂S.

Finding empirical formulae from percentage data

You might have been given the last example in a different form. You could have been told that the compound contained 59.0% of Na and 41.0% of S by mass.

That's not a problem. If you had 100 g of the compound, then the masses of sodium and sulfur would be 59.0 g and 41.0 g respectively. So use those figures in a sum like the last one. Try it, and make sure you get the same answer as before.

Because this topic is covered in my chemistry calculations book (see pages 31 to 33), I can't go any further than this with online help. For reasons that I have explained on another page, all I can do in these cases is to refer you to the book.

Finding empirical formulae from combustion data

This isn't covered explicitly in my book, and so I can give a complete explanation here.

If you burn a compound containing carbon and hydrogen in an excess of air or oxygen, carbon dioxide and water are formed. If you can trap the water and carbon dioxide separately, you can find out what mass of each is formed. From that, you can work out the empirical formula.

How you do this varies slightly depending on whether the compound contains anything else as well as the carbon and hydrogen. We'll look at the two cases separately.

If the compound only contains carbon and hydrogen

A compound which only contains carbon and hydrogen is called a hydrocarbon. Look for this word in the question. You can only use this shorter method if you know that there is nothing else present.

Here's an example:

When 0.78 g of a hydrocarbon was burned in excess air, 2.64 g of carbon dioxide and 0.54 g of water were formed. Find the empirical formula of the hydrocarbon.

The important things to notice is that every mole of CO₂ contains 1 mole of carbon atoms. Every mole of H₂O contains 2 moles of hydrogen atoms.

1 mole of CO₂ weighs 44 g.

No of moles of CO₂ = 2.64/44 = 0.06

Therefore, no of moles of carbon atoms, C = 0.06

1 mole of H₂O weighs 18 g.

No of moles of H₂O = 0.54/18 = 0.03

Each mole of H₂O contains 2 moles of hydrogen atoms.

Therefore, no of moles of hydrogen atoms, H = 0.06

The ratio of the number of moles of C : H is 1 : 1.

The empirical formula is CH.

It has taken several lines to write this down, but it is a simple calculation.

If the compound contains other things as well as carbon and hydrogen

This also applies to cases where you haven't been told that the compound is a hydrocarbon. It might be a hydrocarbon, but you aren't sure.

The most likely cases you will come across will probably contain oxygen as well as carbon and hydrogen.

In these cases, you have to put in extra steps in order to find out how much oxygen (or whatever) is present. The sequence is:

• Use the carbon dioxide figure to work out the number of moles of carbon atoms.

• From this, work out the mass of carbon present.

• Use the water figure to work out the number of moles of hydrogen atoms.

• From this, work out the mass of hydrogen present.

• Add up the carbon and hydrogen masses, and compare them with the original mass of compound. If there is no difference, then you have a hydrocarbon, and the problem disappears. Just look at the ratio of the numbers of moles of carbon and hydrogen atoms.

  If there is a difference, assume the difference is due to oxygen unless you are told differently. Work out the mass of oxygen present.

• Find the number of moles of oxygen atoms present.

• From the mole figures for carbon, hydrogen and oxygen, work out the empirical formula.

This all seems a bit long-winded, but it is simple enough if you follow the stages through in an example.

The question:

0.23 g of a compound containing carbon, hydrogen and possibly oxygen was burned in an excess of air. 0.44 g of carbon dioxide and 0.27 g of water were formed. Work out the empirical formula of the compound.

No of moles of CO₂ = 0.44/44 = 0.01

Therefore, no of moles of carbon atoms, C = 0.01

and mass of carbon = 0.01 x 12 g = 0.12 g

No of moles of H₂O = 0.27/18 = 0.015

Each mole of water contains 2 moles of hydrogen atoms.

Therefore, no of moles of hydrogen atoms, H = 2 x 0.015 = 0.03

and mass of hydrogen = 0.03 x 1 g = 0.03 g

Total mass of carbon and hydrogen in compound = 0.12 + 0.03 g = 0.15 g

Since you started with 0.23 g of compound, there is missing mass which must be oxygen.

Mass of oxygen = 0.23 - 0.15 g = 0.08 g

No of moles of oxygen atoms, O = 0.08/16 = 0.005

The ratio of number of moles is C 0.01 : H 0.03 : O 0.005

Dividing through by the smallest number to give a simple ratio gives:

C 2 : H 6 : O 1

The empirical formula is C₂H₆O.

Converting empirical formulae into molecular formulae

This is really simple! You can do it if you are told either the relative formula mass of the compound or the mass of 1 mole (which is just the relative formula mass expressed in grams).

Example 1

Let's follow on with the example above which resulted in an empirical formula of CH. Suppose you knew that the relative formula mass was 78.

If you add up the relative formula mass of the empirical formula, CH, it comes to 12 + 1 = 13.

The true molecular formula must be some multiple of this - so how many times does 13 go into 78?

Dividing 78 by 13 gives 6, and so the molecular formula must be 6 times bigger than CH - in other words C₆H₆.

Example 2

And let's also follow on with the other example above which resulted in an empirical formula of C₂H₆O. Suppose you knew that the mass of 1 mole was 46 g.

1 mole of the empirical formula, C₂H₆O, would have a mass of

(2 x 12) + (6 x 1) + 16 g = 46 g.

That's the same as the mass of 1 mole that you are given. Therefore the molecular formula must be the same as the empirical formula - C₂H₆O.

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