Chemguide: Support for CIE A level Chemistry
Learning outcomes 2.3.3 and 2.3.5
Molecular and empirical formulae
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.
Note: You may come across a few ionic compounds such as sodium peroxide, Na2O2, or mercury(I) chloride, Hg2Cl2, where the formula normally used isn't the empirical formula. There are good reasons for this which would be unnecessarily confusing to discuss now. Almost all ionic compounds use the empirical formula.
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.
Note: You may find that this time it isn't as easy to spot the ratio. If it isn't immediately obvious, try dividing through by the smallest number. That will almost invariably help.
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.
It has taken several lines to write this down, but it is a simple calculation.
Note: If you are writing down a complete calculation rather than filling in answers in a structured exam paper, it is important to include lots of words. Notice how many words there are in the simple calculation above compared with the amount of numbers. You have to include enough words to make it absolutely clear what you are doing
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:
This all seems a bit long-winded, but it is simple enough if you follow the stages through in an example.
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.
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).
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.
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.
© Jim Clark 2019