Chemguide: Support for CIE A level Chemistry


Learning outcome 1.4(b)

This statement deals with how you work out empirical formulae and molecular formulae from experimental data. It expects you to be able to do it in two ways - from data about the masses or percentages of the various elements combined together, or from data obtained from combustion experiments.

You also have to be able to work out a molecular formula from an empirical formula.

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


Finding 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:

NaS
mass0.46 g0.32 g
number of moles of atoms0.46/230.32/32
= 0.02= 0.01
ratio21

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.

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.


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:

  • 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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© Jim Clark 2010 (last modified March 2014)