Chemguide: Core Chemistry 14 - 16

Empirical formulae

This page shows how you can find formulae of compounds from experimental data. I am assuming you have already read and understood the page about moles.

Explaining what an empirical formula is

The empirical formula tells you the simplest ratio of the various atoms present in a substance. For example, in ethane, C₂H₆, the ratio of the number of carbon to hydrogen atoms is 1:3. The empirical formula is CH₃. For hydrogen peroxide, H₂O₂, where the simplest ratio is 1:1, the empirical formula is HO.

The empirical formula for most molecular substances is virtually never used, unless it happens 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 need to know, or be able to calculate, the relative formula mass.

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.

Relative atomic masses: Na = 23; S = 32

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.

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 suppose that you have calculated the empirical formula of a hydrocarbon as CH₂. And suppose you also knew that the relative formula mass was 42.

If you add up the relative formula mass of the empirical formula, CH₂, it comes to 12 + (2 x 1) = 14.

The true molecular formula must be some multiple of this - so how many times does 14 go into 42?

Dividing 42 by 14 gives 3, and so the molecular formula must be 3 times bigger than CH₂ - in other words C₃H₆.

Example 2

Let's take another example which gave 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.

Empirical formulae experiments

Finding the formula of magnesium oxide

This video shows the experimental method and the calculation. I will repeat the calculation below.

Results

Mass of crucible + lid = 40.19 g

Mass of crucible + lid + magnesium = 40.30 g

Mass of crucible + lid + magnesium oxide = 40.37 g

Now you can use those results to find the masses of magnesium and oxygen.

Mass of magnesium = 40.30 - 40.19 = 0.11 g

Mass of oxygen = 40.37 - 40.30 = 0.07 g

So you have all the information to do the sort of sum we looked at further up the page. The relative atomic masses are: O = 16; Mg = 24.3.

The formula is therefore MgO.

If you do this sort of calculation in an exam, the answers will always work out as some exact ratio. If you do it practically, errors in the experiment will give you the sort of answer above, and you accept that this is a "near enough" 1 : 1 ratio.

You can't do this experiment with a 2 decimal place balance and expect to get a good result. On the screen, the results table shows each result with an error of ± 0.01 g. When you have calculated the results by doing two subtractions, the error in each answer has grown to ± 0.02 g. That means that the mass of magnesium could realistically be anywhere between 0.09 g and 0.13 g, and the oxygen anywhere between 0.05 g and 0.09 g.

Finding the formula of black copper oxide

(There are actually two different copper oxides, a black one and a red one. The familiar one is black copper(II) oxide, but I can't call it that, because that prejudges the result of this experiment!)

This next video is actually produced to show teachers how to do this experiment and therefore gives a lot of detail.

Essentially, you reduce the copper oxide to copper by heating it in a stream of hydrogen taken from a cylinder. You can ignore all the detail about using the cylinder.

Results

Mass of tube = 28.93 g

Mass of tube + copper oxide = 33.03 g

Mass of tube + copper = 32.13 g

Now you can use those results to find the masses of copper and oxygen.

Mass of copper = 32.13 - 28.93 = 3.20 g

Mass of oxygen = 33.03 - 32.13 = 0.90 g

The relative atomic masses are: O = 16; Cu = 63.5.

This is consistent with the sort of result you would get if you did this in the lab. The correct ratio is obviously going to be 1:1, and the formula is CuO.

Finding the formula of hydrated copper(II) sulfate

Blue copper(II) sulfate crystals contain water of crystallisation. This experiment sets out to find out how many molecules of water of crystallisation are present.

In other words, if copper(II) sulfate crystals are CuSO₄.nH₂O, what is the value of n?

Notice that the dish was heated very gently electrically. Strong heat decomposes copper(II) sulfate, and we have to avoid that.

Here are the results again.

Mass of dish = 25.1 g

Mass of dish + hydrated copper(II) sulfate = 29.2 g

Mass of tube + anhydrous copper(II) sulfate = 27.7 g

Now you can use those results to find the masses of anhydrous copper(II) sulfate and water.

Mass of anhydrous copper(II) sulfate = 27.7 - 25.1 = 2.6 g

Mass of water = 29.2 - 27.7 = 1.5 g

The relative atomic masses are: H = 1; O = 16; S = 32; Cu = 63.5.

This time you will first have to calculate the relative formula masses of anhydrous copper(II) sulfate, CuSO₄, and water.

RFM H₂O = (2 x 1) + 16 = 18

RFM CuSO₄ = 63.5 + 32 + (4 x 16) = 159.5

You find the ratio by dividing by the smaller number in the last stage - in other words divide both 0.0163 and 0.0833 by 0.0163.

So the answer is actually remarkably close to the expected value of 5 given the errors introduced by only using a 1 decimal place balance.