Mole calculations


The mole is a unit frequently used by chemists as it measures the amount of substance. Calculations involving moles are very common in chemistry. Mistakes are also very common when students simply try to learn the equations, rather than thinking about what they mean.


Working out the mass from the number of moles

In the same way as the quantities given in a recipe might be inappropriate and need to be changed, we frequently need to use quantities of chemicals other than one mole. Students find this easy to do if the numbers are simple, but often struggle otherwise. It really is easy! Taking CaCO3 as our example. If this is added up, the molar mass is 100 g mol-1. That is, one mole has a mass of 100 g. Clearly, two moles weigh 200 g (that is 2 100 g). This is common sense, but if it helps you can learn the equation:

Mass of substance (in g) = number of moles (in mol) mass of one mole (in g mol-1)

You will be a better chemist if you understand this, rather than memorizing it. If you get stuck when dealing with awkward numbers of moles just think back to the simple example above. If you're not given the formula of the substance, make sure you work it out correctly otherwise your calculation will go wrong from the start.



Work out the mass of the following. Give all your answers to three significant figures. You can check out the answer by clicking in the answer box.

Amount of substance

Mass (in grams)

2 moles of HNO3

0.0312 moles of Cu(OH)2 

5.61 10-4 moles of FeSO4.7H2O

1.7 moles of C7H6O2

0.188 moles of ammonium dichromate

More detailed answers


Working out the number of moles from the mass

This can be done by rearranging the last equation or by choosing an easy example to work with. For example, using the CaCO3 from the last section which had a molar mass of 100 g mol-1. If we have 200 g of CaCO3 (and each mole has a mass of 100 g), it is fairly obvious that we have 2 moles. The calculation is 200/100 = 2. In other words, dividing the mass we have by the mass of one mole gives the number of moles:

Students often make errors when the numbers involved are less straightforward than in the above example. One solution is to replace the numbers you have been given with much easier ones. Think what you would do with the easier numbers, then swap back the original data. Another good check is to ask yourself whether you have less than one mole or more than one mole. This avoids putting down some very silly answers! Don't quote too many significant figures in your answer.


How many moles are there in 0.326 g of barium hydroxide?

Ba(OH)2 = 171. That is, one mole of barium hydroxide has a mass of 171 g. As we have less than 1 gram, we clearly have less than one mole - we're looking for a small answer!

Number of moles = 0.326/171 = 0.00191 mol



Work out the number of moles in each of the following. Give all your answers to three significant figures. You can check out the answer by clicking in the answer box.

Mass of substance

Amount (in mol) 

1000 g of C12H22O11 (sugar)

568 g (1 pint) of water

0.174 g of sodium carbonate

4.28 g of cobalt(II) chloride-6-water

1.00 mg of C16H18O4N2S (penicillin)

More detailed answers

This calculation can only be done with pure substances. A common error is to try and use this equation with aqueous solutions. For example, 20 grams of copper(II) sulphate solution does not contain 20 grams of copper(II) sulphate. Most of it is water. So a much smaller (and unknown) quantity is actually the copper(II) sulphate. Without the actual mass, we cannot use the above equation. If we know the concentration of the solution we can use a concentration calculation to work out the number of moles of copper(II) sulphate which it contains.


Working with reacting masses - using equations

We usually work out the number of moles of a substance so that we can then work out the number of moles of another substance which is reacting with the first. In order to do this we need to look at the equation for the reaction. Again, don't write down too many significant figures, but do use them all in your calculations. This type of calculation is important if we are planning the quantities of reactants to use in an experiment.

Example 1

What mass of iron reacts with 10.0 g of sulphur? [Fe = 56, S = 32]

First we need the correctly balanced equation:

Fe(s) + S(s) FeS(s)

The equation shows that one mole of sulphur reacts with one mole of iron. We always need the same number of moles of iron and sulphur. We can work out the number of moles of sulphur from the mass:

Number of moles of sulphur = 10.0/32 = 0.313 mol

\ Number of moles of iron = 0.313 mol (the same as the sulphur)

\ Mass of iron = 0.313 56 = 17.5 g


Example 2

What mass of magnesium oxide is made when 250 g of oxygen reacts with excess magnesium? It is important to have an excess of magnesium so that all of the oxygen reacts. [Mg =24, O =16].

2Mg(s) + O2(g) 2MgO(s)

The equation shows that 1 mole of oxygen produces two moles of magnesium oxide. There will always be twice as many moles of magnesium oxide as oxygen (or half as many moles of oxygen compared to magnesium oxide).

Number of moles of oxygen = 250/32 = 7.81 mol

\ Number of moles of magnesium oxide = 2 7.81 = 15.6 mol (twice as many)

MgO = 40

\ Mass of magnesium oxide = 15.6 40 = 625 g



1. What mass of potassium hydrogencarbonate is needed to make 100 g of potassium carbonate on thermal decomposition?



2. What mass of zinc sulphate-7-water could be made from 1.00 g of zinc metal reacting with excess sulphuric acid?



There is another way of carrying out the above calculations which you might find easier. You can see how it is done here.


Other calculations

When working with gases we generally prefer to deal with volumes. Molar volume calculations first start with a moles calculation.

We have already seen above that aqueous solutions are not pure substances. This means that the equations on this page do not apply to aqueous solutions. It is very important to be able to work out the number of moles in an aqueous solution. You can check out the appropriate calculations here.

Another big source of mole calculations is the titration calculation. For even more moles see gravimetric work.