Higher tier only

Make sure you know what is meant by the terms A_{r}, M_{r} and you know how to calculate the mass of
1 mole of a substance before reading this page!

One of the main uses of moles in GCSE chemistry is in calculating the masses of
reactants
and products used up or produced in chemical reactions. The best way to learn how to do these calculations is by completing lots of
examples!

Example 1. Calculate the mass of water produced by burning 25g of hydrogen in air.

AStep 1- write a balanced symbolic equation for the reaction:

Have you have ever wondered what the numbers shown in green represent in the symbolic equation. Some people might read them as meaning 2 molecules of hydrogen react with 1 molecule of oxygen to make 2 molecules of hydrogen oxide or water.

However in mole calculations you will likely be calculating the masses of one or more of the reactants or products produced in a chemical reaction and so weighing
out 2 molecules of hydrogen and 1 molecule of oxygen is going to be a little tricky! So scale the
whole equation up; scale up by a factor of 6 x10^{23}; that is the number of particles present in 1 mole
of any substance. That means we now have 2 moles of hydrogen
reacting with 1 mole of oxygen to give 2 moles of water and not 2 molecules reacting with 1 molecule to form 2 molecules of water! The numbers in front of the substances
in symbolic equations tell you the number of moles that are reacting or the number of moles of product that are produced.

Remember that moles are a measure of the amount or mass of a substance. So:

- The M
_{r}of hydrogen (H_{2}) will be 2; so 1 mole of hydrogen will have a mass of 2 grams. - The M
_{r}of oxygen (O_{2}) will be 32; so 1 mole of oxygen will have a mass 32 grams.

Example 2: Calculate the mass of carbon dioxide produce by burning 750g of methane in air.

Methane (CH_{4}) is the gas that is burned in a Bunsen burner and it is also the gas that is used at home for
cooking and heating. It burns to release lots of heat energy according to the equations below:

A balanced symbolic equation is shown below for this reaction. We can see that 1 mole of methane produces 1 mole of carbon dioxide gas.

Recall that the A_{r} of carbon=12 A_{r} of hydrogen=1 A_{r} of oxygen=16

So 1 mole of methane (CH_{4}) has a mass of 16g while 1 mole of carbon dioxide (CO_{2}) has a mass 44g.

So we can summarise this as:

Example 3: Calculate the mass of copper metal obtained by the reduction of 100g of copper oxide

Black copper oxide powder was placed in a glass tube and heated gently using a Bunsen burner, as shown below. If hydrogen gas is then fed into the glass tube it will reduce the black copper oxide to brown metallic copper. Word and symbolic equations are shown below for this reduction reaction:

As before we need to start with the balanced symbolic equation for the reaction:

Calculate the relative formula mass (M_{r}) of copper oxide using the relative atomic masses (A_{r}) from the periodic table and we have:

A_{r} of Cu=63.5 A_{r} of O=16.

So 1 mole copper oxide has a mass of 79.5g while 1 mole of copper metal has a mass of 63.5g.

or

From the equations above you should see that 1 mole of copper oxide will produce 1 mole of copper and as above calculate the amount of copper metal obtained from 1g of copper oxide, which is easily done by simply dividing each side of the equation above by 79.5. This gives:

or

Example 4: How much iron can be produced from 150 tonnes of iron oxide?

In a blast furnace iron oxide is reduced to iron, the reaction is given by the equation below.

A
From the symbolic equation we can see that 1 mole of iron oxide will produce 2 moles of iron. So now it is
simply a matter of scaling up from grams to tonnes.

Now 1000kg is 1 tonne. So we have:

I have drawn out these calculations to show you step by step how to carry out mole calculations. With practice you will be able to complete mole calculations in a few steps, but until you are confident with what you are doing go slow and make sure you are clear about what you are doing in each step.