Higher tier only
Mole calculations
Make sure you know what is meant by the terms Ar, Mr 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.
Ar of H=1 Ar of O=16
Step 1- write a balanced symbolic equation for the reaction:
hydrogen(s) + oxygen(g) → hydrogen oxide(l)
2H2(g) + O2(g) → 2H2O(l)
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 x1023; 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 Mr of hydrogen (H2 ) will be 2; so 1 mole of hydrogen will have a mass of 2 grams.
- The Mr of oxygen (O2) will be 32; so 1 mole of oxygen will have a mass 32 grams.
From the balanced symbolic equation above we can see that 2 moles of hydrogen react with 1 mole of oxygen; well we already know that 1 mole of hydrogen (H2) has a mass of 2g; so 2 moles will have a mass of 4g.
This means that 4g of hydrogen reacts with 1 mole or 32g of oxygen to produce 2 moles of water. The Mr of
water is (2x Ar H + Ar of Oxygen) 18; so 1 mole of water has a mass of 18g, in our
equation above we produce 2 moles of water that is 36g of water. This is outlined below:
2H2(g) + O2(g) → 2H2O(l)
4g + 32g → 36g
The original question was how much water is produced by burning 25g of hydrogen in air.
Well we can use or equation to work this out:
2H2(g) + O2(g) → 2H2O(l)
however the questions asks nothing about oxygen so we can simply ignore it or simply remove it from the equation,
the question we have to answer only asks about hydrogen and water. So we have:
2H2(g) → 2H2O(l)
4g → 36g
So from our equation we can see that that 4g of hydrogen will produce 36g of water.
Now for the moment until you get to grips with mole calculations, calculate how much water 1g
of hydrogen would produce. Well if 4g of hydrogen produces 36g of water, then divide
by 4 to get what 1g of hydrogen would produce:
2H2(g) → 2H2O(g)
4g → 36g
dividing by 4:
1g of hydrogen gives → 9g of water
so if 1 gram of hydrogen produces 9g of water. So to find out how much water is produced from burning 25g of hydrogen then simply multiply by 25 to find out how much water is produced.
2H2(g) → 2H2O(g)
1g → 9g
25g → 9g x 25 = 225g of water
This method might seem long winded but once you get into the swing of it, you will be carrying out mole
calculations in no time at all.
Example 2: Calculate the mass of carbon dioxide produce by burning
750g of methane in air.
Methane (CH4) 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.
methane(g) + oxygen(g) → carbon dioxide(g) + hydrogen oxide(g)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
the questions does not ask about oxygen or water, so remove them from the equation. So we have 1 mole of methane produces 1 mole of carbon dioxide as shown in the equation below:
Recall that the Ar of carbon=12 Ar of hydrogen=1 Ar of oxygen=16
So 1 mole of methane (CH4) has a mass of 16g while 1 mole of carbon dioxide (CO2) has a mass 44g.
CH4(g) → CO2(g)
1 mole of methane = 16g → 1 mole of CO2 = 44g
So we can summarise this as:
CH4(g) → CO2(g)
16g of methane produces → 44g of carbon dioxide
divide both sides of the above equation by 16 to find out what 1g of methane (CH4), would produce:
1g of methane produces → 44g /16 = 2.75g of carbon dioxide
so if 1 gram of methane produces 2.75g of carbon dioxide, then simply multiply by 750 to calculate how much carbon dioxide would be produced by burning 750g of methane:
2.75 x 750 = 2062.5g of carbon dioxide gas, or 2.062 kg of carbon dioxide gas.
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:
copper oxide(s) + hydrogen(g) → copper(s) + hydrogen oxide(g)
CuO(s) + H2(g) → Cu(s) + H2O(g)
From the balanced symbolic equation we can see that 1 mole of copper oxide produces 1 mole of copper metal. The
question does not
ask about the hydrogen or hydrogen oxide (water) so simply remove them from the equation to simplify things a bit! We now have:
1 mole CuO(s)→ 1 mole Cu(s)
Calculate the relative formula mass (Mr) of copper oxide using the relative atomic masses (Ar) from the periodic table and we have:
Ar of Cu=63.5 Ar 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.
1 mole CuO(s)→ 1 mole Cu(s)
or
79.5g CuO(s)→ 63.5g Cu(s)
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:
79.5g/79.5 CuO(s)→ 63.5g/79.5 Cu(s)
or
1g CuO(s)→ 0.8g Cu(s)
So if 1g of copper oxide produces 0.8g of copper, then 100g of copper oxide will simply produce 80g of copper.
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.
Ar of Fe = 56 Ar C =12 Ar O =16
iron oxide(s) + carbon monoxide(g) → iron(s) + carbon dioxide(g)
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
As before to calculate how much iron will be formed calculate the mass of 1 mole of each of the reactants and products and scale up by the number of moles
of each substance present. The equation below has the masses of the number of moles for each of the reactants and products present.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
160g + 84g → 112g + 132g
The question asks nothing about carbon monoxide or carbon dioxide so simply remove these from our original equation, this gives:
Fe2O3(s) → 2Fe(s)
160g of iron oxide will produce → 112g of iron
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:
Fe2O3(s) → 2Fe(s)
160g of iron oxide → 112g iron or scale up by x1000 to covert grams to kilograms
160kg → 112kg or scale up by x1000 to convert kg to tonnes
160tonnes → 112tonnes
as before work out how much iron you will get from 1 tonne of
iron oxide:
160 tonnes → 112 tonnes so divide by 160 to calculate what 1 tonne will produce
1 tonne of iron oxide → 0.75 tonnes of iron.
so simply multiply the answer by 750 to calculate the mass of iron obtained from 750 tonnes of iron oxide;
0.7 x 750 = 525 tonnes of iron.
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.
Practice questions
Next
