Higher tier only

One of the main uses of moles in GCSE chemistry is in calculating the masses of reactants
and products used in chemical reactions. The best way to do this is by completing lots of
examples!

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

A_{r} of H=1 A_{r} of O = 16

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

##### hydrogen_{(s)} + oxygen_{(g) } → hydrogen oxide_{(l) }

##### 2H_{2(g)} + O_{2}_{(g) } → 2H_{2}O_{(l) }

Have you have ever wondered what the numbers shown in green are 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 we are looking to calculate masses and 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 means we now have 2 moles of hydrogen
reacting with 1 mole of oxygen to give 2 moles of water. The numbers in front of the substances
in symbolic equations tell you the number of moles that are reacting.

Remember that moles are a measure of the amount 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.
So from the equation we can see that we need 2 moles of hydrogen, well 1 mole is 2g, so 2 moles will be 4g.
This 4g of hydrogen reacts with 1 mole, 32g of oxygen to produce 2 moles of water. The M_{r} of
water is (2x A_{r} H + A_{r} of Oxygen) 18, so 1 mole of water is 18g, in our
equation we produce 2 moles of water, 36g.

##### 2H_{2(g)} + O_{2}_{(g) } → 2H_{2}O_{(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:
##### 2H_{2(g)} + O_{2}_{(g) } → 2H_{2}O_{(l) }

but the questions askes nothing about oxygen so we can simply ignore it or simply remove it from the equation,
it only askes about hydrogen and water. So we have:
##### 2H_{2(g)} → 2H_{2}O_{(l) }

##### 4g → 36g

So from our equation we know that 4g of hydrogen 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 produce 36g of water, then divide
by 4 to get what 1g of hydrogen would produce:
##### 2H_{2(g)} → 2H_{2}O_{(g) }

##### 4g → 36g

dividing by 4:
##### 1g hydrogen gives → 9g of water

so if 1 gram of hydrogen produces 9g of water, then multiply by 25 to find out how much 25g of hydrogen would produce.
##### 2H_{2(g)} → 2H_{2}O_{(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 quickly.
Example 2: Methane burns in air to produce carbon dioxide and water. A balanced
symbolic equation is shown below for this reaction. Calculate the mass of carbon dioxide produce by burning
750g of methane in air.

##### methane_{(g)} + oxygen_{(g) } → carbon dioxide_{(g)} + hydrogen oxide_{(l) }

##### CH_{4(g)} + 2O_{2}_{(g) } → CO_{2(g)} + 2H_{2}O_{(l) }

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:
##### CH_{4(g)} → CO_{2(g)}

##### 1 mole = 16g → 1 mole = 44g

##### CH_{4(g)} → CO_{2(g)}

##### 16g of methane produces → 44g of carbon dioxide

divide 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 methane, then simply multiply by 750 to calculate what 750g of methane would produce:
##### 2.75 x 750 = 2062.5g of carbon dioxide gas, or 2.062 Kg of carbon dioxide gas.

Example 3:

e.g. In a blast furnace iron oxide is reduced to iron, the reaction is given by the equation below.
How much iron can be produced from 150 tonnes of iron oxide?

A_{r} of Fe = 56 A_{r} C =12 A_{r} O =16

##### iron oxide_{(s)} + carbon monoxide_{(g) } → iron_{(s)} + carbon dioxide_{(g) }

##### Fe_{2}O_{3(s)} + 3CO_{(g)} → 2Fe_{(s)} + 3CO_{2(g)}

As before 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:
##### Fe_{2}O_{3(s)} + 3CO_{(g)} → 2Fe_{(s)} + 3CO_{2(g)}

##### 160g + 84g → 112g + 132g

The question askes nothing about carbon monoxide or carbon dioxide so remove these from our original equation:
##### Fe_{2}O_{3(s)} → 2Fe_{(s)}

##### 160g of iron oxide will produce → 112g of iron

So now it is simply a matter of scaling up from grams to tonnes. Now 1000Kg is 1 tonne. So we have:
##### Fe_{2}O_{3(s)} → 2Fe_{(s)}

##### 160g → 112g or scale up by x1000 to covert grams to kilograms

##### 160Kg → 112Kg or scale up by x1000 to covert Kg to tonnes

##### 160tonnes → 112tonnes

as before work out how much iron you will get from 1 tonne of iron oxide:
##### 160tonnes → 112tonnes so divide by 160 to calculate what 1 tonne will produce

##### 1 tonnes → 0.75 tonnes

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.

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