More on Moles (part 2 of 3)

Part 2 of 3 on moles

Today’s blog post will be the second part of a three-part series on a Chemistry topic called moles.lorenzo avagadro and his mole I wrote the first part called ‘What is a Mole?’ recently and I would suggest reading that first, but essentially I explained that a mole is a unit of matter in which a known number of atoms or molecules is thought to be contained (specifically, 6.02×1023 atoms/molecules which is also known as Avogadro’s number).

This number was chosen because a mole of any substance will end up having the same mass as its mass number, which is also the number of protons and neutrons in the nucleus. So, you can find the molar mass of any element by looking up the mass number in the periodic table.

In this post I will be focussing on how to apply this concept to complete calculate expected masses from chemical reactions as in the question below:

Aluminium is extracted from aluminium oxide as shown. Calculate the mass of aluminium that can be formed from 1020 g of aluminium oxide.

2 Al2O3 → 4 Al + 3 O2


Calculating an expected mass

So let’s start with the first question and try to understand what information we have and what we are trying to find out.

Aluminium is extracted from aluminium oxide as shown. Calculate the mass of aluminium that can be formed from 1020 g of aluminium oxide.

2 Al2O34 Al + 3 O2


Step 1: Identify the known and target substance

I’m referring to the known substance as the one we know the information the mass of. We have been given the mass of the aluminium oxide as 1020g. I’ve coloured it pink in the equation and in the question

This is not the molar mass!!

 So often people confuse this. The molar mass is the mass that one mole would have. This is just the mass we have. We don’t know how many moles there are yet, but we are going to find out.

The target substance is the substance we are being asked to calculate the expected mass of. In this case, this is aluminium. I have highlighted this in green.


Step 2: work out the molar mass of the target and known substance

Using the periodic table and the chemical formulae of the target and known substances we can calculate how many grams one mole would be. Ignore the big numbers on the left of the formulae in the balanced equation. Just find the mass of one mole of each substance. Eg. Al not 4Al.

We can see from the mass number that the molar mass of aluminium (Al) is 27g/mol.

For aluminium oxide (Al2O3), we can see it is:

(2 x 27g) + (3 x 16g) = 54g + 48g = 102g/mol

If I had 102g of aluminium, I would have one mole.


Step 3: Calculate the number of moles we have of the known substance

They have been nice here. We have 1020g of aluminium oxide. We know that one mole is 102g, so clearly we have 10 moles. However, sometimes they are not so nice, so it is important to know how to calculate the number of moles if you were given awkward numbers. We easily can see here that this is 10 moles, but let’s see if we can work out what sum we did.

10mol = 1020g ÷ 102g/mol

Or Moles = mass ÷ molar mass

We can use this to make a formula triangle. The right hand side of the equation shows us that mass would go at the top and molar mass at the bottom. The left hand side of the equation shows us that moles goes in the gap left behind at the bottom.


Step 4: Use the balanced equation to work out the number of moles expected of the target substance

The equation below shows us that if we had 2 moles of aluminium oxide then we could expect to get 4 moles of aluminium.

2 Al2O34 Al + 3 O2

I need to make two things clear here

  1. In reality we will not get this much for a couple of reasons. Some of the products could get lost in the process or maybe not all of the reactants managed to react. We should really call the expected number of moles the theoretical yield.
  2. This is if we had 2 moles of aluminium oxide. So often students here record we have 4 moles of the target substance and complete the calculation from that. You need to keep the word ‘if’ needs in your mind when looking at moles.

Now… we know from earlier that we have 10 moles of aluminium oxide, so if every 2 moles of aluminium oxide produces 4 moles of aluminium, we can see the ratio is 1:2 and so we can expect 20 moles of aluminium in this case.


Step 5: Converting the moles of the target substance into a mass. THE FINAL STEP!

We’ve nearly made it! Going back to the previous formula triangle, we can calculate the mass of aluminium as follows:

mass    = moles x molar mass

= 20 moles x 27g = 540g

So, there you are. The maximum theoretical yield of aluminium expected in this example is 540g.

I hope you found that helpful and look out for the next blog post on performing a titration calculation. See you there!




A bit about the author, Paul H:

Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a full-time tutor, providing online tuition using a variety of hi-tech resources to provide engaging and interesting lessons.  He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE, plus teaches Physics, Maths, and Chemistry to A-Level across all the major Exam Boards.

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