More on Moles (part 2 of 3 on moles)
Today’s blog post will be the second part in a three-part series on a Chemistry topic called moles. I wrote the first part recently called ‘What is a Mole?’. 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, 6.02x1023 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 amount 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.
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.
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.
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.
Two things need to be made clear here
In reality we will not get this much for a couple of reasons. Some of the product 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.
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. The word ‘if’ needs to be kept in your mind when looking at moles.
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!
