The chapter on moles calculation can be very confusing. With so many different formulas, the common complaint I hear from students is… “How/Where do I start?”
Well, whether you are doing combined chemistry or pure chemistry, the surprising fact is this – most mole calculation questions can be solved in 3 steps! These same steps can be used repeatedly, over and over again, no matter how they twist and turn the question!
Before I get to the 3 Step Method, here’s a nifty flowchart you can use if you haven’t already memorised the formulas!
The 3 Step Method
Step 1 – Find the number of moles
The first thing you want to do when you come across ANY mole calculation question is quickly look through the given information and calculate the number of moles of substances given!
If you are given the mass of a reactant, divide by its Ar/Mr to find the number of moles.
If given the volume of gas of a reactant, divide by 24dm^3 to find the number of moles.
If given the volume and concentration of a reactant, multiply them together to obtain the number of moles!
(conc = mol ÷ volume)
E.g. A student adds 48g of Mg to a solution of CuSO4. Find the mass of Cu obtained at the end of the reaction.
The only information we have here is the mass of Mg. Hence, use the mass of Mg to find the number of moles of Mg first!
Moles = 48g ÷ 24(Ar of Mg) = 2 moles
Step 2 – Mole ratio
Once you have the number of moles, we move on to the next step – which is to use the mole ratio to find the number of moles of whatever substance you are looking for!
known : unknown
For instance, Mg + CuSO4 → MgSO4 + Cu
The mole ratio of Mg(known) to Cu(what you want to find) is 1 : 1.
Since we calculated 2 moles of Mg in step 1, there will be 2 moles of Cu formed.
Step 3 – Finding the unknown (mass)
The last step of the 3 step method is to find the unknown… in this case unknown mass. Since we already know the number of moles of Cu produced, we can use the mass formula in the flowchart to find the unknown mass!
Mass of Cu = 2 moles × 64 (Ar of Cu) = 128g
How Questions Can Become More Complex
Ok. Before some of you cry foul and call me a cheat, I must admit – that was an oversimplification of the chapter on moles…
As some of you would have suspected, there are a group of teachers who will spend most of their free time, thinking of ways to make questions more… interesting!
So how can we solve these more complex problems? Well, I’ve grouped these more complex problems into 3 main groups.
Type 1 – 3 Step Method (Direct)
The first type of question is exactly the same as what I describe above. This is the most basic/simple type of question. It usually comes out ONLY in combined chemistry papers…
So if you take PURE chemistry… your chances of getting a question like this are close to…. almost 0. =)
Type 2 – Excess/Limiting
This type of question is slightly more complex.
You are given information on BOTH reactants. So the question is… which should you use in the mole ratio step? To determine which reactant to use, you will need to establish which reactant is in excess and which is limiting.
The way to do this is as follows:
a. Find the number of moles of both reactants.
b. Choose 1 reactant, either one is fine, use the mole ratio and determine if the other reactant is excess or limiting.
Perhaps an example would paint a clearer picture…
e.g. CH4 + 2O2 → CO2 + 2H2O
In the above reaction, 10g of methane, CH4, reacts with 5g of oxygen to form carbon dioxide and water. Find the volume of carbon dioxide gas produced.
As we are given information on BOTH reactants, we need to first figure out which is excess and which is limiting!
Number of moles of methane = 10g ÷ 16 (Mr of methane) = 0.625moles
Number of moles of oxygen = 5g ÷ 32 (Mr of oxygen) = 0.15625moles
Choose 1 reactant (up to you!) – I will choose oxygen.
From the mole ratio, methane : oxygen is 1 : 2. This means that 0.15625 moles of oxygen will require 0.078125 moles of methane. As we have 0.625moles of methane, we have MORE than required hence methane is in EXCESS.
Once we know which reactant is limiting, we will use the number of moles of the limiting reactant and go back to the 3 Step Method to solve for unknown!
Mole ratio of oxygen : carbon dioxide is 2 : 1. Hence 0.15625 moles of oxygen will produce 0.078125 moles of carbon dioxide.
Volume of carbon dioxide gas is = 0.078125 x 24dm3 = 1.875dm3.
Type 3 – Reactants + Products
The last type of question is when information is given about both reactants and products. If you encounter questions like that, ALWAYS use the information on the products to calculate the number of moles and carry out the 3 step method. The information on the reactants is probably redundant.
Let me give you an analogy…
If you have 1000 eggs and you only managed to bake a 1kg cake. How much sugar do you need?
(For example purposes… lets assume that 100g of sugar and 10 eggs will get us a 500g cake… I know… I’m no baker!)
Since we only managed to bake a 1kg cake, 1000g, then the mass of sugar needed is only 200g. (Notice how we didn’t make use of the given information on the eggs!)
Congratulations! You made it to the end of a very very long blog post.
One of the fastest ways to improve is to practice more questions. So, go grab your questions on moles, and test out the 3 step method! As always, if you have any questions, feel free to contact me via the contact page here, leave a comment in the box below! I promise I will try to reply ASAP!