calculations for stock solutions...

If you work in a lab you will have to make up stock solutions of chemicals all the time - on Tuesday you will be getting some more practise at this, and I have already written a post that goes over how to correctly go about dissolving your chemical and making it up to the correct volume (see previous post on making up stock solutions). 

But what about how you calculate how much chemical you want to weigh out in the first place?

Here is a simple example....

Imagine you want to make 500 mls of a NaCl solution that has a concentration of 2 M.

This is how I would go about figuring out how much NaCl I would need to weigh out if I was preparing this solution:

First, I would write down a couple of my all time favourite equations that are going to help me with this problem:

Ok, so my aim is to figure out what mass of NaCl I need to weigh out - you can find mass in the equation on the right. But in order to figure out the mass, I need to know the other two things in this equation - the molecular mass and the number of moles.

The molecular mass is easy. To find this out I just have to look at the periodic table and find the relative atomic mass of Sodium and the relative atomic mass of Chlorine and then add these two numbers together (remember, the relative atomic mass is the mass of 1 mole of an atom, and the molecular mass is the mass of 1 mole of a molecule). So, for NaCl the molecular mass will be 

23 + 35.5 = 58.5

But to find out how many moles you have you need to use the equation on the left - you know the concentration you want (2 M) and you know the volume you want (500 ml) so you can put these into this equation to calculate the number of moles this solution would contain:

Concentration (in M) = Moles / Volume (in L)

2 M = Moles / 0.5 L    

- note I have changed the volume to litres 

because you have to use litres in this equation as the 

definition of Molarity is the number of moles per litre

Then rearrange the equation to find the number of moles:

Moles =  2 M x 0.5 L 

Moles =  1

Great, so now you know that your solution will contain 1 mole of NaCl (i.e. 6.02 x 10²³ NaCls - see my previous post on moles if you don't get this).

Now you can go back to the right hand equation above and solve for mass:

Moles =  Mass / Molecular Mass

1 =  Mass / 58.5

Mass = 1 x 58.5

Mass = 58.5 g

So that is how you can go about figuring how much of a chemical you need to weigh out if you want to make a stock solution of a particular concentration and volume :)

Hope that helps