Why is memorising the molarity formula a bad idea?

OK,  so maybe you’re here for the molarity formula, in which case here it is:

Molarity = the number of moles / volume in liters

The molarity of a solution is the number of moles present, divided by the volume of the solution in liters.

That is it, that is all there is you need to know or is it? Is there more to this than meets the eye? If you really want to understand molarity then you need to understand moles and why they are important, and how they relate to molarity, so below I will cover:

  1. What is a Mole?
  2. Why are Moles important?
  3. How do I know how many atoms, ions or molecules I have?
  4. What is Molarity?
  5. Summary

And I would argue that just remembering the formula is a bad idea because under the pressure of exams, or in the lab, you may forget it, or get confused. However, if you understand moles and molarity, and how they are related, then you are less likely to forget.

What is a mole?

A mole is a certain number of atoms, ions or molecules.  In the previous sentence, the keyword is number – a mole is not a weight or an amount of something, it is a number, it is a count.

moles and molarity

This is NOT the type of moles I am talking about…

One mole of something is actually equal to 6.0221415 × 1023 of that something, and that, I think we can agree, is a very big number, but interestingly we don’t need to worry about it. All we need to know is that a mole is a count of something. So, 1 mole of atoms would contain 6.0221415 × 1023  atoms, 1 mole of ions would contain 6.0221415 × 1023  ions, and, you guessed it, 1 mole of molecules would contain 6.0221415 × 1023  molecules. (By the way, 6.0221415 × 1023 is the Avogadro constant and is derived from the number of atoms on 12 g of carbon-12.)

Why are moles important?

Moles are important because atoms, ions, and molecules will react together in 1:1, or 1:2, 1:3, etc. configurations. That is, 1 atom of sodium will react with 1 atom of chlorine to make 1 molecule of sodium chloride.

moles and molarity reaction

A one-to-one (1:1) reaction – one molecule (or atom) of A reacts with one molecule (or atom) of B.

So, if we want to make 10 molecules of sodium chloride, we would take 10 atoms of sodium and add 10 atoms of chlorine. We would count out the 10 atoms of sodium, and 10 atoms of chlorine, and combine them. And here is the key point, we wouldn’t weigh out 10 g of sodium and 10 g of chlorine because sodium atoms and chlorine atoms weigh different amounts, and so 10 g of sodium would contain more atoms than 10 g of chlorine as a sodium atom weighs less than a chlorine.

molarity and moles reaction

A one-to-one (1:1) reaction – many molecule (or atom) of A reacts with many molecule (or atom) of B. The ratio of A and B reacting is still 1:1.

If we mixed 10 g of sodium with 10 g of chlorine, we wouldn’t get 10 g of sodium chloride, and I’ll explain why, and what we would get, below.

How do I know how many atoms, ions or molecules I have?

This is where atomic weights, in the case of atoms, or molecular weight, in the case of molecules, comes in. (Ions can have an atomic weight if the ion is a charged atom, or a molecular weight if the ion is a charged molecule.)

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The atomic or molecular weight of something is how much 6.0221415 × 1023 atoms would weigh, and the molecular weight of a molecule is how many 6.0221415 × 1023 molecules would weigh.  It is the weight of one mole. It is the weight of 6.0221415 × 1023 atoms, ions or molecules. That is it…

If you counted out 6.0221415 × 1023 atoms of something you would have 1 mole, and the weight of that pile of atoms would be the atomic weight.

So, if something has a molecular weight of 10 g/mol (grams per mole) and you weigh out 10 g, you will have 1 mole which is 6.0221415 × 1023 molecules.  If you weigh out 5 g, as 5 is half of 10, you have 0.5 moles. 20 g would give you 2 moles. And this brings me to the first thing you need to remember:

The molecular weight of something is the number of grams that make one mole

Let’s go back to sodium and chlorine. The atomic weight of sodium is 22.9898 g/mol, and for chlorine, it is 35.453 g/mol. Therefore if I weigh out 22.9898 g of sodium I would have 6.0221415 × 1023 atoms of sodium, and 35.453 g of chlorine would give me 6.0221415 × 1023 atoms of chlorine.

Therefore to react sodium and chlorine 1:1 I would have to weigh out 22.9898 g of sodium, and 35.453 g of chlorine (these two weights would contain the same number of atoms). If I didn’t want to make that much, i.e. 1 mole of sodium chloride, I could go for an equal fraction of each. So, say I wanted to make 0.1 moles of sodium chloride then I would take 22.9898 x 0.1 = 2.29898 g of sodium, and 35.453 x 0.1 = 3.5453 g of chlorine.

In the case above where we had 10 g of sodium and 10 g of chlorine things are a bit different. Yes, the weights are 1:1, but what we need is the atoms to be 1:1, and we know a chlorine atom weighs more than a sodium atom.

Well in the case of sodium we would have 10 / 22.9898 = 0.435 moles, and for chlorine, 10 / 35.453 = 0.282 moles. As sodium reacts with chlorine 1:1 then we could react all 0.282 moles of chlorine with the 0.435 moles of sodium, which would leave 0.435 – 0.282 = 0.153 moles of sodium unreacted (or 0.153 x 22.9898 =  3.517 g of sodium). The reaction would yield 0.282 moles of sodium chloride, which has a molecular weight of 22.9898 + 35.453 = 58.4428 g/mol, so we would have 0.282 x 58.4428 = 16.480 g of sodium chloride, plus 3.517 g of sodium left over. We certainly wouldn’t have 20 g of sodium chloride.

Reaction - Moles and Molarity

Moles are important because by working in moles we keep the number of atoms or molecules present equal. For example, say B weighs 2 g each, and A weighs 1 g. If I weigh out 4 g of A, I will have 4 As, but if I weigh out 4 g of B I will only have 2. If I react A with B, and they combine 1:1, then the 2 B will combine with 2 A, and this will leave 2 A leftover.

What is molarity?

Molarity is just a concentration. It is the number of moles per volume of liquid. It is moles per liter. So, 1 mole of something made up to 1 liter, would be a 1 Molar solution (we write that as 1 M). And this brings me to the second thing to remember:

1 mole made up to one liter gives a 1 Molar solution

Now above we said that the molecular weight of something is the number of grams that make one mole, so we can replace “1 mole” in the above with molecular weight, and this gives us the third thing you need to remember:

The molecular weight of something made up to one liter gives a 1 Molar solution

So using our example above of a substance with a molecular weight of 10 g/mol if we took 10 g, dissolved it in water, and made the volume up to 1 liter, then we would have a 1 M solution.

Some more worked examples of this:

  • If we took 5 g of the same substance and made it up to 1 liter with water, we would have a 0.5 M solution.
  • If we took 20 g of the same substance and made it up to 1 liter with water, we would have a 2.0 M solution.
  • If we took 5 g of the same substance and made it up to 0.5 liter with water, we would have a 1.0 M solution.

In the above examples, all have to do is take the amount of substance, say 5 g, and divide that by the molecular weight – 10 g/mol, to give the number of moles, which is 0.5 moles. Next, look at the volume, if it is 1 liter, then you have 0.5 moles per 1 liter, or 0.5 mol/l. And as mol/l is Molar, then we have 0.5 M.

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So for 5 g of substance in 0.5 liters. We have 5 g of a substance with a molecular weight of 10 g/mol, hence we have 0.5 moles (5/10 = 0.5). This 0.5 moles is in 0.5 liters, so to get to 1 liter we need to double the volume (2 x 0.5), and as we have multiplied the volume by 2, we have to do the same to the moles, so we get 0.5 x 2 = 1. Hence, the answer is 1 M. We are just dividing the number of moles by the volume in liters, for example, 0.5 moles in 0.25 l is 0.5 / 0.25 = 2 M. And again, 0.1 moles in 0.4 liters is 0.1 / 0.4 = 0.25 M.

There is a formula you can use to work this out, and it can be derived from what we have discussed above, and the formula is:

M = g / (MW x v)

Where: M = molarity (M); g = mass in grams; MW = molecular weight (grams per mole); v = volume (liter)

In the lab you may be asked to make up a certain volume of a solution at a given molarity. For this you need to rearrange the above equation so that you can solve for grams (g):

g = M x MW x v

However, it is tricky to remember these two equations and under pressure, in an exam or in the lab, you may get things confused.

Is it M = MW / (g x v) or M = g / (MW x v)?

But I think you are less likely to forget:

The molecular weight of something made up to one liter gives 1 Molar solution

and by remembering the above you have molarity covered, and you can quickly and easily derive the equations to answer the questions.

Phew…  That turned out to be a much longer post than I thought it would be. However, an understanding of moles and molarity is critical, so it is worth the effort of getting a good understanding of this material. As I said above, I think the key is:

The molecular weight of something made up to one liter gives 1 Molar solution.

Summary

So, you could just remember the molarity formula:

Molarity = the number of moles / volume in liters

But, by know of what moles are, why they are important, and how they relate to molarity, you get a much better understanding as to what the formula means, and you are less likely to forget it under the pressure of exams or in the lab.

If you would like a copy of this post for your notes then grab the free Maths4Biosciences eBook. Besides the material above it also has sections on percentage solutions, dilutions, and how to draw a graph in the lab.

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2 thoughts on “Why is memorising the molarity formula a bad idea?

  1. rhymingchemist

    I’m not sure I agree. It is worth memorising the molarity equation and then also striving to understand the underlying science. Such understanding will come to some students quite quickly while for others it may emerge more slowly, over a series of lessons and activities. But by memorising definitions and equations, these mental building blocks can be recalled easily, which frees up cognitive capacity for deeper understanding of related points.

    Reply
    1. Nick Post author

      Sorry for the delayed reply (I’ve been a bit busy). I can see your point, and in a way, I agree with what you say. However, the point I was trying to make is that some students feel that memorising the formula is all they need to do and that they don’t need to develop a deeper understanding.

      Reply

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