# What is a mole and why is it useful?

Molecules are very small, much smaller than we can see with our eyes. For this reason, we can’t simply just count the number of molecules in a sample by looking at it. Even if we could, there are so many molecules in a given sample (there are more molecules of water in a teaspoon of water than there are teaspoons of water in the world!) that in order to simplify the maths of day to day chemistry, the unit of a mol was created. A mol was decided to be 6.022 *10^{23} of the smallest particle present, which could be an atom, a molecule, an ion etc.

Eg, How many molecules are present in one mol of CaCO_{3}?

Answer: (1 mol) * (6.022 * 10^{23})= 6.022 * 10^{23} molecules.

This is however, only useful on a molecular scale. We can only measure the general properties of a sample, for example how much it weighs or its volume. Therefore we need a way of converting between the two units of measurement – the one we can see and the one which tells us what is going on at a molecular level.

Eg, How many moles are present in one gram of CaCO_{3}?

First of all we need to work out how many grams one mole of CaCO_{3} weighs. This is done by adding the individual weights of *each* atom present in the molecule (also known as the molar mass).

(Ca = 40 g/mol) + (C = 12 g/mol) + 3(O = 16 g/mol) = 100 g/mol

What this means is that every mol of CaCO_{3} weighs 100 g. If we now look at the units, a conversion factor has been created between what we can see and the molecular level: Mass of sample / molar mass = g / (g/mol) = mol

Answer: 1 g / (100 g/mol) = 0.01 mol

Why is using moles important?

Eg, How many grams of CaCl_{2} would be produced according to the following equation: CaCO_{3} + 2HCl => CaCl_{2} + CO_{2} + H_{2}O.

It would be very hard to work out how many grams of CaCl_{2} would be produced if we simply put 1 g of CaCO_{3} into a cubic metre of concentrated HCl - we wouldn’t know how many reactions would take place. However if we look at the equation, it says that for every molecule of CaCO_{3} that reacts, 2 molecules of HCl react with it to produce one molecule of CaCl2. In other words, every molecule of CaCO3 goes on to produce one molecule of CaCl2 (It should be noted that this is an idealised case and very rarely occurs in real life due to other factors). Therefore, if we calculate the number of moles of CaCO3 added to the reaction, we know the number of moles of CaCl2 produced, and using a method similar to the 2^{nd} example, the mass of CaCl2 produced in the reaction can be predicted

mol CaCO_{3} added => 0.01 mol CaCl_{2} produced

Molar mass of CaCl_{2} = 40 + 2*35.5 = 111 g/mol

0.01 mol * 111 g/mol = 1.11 g CaCl_{2} produced.