The mole and Avogadro's number

Counting by weighing

A shopkeeper does not count grains of rice one by one. A shopkeeper weighs them. A chemist is in the same position, but the grains are atoms and molecules, and there are very many of them. A chemist needs a way to count a huge number of particles by weighing.

The mole

The mole is the unit for the amount of a substance. One mole is the amount that has the same number of particles as there are atoms in exactly $12\ \text{g}$ of carbon-12. This number is

This is called the Avogadro constant (or Avogadro's number), after the Italian scientist Amedeo Avogadro. It is written $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.

One mole can be a mole of atoms, of molecules or of ions. The word is like "dozen". A dozen is $12$ of anything. A mole is $6.022 \times 10^{23}$ of anything.

How large is this number? If you had a mole of grains of sand, they would cover a large country to a depth of many metres. A mole of rupee coins would be more than all the money in the world. It is a number that fits only for very small particles. A mole of water is only $18\ \text{g}$, about a tablespoon, and it holds $6.022 \times 10^{23}$ molecules.

Molar mass

The molar mass is the mass of one mole of a substance. Its unit is $\text{g/mol}$. And here is the useful fact:

The molar mass in g/mol has the same number as the atomic or molecular mass in u.

So $12\ \text{g}$ of carbon, $18\ \text{g}$ of water and $58.5\ \text{g}$ of salt each have $6.022 \times 10^{23}$ particles, though the masses are different. This is how we count by weighing.

Why does this work? One atom of carbon is $12\ \text{u}$, and one atom of oxygen is $16\ \text{u}$. So oxygen atoms are $\dfrac{16}{12}$ as heavy as carbon atoms. If we take the same number of each, the masses are also in the ratio $\dfrac{16}{12}$. And $12\ \text{g}$ of carbon and $16\ \text{g}$ of oxygen are in exactly that ratio. So $16\ \text{g}$ of oxygen atoms has the same number of atoms as $12\ \text{g}$ of carbon atoms. The mole is a clever way to link the world of single atoms (in u) with the world of the laboratory (in g).

The three formulas

Let $n$ be the number of moles, $m$ the mass in grams, $M$ the molar mass in g/mol, and $N$ the number of particles.

These two give a third: $N = \dfrac{m}{M} \times N_A$.

Rearranged: $m = n \times M$. To find a mass from the moles, multiply by the molar mass.

A worked example

How many moles and how many molecules are there in $36\ \text{g}$ of water?

• The molar mass of water is $18\ \text{g/mol}$.
• $n = \dfrac{36\ \text{g}}{18\ \text{g/mol}} = 2\ \text{mol}$.
• $N = 2\ \text{mol} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 1.204 \times 10^{24}$ molecules.

More worked examples

1. From moles to mass. What is the mass of $0.25\ \text{mol}$ of $\text{CO}_2$? The molar mass is $44\ \text{g/mol}$. $m = 0.25\ \text{mol} \times 44\ \text{g/mol} = 11\ \text{g}$.

2. Atoms in a molecule. How many atoms are there in $18\ \text{g}$ of water? $18\ \text{g}$ is $1\ \text{mol}$, which is $6.022 \times 10^{23}$ molecules. Each molecule has $3$ atoms ($2\ \text{H}$ and $1\ \text{O}$). So the number of atoms is $3 \times 6.022 \times 10^{23} = 1.807 \times 10^{24}$.

3. From the number of particles. What is the mass of $3.011 \times 10^{23}$ molecules of oxygen, $\text{O}_2$? $n = \dfrac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.5\ \text{mol}$. $m = 0.5\ \text{mol} \times 32\ \text{g/mol} = 16\ \text{g}$.

4. The mass of one molecule. The mass of one molecule of water is the molar mass divided by the Avogadro constant: $\dfrac{18\ \text{g}}{6.022 \times 10^{23}} = 2.99 \times 10^{-23}\ \text{g}$. This is a very small mass.

You should find that $40\ \text{g}$ of oxygen gas is $1.25\ \text{mol}$, and that the same $40\ \text{g}$ of hydrogen gas would be $20\ \text{mol}$.

A table to help you

Practise with new numbers

Try it in Python

The program changes three masses into moles and into numbers of particles.

Common mistakes

• Using the atomic mass of an atom for a molecule. Use $32\ \text{g/mol}$ for $\text{O}_2$, not $16\ \text{g/mol}$.
• Mixing up atoms and molecules. $1\ \text{mol}$ of $\text{O}_2$ has $6.022 \times 10^{23}$ molecules and twice as many atoms.
• Forgetting the unit. The molar mass is in g/mol, and the mass is in g.
• Making an error with the power of ten. $6.022 \times 10^{23}$ times $2$ is $1.204 \times 10^{24}$, not $12.04 \times 10^{23}$ written wrongly.

Remember these

• One mole has $6.022 \times 10^{23}$ particles (the Avogadro constant).
• The molar mass in g/mol is equal to the atomic or molecular mass in u.
• $n = m \div M$, and $N = n \times N_A$.
• The mole links the mass that we weigh to the number of particles that we cannot count.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the mass of one mole of ammonia, $\text{NH}_3$?

2. How many particles are there in $0.75\ \text{mol}$ of any substance?

3. How many moles are there in $27\ \text{g}$ of aluminium, $\text{Al} = 27$?

4. What is the mass of $5\ \text{mol}$ of oxygen gas, $\text{O}_2$?