Strategy: the mole map

One map for every mole problem

A mole problem looks hard because it asks for something strange: the number of atoms in a spoonful, or the mass of a single molecule. But it always has the same three places.

• Mass to moles: divide by the molar mass, $M$, in g/mol.
• Moles to particles: multiply by $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$.
• The other way: do the opposite step.

The moles are always in the middle. You never go from mass straight to particles. Always stop at moles. This is like a railway map: to go from one town to another, you may have to change at a junction. Moles are the junction.

The four-step method

1. Write what you have and what you want.
2. Go to moles.
3. Move along the map to what you want.
4. Count the atoms last. A molecule of $\text{CO}_2$ has $3$ atoms. Multiply the molecules by $3$ at the end.

Example. How many atoms are there in $4.4\ \text{g}$ of carbon dioxide?

1. Have: $4.4\ \text{g}$ of $\text{CO}_2$. Want: atoms.
2. Moles: $\dfrac{4.4\ \text{g}}{44\ \text{g/mol}} = 0.1\ \text{mol}$.
3. Molecules: $0.1\ \text{mol} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 6.022 \times 10^{22}$.
4. Atoms: $3 \times 6.022 \times 10^{22} = 1.807 \times 10^{23}$.

More examples

From the number of particles to the mass. What is the mass of $1.2044 \times 10^{24}$ molecules of ammonia, $\text{NH}_3$?

1. Have: the number of molecules. Want: the mass.
2. Moles: $\dfrac{1.2044 \times 10^{24}}{6.022 \times 10^{23}} = 2\ \text{mol}$.
3. Mass: $2\ \text{mol} \times 17\ \text{g/mol} = 34\ \text{g}$.

From the mass to the atoms of one kind. How many oxygen atoms are there in $49\ \text{g}$ of sulphuric acid, $\text{H}_2\text{SO}_4$ (molar mass $98\ \text{g/mol}$)?

1. Moles: $\dfrac{49\ \text{g}}{98\ \text{g/mol}} = 0.5\ \text{mol}$.
2. Molecules: $0.5\ \text{mol} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 3.011 \times 10^{23}$.
3. Oxygen atoms: each molecule has $4$, so $4 \times 3.011 \times 10^{23} = 1.204 \times 10^{24}$.

The mass of one particle

Turn the map round. One mole of molecules has the mass $M$, and holds $N_A$ molecules. So one molecule has the mass $\dfrac{M}{N_A}$. For water: $\dfrac{18\ \text{g}}{6.022 \times 10^{23}} = 3.0 \times 10^{-23}\ \text{g}$.

You should find that the step from moles to molecules is always the same giant jump, whatever the substance.

Check by size

• A small mass must give a small number of moles. $4.4\ \text{g}$ of a gas of molar mass $44$ cannot be more than one mole.
• A number of particles of the order $10^{23}$ means about a mole. A number like $10^{10}$ is a very small amount.
• If an answer for atoms is smaller than the answer for molecules, you divided when you should have multiplied.
• The mass of one molecule must be a very small number of grams, of the order of $10^{-23}$.

Practise with new numbers

Try it in Python

This program follows the map for $4.4\ \text{g}$ of carbon dioxide.

Common mistakes

• Going from mass straight to particles. Always stop at moles.
• Forgetting to count the atoms in a molecule. The last step is easy to miss.
• Using the wrong molar mass. Add the atomic masses of all the atoms in the formula first.
• Mistakes with the power of ten. $6.022 \times 10^{23}$ times $3$ is $1.807 \times 10^{24}$, and not $18.07 \times 10^{23}$ written in a wrong place.

Remember these

• The map: mass $\leftrightarrow$ moles $\leftrightarrow$ particles. Moles are always in the middle.
• Mass to moles: divide by $M$. Moles to particles: multiply by $N_A$.
• Count the atoms in a molecule at the end.
• The mass of one molecule is $M \div N_A$.

Check yourself

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

1. How many moles are there in $14\ \text{g}$ of nitrogen gas, $\text{N}_2$?

2. How many molecules is that?

3. How many nitrogen atoms are there in $14\ \text{g}$ of nitrogen gas?

4. What is the mass of $1.5055 \times 10^{23}$ molecules of water?