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Stoichiometry & The Mole

Chemists cannot count atoms one by one, so they count them in moles. This note builds the mole idea from relative masses up to gas volumes, concentrations and yield, so the numbers in an equation finally mean something you can weigh out.

MYP 5ChemistryQuantitative ChemistryCriteria A · C~13 min read

Written by Fayek Jawad Haider · MYP student

Relative atomic & molecular mass

Atoms are far too light to weigh individually, so we compare their masses instead. The relative atomic mass (\(A_\text{r}\)) of an element compares the mass of its atoms to one twelfth of a carbon-12 atom. It has no units because it is a ratio.

Relative molecular mass (\(M_\text{r}\))
The sum of the relative atomic masses of every atom shown in a formula. Also called relative formula mass for ionic compounds.

To find \(M_\text{r}\), just add up the \(A_\text{r}\) values. For water, H2O: \((2 \times 1) + 16 = 18\). For carbon dioxide, CO2: \(12 + (2 \times 16) = 44\).

The mole and Avogadro

A mole is simply a fixed number of particles, in the same way a dozen is 12. The catch is that atoms are tiny, so the number is enormous.

The mole
The amount of substance that contains \(6.02 \times 10^{23}\) particles. This value is the Avogadro constant.

The clever part: the mass of one mole of a substance in grams is numerically equal to its \(A_\text{r}\) or \(M_\text{r}\). This is the molar mass (\(M\)), in grams per mole. So one mole of carbon (\(A_\text{r} = 12\)) weighs 12 g, and one mole of water weighs 18 g. That single fact links the invisible world of atoms to a mass you can put on a balance.

Moles, mass and \(n=m/M\)

The three quantities, number of moles (\(n\)), mass (\(m\)) and molar mass (\(M\)), are tied together by one equation you will use constantly:

\[ n = \frac{m}{M} \]

Here \(n\) is in mol, \(m\) is in grams and \(M\) is in g/mol. Rearranged, \(m = n \times M\) and \(M = m / n\).

Worked example

How many moles are there in 22 g of carbon dioxide, CO2?

1
Find the molar mass: \(M(\text{CO}_2) = 12 + (2 \times 16) = 44\) g/mol.
2
Substitute into \(n = m/M\): \(n = \dfrac{22}{44}\).
3
Divide: \(22 \div 44 = 0.5\).
\(n = 0.5\) mol of CO2

Check the units first

Before you divide, make sure mass is in grams. A tonne or a milligram will give an answer that is out by a factor of a thousand or a million. Units catch more errors than arithmetic does.

Molar gas volume

Gases are awkward to weigh, so for gases we count moles by volume instead. At room temperature and pressure (rtp), one mole of any gas occupies the same volume: 24 dm3 (which is 24 000 cm3). This works because gas particles are so spread out that their own size barely matters.

\[ n = \frac{V}{24} \quad\text{(}V\text{ in dm}^3\text{ at rtp)} \]

So 48 dm3 of oxygen at rtp is \(48 \div 24 = 2\) mol, whatever the gas happens to be. Note this only holds at rtp; change the temperature or pressure and the volume changes.

Concentration

For solutions, the amount of substance dissolved is measured as concentration: moles of solute per cubic decimetre of solution (1 dm3 = 1 litre = 1000 cm3).

\[ c = \frac{n}{V} \quad\text{(}c\text{ in mol/dm}^3,\; V\text{ in dm}^3\text{)} \]
Worked example

0.25 mol of sodium chloride is dissolved to make 500 cm3 of solution. Find the concentration.

1
Convert the volume to dm3: \(500 \div 1000 = 0.5\) dm3.
2
Substitute into \(c = n/V\): \(c = \dfrac{0.25}{0.5}\).
\(c = 0.5\) mol/dm3

Percentage yield

In theory an equation tells you exactly how much product you should make. In practice you always get less, reactions may not finish, some product is lost, or side reactions occur. Percentage yield compares what you actually got to what was theoretically possible.

\[ \text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \]
Worked example

A reaction should produce 8.0 g of product, but only 6.0 g is collected. Calculate the percentage yield.

1
Identify the values: actual yield = 6.0 g, theoretical yield = 8.0 g.
2
Substitute: \(\dfrac{6.0}{8.0} \times 100\).
3
Calculate: \(0.75 \times 100 = 75\).
Percentage yield = 75%

Where this is assessed

Mole calculations are graded under Criterion A (applying the right relationship) and, when you handle experimental data and yields, Criterion C (Processing and evaluating). Always quote units and a sensible number of significant figures.

Check yourself

1. What is the mass of 2 mol of water, H2O? +

\(M(\text{H}_2\text{O}) = (2\times 1) + 16 = 18\) g/mol. Then \(m = n \times M = 2 \times 18 = \mathbf{36\text{ g}}\).

2. What volume does 3 mol of hydrogen gas occupy at rtp? +

At rtp one mole occupies 24 dm3, so \(V = n \times 24 = 3 \times 24 = \mathbf{72\text{ dm}^3}\).

3. A reaction has a theoretical yield of 50 g but produces 40 g. What is the percentage yield? +

\(\dfrac{40}{50} \times 100 = 0.8 \times 100 = \mathbf{80\%}\).

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