Physical Chemistry / Quantitative and practical chemistry

Moles and chemical calculations

The mole, molar mass, reacting masses, limiting reagents, percentage yield and gas volumes.

GCSEA-Level

In a nutshell

Atoms are far too small to count one by one, so chemists count them in batches called moles. One mole is about 6.02\times10^{23} particles, the Avogadro constant, chosen so that one mole of a substance has a mass in grams equal to its relative formula mass. One mole of carbon-12 atoms weighs 12 g; one mole of water weighs about 18 g. Using moles, a balanced equation becomes a recipe: it tells you what mass of each reactant you need and how much product to expect. Moles also let you work out which reactant runs out first, how efficient a reaction was (the percentage yield) and what volume of gas will be produced. Every quantitative calculation in chemistry, from medicine doses to fertiliser production, rests on this idea.

A mole is an amount of substance containing 6.02\times10^{23} particles (the Avogadro constant). The mass of one mole, in grams, equals the relative formula mass M_\text{r}: add up the relative atomic masses of all the atoms in the formula.

\text{moles} = \frac{\text{mass (g)}}{M_\text{r}}

Example: M_\text{r}(\mathrm{CO_2}) = 12 + 2 \times 16 = 44, so 88 g of CO_2 is 2 mol.

Reacting masses

A balanced equation gives the ratio of moles. For \mathrm{2Mg + O_2 \rightarrow 2MgO}, 2 mol Mg makes 2 mol MgO. What mass of MgO forms from 6.0 g Mg?

1. Moles of Mg = 6.0/24 = 0.25 mol. 2. Ratio 1:1, so 0.25 mol MgO. 3. Mass = 0.25 \times 40 = 10 g.

Limiting reactant

The reactant used up first is the limiting reactant; it sets how much product forms. The other is in excess.

Percentage yield

Less product than expected is usually obtained: the reaction may be reversible, product is lost during separation, or side reactions occur.

\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100

Gas volumes

At room temperature and pressure, one mole of any gas occupies about 24 dm^3. So 0.50 mol of hydrogen takes up 12 dm^3.

The mole is the SI unit of amount of substance: exactly 6.022\,140\,76\times10^{23} specified entities (the Avogadro constant, N_\text{A}). The molar mass M (g mol^{-1}) is numerically equal to the relative formula mass. Three relationships cover most calculations:

n = \frac{m}{M}, \qquad n = c\,V, \qquad pV = nRT

For the ideal gas equation use SI units: p in Pa, V in m^3, T in K, R = 8.314 J K^{-1} mol^{-1}. Convert cm^3 to m^3 by multiplying by 10^{-6}.

Worked example: limiting reagent and yield

5.00 g of aluminium reacts with 10.0 g of chlorine: \mathrm{2Al + 3Cl_2 \rightarrow 2AlCl_3}.

n(\mathrm{Al}) = 5.00/27.0 = 0.185 mol; n(\mathrm{Cl_2}) = 10.0/71.0 = 0.141 mol.
0.185 mol Al would need 0.185 \times 3/2 = 0.278 mol Cl_2, which is not available, so chlorine is limiting.
Theoretical n(\mathrm{AlCl_3}) = 0.141 \times 2/3 = 0.0939 mol, mass = 0.0939 \times 133.5 = 12.5 g.
If 10.0 g is collected, percentage yield = 10.0/12.5 \times 100 = 80\%.

Empirical and molecular formulae

Divide the mass (or percentage) of each element by its A_\text{r}, then by the smallest result, to get the simplest whole-number ratio. A compound with 40.0% C, 6.7% H and 53.3% O gives a ratio of 3.33 : 6.7 : 3.33, i.e. CH_2O. If M_\text{r} = 180, the molecular formula is C_6H_{12}O_6.

Atom economy

\text{atom economy} = \frac{M_\text{r}(\text{desired product})}{\sum M_\text{r}(\text{all products})} \times 100

Yield measures how well a reaction was carried out; atom economy measures how wasteful the reaction is by design. Addition reactions have 100% atom economy; substitutions and eliminations produce by-products and score lower. Green chemistry aims to maximise both.

Always keep three significant figures through intermediate steps and round only at the end.

Key points

One mole contains 6.022\times10^{23} particles; its mass in grams equals M_\text{r}.
n = m/M, n = cV and pV = nRT link moles to mass, solution volume and gas volume.
Use the balanced equation's mole ratio; the limiting reagent sets the theoretical yield.
Percentage yield = actual ÷ theoretical × 100; atom economy measures waste built into the equation.
At room temperature and pressure a mole of any gas occupies about 24 dm^3.

Check yourself

5 questions

1How many moles are in 36 g of water (M_\text{r} = 18)?
n = m/M_\text{r} = 36/18 = 2 mol.
24 mol of H_2 reacts with 1 mol of O_2 to form water (\mathrm{2H_2 + O_2 \rightarrow 2H_2O}). Which is the limiting reactant?
1 mol O_2 needs only 2 mol H_2, so oxygen is used up first and 2 mol H_2 remains in excess.
3A reaction should produce 20 g of product but gives 15 g. What is the percentage yield?
15/20 \times 100 = 75\%.
4What volume does 0.100 mol of an ideal gas occupy at 300 K and 100 kPa?
V = nRT/p = 0.100 \times 8.314 \times 300/100\,000 = 2.49\times10^{-3} m^3 = 2.49 dm^3.
5A reaction has a 95% yield but a 40% atom economy. What does this tell you?
Yield reflects practical efficiency; atom economy is fixed by the equation and shows how much mass goes to unwanted products.

Recent preprints on this topic

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Matched by keyword from ChemRxiv via Crossref. Not peer-reviewed; read critically.