Science · Chemistry

The Mole and Chemical Calculations: Counting the Uncountable

Atoms are far too small to count one by one. The mole lets chemists count them by weighing, and turns every equation into a recipe you can calculate.

EDUSAMBAM Editorial Team | 18 min read | Updated 21 October 2026

A chemist cannot count atoms one at a time, because even a speck of salt contains more than a billion billion of them. The solution is to count in very large groups, in the same way a shop counts eggs by the dozen. The chemist’s group is called the mole. Once you understand it, a balanced equation becomes a recipe that tells you exactly how much to weigh out and how much product to expect.

1.Relative Atomic Mass and Relative Formula Mass

Atoms are so light that grams are not a convenient unit. Instead we compare each atom with carbon-12. The relative atomic mass (Ar) of an element is the average mass of its atoms compared with one-twelfth of the mass of a carbon-12 atom, taking into account its isotopes. It has no units. For example, Ar of hydrogen is 1, carbon 12, oxygen 16, sodium 23, and chlorine 35.5 (an average, because chlorine has two common isotopes).

The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of all the atoms in its formula. For molecules it is called the relative molecular mass. Use the Ar values given in your exam or textbook.

Compound	Working	Mr
Water, H₂O	2 × 1 + 16	18
Calcium carbonate, CaCO₃	40 + 12 + 3 × 16	100
Magnesium hydroxide, Mg(OH)₂	24 + 2 × (16 + 1)	58
Ammonium sulfate, (NH₄)₂SO₄	2 × (14 + 4 × 1) + 32 + 4 × 16	132
Copper(II) sulfate, CuSO₄	64 + 32 + 4 × 16	160

Notice how brackets work: the number after the bracket multiplies everything inside it.

2.What Is a Mole?

A mole (symbol mol) is the amount of substance that contains 6.02 × 10²³ particles. This number is called the Avogadro constant. The particles may be atoms, molecules or ions, so you must say which: one mole of oxygen atoms (O) is not the same as one mole of oxygen molecules (O₂).

The number was chosen so that one mole of any substance has a mass in grams equal to its relative mass (Ar or Mr). This mass is called the molar mass, with the unit g/mol. So one mole of carbon is 12 g, one mole of water is 18 g, and one mole of sodium chloride (Mr 58.5) is 58.5 g.

Figure 1. One mole of different substances. The masses are very different, but each contains exactly the same number of particles, 6.02 × 10²³.

amount in moles (n) = mass (g) ÷ molar mass (g/mol)

number of particles = amount (mol) × 6.02 × 10²³

Worked example 1: mass to moles to particles

Question: How many moles, and how many molecules, are there in 36 g of water?

Molar mass of H₂O = 18 g/mol.
n = 36 ÷ 18 = 2.0 mol.
Molecules = 2.0 × 6.02 × 10²³ = 1.20 × 10²⁴ molecules.

Answer: 2.0 mol, or about 1.2 × 10²⁴ molecules.

3.The Mole Map

The mole connects four quantities. Whenever you are stuck in a calculation, first convert what you are given into moles, do the chemistry in moles, and then convert to what the question asks for.

Figure 2. The mole sits at the centre. Mass, gas volume, particle number and solution concentration all convert through moles. The gold links pulse to remind you that every conversion goes through the mole.

Gas volumes. Equal numbers of molecules of different gases take up the same volume at the same temperature and pressure (Avogadro’s law). At room temperature and pressure (rtp, about 20 °C and 1 atmosphere), one mole of any gas occupies about 24 dm³ (24 000 cm³). At standard temperature and pressure (0 °C and 1 atmosphere), it is about 22.4 dm³. Use the value your syllabus specifies; this article uses 24 dm³ at rtp.

Concentration. For a solution, n = c × V, where c is concentration in mol/dm³ and V is volume in dm³ (divide cm³ by 1000).

4.Percentage Composition and Formulae

The percentage by mass of an element in a compound is: (Ar × number of atoms ÷ Mr) × 100.

Worked example 2: percentage of nitrogen in a fertiliser

Question: What is the percentage of nitrogen in ammonium sulfate, (NH₄)₂SO₄?

Mr = 132 (from the table above).
Two nitrogen atoms: 2 × 14 = 28.
Percentage N = 28 ÷ 132 × 100 = 21.2%.

Answer: about 21%. Farmers compare fertilisers this way.

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of atoms in one molecule, and is a whole-number multiple of the empirical formula. For example, glucose has the molecular formula C₆H₁₂O₆ and the empirical formula CH₂O.

Worked example 3: finding an empirical and molecular formula

Question: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and its Mr is 56. Find its empirical and molecular formulae.

Take 100 g: 85.7 g C and 14.3 g H.
Moles: C = 85.7 ÷ 12 = 7.14; H = 14.3 ÷ 1 = 14.3.
Divide by the smallest (7.14): C = 1; H = 2.0. Ratio 1 : 2.
Empirical formula = CH₂, with mass 12 + 2 = 14.
Multiplier = 56 ÷ 14 = 4, so the molecular formula is C₄H₈.

Answer: empirical CH₂; molecular C₄H₈ (butene).

Worked example 4: water of crystallisation

Question: When 2.50 g of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated, 1.60 g of anhydrous CuSO₄ remains. Find x.

Mass of water lost = 2.50 − 1.60 = 0.90 g.
Moles of H₂O = 0.90 ÷ 18 = 0.050 mol.
Moles of CuSO₄ = 1.60 ÷ 160 = 0.010 mol.
Ratio H₂O : CuSO₄ = 0.050 : 0.010 = 5 : 1, so x = 5.

Answer: CuSO₄·5H₂O.

5.Reacting Masses and Gas Volumes

A balanced equation gives the ratio of moles, not grams. The method is always the same.

Write the balanced equation.
Convert the given mass (or volume) to moles.
Use the ratio in the equation to find moles of the substance you want.
Convert moles to mass, volume or whatever is asked.

Worked example 5: mass of product

Question: What mass of carbon dioxide forms when 10.0 g of calcium carbonate is heated strongly? CaCO₃ → CaO + CO₂ (Mr CaCO₃ = 100; Mr CO₂ = 44)

Moles of CaCO₃ = 10.0 ÷ 100 = 0.100 mol.
Ratio 1 : 1, so moles of CO₂ = 0.100 mol.
Mass of CO₂ = 0.100 × 44 = 4.4 g.
Volume of CO₂ at rtp (extra) = 0.100 × 24 = 2.4 dm³.

Answer: 4.4 g (or 2.4 dm³ of gas).

Worked example 6: gas volume from a reaction

Question: 2.4 g of magnesium reacts with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen forms at rtp?

Moles of Mg = 2.4 ÷ 24 = 0.10 mol.
Ratio Mg : H₂ = 1 : 1, so moles of H₂ = 0.10 mol.
Volume = 0.10 × 24 = 2.4 dm³.

Answer: 2.4 dm³ (2400 cm³) of hydrogen.

6.Limiting Reactant

In many reactions one reactant runs out first. It is the limiting reactant; it decides how much product forms. The other reactant is in excess, and some is left over. A sandwich analogy makes the idea clear.

Figure 3. The cheese is limiting because each sandwich needs one slice and there are only two. The leftover bread (pulsing) is in excess.

Worked example 7: finding the limiting reactant

Question: 4.8 g of magnesium burns in 2.4 g of oxygen: 2Mg + O₂ → 2MgO. Which is limiting, and what mass of MgO forms? (Ar: Mg 24, O 16; Mr MgO = 40)

Moles of Mg = 4.8 ÷ 24 = 0.20 mol. Moles of O₂ = 2.4 ÷ 32 = 0.075 mol.
The equation needs 2 mol Mg for every 1 mol O₂. For 0.20 mol Mg we would need 0.10 mol O₂.
Only 0.075 mol O₂ is available, so oxygen is limiting.
Moles of MgO = 2 × 0.075 = 0.15 mol.
Mass of MgO = 0.15 × 40 = 6.0 g.

Answer: oxygen is limiting; 6.0 g of magnesium oxide forms.

7.Percentage Yield and Atom Economy

The theoretical yield is the maximum mass of product predicted by the calculation. The actual yield is what you really obtain, which is usually less because of incomplete reactions, side reactions and losses when separating the product.

percentage yield = (actual yield ÷ theoretical yield) × 100

Worked example 8: percentage yield

Question: In Worked example 7, only 4.5 g of magnesium oxide is collected. What is the percentage yield?

Theoretical yield = 6.0 g.
Percentage yield = 4.5 ÷ 6.0 × 100 = 75%.

Atom economy measures how much of the reactants’ mass ends up in the useful product, which matters for saving resources and reducing waste:

atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100

For CaCO₃ → CaO + CO₂, if CaO is the desired product, atom economy = 56 ÷ (56 + 44) × 100 = 56%. Reactions that give only one product have an atom economy of 100%, such as the addition reactions of alkenes.

8.Common Mistakes to Avoid

Using grams instead of moles in the equation ratio. The ratio is in moles.
Forgetting to balance the equation first. A wrong ratio gives a wrong answer.
Using Mr of the wrong particle. Oxygen gas is O₂ (Mr 32), not O (16).
Mixing cm³ and dm³. Divide cm³ by 1000 for dm³.
Rounding too early. Keep extra digits until the final answer.
Forgetting units and significant figures. Write g, mol, dm³ and keep sensible precision.

9.Check Your Understanding

Try each question first, then tap it to reveal the answer.

1. Calculate the Mr of H₂SO₄. (Ar: H 1, S 32, O 16)

Answer: 2 × 1 + 32 + 4 × 16 = 98.

2. How many moles are in 11 g of CO₂? (Mr = 44)

Answer: 11 ÷ 44 = 0.25 mol.

3. What volume do 0.50 mol of any gas occupy at rtp?

Answer: 0.50 × 24 = 12 dm³.

4. Find the concentration of a solution containing 0.20 mol in 500 cm³.

Answer: 0.20 ÷ 0.500 = 0.40 mol/dm³.

5. What is the empirical formula of C₂H₄O₂?

Answer: CH₂O.

6. In 2H₂ + O₂ → 2H₂O, what mass of water forms from 4 g of hydrogen? (Mr H₂ = 2, H₂O = 18)

Answer: Moles H₂ = 2.0; moles H₂O = 2.0; mass = 2.0 × 18 = 36 g.

7. An experiment gives 3.0 g of product but the theoretical yield is 4.0 g. Find the percentage yield.

Answer: 3.0 ÷ 4.0 × 100 = 75%.

10.Summary

Ar and Mr compare masses with carbon-12; Mr is the sum of Ar values in the formula.
One mole contains 6.02 × 10²³ particles and has a mass in grams equal to Mr (the molar mass).
n = m ÷ Mr; for gases at rtp n = V ÷ 24 dm³; for solutions n = c × V.
Equations give mole ratios: convert to moles, use the ratio, convert back.
Empirical formula is the simplest ratio; molecular formula is a multiple of it.
The limiting reactant controls the amount of product.
Percentage yield = actual ÷ theoretical × 100; atom economy shows how efficiently atoms are used.

For teachers and parents

Suggested use: let students weigh out one mole of several substances (for example 12 g of carbon, 18 g of water, 58.5 g of salt) to make the idea tangible. Use a table of Ar values in every question so students practise reading it. Encourage a standard layout for every calculation: equation, moles known, ratio, moles wanted, answer with units. Parents can help by checking units and rounding. Please follow your own syllabus for rtp and standard values and for the exact Ar values to use.

11.Sources and Further Reading

These reputable sources were consulted to verify the facts in this article:

Encyclopaedia Britannica: Overview of the mole and Avogadro’s number. Mole (chemistry).
Royal Society of Chemistry: Teaching resources for school chemistry. rsc.org.
American Chemical Society: Educational chemistry resources. acs.org.
OpenStax: Free peer-reviewed textbook with a chapter on stoichiometry. Chemistry 2e.
LibreTexts Chemistry: Open educational chemistry library. chem.libretexts.org.

Copyright note: This article is original EDUSAMBAM educational writing. The diagrams are original inline SVG schematics created for this page and are not copied from any source. Links open external websites; the sources were used to check facts and are suggested for further study.

Last updated: 21 October 2026. Written for students up to age 18 and checked against standard school-level chemistry references. This is general education, not a substitute for your teacher, textbook or syllabus.