ChemistryISC Class 11 · Some Basic Concepts of Chemistry

Mole Concept: Moles, Avogadro Number and Empirical Formula

Updated 17 October 2026 · Fact-checked

One mole contains 6.022 × 10²³ entities (Avogadro number). Moles are found from n = mass/molar mass = N/N_A = V/22.4 L (a gas at STP). An empirical formula comes from percentage composition by dividing by atomic masses and taking the simplest whole-number ratio; the molecular formula is a multiple of it.

Moles connect mass, particles and gas volume; for empirical formula, convert % to grams, divide by atomic mass, divide by the smallest, then clear fractions.

Illustration drawn for QuestionBuddy.

Understand Mole Concept and Empirical Formula

The mole. A mole is the amount of substance containing as many elementary entities as there are atoms in 12 g of carbon-12. The number of entities per mole is Avogadro's number, N_A = 6.022 × 10²³ mol⁻¹. The mass of one mole is the molar mass (g mol⁻¹), numerically equal to the relative atomic or molecular mass.

Conversions. n = m/M = N/N_A. For gases at STP (273 K and 1 atm) one mole occupies 22.4 L. Always count what is asked: moles of molecules, atoms, ions or electrons. For example, in 1 mol of CO₂ there are 1 mol of molecules, 3 mol of atoms and 22 mol of electrons.

Average atomic mass. For an element with isotopes, the atomic mass is the weighted mean: Σ(abundance × isotopic mass)/100.

Percentage composition. % of an element = (mass of the element in one mole of compound/molar mass) × 100. For a hydrate, include the water of crystallisation in the molar mass.

Empirical formula. Steps: (1) take the percentages as masses in 100 g; (2) divide by atomic masses to get moles; (3) divide all by the smallest; (4) if needed multiply by 2, 3 ... to clear fractions (for example 1.5 → multiply by 2). The molecular formula is (empirical formula)ₙ with n = molar mass/empirical formula mass.

Combustion analysis. For a compound of C and H (or C, H, O), the mass of carbon = (12/44) × mass of CO₂ and of hydrogen = (2/18) × mass of H₂O; oxygen is found by difference.

Mole Concept and Empirical Formula: formulas and key facts

Moles: $n = m / M ∣ n = N / N_{A} ∣ n = V /22.4 (S T P)$; V in litres of gas at STP.
Average atomic mass: $Σ (ab u n d an ce \times ma ss) /100$; Weighted mean of isotopes.
Percentage composition: Include any water of crystallisation.
Molecular formula: $n = m o l a r ma ss / e m p i r i c a l f or m u l ama ss$; Molecular formula = (empirical formula) × n.

Mole Concept and Empirical Formula in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Water in blue vitriol

Suppose a lab technician has 24.95 g of hydrated copper sulphate, CuSO₄·5H₂O (Cu = 63.5, S = 32, O = 16, H = 1).

Find the percentage of water and the mass of anhydrous salt left on strong heating.

Try it first, then show the solutionSolution

M = 63.5 + 32 + 64 + 5 × 18 = 249.5 g mol⁻¹.
n = 24.95/249.5 = 0.1 mol; water = 0.5 mol = 9.0 g, so the percentage of water = 9.0/24.95 × 100 = 36.1%.
Anhydrous CuSO₄ = 0.1 × 159.5 = 15.95 g.

Answer: 36.1% water; 15.95 g anhydrous salt.

Memory hook: Always include the water of crystallisation in the molar mass.

Tricks to remember Mole Concept and Empirical Formula

Short memory aids. Say each one aloud once, then use it on a question the same day.

Everything goes through the mole
Convert mass, volume or number into moles first; then compare moles.
Use it when: Any stoichiometry question.
Divide, divide, clear
Divide by atomic mass, divide by the smallest, clear fractions.
Use it when: Empirical formula steps.

Mole Concept and Empirical Formula: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Molecules in a mole · 1 mark · MCQBoard level

The number of molecules in 18 g of water is:
(a) 3.011 × 10²³
(b) 6.022 × 10²³
(c) 6.022 × 10²²
(d) 1.8 × 10²⁴

Try it first, then show the solutionSolution

M(H₂O) = 18 g mol⁻¹, so 18 g is 1 mol, containing 6.022 × 10²³ molecules.

Answer: (b) 6.022 × 10²³

Example 2: Moles, molecules, atoms and volume · 3 marks · Short answerBoard level

For 6.6 g of CO₂ (C = 12, O = 16) find the number of moles, the number of molecules, the number of atoms and the volume at STP.

Try it first, then show the solutionSolution

M = 44 g mol⁻¹, so n = 6.6/44 = 0.15 mol. (0.75 mark)
Molecules = 0.15 × 6.022 × 10²³ = 9.03 × 10²². (0.75 mark)
Atoms = 3 × 9.03 × 10²² = 2.71 × 10²³. (0.75 mark)
V = 0.15 × 22.4 = 3.36 L. (0.75 mark)

Answer: 0.15 mol; 9.03 × 10²² molecules; 2.71 × 10²³ atoms; 3.36 L.

Example 3: Empirical to molecular formula · 4 marks · Short answerHarder board level

A liquid contains 54.5% C, 9.1% H and 36.4% O by mass, and its molar mass is 88 g mol⁻¹. Find its empirical and molecular formulae. (C = 12, H = 1, O = 16)

Try it first, then show the solutionSolution

Moles in 100 g: C 54.5/12 = 4.54; H 9.1/1 = 9.1; O 36.4/16 = 2.275. (1 mark)
Dividing by 2.275: C 2 : H 4 : O 1, so the empirical formula is C₂H₄O. (1 mark)
Empirical formula mass = 24 + 4 + 16 = 44. (1 mark)
n = 88/44 = 2, so the molecular formula is C₄H₈O₂. (1 mark)

Answer: Empirical formula C₂H₄O; molecular formula C₄H₈O₂.

Example 4: Combustion analysis · 4 marks · Short answerHarder board level

Complete combustion of 0.58 g of a hydrocarbon gives 1.76 g of CO₂ and 0.90 g of H₂O. The molar mass of the compound is 58 g mol⁻¹. Find its molecular formula.

Try it first, then show the solutionSolution

Mass of C = (12/44) × 1.76 = 0.48 g; mass of H = (2/18) × 0.90 = 0.10 g. (1 mark)
Total = 0.58 g, so it is a hydrocarbon. Moles: C 0.48/12 = 0.04; H 0.10/1 = 0.10. (1 mark)
Ratio C : H = 0.04 : 0.10 = 2 : 5, so the empirical formula is C₂H₅ (mass 29). (1 mark)
n = 58/29 = 2, so the molecular formula is C₄H₁₀. (1 mark)

Answer: C₄H₁₀ (empirical formula C₂H₅).

Example 5: A mixture of gases · 5 marks · Long answerHOTS / competency

A gas mixture contains 5.6 L of CO₂ measured at STP and 3.2 g of CH₄ (C = 12, H = 1, O = 16). Find the total number of moles, the total mass, the number of carbon atoms and the total number of atoms.

Try it first, then show the solutionSolution

n(CO₂) = 5.6/22.4 = 0.25 mol; n(CH₄) = 3.2/16 = 0.20 mol. Total = 0.45 mol. (1 mark)
Mass = 0.25 × 44 + 3.2 = 11 + 3.2 = 14.2 g. (1 mark)
Carbon atoms = 0.45 × 6.022 × 10²³ = 2.71 × 10²³. (1 mark)
Atoms: CO₂ gives 0.25 × 3 = 0.75 mol, CH₄ gives 0.20 × 5 = 1.00 mol; total 1.75 mol. (1 mark)
Number = 1.75 × 6.022 × 10²³ = 1.05 × 10²⁴ atoms. (1 mark)

Answer: 0.45 mol; 14.2 g; 2.71 × 10²³ C atoms; 1.05 × 10²⁴ atoms.

Where students lose marks in Mole Concept and Empirical Formula

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong molar mass
Using atomic mass (N = 14) for a diatomic gas.
Do this: Write the formula of the species (N₂ = 28) before computing.
Usually costs: 1 to 2 marks.
Mistake: Not clearing fractions in the empirical formula
Writing Fe₁O₁.₅ or rounding 1.5 to 1 or 2.
Do this: Multiply the ratio by 2 when it is x.5; by 3 for x.33.
Usually costs: 1 to 2 marks.
Mistake: Mixing molecules and atoms
Giving the number of molecules when atoms are asked.
Do this: Multiply by the atoms per molecule.
Usually costs: 1 mark.

Exam tips

Write the formula and molar mass first, with the atomic masses used.
Show each conversion step: mass → moles → number.
Check combustion data: C + H masses should add up to the sample mass for a hydrocarbon.
State the conditions for 22.4 L (STP).

Practice questions on Mole Concept and Empirical Formula

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The mass of one mole of nitrogen molecules (N = 14) is:
(a) 14 g
(b) 28 g
(c) 6.022 × 10²³ g
(d) 22.4 g
Hint
Nitrogen gas is N₂.
Answer
(b) 28 g
Q2 · 2 marks · Short answerBoard level
Find the mass of 3.011 × 10²³ molecules of NH₃ (N = 14, H = 1).
Hint
Convert molecules to moles first.
Answer
n = 3.011 × 10²³/6.022 × 10²³ = 0.5 mol; mass = 0.5 × 17 = 8.5 g.
Q3 · 3 marks · Short answerHarder board level
An oxide of iron contains 69.9% iron and 30.1% oxygen. Find its empirical formula. (Fe = 56, O = 16)
Hint
Divide by atomic mass and clear the fraction.
Answer
Fe 69.9/56 = 1.248; O 30.1/16 = 1.881. Ratio 1 : 1.507 ≈ 2 : 3, so Fe₂O₃.
Q4 · 3 marks · Short answerHarder board level
How many electrons are present in 1.8 g of water? (H = 1, O = 16, N_A = 6.022 × 10²³)
Hint
A water molecule has 10 electrons.
Answer
n = 1.8/18 = 0.1 mol of molecules, so electrons = 0.1 × 10 = 1.0 mol = 6.022 × 10²³.
Q5 · 3 marks · Short answerHarder board level
Chlorine has two isotopes: ³⁵Cl (75%) and ³⁷Cl (25%). Find its average atomic mass.
Hint
Weighted mean.
Answer
(75 × 35 + 25 × 37)/100 = (2625 + 925)/100 = 35.5.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student studies a hydrate of sodium carbonate, Na₂CO₃·xH₂O (Na = 23, C = 12, O = 16, H = 1). A 2.86 g sample loses 1.80 g of water on heating.
(i) Find the mass of anhydrous Na₂CO₃. (1 mark)
(ii) Find its moles. (1 mark)
(iii) Find the moles of water lost. (1 mark)
(iv) Find x. (1 mark)
Hint
The ratio of moles of water to moles of salt gives x.
Answer
(i) 2.86 − 1.80 = 1.06 g. (ii) M = 106, so n = 0.01 mol. (iii) 1.80/18 = 0.10 mol. (iv) x = 0.10/0.01 = 10, so Na₂CO₃·10H₂O.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is Avogadro's number?Show answer
6.022 × 10²³ mol⁻¹.
2.Volume of 1 mol of gas at STP?Show answer
22.4 L.
3.How is the empirical formula found?Show answer
Divide percentages by atomic masses, then by the smallest, then clear fractions.
4.What is the molecular formula in terms of the empirical formula?Show answer
(Empirical formula) × n, n = M/empirical mass.

Mole Concept and Empirical Formula: frequently asked questions

Is the mole a unit of mass?

No. The mole measures the amount of substance, that is, a number of entities. The mass of one mole is the molar mass.

Why can two compounds have the same empirical formula?

Because different molecules, such as CH₂O (formaldehyde) and C₆H₁₂O₆ (glucose), may have atoms in the same ratio but different numbers of atoms.

Is STP always 22.4 L?

The value 22.4 L is for 273 K and 1 atm. At 273 K and 1 bar, the molar volume is a little larger (22.7 L); use the value stated in the question.