Form Three Chemistry Notes
Chapter One: The Mole Concept · Ordinary Level (Form Three) · Study Notes. Read this chapter free, then continue every chapter in the Click Online Academy app.
All Form Three Chemistry Chapters
- Chapter 1The Mole ConceptRead below · Free
- Chapter 2Volumetric AnalysisContinue in the app
- Chapter 3Ionic Theory and ElectrolysisContinue in the app
- Chapter 4Compounds of MetalsContinue in the app
- Chapter 5Non-metals and their CompoundsContinue in the app
- Chapter 6Extraction of IronContinue in the app
Learning Outcomes
By the end of this chapter, you should be able to:
- define the mole and state Avogadro’s constant;
- distinguish between relative atomic mass, relative molecular/formula mass and molar mass;
- calculate the number of moles from mass, and mass from number of moles;
- calculate the number of particles in a given amount of substance;
- use the molar volume of gases at STP in calculations;
- apply the mole concept to simple stoichiometric problems.
Key Terms and Concepts
| Term | Meaning |
|---|---|
| Mole | The amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of carbon-12. Symbol: mol. |
| Avogadro’s constant (L or NA) | The number of particles in one mole: 6.02 × 1023 mol−1. |
| Relative atomic mass (Ar) | The average mass of an atom of an element compared with 1/12 of the mass of an atom of carbon-12. It has no units. |
| Relative molecular / formula mass (Mr) | The sum of the relative atomic masses of all the atoms in a molecule or formula unit. No units. |
| Molar mass (M) | The mass of one mole of a substance. Unit: g mol−1. Numerically equal to Ar or Mr. |
| Molar volume | The volume occupied by one mole of any gas at stated conditions. At STP it is 22.4 dm3 (22.4 L). |
| STP | Standard temperature and pressure: 0 °C (273 K) and 1 atmosphere (101.325 kPa). |
| Stoichiometry | The quantitative relationship between reactants and products in a chemical reaction, based on the balanced equation and the mole concept. |
1. What is the Mole?
Atoms and molecules are extremely small. Counting them one by one is impossible, so chemists use a convenient counting unit called the mole.
Just as a dozen means 12 and a gross means 144, one mole means 6.02 × 1023 particles. This huge number is called Avogadro’s constant (or Avogadro’s number).
The official definition is:
One mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 grams of the carbon-12 isotope.
Those elementary entities can be atoms, molecules, ions, electrons, formula units, etc.
| Substance | 1 mole contains | Mass of 1 mole |
|---|---|---|
| Carbon (C) | 6.02 × 1023 atoms | 12 g |
| Iron (Fe) | 6.02 × 1023 atoms | 56 g |
| Hydrogen (H2) | 6.02 × 1023 molecules | 2 g |
| Oxygen (O2) | 6.02 × 1023 molecules | 32 g |
| Water (H2O) | 6.02 × 1023 molecules | 18 g |
| Sodium chloride (NaCl) | 6.02 × 1023 formula units | 58.5 g |
TEST YOUR UNDERSTANDING
Q1. Define the mole.
Q2. What is the value of Avogadro’s constant?
2. Molar Mass and Relative Masses
Relative atomic mass (Ar) is the average mass of atoms of an element compared with 1/12 the mass of a carbon-12 atom. It has no unit.
Relative molecular mass (Mr) or relative formula mass is the sum of the relative atomic masses of all the atoms shown in the formula of a molecule or ionic compound. It also has no unit.
Molar mass (M) is the mass of one mole of a substance. Its unit is g mol−1. Numerically, molar mass is the same as Ar or Mr.
Calculating relative formula mass / molar mass
Add the relative atomic masses of every atom in the formula.
- CO2: C = 12, O = 16 × 2 → Mr = 12 + 32 = 44 → molar mass = 44 g mol−1
- H2SO4: (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 → 98 g mol−1
- CaCO3: 40 + 12 + (3 × 16) = 100 → 100 g mol−1
- Na2SO4: (2 × 23) + 32 + (4 × 16) = 46 + 32 + 64 = 142 → 142 g mol−1
TEST YOUR UNDERSTANDING
Q3. Calculate the molar mass of (a) NH3, (b) Mg(OH)2. (N=14, H=1, Mg=24, O=16)
3. Mole Calculations
The fundamental relationship is:
Number of moles (n) = mass (m) / molar mass (M)
or rearranged:
- mass = n × M
- M = m / n
Number of particles
Number of particles = number of moles × Avogadro’s constant
N = n × 6.02 × 1023
Worked examples
Example 1. How many moles are in 11.5 g of sodium? (Ar Na = 23)
n = 11.5 / 23 = 0.5 mol
Example 2. What mass of water contains 0.25 mol of H2O? (M = 18 g mol−1)
m = 0.25 × 18 = 4.5 g
Example 3. How many molecules are in 9 g of water?
n = 9 / 18 = 0.5 mol
N = 0.5 × 6.02 × 1023 = 3.01 × 1023 molecules
Example 4. Calculate the number of moles of oxygen atoms in 16 g of O2.
n(O2) = 16 / 32 = 0.5 mol of molecules
Each O2 molecule contains 2 O atoms → n(O atoms) = 0.5 × 2 = 1 mol
TEST YOUR UNDERSTANDING
Q4. Calculate the number of moles in 10 g of CaCO3. (Ca=40, C=12, O=16)
Q5. How many atoms are present in 0.1 mol of iron?
Q6. What mass of copper contains the same number of atoms as 12 g of carbon? (Cu=63.5, C=12)
4. Moles and Gases
Avogadro’s law states that equal volumes of all gases, measured at the same temperature and pressure, contain the same number of molecules.
At STP (0 °C and 1 atm), one mole of any gas occupies 22.4 dm3 (22.4 litres). This is the molar volume of a gas at STP.
n = volume of gas at STP (dm3) / 22.4
Worked examples
Example 5. How many moles of nitrogen gas are in 5.6 dm3 of N2 at STP?
n = 5.6 / 22.4 = 0.25 mol
Example 6. What volume does 8 g of oxygen gas occupy at STP? (O2 = 32)
n = 8 / 32 = 0.25 mol
Volume = 0.25 × 22.4 = 5.6 dm3
TEST YOUR UNDERSTANDING
Q7. Calculate the volume occupied by 0.5 mol of any gas at STP.
Q8. How many molecules are present in 11.2 dm3 of hydrogen gas at STP?
5. Applications of the Mole Concept
The mole concept is the foundation of quantitative chemistry. It is used to:
- convert between mass and amount of substance;
- find the number of atoms, molecules or ions in a sample;
- determine empirical and molecular formulae from percentage composition;
- carry out stoichiometric calculations from balanced equations (mass–mass, mass–volume, volume–volume);
- prepare solutions of known concentration (molarity) — a key skill for volumetric analysis in the next chapter;
- calculate percentage purity and water of crystallisation.
Simple stoichiometry example
Consider the reaction: 2Mg + O2 → 2MgO
From the equation, 2 moles of Mg react with 1 mole of O2.
If 12 g of magnesium reacts completely, how many grams of oxygen are needed? (Mg=24, O=16)
n(Mg) = 12 / 24 = 0.5 mol
Mole ratio Mg : O2 = 2 : 1 → n(O2) = 0.25 mol
Mass of O2 = 0.25 × 32 = 8 g
TEST YOUR UNDERSTANDING
Q9. From the equation 2H2 + O2 → 2H2O, how many moles of water are formed when 4 g of hydrogen reacts completely? (H=1)
Chapter Summary
- The mole is the SI unit of amount of substance; 1 mol contains 6.02 × 1023 particles (Avogadro’s constant).
- Molar mass (g mol−1) is numerically equal to relative atomic or formula mass.
- Key equations: n = m / M ; N = n × 6.02 × 1023 ; for gases at STP, n = V / 22.4.
- The mole concept links mass, number of particles and volume of gases and is essential for stoichiometry and volumetric analysis.
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Continue in the appEnd-of-Topic Questions
Attempt all questions in your exercise book before checking the suggested answers. (New Syllabus 2027)
- Section A: Multiple choice. Choose the correct answer. (i) One mole of any substance contains: A. 6.02 × 1023 particles B. 12 particles C. 22.4 particles D. 6.02 particles. (ii) The molar mass of CO2 is: A. 28 g mol−1 B. 44 g mol−1 C. 32 g mol−1 D. 16 g mol−1. (iii) The volume occupied by 1 mole of any gas at STP is: A. 22.4 cm3 B. 22.4 dm3 C. 24 dm3 D. 11.2 dm3. (iv) Number of moles = mass ÷ : A. Avogadro’s constant B. molar mass C. relative atomic mass D. density. (v) 0.5 mol of oxygen gas (O2) contains how many molecules? A. 3.01 × 1023 B. 6.02 × 1023 C. 1.204 × 1024 D. 0.5. (5 marks)
- Match Column A with Column B: (i) Avogadro’s constant (ii) Molar mass of water (iii) Molar volume at STP (iv) Relative atomic mass of carbon (v) 1 mole of NaCl. Column B: A. 12 B. 6.02 × 1023 formula units C. 18 g mol−1 D. 22.4 dm3 E. 6.02 × 1023 mol−1. (5 marks)
- (a) Define the mole. (b) State the value of Avogadro’s constant. (4 marks)
- Calculate the molar mass of (a) H2SO4, (b) Ca(OH)2. (H=1, S=32, O=16, Ca=40) (4 marks)
- How many moles are present in 25 g of CaCO3? (Ca=40, C=12, O=16) (3 marks)
- Calculate the number of molecules in 4.4 g of CO2. (C=12, O=16) (4 marks)
- What volume does 8 g of oxygen gas occupy at STP? (O=16) (3 marks)
- From the equation 2Mg + O2 → 2MgO, calculate the mass of magnesium oxide formed when 6 g of magnesium burns completely in oxygen. (Mg=24, O=16) (6 marks)
- Explain why chemists prefer to use the mole rather than counting individual atoms or molecules. (4 marks)
Suggested Answers
- (i) A. (ii) B. (iii) B. (iv) B. (v) A.
- (i) E. (ii) C. (iii) D. (iv) A. (v) B.
- (a) The amount of substance containing as many elementary entities as there are atoms in 12 g of carbon-12. (b) 6.02 × 1023 mol−1.
- (a) 2 + 32 + 64 = 98 g mol−1. (b) 40 + 32 + 2 = 74 g mol−1.
- M = 100 g mol−1; n = 25 / 100 = 0.25 mol.
- n = 4.4 / 44 = 0.1 mol; molecules = 0.1 × 6.02 × 1023 = 6.02 × 1022.
- n = 8 / 32 = 0.25 mol; V = 0.25 × 22.4 = 5.6 dm3.
- n(Mg) = 6 / 24 = 0.25 mol; mole ratio Mg : MgO = 2 : 2 = 1 : 1 → 0.25 mol MgO; mass = 0.25 × 40 = 10 g.
- Atoms and molecules are far too small and numerous to count individually; the mole provides a practical unit that links measurable mass to a known number of particles.
These are suggested answers. Credit is also given to correct points expressed in your own words. Use a calculator where necessary and show working.
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