IB MYP 4-5 Chemistry -The mole concept and Avogadro’s constant- Study Notes - New Syllabus
IB MYP 4-5 Chemistry -The mole concept and Avogadro’s constant- Study Notes
Key Concepts
- The Mole Concept and Avogadro’s Constant
 
The Mole Concept and Avogadro’s Constant
The Mole Concept and Avogadro’s Constant
The mole is the standard scientific unit used to measure the amount of substance. It allows chemists to count particles (atoms, ions, or molecules) by weighing them.
Key Idea: One mole of any substance contains the same number of particles — this number is known as Avogadro’s constant.
Avogadro’s Constant (\( \mathrm{N_A} \))
Avogadro’s constant (\( \mathrm{N_A} \)) is the number of particles (atoms, molecules, or ions) in one mole of a substance.
\( \mathrm{N_A = 6.022 \times 10^{23}\ particles\ mol^{-1}} \)
Meaning:
- 1 mole of atoms = \( \mathrm{6.022 \times 10^{23}} \) atoms.
 - 1 mole of molecules = \( \mathrm{6.022 \times 10^{23}} \) molecules.
 - 1 mole of ions = \( \mathrm{6.022 \times 10^{23}} \) ions.
 
Example: 1 mole of water (\( \mathrm{H_2O} \)) contains:
- \( \mathrm{6.022 \times 10^{23}} \) molecules of water
 - \( \mathrm{(6.022 \times 10^{23}) \times 3 = 1.807 \times 10^{24}} \) atoms (2 H + 1 O)
 
Relationship Between Mass, Moles, and Molar Mass
Formula:
\( \mathrm{n = \dfrac{m}{M}} \)
Where:
- \( \mathrm{n} \) = number of moles
 - \( \mathrm{m} \) = mass of the substance (in grams)
 - \( \mathrm{M} \) = molar mass (g/mol)
 
Rearranged Forms:
- \( \mathrm{m = nM} \)
 - \( \mathrm{M = \dfrac{m}{n}} \)
 
Relationship Between Number of Particles and Moles
Formula:
\( \mathrm{n = \dfrac{N}{N_A}} \)
Where:
- \( \mathrm{N} \) = number of particles (atoms, molecules, or ions)
 - \( \mathrm{N_A} \) = Avogadro’s constant
 - \( \mathrm{n} \) = number of moles
 
Rearranged Forms:
- \( \mathrm{N = n \times N_A} \)
 - \( \mathrm{n = \dfrac{N}{N_A}} \)
 
Relationship Between Volume and Moles (for Gases)
At standard temperature and pressure (STP: 0°C and 1 atm), 1 mole of any gas occupies:
\( \mathrm{22.4\ dm^3 = 22,400\ cm^3} \)
Formula:
\( \mathrm{n = \dfrac{V}{V_m}} \)
Where:
- \( \mathrm{n} \) = number of moles
 - \( \mathrm{V} \) = volume of gas (in dm³)
 - \( \mathrm{V_m} \) = molar volume = \( \mathrm{22.4\ dm^3\ mol^{-1}} \) at STP
 
Molar Mass and Relative Atomic Mass
The molar mass (\( \mathrm{M} \)) of a substance is the mass of one mole of that substance.
\( \mathrm{M = Relative\ Atomic\ or\ Molecular\ Mass\ (in\ g/mol)} \)
Examples:
- 1 mole of \( \mathrm{H_2} \) = 2 g
 - 1 mole of \( \mathrm{O_2} \) = 32 g
 - 1 mole of \( \mathrm{H_2O} \) = 18 g
 
Summary Table of Relationships
| Quantity | Symbol | Formula | Units | 
|---|---|---|---|
| Number of moles | \( \mathrm{n} \) | \( \mathrm{n = \dfrac{m}{M}} \) | mol | 
| Number of particles | \( \mathrm{N} \) | \( \mathrm{N = n \times N_A} \) | — | 
| Mass | \( \mathrm{m} \) | \( \mathrm{m = n \times M} \) | g | 
| Volume (for gases at STP) | \( \mathrm{V} \) | \( \mathrm{n = \dfrac{V}{22.4}} \) | dm³ | 
Example
Calculate the number of moles in 12 g of carbon.
▶️ Answer / Explanation
Step 1: Molar mass of carbon = 12 g/mol.
Step 2: \( \mathrm{n = \dfrac{m}{M} = \dfrac{12}{12} = 1\ mol} \)
Final Answer: 1 mole of carbon contains \( \mathrm{6.022 \times 10^{23}} \) atoms.
Example
Find the number of water molecules in 36 g of water (\( \mathrm{H_2O} \)).
▶️ Answer / Explanation
Step 1: Molar mass of water = 18 g/mol.
Step 2: \( \mathrm{n = \dfrac{36}{18} = 2\ mol} \)
Step 3: \( \mathrm{N = n \times N_A = 2 \times 6.022 \times 10^{23}} \)
Step 4: \( \mathrm{N = 1.204 \times 10^{24}\ molecules} \)
Final Answer: 36 g of water contains \( \mathrm{1.204 \times 10^{24}} \) molecules.
Example
At STP, 11.2 dm³ of nitrogen gas (\( \mathrm{N_2} \)) is collected. Calculate:
- (a) The number of moles of nitrogen gas
 - (b) The number of nitrogen molecules
 - (c) The number of nitrogen atoms
 
▶️ Answer / Explanation
(a) Using \( \mathrm{n = \dfrac{V}{22.4}} \)
\( \mathrm{n = \dfrac{11.2}{22.4} = 0.5\ mol} \)
(b) Number of molecules \( \mathrm{= n \times N_A} \)
\( \mathrm{N = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\ molecules} \)
(c) Each molecule of \( \mathrm{N_2} \) contains 2 atoms:
\( \mathrm{Atoms = 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23}} \)
Final Answer:
- (a) 0.5 mol of \( \mathrm{N_2} \)
 - (b) \( \mathrm{3.011 \times 10^{23}} \) molecules
 - (c) \( \mathrm{6.022 \times 10^{23}} \) atoms
 
