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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

IB MYP 4-5 Chemistry Study Notes – All topics

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

QuantitySymbolFormulaUnits
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
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