What is the mole concept in the context of IB MYP Science?

In the IB MYP Science curriculum, the mole concept is a fundamental principle that helps students understand the amount of substance in chemistry. It defines a mole as a unit that measures the amount of substance, specifically 6.022 x 10^23 particles, which could be atoms, molecules, ions, or electrons. This concept is crucial for balancing chemical equations and understanding chemical reactions.

How is the Avogadro Constant used in chemical kinetics?

The Avogadro Constant, which is 6.022 x 10^23 mol^-1, is used in chemical kinetics to relate the macroscopic scale of substances to the microscopic scale of atoms and molecules. It allows scientists and students to calculate the number of particles involved in chemical reactions, which is essential for understanding reaction rates and mechanisms in the study of chemical kinetics.

Why is the mole concept important for understanding chemical reactions?

The mole concept is vital for understanding chemical reactions because it provides a bridge between the atomic scale and the macroscopic scale. By using moles, students can quantify the exact amounts of reactants and products involved in a reaction, allowing for precise calculations and predictions of reaction outcomes. This is particularly important in the IB MYP Science curriculum for mastering stoichiometry and reaction balancing.

How do you calculate the number of moles from a given mass?

To calculate the number of moles from a given mass, you need to use the formula: number of moles = mass (g) / molar mass (g/mol). The molar mass is the mass of one mole of a substance and can be found on the periodic table as the atomic or molecular weight. This calculation is a key skill in the IB MYP Science curriculum for solving problems related to the mole concept.

What is the significance of Avogadro's number in the IB MYP curriculum?

Avogadro's number, 6.022 x 10^23, is significant in the IB MYP curriculum because it provides a consistent way to count particles at the atomic level. It is essential for converting between the number of particles and the amount of substance in moles, which is a fundamental aspect of understanding chemical reactions, stoichiometry, and the behavior of gases in chemical kinetics.

Why is "Using ratios of masses instead of moles in a reaction." a common mistake in The Mole Concept and the Avogadro Constant?

Why it happens: Masses are given in the question. How to avoid it: Convert mass to MOLES first; use the equation's mole ratio; convert back to mass at the end.

Why is "Writing molar mass without units or as a plain number." a common mistake in The Mole Concept and the Avogadro Constant?

Why it happens: It looks similar to Aᵣ. How to avoid it: Molar mass is in g/mol. Aᵣ has no units. Mᵣ has no units. Numerically equal but conceptually different.

Why is "Writing Nₐ as 6.022 × 10⁻²³." a common mistake in The Mole Concept and the Avogadro Constant?

Why it happens: Sign of exponent slip. How to avoid it: Avogadro's constant is HUGE (6 × 10²³, not 6 × 10⁻²³). Memorise as 6 followed by 23 zeros (roughly).

Chemical kineticsIB MYP Sciences (Years 1-5)

The Mole Concept and the Avogadro Constant

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The Mole and Avogadro's Constant — IB MYP Sciences: counting atoms by weighing

Atoms are far too small to count one by one. The 'mole' is a chemist's standard count — 6.022 × 10²³ particles. This MYP Sciences note covers what a mole is, how to convert between mass, moles and particles, and why this lets us connect the microscopic and macroscopic worlds.

What you’ll learn

Mapped to the IB MYP Sciences subject guide (2026 onwards).

MYP Sciences A — Define the mole and state Avogadro's constant.
MYP Sciences A — Calculate molar mass (Mᵣ) of compounds from atomic masses.
MYP Sciences C — Convert between mass, moles and number of particles.
MYP Sciences C — Use balanced equations and moles to calculate reacting masses.

What is a mole?

A mole is a unit of counting — like a 'dozen' but much bigger.

Atoms are tiny: a teaspoon of water contains about 1.7 × 10²³ molecules — far too many to count one at a time. Chemists invented a useful unit of counting:

1 mole = 6.022 × 10²³ particles (atoms, molecules, ions or electrons).

This number is Avogadro's constant (Nₐ). It was chosen because exactly 1 mole of carbon-12 atoms has a mass of 12 grams — a beautifully clean link between atomic-mass units and grams.

Useful for any kind of particle:

1 mol of Na atoms = 6.022 × 10²³ atoms = 23 g (relative atomic mass of Na is 23).
1 mol of H₂O molecules = 6.022 × 10²³ molecules = 18 g (Mᵣ = 2 + 16 = 18).
1 mol of electrons = 6.022 × 10²³ electrons.

1 mol = 6.022 × 10²³ particles (Nₐ).
1 mol of C-12 atoms = 12 g exactly.
Works for atoms, molecules, ions, electrons — anything you can count.

Molar mass and the n = m/M relationship

Mass divided by molar mass = number of moles.

Molar mass (M) of a substance is the mass of 1 mole, in grams. Numerically equal to the relative atomic or formula mass (Aᵣ or Mᵣ).

Examples:

Carbon (C): Aᵣ = 12 → M = 12 g/mol.
Oxygen molecule (O₂): Mᵣ = 32 → M = 32 g/mol.
Water (H₂O): Mᵣ = 18 → M = 18 g/mol.
Sodium chloride (NaCl): Mᵣ = 58.5 → M = 58.5 g/mol.

The key equation:

$n = \frac{m}{M}$

where n = moles, m = mass (g), M = molar mass (g/mol).

Cover the unknown to find it: cover n for n = m/M; cover m for m = n × M; cover M for M = m/n.

Worked example. How many moles in 36 g of water?

n = m/M = 36 / 18 = 2 moles. (And 2 mol × 6.022 × 10²³ ≈ 1.2 × 10²⁴ molecules.)

Worked example. What is the mass of 0.25 mol of CO₂?

Mᵣ(CO₂) = 12 + 16 + 16 = 44. m = n × M = 0.25 × 44 = 11 g.

Molar mass = mass of 1 mol = Aᵣ or Mᵣ in grams.
n = m / M.
Rearrange: m = n × M, or M = m / n.

Using moles in reactions

A balanced equation tells you the mole ratio.

A balanced chemical equation tells you the RATIO of moles of reactants to products. Combined with the n = m/M formula, you can answer questions like 'how much of A do I need to react with B?'

Worked example. What mass of magnesium oxide (MgO) is produced from 2.4 g of magnesium burning?

Equation: 2Mg + O₂ → 2MgO. Mole ratio: 2 Mg : 2 MgO = 1:1.
Moles of Mg = 2.4 / 24 = 0.10 mol.
By the 1:1 ratio, moles of MgO = 0.10 mol.
Mass of MgO = n × M = 0.10 × 40 = 4.0 g.

This is called a mass-to-mass calculation or stoichiometry. It's how chemists work out exact amounts in industry, in research, and in scaling up reactions safely.

Balanced equation = mole ratio.
Convert mass → moles → use ratio → moles of product → mass.
Stoichiometry: how chemists scale reactions.

How it’s examined

Criterion C is the main test ground — convert mass to moles, use ratios in a balanced equation, find mass of product. Criterion A tests definitions and the formula n = m/M.

Sources: IB MYP Sciences guide (IBO, official subject guide). Last reviewed 2026-05-25.

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Step-by-step worked examples — The Mole Concept and the Avogadro Constant

Step-by-step solutions to past-paper-style questions on the mole concept and the avogadro constant, written exactly the way a tutor would explain them at the board.

1Moles from mass
Getting started• n = m/M
Question
How many moles are in 50 g of CaCO₃? (Mᵣ = 100)
Step-by-step solution
1. Step 1
  n = m / M = 50 / 100.
2. Step 2
  n = 0.50 mol.
Answer
0.50 mol.
2Mass from moles
Getting started• m = nM
Question
What is the mass of 0.30 mol of NaCl? (Mᵣ = 58.5)
Step-by-step solution
1. Step 1
  m = n × M = 0.30 × 58.5.
2. Step 2
  m = 17.55 g ≈ 17.6 g.
Answer
≈ 17.6 g.
3Counting particles
Building confidence• Avogadro
Question
How many water molecules are in 9.0 g of water? (Mᵣ(H₂O) = 18)
Step-by-step solution
1. Step 1
  Moles = 9.0 / 18 = 0.50 mol.
2. Step 2
  Particles = 0.50 × 6.022 × 10²³ = 3.011 × 10²³ molecules.
Answer
≈ 3.0 × 10²³ molecules.
4Mass-to-mass stoichiometry
Stretch• stoichiometry
Question
Calculate the mass of MgO produced from completely burning 6.0 g of magnesium. (Aᵣ Mg = 24, Mᵣ MgO = 40.)
Step-by-step solution
1. Step 1
  Equation: 2Mg + O₂ → 2MgO. Ratio Mg:MgO = 1:1.
2. Step 2
  Moles of Mg = 6.0 / 24 = 0.25 mol.
3. Step 3
  Moles of MgO = 0.25 mol (1:1 ratio).
4. Step 4
  Mass of MgO = 0.25 × 40 = 10 g.
Answer
10 g of MgO.

Key Formulae — The Mole Concept and the Avogadro Constant

The formulae you need to memorise for the mole concept and the avogadro constant on the IB MYP Sciences paper, with every variable defined in plain English and a note on when to use it.

Moles from mass
$n = m / M$
$n$
amount in moles (mol)
$m$
mass (g)
$M$
molar mass (g/mol)
When to use
Anywhere you need to swap between mass and moles.
Number of particles
$N = n × Nₐ$
$N$
number of particles
$n$
moles
$Nₐ$
Avogadro's constant (6.022 × 10²³ /mol)
When to use
When you need a number of atoms or molecules.

Key Definitions and Keywords — The Mole Concept and the Avogadro Constant

Definitions to memorise and the exact keywords mark schemes credit for the mole concept and the avogadro constant answers — sharpened from recent examiner reports for the 2026 IB MYP Sciences sitting.

Mole (mol)
Examiner keyword
The chemist's unit of amount: 1 mol contains exactly 6.022 × 10²³ particles.
Avogadro's constant (Nₐ)
Examiner keyword
The number of particles in 1 mole: 6.022 × 10²³ per mol.
Molar mass (M)
Examiner keyword
Mass of 1 mole of a substance, in g/mol. Numerically equal to Aᵣ or Mᵣ.
RTP (room temperature and pressure)
~25 °C, 1 atm. 1 mol of any gas at RTP occupies ~24 dm³.

Common Mistakes and Misconceptions — The Mole Concept and the Avogadro Constant

The traps other students keep falling into on the mole concept and the avogadro constant questions — taken from recent IB MYP Sciences examiner reports and mark schemes — and how to avoid them.

✕Using ratios of masses instead of moles in a reaction.
Why it happens
Masses are given in the question.
How to avoid it
Convert mass to MOLES first; use the equation's mole ratio; convert back to mass at the end.
✕Writing molar mass without units or as a plain number.
Why it happens
It looks similar to Aᵣ.
How to avoid it
Molar mass is in g/mol. Aᵣ has no units. Mᵣ has no units. Numerically equal but conceptually different.
✕Writing Nₐ as 6.022 × 10⁻²³.
Why it happens
Sign of exponent slip.
How to avoid it
Avogadro's constant is HUGE (6 × 10²³, not 6 × 10⁻²³). Memorise as 6 followed by 23 zeros (roughly).

Practice questions

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