Mole and Avogadro Constant
Learn and apply Mole and Avogadro Constant in the published Chemistry course sequence.
Continue where you stopped
The core idea
On this page
Mole and Avogadro Constant: Orientation
The mole is the bridge between the microscopic (particles) and macroscopic (mass). Most stoichiometry marks start with one of two conversions: particles ↔ moles (N = nN_A) or mass ↔ moles (n = m/M).
Build this on Solution Concentration and Dilution and keep the Stoichiometry hub open so unit conversions and mole logic stay coherent.
Definitions (Must Know)
A. Mole
The mole is an amount of substance that contains Avogadro’s constant number of entities.
B. Avogadro constant, N_A
Avogadro’s constant, N_A, is the number of entities in 1 mol: N_A = 6.02 × 10²³ mol⁻¹
C. Molar mass, M
The molar mass, M, is the mass of 1 mol of a substance (units: g mol⁻¹).
D. Formula unit
A formula unit is the simplest ratio of ions in an ionic compound, shown by its chemical formula (e.g. NaCl).
E. Relative masses
- Relative isotopic mass is the mass of one atom of an isotope relative to 1/12 of the mass of one carbon-12 atom.
- Relative atomic mass, Aᵣ, is the weighted mean mass of an atom of an element relative to 1/12 of the mass of one carbon-12 atom.
- Relative molecular mass, Mᵣ, is the sum of the relative atomic masses of all atoms in one molecule.
- Relative formula mass is the corresponding sum for a formula unit of an ionic or giant substance.
Relative masses have no units because they are ratios. Molar mass has units of g mol⁻¹.
Detailed Explanations
A. Particles ↔ moles (workflow)
- Decide the entity (atoms / molecules / ions / formula units).
- Use N = nN_A (or rearrange to n = N/N_A).
- Write the final answer with the entity stated.
Mini example:
- 0.250 mol of CO₂ contains N = 0.250 × 6.02 × 10²³ = 1.51 × 10²³ molecules.
B. Mass ↔ moles (workflow)
- Find M in g mol⁻¹ (sum of Aᵣ values).
- Use n = m/M (or m = nM).
- Keep units until the final line.
Mini example:
- For Al, M = 27.0 g mol⁻¹, so 2.70 g is n = 2.70/27.0 = 0.100 mol.
C. “Atoms in a compound” questions (don’t lose the entity mark)
Because N = nN_A counts entities, you must convert “molecules” into “atoms” when asked.
Two safe methods:
- count molecules, then multiply by atoms per molecule
- convert to moles of atoms first (then multiply by N_A)
D. Formula units and ions (common in data questions)
- 1 mol of NaCl contains 1 mol of Na⁺ and 1 mol of Cl⁻ (once dissolved).
- 1 mol of Al₂(SO₄)₃ contains 2 mol of Al³⁺ and 3 mol of SO₄²⁻ (once dissolved).
E. Isotopic abundance and relative atomic mass
Aᵣ is a weighted mean, so a more abundant isotope contributes more strongly. If chlorine is 75.8% chlorine-35 and 24.2% chlorine-37:
Aᵣ(Cl) = ((35)(75.8) + (37)(24.2))/100 = 35.5
The answer lies between 35 and 37 and is closer to 35 because chlorine-35 is more abundant. In a mass spectrum, use the stated peak abundances in the same way; do not simply average the isotope masses.
F. Balance first, then use the mole ratio
Chemical formulae tell you which substances are present. Coefficients tell you the reacting amounts. For methane combustion:
CH₄ + 2O₂ → CO₂ + 2H₂O
The equation shows that 1 mol of CH₄ reacts with 2 mol of O₂. Balance by changing coefficients only—never alter a chemical formula to make the atom count fit.
Worked Examples
Modelled example 1
Count Carbon Dioxide Molecules
Problem
Study the worked solution
Choose the entity relationship
Method
Use N = nN_A and retain molecules as the named entity.Reason
One mole contains Avogadro’s constant of the specified entities.Working
N = (0.250)(6.02 × 10²³ mol⁻¹)Calculate and label
Method
Evaluate to three significant figures and state the entity.Reason
The numerical count is ambiguous unless molecules, atoms, ions or formula units are named.Working
N = 1.51 × 10²³ CO₂ molecules.
Guided practice 2
Convert Aluminium Mass to Amount
Problem
Try this before viewing the solution
Hints
Hint 1: identify molar mass
Hint 2: choose the quotient
View solution step by step
State molar mass
Method
Attach molar-mass units to the relative atomic mass value.Reason
A mass-to-amount calculation requires M in g mol⁻¹.Working
M(Al) = 27.0 g mol⁻¹Divide mass by molar mass
Method
Use n = m/M.Reason
The gram units cancel, leaving moles.Working
n = 2.70/27.0 = 0.100 mol
Common misconception 3
Distinguish Molecules from Oxygen Atoms
Learner claim
A learner says 0.250 mol of CO₂ contains 1.51 × 10²³ oxygen atoms because N = nN_A. Identify the first error and calculate the oxygen-atom count.
Diagnose the entity first
View solution step by step
Convert to moles of oxygen atoms
Method
Multiply the carbon dioxide amount by two.Reason
Every CO₂ molecule contains two oxygen atoms.Working
n(O\ atoms) = 2(0.250) = 0.500 molCount oxygen atoms
Method
Apply Avogadro’s constant to the oxygen-atom amount.Reason
N = nN_A now counts the entity actually requested.Working
N = (0.500)(6.02 × 10²³) = 3.01 × 10²³ oxygen atoms.
Examiner practice 4
Count Chloride Ions from a Compound Mass
Problem
A 4.75 g sample of anhydrous MgCl₂ dissolves completely. Calculate the number of chloride ions produced. Use Aᵣ(Mg) = 24.3, Aᵣ(Cl) = 35.5 and N_A = 6.02 × 10²³ mol⁻¹. [4 marks]
Try this before viewing the solution
View solution step by step
Calculate molar mass
1 markMethod
Add one magnesium and two chlorine relative masses.Reason
The formula MgCl₂ fixes the atom ratio in one mole of formula units.Working
M = 24.3 + 2(35.5) = 95.3 g mol⁻¹Find compound amount
1 markReason
Dividing sample mass by molar mass gives moles of MgCl₂ formula units.Working
n(MgCl₂) = 4.75/95.3 = 4.98 × 10⁻² molConvert to chloride amount
1 markReason
Each formula unit releases two chloride ions on complete dissolution.Working
n(Cl⁻) = 2(4.98 × 10⁻²) = 9.97 × 10⁻² molCount chloride ions
1 markMethod
Multiply the chloride amount by Avogadro’s constant.Reason
The result must name chloride ions rather than formula units.Working
N = (9.97 × 10⁻²)(6.02 × 10²³) = 6.00 × 10²² chloride ions.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Credit the compound amount and chloride multiplier as separate decisions.
Challenge 5
Infer Molar Mass from a Molecule Count
Problem
A 2.20 g sample contains 3.01 × 10²² molecules. Calculate its molar mass and identify it as N₂, CO₂ or SO₂. Use N_A = 6.02 × 10²³ mol⁻¹.
Try this before viewing the solution
Hints
Hint 1: start from the molecule count
Hint 2: recover molar mass
View solution step by step
Convert molecules to amount
Method
Divide the molecule count by Avogadro’s constant.Reason
The mass is for the same sample, so its molecular count must first be expressed in moles.Working
n = (3.01 × 10²²)/(6.02 × 10²³) = 0.0500 molInfer molar mass
Reason
Molar mass is sample mass divided by sample amount.Working
M = 2.20/0.0500 = 44.0 g mol⁻¹Identify the molecule
Method
Match 44.0 g mol⁻¹ to carbon dioxide.Reason
Mᵣ(CO₂) = 12.0 + 2(16.0) = 44.0, unlike N₂ or SO₂.Working
The sample is CO₂.
Mind Stretchers
Mind stretcher 1Extension
1.20 × 10²⁴ chloride ions are produced when an ionic solid dissolves. How many moles of chloride ions is this?
Show Hint
Convert the entity count to moles first. Keep the entity as chloride ions; no formula multiplier is needed.
Show Answer
Mark scheme:
- n = N/N_A = (1.20 × 10²⁴)/(6.02 × 10²³) = 1.99 mol ≈ 2.00 mol
Mind stretcher 2: Counting every ion releasedExtension
Question. A sample contains 0.150 mol of Al₂(SO₄)₃. It dissolves completely. Calculate the total number of ions produced.
Show Hint
One formula unit produces two aluminium ions and three sulfate ions.
Show Answer
Each formula unit gives five ions, so the amount of ions is 5(0.150) = 0.750 mol. The number is 0.750(6.02 × 10²³) = 4.52 × 10²³ ions.