Chemistry • Year 11 • Module 2 • Lesson 1

The Mole Concept

Apply N = n × N_A to real data, interpret a scale-of-matter table, and reason about moles in context before Band 5–6 questions.

Apply · Data & Reasoning

1. Interpret a data table — comparing moles across substances

The table below shows six samples of different substances. The amount in moles and the substance formula are given. 9 marks

Sample	Substance	n (mol)	Elementary entity
A	Carbon atoms (C)	1.0	atom
B	Water molecules (H₂O)	0.50
C	Sodium chloride (NaCl) — formula units	2.0	formula unit
D	Helium atoms (He)	0.25
E	Oxygen gas (O₂)	3.0
F	Glucose molecules (C₆H₁₂O₆)	0.10

1.1 Complete the missing cells in the table above. Use N_A = 6.022 × 10²³ mol⁻¹. Show one worked example of your calculation in the space below. 7 marks (1 per row for N; 1 worked example)

1.2 Using your completed table, identify which sample (A–F) contains the greatest number of entities, and which contains the fewest. Explain why the sample with the most moles does not necessarily contain the most entities. 2 marks

Stuck? Use N = n × N_A for each row. Remember the elementary entity depends on the substance: atoms for elements, molecules for molecular compounds, formula units for ionic compounds.

2. Interpret graph — scale of Avogadro’s number

A science journalist compared Avogadro's number (N_A = 6.022 × 10²³) to other large quantities in nature. The bar chart below plots the base-10 logarithm of each quantity so they can be compared on the same axis. 8 marks

Figure 2. Log-scale comparison of large natural quantities. Illustrative data; order-of-magnitude values.

2.1 Using the graph, identify which quantity is closest in magnitude to Avogadro's number and estimate by how many orders of magnitude it differs from N_A. 2 marks

2.2 The quantity “seconds since the Big Bang” is approximately 4 × 10¹⁷. Explain why this value is NOT comparable to Avogadro's number, and how many times larger N_A is. Show your calculation. 3 marks

2.3 One litre of water contains approximately 3.34 × 10²⁵ molecules. Calculate the number of moles of water molecules in 1 L of water. Show full working. 3 marks

Stuck? For 2.3, use n = N ÷ N_A. Remember to state the formula, substitute, and calculate with correct units.

3. Compare N and n across five features

Complete the two-column table below. For each feature, write a concise description that distinguishes the two symbols. 10 marks (1 per cell)

Feature	N (number of particles)	n (amount in moles)
What it represents
Units
Typical scale
Symbol convention
Example value for 2 mol of carbon

Stuck? Revisit the “N vs n” callout in the lesson and the formula reference panel.

4. Predict and justify — a laboratory scenario

A Year 11 chemistry student is preparing two solutions for an experiment. Solution X contains 1.5 mol of NaCl dissolved in water. Solution Y contains 9.033 × 10²³ formula units of NaCl dissolved in water. The student claims that Solutions X and Y contain different amounts of NaCl and should behave differently in the experiment.

5 marks

4.1 Calculate the number of moles of NaCl in Solution Y. Show all working. 3 marks

4.2 Is the student’s claim correct? Justify your answer by comparing the number of moles in each solution and predicting whether the two solutions would behave identically or differently. 2 marks

Stuck? Convert N to n using n = N ÷ N_A, then compare n for both solutions.

Answers — Do not peek before attempting

Q1.1 — Completed table (7 marks)

A: entity = atom; N = 1.0 × 6.022 × 10²³ = 6.022 × 10²³ atoms.

B: entity = molecule; N = 0.50 × 6.022 × 10²³ = 3.011 × 10²³ molecules.

C: entity = formula unit (given); N = 2.0 × 6.022 × 10²³ = 1.204 × 10²⁴ formula units.

D: entity = atom; N = 0.25 × 6.022 × 10²³ = 1.506 × 10²³ atoms.

E: entity = molecule; N = 3.0 × 6.022 × 10²³ = 1.807 × 10²⁴ molecules.

F: entity = molecule; N = 0.10 × 6.022 × 10²³ = 6.022 × 10²² molecules.

Worked example (Sample C): Known: n = 2.0 mol, N_A = 6.022 × 10²³ mol⁻¹. Formula: N = n × N_A. Substitute: N = 2.0 × 6.022 × 10²³ = 1.204 × 10²⁴ formula units. [1 mark for a clear worked example]

Q1.2 — Greatest and fewest (2 marks)

Greatest: Sample E (3.0 mol O₂) with N = 1.807 × 10²⁴ molecules [1]. Fewest: Sample F (0.10 mol glucose) with N = 6.022 × 10²² molecules [1]. The sample with most moles always has the most entities because N = n × N_A and N_A is a fixed constant — so N is directly proportional to n. (Note: the type of particle does not affect N; only n matters.)

Q2.1 — Closest quantity to N_A (2 marks)

Stars in the observable universe (~10²²) is the closest in log scale to N_A (~6 × 10²³) among the comparison quantities [1]. N_A is approximately 1.78 orders of magnitude (about 60 times) larger than the number of stars [1]. (Accept answers in the range 1 to 2 orders of magnitude.)

Q2.2 — Seconds since Big Bang vs N_A (3 marks)

Seconds since Big Bang ≈ 4 × 10¹⁷. This is much smaller than N_A = 6.022 × 10²³ [1]. Ratio: N_A ÷ (4 × 10¹⁷) = 6.022 × 10²³ ÷ 4 × 10¹⁷ ≈ 1.5 × 10⁶ [1]. N_A is approximately 1.5 million times larger than the number of seconds since the Big Bang. They differ by about 6 orders of magnitude, meaning they are emphatically not comparable [1].

Q2.3 — Moles of water in 1 L (3 marks)

Known: N = 3.34 × 10²⁵ molecules; N_A = 6.022 × 10²³ mol⁻¹ [1].

Formula: n = N ÷ N_A [1].

n = 3.34 × 10²⁵ ÷ 6.022 × 10²³ = 55.5 mol [1].

Q3 — Comparing N and n

What it represents: N = the actual count of individual particles (atoms, molecules, etc.). n = the amount of substance grouped into moles.

Units: N = no units (dimensionless). n = mol.

Typical scale: N = very large numbers, typically 10²⁰–10²⁵. n = small numbers, typically 0.001–10 for lab amounts.

Symbol convention: N = capital letter (particle number). n = lowercase letter (amount in moles).

Example (2 mol C): N = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ atoms. n = 2 mol.

Q4.1 — Moles in Solution Y (3 marks)

Known: N = 9.033 × 10²³ formula units; N_A = 6.022 × 10²³ mol⁻¹ [1].