Chemistry • Year 11 • Module 1 • Lesson 1

Properties and Classification of Matter

Apply your classification skills to real measurement data, case studies and structured comparisons at Band 4–5 level.

Apply · Data & Reasoning

1. Interpret measurement data — classifying five mystery substances

A student measured properties of five laboratory samples using standard equipment. The table below records the results. 10 marks

Sample Appearance Melting point (°C) Boiling point changes with concentration? Can be decomposed chemically?
W Clear colourless liquid, uniform 0 — sharp, fixed No Yes — by electrolysis to H2 and O2
X Shiny orange solid, uniform 1085 — sharp, fixed No No
Y Clear liquid, uniform Varies with salt concentration Yes No (only by physical means)
Z White solid granules mixed with black powder; two distinct regions visible No single fixed point n/a Not as a unit (components retain individual properties)
V Yellow-green gas, uniform −101 — sharp, fixed No No

1.1 Complete the table below by classifying each sample and providing a one-sentence justification using only the data above. 7 marks

SampleClassificationJustification using the data
W
X
Y
Z
V

1.2 For samples W and Y: both are clear colourless (or near-colourless) liquids that look uniform. Identify the one piece of data in the table that most clearly distinguishes them, and explain what it tells you about each. 2 marks

1.3 Why can appearance alone never definitively determine whether a substance is a pure substance or a homogeneous mixture? Use samples from the table to support your answer. 1 mark

Stuck? Use the decision framework from Card 3: fixed composition? → can be broken down chemically? → is the mixture uniform?

2. Interpret graph — melting point as evidence of purity

A student heated three different 5 g samples (P, Q, R) in a capillary tube apparatus and recorded temperature vs time. The graph below shows the heating curves. 8 marks

20 36 52 68 84 100 0 50 100 150 200 250 300 Time (s) Temperature (°C) P: 80 °C plateau Q: no plateau (mixture) R: 63 °C plateau Sample P Sample Q Sample R

Figure 2. Heating curves for 5 g samples P, Q, R heated at a constant rate. Illustrative data.

2.1 For each sample, state whether it is most likely a pure substance or a mixture, and justify your answer using the shape of the heating curve. 3 marks (1 per sample)

2.2 Read the melting point of Sample P from the graph. Using the decision tree from the lesson, state whether Sample P is an element or compound, and explain what additional information you would need to determine this. 2 marks

2.3 A student concludes: “Sample Q is a heterogeneous mixture because its temperature keeps rising.” Identify one flaw in this reasoning and write a more accurate conclusion. 2 marks

2.4 Samples P and R both show flat plateaus. Suggest why their melting point temperatures differ. 1 mark

Stuck? Revisit the “exam trap” note: a sharp fixed melting point is a key indicator of a pure substance. The shape of the curve (plateau vs. gradual rise) is the evidence.

3. Compare element, compound and mixture across five features

Complete the three-column table below. For each feature, write a concise description that contrasts all three types. 10 marks)

FeatureElementCompoundMixture
Number of atom types
Is composition fixed?
Can be separated by physical means?
Example formula or name
Australian real-world example
Stuck? Revisit Card 2 (Pure Substances) and Card 3 (Mixtures) in the lesson. For Australian examples, think of minerals from Australian mines, ocean water, alloys used in industry.

4. Predict and justify — the Broken Hill ore body

The Broken Hill silver–lead–zinc ore body in far-western NSW contains several minerals mixed together, including galena (PbS), sphalerite (ZnS) and gangue (waste rock). Miners extract the ore and process it in a concentrator to separate the different minerals.

5 marks

4.1 Classify the raw ore body (PbS, ZnS, gangue mixed together) as one of: element, compound, homogeneous mixture, or heterogeneous mixture. Justify your classification using two pieces of evidence from the description. 3 marks

4.2 After smelting, a silver–lead alloy is produced. Predict the classification of this alloy and explain why its classification differs from that of galena (PbS). 2 marks

Stuck? Use the classification decision framework: fixed composition? → chemical bonds in fixed ratio? → uniform appearance?
Answers — Do not peek before attempting

Q1.1 — Classification table

W: Compound — sharp fixed melting point (pure substance); can be decomposed by electrolysis into two different elements (H2 and O2), so it is chemically breakable → compound. (This is water, H2O.) X: Element — sharp fixed melting point (pure substance); cannot be decomposed chemically; contains only one type of atom (copper, Cu). Y: Homogeneous mixture — boiling point changes with concentration (variable composition); looks uniform (one visible phase) → homogeneous mixture. (This is a salt solution.) Z: Heterogeneous mixture — two distinct visible regions (white solid + black powder), no single fixed melting point → heterogeneous mixture. V: Element — sharp fixed melting point (pure substance); cannot be decomposed chemically → element. (This is chlorine gas, Cl2.)

Q1.2 — Distinguishing W and Y (2 marks)

The most useful distinguishing piece of data is whether the boiling point changes with concentration [1]. Sample W has a fixed melting point of 0°C and a fixed boiling point, indicating it is a pure substance — its composition does not vary. Sample Y has a boiling point that changes with salt concentration, confirming its composition is variable and therefore it is a mixture [1].

Q1.3 — Appearance alone insufficient (1 mark)

Appearance alone cannot determine purity because a homogeneous mixture looks just as uniform as a pure substance [1]. Samples W (pure compound) and Y (homogeneous mixture) both appear as clear uniform liquids, yet they have fundamentally different classifications based on composition data. The appearance “uniform” refers to visual phase, not chemical identity.

Q2.1 — Classification from heating curves (3 marks)

Sample P — pure substance. The curve shows a horizontal flat plateau at 80°C, indicating all energy is being used to melt the substance at a fixed temperature. A sharp fixed melting point is characteristic of a pure substance [1]. Sample Q — mixture. The curve shows a gradual, continuous rise with no flat plateau. In a mixture, the melting occurs over a temperature range because different components melt at different temperatures, and the composition shifts as melting proceeds [1]. Sample R — pure substance. Like P, it shows a sharp plateau (at 63°C), indicating a pure substance with a distinct fixed melting point [1].

Q2.2 — Melting point of P; element or compound? (2 marks)

The melting point of Sample P reads as 80°C from the plateau on the graph [1]. Knowing that P is a pure substance, the heating curve cannot distinguish between an element and a compound. Additional information needed: the formula or composition of the substance (whether it contains one or two or more element types). If the formula has only one element symbol — element; if it has two or more different element symbols — compound [1]. Accept also: carrying out a chemical decomposition test (e.g. electrolysis) to determine if it can be broken down into simpler substances.

Q2.3 — Flaw in student reasoning (2 marks)

Flaw: the absence of a flat plateau indicates a mixture, but the type of mixture (heterogeneous vs homogeneous) cannot be determined from a heating curve alone. The student incorrectly uses a thermal property to determine the uniformity type [1]. More accurate conclusion: Sample Q is most likely a mixture (because it has no sharp melting point), but whether it is homogeneous or heterogeneous requires observing the sample directly — heating curve data alone cannot distinguish the two [1].

Q2.4 — Why plateaus differ (1 mark)

Samples P and R are different pure substances (with different chemical identities), so they have different melting points. Each pure substance has a characteristic, fixed melting point that is unique to that substance — this is one of the properties that defines and identifies it [1].

Q3 — Compare and contrast table

Atom types: Element — one type only. Compound — two or more types, bonded in fixed ratio. Mixture — two or more types, physically combined, not bonded. Fixed composition: Element — yes. Compound — yes (always the same ratio). Mixture — no (can vary). Physical separation: Element — no (no components to separate). Compound — no (would require breaking chemical bonds, i.e. chemical means). Mixture — yes (e.g. filtration, distillation, magnetic separation). Example formula: Element — Fe, Cu, O2, Cl2. Compound — NaCl, H2O, CO2. Mixture — ocean water, air, granite. Australian example: Element — copper ore (Cu) from Mount Isa mine. Compound — iron oxide (Fe2O3) ore at Pilbara, WA. Mixture — ocean water off Sydney coast (homogeneous); Broken Hill ore body (heterogeneous). Accept any valid Australian examples.

Q4.1 — Broken Hill ore body (3 marks)

The raw ore body is a heterogeneous mixture [1]. Evidence 1: multiple different mineral components are present (PbS, ZnS and gangue), so composition is not fixed and varies from location to location in the ore body [1]. Evidence 2: different minerals are described as “mixed together” with distinct visible regions (different coloured minerals), indicating non-uniform composition with multiple phases — the defining feature of a heterogeneous mixture [1]. Accept also: the minerals can be physically separated in a concentrator (flotation process), consistent with a mixture.

Q4.2 — Silver–lead alloy vs galena (2 marks)

The silver–lead alloy is a homogeneous mixture [1]. It differs from galena (PbS) because galena is a compound — two different elements (Pb and S) are chemically bonded in a fixed 1:1 ratio; it has a definite composition that cannot vary and can only be separated by chemical means. The alloy is a physical blend of silver and lead with no chemical bonds between Ag and Pb atoms; its composition can vary (different Ag:Pb ratios are possible) and the metals can in principle be separated by physical processes such as fractional distillation [1].