Pure substances vs mixtures — elements, compounds, homogeneous - heterogeneous: Exercises

This page is a ladder. Each rung is harder than the last, and every problem tells you its level:

L1 Recognition — can you name the category?
L2 Application — can you use a rule on numbers or descriptions?
L3 Analysis — can you reason from evidence to a conclusion?
L4 Synthesis — can you combine several ideas at once?
L5 Mastery — can you handle a tricky, real-world, edge-case scenario?

Try each one on paper first. Then open the collapsible Solution to check yourself. Everything here builds on the parent topic.

Intuition The one map you need

Before the questions, hold this decision tree in your head. Every problem below is just a walk down these branches.

Read the flowchart top to bottom. You start at "a sample of matter". The first diamond asks "is the composition fixed everywhere?" — and here everywhere means "in every tiny region you could sample, right down to the molecular scale." If yes, you are on the pure substance branch; a second question, "is there only one kind of atom?", splits that into element (one kind) or compound (two or more kinds bonded in a fixed ratio). If the composition is not fixed, you are on the mixture branch; the question "does it look the same everywhere at the scale of the eye and molecules?" splits it into homogeneous (uniform) or heterogeneous (visible different regions).

Two questions decide everything: "is the composition fixed?" (pure vs mixture) and then either "one type of atom?" (element vs compound) or "uniform to the eye/molecular scale?" (homogeneous vs heterogeneous).

The figure below re-draws exactly this logic as a labelled blueprint chart — use it as the master key you return to for every problem on this page. Notice the amber diamonds are the two deciding questions, and the cyan boxes are the four possible answers (element, compound, homogeneous, heterogeneous). The yes/no arrows are the paths you walk.

Figure — Pure substances vs mixtures — elements, compounds, homogeneous - heterogeneous

Figure 1 — The classification decision map: two amber questions ("fixed composition?" then either "one kind of atom?" or "uniform everywhere?") route any sample to one of four cyan outcomes.

Level 1 — Recognition

Problem 1.1

Classify each as element, compound, or mixture: (a) Helium gas (He) (b) Table salt (NaCl) (c) Italian salad dressing (oil + vinegar) (d) Pure copper wire (Cu) (e) Carbon dioxide (CO₂)

Recall Solution 1.1

Walk the tree in Figure 1: first ask "fixed composition?", then split.

(a) He — element. One type of atom (2 protons). Cannot be broken by chemistry.
(b) NaCl — compound. Two different elements chemically bonded in a fixed 1:1 ratio.
(c) oil + vinegar — mixture (heterogeneous — the layers separate). Variable ratio, no chemical bonds between oil and vinegar.
(d) Cu — element. One type of atom (29 protons).
(e) CO₂ — compound. Two elements (C and O) bonded in a fixed 1:2 ratio.

Problem 1.2

For each, say whether the mixture is homogeneous or heterogeneous: (a) Air (b) Sand in water (c) Filtered seawater (d) Chocolate-chip cookie

Recall Solution 1.2

Ask: "if I zoom into any tiny region, is the ratio the same?" (the right-hand branch of Figure 1)

(a) Air — homogeneous. Gases mix at the molecular level; every breath is ~78% N₂.
(b) Sand in water — heterogeneous. You can see two distinct phases; sand settles.
(c) Filtered seawater — homogeneous. Salt is dissolved molecule/ion-by-ion; uniform throughout.
(d) Cookie — heterogeneous. Chips sit in specific spots; a bite of dough ≠ a bite of chip.

Level 2 — Application

Problem 2.1

A sample labelled "water" contains 2.02 g of hydrogen and 16.00 g of oxygen. Pure water has a hydrogen-to-oxygen mass ratio of 0.126 : 1 (Law of Definite Proportions). Is this sample pure water?

Recall Solution 2.1

What we do: compute the actual mass ratio and compare to the fixed value. Why: a compound has one fixed ratio; a match means "consistent with pure compound," a mismatch means "mixture or impure." $ratio = \frac{2.02}{16.00} = 0.1263$ This equals $0.126$ to three significant figures. Yes — consistent with pure water. The 2:1 atom ratio (H₂O) forces this mass ratio, no matter where the water came from.

Problem 2.2

Seawater is about 3.5% salt by mass. Physiological saline is 0.9% salt by mass. Both are called "saltwater." Are they the same substance? What does this tell you about their classification?

Recall Solution 2.2

What we do: compare the compositions. $3.5% \neq = 0.9%$ — the ratio of salt to water is different. Why this matters: a compound is locked to one composition. Because saltwater comes in a range of compositions and still counts as "saltwater," it cannot be a compound. It is a homogeneous mixture (a solution). The variability is the fingerprint of a mixture.

Problem 2.3

Brass is roughly 70% Cu and 30% Zn by mass. Estimate its density given $ρ_{Cu} = 8.96 g/cm^{3}$ and $ρ_{Zn} = 7.14 g/cm^{3}$ .

Recall Solution 2.3

What we do — and why densities don't simply mass-average. Density $ρ$ is mass divided by volume, $ρ = m / V$ . When two metals mix, it is the volumes that add (each atom still takes up its own room in the lattice), so the natural quantity to add is specific volume $1/ ρ$ (volume per gram), weighted by mass fraction.

Let $w_{Cu} = 0.70$ and $w_{Zn} = 0.30$ be the mass fractions. Take a 1 g sample. Its total volume is the sum of the volume each component contributes: $V_{total} = \frac{w _{Cu}}{ρ _{Cu}} + \frac{w _{Zn}}{ρ _{Zn}} = \frac{0.70}{8.96} + \frac{0.30}{7.14} = 0.07813 + 0.04202 = 0.12015 cm^{3} .$ Since we used 1 g, the mixture density is $ρ = \frac{1 g}{0.12015 cm ^{3}} = 8.32 g/cm^{3} .$ Why this is the honest answer, not a straight mass-average of densities. A mass-weighted average of the densities ( $0.70 \times 8.96 + 0.30 \times 7.14 = 8.41$ ) is only an approximation that happens to be close here because the two densities are similar. The correct rule — averaging specific volumes by mass, because volumes are what physically add — gives $8.32 g/cm^{3}$ . Either way, the value lands in the measured range for brass (~8.3–8.5 g/cm³), and the point stands: brass's density is a blend of its components' values. That blendability (properties are weighted averages) is the fingerprint of a solid solution (homogeneous mixture), not a compound with brand-new properties.

Level 3 — Analysis

Problem 3.1

You have an unknown clear liquid. You boil it completely dry and find a solid residue left behind. Before boiling, the liquid looked uniform. What can you conclude, and why?

Recall Solution 3.1

Reasoning chain:

It looked uniform → homogeneous.
Boiling (a physical method) separated it into two parts (vapour + residue) → separable physically → mixture, not a pure substance.
Uniform + separated by physical means → a homogeneous mixture (solution) of a dissolved solid in a liquid.

Why boiling is decisive: physical methods only separate mixtures. A pure compound would boil away entirely (or decompose chemically), never leaving a different residue by simple evaporation. See Separation techniques in chemistry.

Problem 3.2

Two clear liquids are mixed. After a minute they form two distinct layers. Explain, in energy terms, why this happens instead of them staying mixed.

Recall Solution 3.2

What we invoke — the full Gibbs criterion, not just energy. Whether things stay mixed is decided by the Gibbs free energy of mixing: $Δ G_{mix} = Δ H_{mix} - T Δ S_{mix},$ where $Δ H_{mix}$ is the enthalpy (the heat/bond-energy change), $T$ is the absolute temperature, and $Δ S_{mix}$ is the entropy change (how much more "spread out / disordered" the arrangement becomes). Mixing happens spontaneously only when $Δ G_{mix} < 0$ . Why both terms matter:

Mixing always increases entropy ( $Δ S_{mix} > 0$ ): scattering two liquids into each other creates more possible arrangements. The $- T Δ S_{mix}$ term therefore pushes toward mixing. This is why most things mix at all.
But for a polar + nonpolar pair (oil + water), forcing them together breaks favourable polar–polar attractions and forms weak, costly polar–nonpolar contacts, so $Δ H_{mix}$ is large and positive.
When that positive $Δ H_{mix}$ outweighs the favourable $- T Δ S_{mix}$ , the total $Δ G_{mix} > 0$ → mixing is not spontaneous → the liquids re-separate into layers even after shaking.

Result: distinct phases with a visible boundary → heterogeneous mixture. Contrast with sugar in water, where $Δ H_{mix}$ is small and entropy wins, so $Δ G_{mix} < 0$ and it stays uniform.

Problem 3.3

A 1 μm (micrometre, one-millionth of a metre) dust particle sits in water at 20 °C. Using Stokes' Law, its settling speed is $v = \frac{2 r ^{2} ( ρ _{p} - ρ _{f} ) g}{9 η},$ where $r$ is the particle radius, $ρ_{p} - ρ_{f}$ is how much denser the particle is than the fluid, $g$ is gravity, and $η$ (the Greek letter eta) is the viscosity — a measure of how "thick" or resistant-to-flow the fluid is (water is thin, honey is thick). Use $r = 1 \times 1 0^{- 6} m$ , $ρ_{p} - ρ_{f} = 1000 kg/m^{3}$ , $g = 9.8 m/s^{2}$ , $η = 1 \times 1 0^{- 3} Pa\cdots$ . Compute $v$ in mm/hr, and say what it implies for classification.

Recall Solution 3.3

What we do: plug in and convert units. $v = \frac{2 ( 1 0 ^{- 6} ) ^{2} ( 1000 ) ( 9.8 )}{9 ( 1 0 ^{- 3} )} = \frac{2 \times 1 0 ^{- 12} \times 9800}{9 \times 1 0 ^{- 3}} \approx 2.18 \times 1 0^{- 6} m/s$ Convert to mm/hr — multiply by $1000 mm/m \times 3600 s/hr$ : $v \approx 2.18 \times 1 0^{- 6} \times 3.6 \times 1 0^{6} \approx 7.8 mm/hr .$

Why this calculation decides the classification. The whole homogeneous-vs-heterogeneous split is a question of scale of uniformity. A true (homogeneous) solution stays perfectly mixed because thermal jiggling completely overwhelms gravity for molecule-sized particles — nothing ever settles. Stokes' Law tells us how fast a given size of particle sinks under gravity. If the settling speed is negligible (particles never fall out), the sample can act homogeneous; if the speed is appreciable and visible (like the $7.8 mm/hr$ here), the particles physically separate into a bottom layer over minutes-to-hours. That visible separation is exactly what "heterogeneous" means. So computing $v$ turns a fuzzy "does it look uniform?" into a concrete number.

Interpretation: $7.8 mm/hr$ is slow but visible over minutes-to-hours → the particle forms a settling layer → the system is a heterogeneous suspension, not a true solution.

The figure below plots this Stokes speed against particle radius, so you can see the whole picture: tiny (molecular) particles settle immeasurably slowly (homogeneous), while micrometre-and-larger particles cross into visible-settling (heterogeneous) territory. The amber dot marks the exact $1 μ m$ case from this problem.

Figure 2 — Stokes settling speed vs particle radius (log–log). Below the dashed 1 mm/hr line settling is effectively invisible (behaves homogeneous); above it particles form visible layers (heterogeneous). Amber dot = the 1 μm particle of Problem 3.3 at ~7.8 mm/hr.

Level 4 — Synthesis

Problem 4.1

You are given a bottle of clear liquid. Design a two-test procedure that tells you whether it is (i) an element/compound (pure), (ii) a homogeneous mixture, or (iii) a heterogeneous mixture — and explain what each outcome means.

Recall Solution 4.1

Test 1 — Look / let it stand / centrifuge (checks uniformity).

If it separates into visible layers or particles settle → heterogeneous. Done.
If it stays perfectly uniform → go to Test 2.

Test 2 — Evaporate/distil a sample (checks fixed vs variable composition).

If it boils away with no residue and a single sharp boiling point → likely a pure substance.
If it leaves a residue or the boiling temperature drifts as it evaporates → composition is changing → homogeneous mixture.

Why this order works: Test 1 uses the cheap uniformity check to catch heterogeneous cases; Test 2 uses a physical separation (only mixtures respond) plus the "fixed boiling point" signature of pure substances. Together they land you in exactly one branch of the decision tree.

Problem 4.2

A student claims: "Air is a compound because it always contains N₂ and O₂." Give three independent pieces of evidence that air is actually a homogeneous mixture.

Recall Solution 4.2

Variable composition. Humidity and CO₂ vary daily; a compound's composition is fixed. Air's isn't.
Physical separability. Fractional distillation of liquid air separates N₂, O₂, Ar by boiling point — a physical method. Compounds need chemical methods (see Separation techniques in chemistry).
Additive properties. Air's density is the weighted average of its components' densities; the N₂ and O₂ keep their own chemical identities (each behaves as itself). A compound would show brand-new properties unlike any component.

Problem 4.3

Carbon dioxide is 27.3% carbon by mass ( $C = 12.01$ , $O = 16.00$ ). Verify this from the formula CO₂, and explain how this single fixed number distinguishes a compound from a mixture.

Recall Solution 4.3

What we do: compute the mass fraction of carbon in one CO₂ formula unit. $% C = \frac{12.01}{12.01 + 2 ( 16.00 )} \times 100% = \frac{12.01}{44.01} \times 100% = 27.29%.$ This matches 27.3%. Why it distinguishes: every pure CO₂ sample gives exactly this percentage — that's the Law of Definite Proportions. A mixture of C and O₂ could be 5% carbon, 50% carbon, anything. A single, reproducible mass percent is the signature of a compound. This links directly to Stoichiometry and the mole concept.

Level 5 — Mastery

Problem 5.1

Bronze (Cu + Sn), ozone (O₃), a colloid of milk, and steam (H₂O gas). For each, give the full classification (element / compound / homogeneous mixture / heterogeneous mixture) and defend the trickiest call.

Recall Solution 5.1

Bronze — homogeneous mixture (solid solution / alloy). Variable Cu:Sn ratio, metallic atoms blended in a lattice, physically separable in principle.
Ozone (O₃) — element. Three atoms, but all oxygen → still one type of atom. (It's an allotrope of oxygen — same element, different arrangement.)
Milk — heterogeneous mixture (colloid). Fat droplets and protein micelles are suspended; under a microscope you see distinct particles. It only looks uniform to the naked eye.
Steam — compound (in the gas phase). Still H₂O molecules, fixed 2:1 atom ratio. Changing phase doesn't change classification (see Phase diagrams and phase changes).

Trickiest call — milk. By eye it looks like a uniform white liquid (tempting "homogeneous"). But the defining test is scale of uniformity: its particles are ~0.1–1 μm — big enough to scatter light (why milk is opaque) and to be counted as separate phases. That scattering (the Tyndall effect) is the tell → heterogeneous colloid.

Problem 5.2

An unknown white solid melts sharply at exactly 801 °C every time and, when melted and electrolysed, yields a soft metal and a yellow-green gas. A second white solid melts over a range (e.g. 120–140 °C) and yields nothing when electrolysed but can be separated by dissolving and re-crystallising. Classify each and justify with all the evidence.

Recall Solution 5.2

First solid — a compound (it's NaCl):

Sharp, reproducible melting point (801 °C) → pure substance (mixtures melt over a range).
Electrolysis (a chemical method) breaks it into a metal (Na) and a gas (Cl₂) → it was made of two different elements bonded → compound.

Second solid — a homogeneous mixture:

Melts over a range (120–140 °C) → its components melt at different temperatures, so no single sharp melting point → not a pure substance → mixture.
Electrolysis yields nothing → there are no ionic bonds to break apart into elements → it was never a single bonded compound; it's just separate substances sitting together.
Separable by dissolving and re-crystallising (a physical method) → confirms mixture, because physical methods only separate mixtures, never compounds.
It appeared uniform and dissolved cleanly (no visible phases or settling) → uniform at the molecular scale → homogeneous mixture of two (or more) solids.

The unifying logic: sharp melting point + chemical-only separation (electrolysis works) = compound; melting range + physical separation works + electrolysis does nothing = homogeneous mixture. The melting behaviour and the type of separation that works are the two most reliable diagnostics — exactly the two questions in Figure 1.

Recall Quick self-test (cloze)

A pure substance made of one type of atom is an element. A pure substance made of two or more elements bonded in a fixed ratio is a compound. A mixture uniform down to the molecular scale is homogeneous. The law that fixes a compound's mass ratios is the Law of Definite Proportions. Ozone O₃ is classified as an ::: element (an allotrope of oxygen) Milk is classified as a ::: heterogeneous mixture (a colloid)

Where this connects

The why behind fixed ratios: Atomic theory and structure and Chemical bonding basics.
Turning mass percents into amounts: Stoichiometry and the mole concept.
Splitting mixtures in the lab: Separation techniques in chemistry.
Concentration of solutions: Solutions and concentration units and Coligative properties.