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

You have met the four boxes in the parent note: element, compound, homogeneous mixture, heterogeneous mixture. Knowing the definitions is one thing; classifying anything the exam throws at you is another. This page walks the entire "space of cases" and works one full example for every corner.

Before a single word: three vocabulary anchors, built from zero.

Definition The three questions that decide everything

To classify any lump of matter, ask up to three yes/no questions, always in this order:

Q1 — Is the composition fixed? "If I take any two samples, is the ratio of ingredients forced to be identical?" Fixed → pure substance (go to Q2). Variable (I can choose the ratio) → mixture (go to Q3).
Q2 — (only if pure) Can a chemical change break it into simpler substances? No → element. Yes → compound.
Q3 — (only if a mixture) Does it look the same everywhere? "If I zoom in with a microscope, do all regions look identical?" Yes, uniform → homogeneous. No, distinct patches → heterogeneous.

So: Q1 first, always. Q2 is for pure substances, Q3 is for mixtures — you never need both. Every example below is labelled with the Q-numbers it uses, so the framework stays consistent throughout the page.

Definition Chemical change vs physical change — what Q2 actually means

Q2 asks whether a chemical change can split the substance. These two words must be pinned down before we use them, because the whole element/compound/mixture split rides on the difference:

A physical change rearranges which particles sit where but leaves each molecule intact — boiling, filtering, evaporating, dissolving, melting. No bonds inside molecules break. This is what separates mixtures.
A chemical change (chemical reaction) breaks and remakes the bonds inside molecules, turning substances into genuinely different substances — electrolysis, burning, reacting with acid. New substances appear that were not there before. This is what decomposes compounds.

The tell-tale sign: after a physical change you can, in principle, get the original pieces back by another physical step; after a chemical change the original molecules are gone and only a reaction can restore them. So "can chemistry break it?" precisely means "does splitting it require breaking bonds inside molecules?"

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

Figure 1 — what it shows (for readers who can't see it): a top-to-bottom flowchart. The top box "Any matter" flows into Q1 (composition fixed?). A "fixed" arrow goes left to the box PURE SUBSTANCE, which then flows into Q2 (chemistry breaks it?), splitting into ELEMENT (arrow "no") and COMPOUND (arrow "yes"). A "variable" arrow goes right to the box MIXTURE, which flows into Q3 (looks the same?), splitting into HOMOGENEOUS (arrow "uniform") and HETEROGENEOUS (arrow "patchy"). Every piece of matter walks these arrows to exactly one of the four leaf boxes — there is no fifth box.

The scenario matrix

Here is the full space of "case classes" this topic can test. Each row is a distinct kind of situation; the last column names the worked example that nails it.

#	Case class	What makes it tricky	Worked example
A	Pure element, single atoms	is a lone gas an element?	Ex 1 — Neon
B	Pure element, molecules (diatomic)	"two atoms" fools you into "compound"	Ex 1 — Oxygen $O_{2}$
C	Pure compound, fixed ratio	new properties hide the ingredients	Ex 2 — Water
D	Homogeneous mixture, variable ratio	uniform but not fixed	Ex 3 — Saltwater
E	Heterogeneous mixture, visible phases	settles / layers	Ex 4 — Sand in water
F	Degenerate: one component only	a mixture of "just water" is pure	Ex 5 — trick sample
G	Limiting case: infinitely dilute	when does a mixture stop being a mixture?	Ex 5
H	Separation test (physical vs chemical)	filter vs electrolysis confirms the box	Ex 6 — the separation test
I	Real-world word problem	messy language, extract the chemistry	Ex 7 — the smoothie
J	Exam twist: same formula, two answers	$O_{3}$ vs $O_{2}$ , air vs oxygen	Ex 8
K	Quantitative: Law of Definite Proportions	prove "fixed ratio" with numbers	Ex 9 — mass ratio
L	Boundary: alloy = solid solution	a metal that is a mixture	Ex 10 — brass

The rest of the page fills these cells in order.

Ex 1 — Elements: single atoms and molecules (cells A, B) · uses Q1, Q2

Forecast: guess now. Many students say " $O_{2}$ has two atoms, so it's a compound." Write down your guess before reading on.

Q1 (composition fixed?) for neon. Why this step? Q1 is always the first fork; pure-vs-mixture comes before anything else. Neon everywhere is neon — 10 protons per atom, no choice of ratio. Fixed → pure substance.
Q2 (can a chemical change break it into simpler substances?) for neon. Why this step? For a pure substance, Q2 splits element vs compound. There are no bonds inside a neon atom to break by any reaction — one type of atom only. → element.
Now $O_{2}$ : Q1 then Q2. Why this step? Q1: fixed (always $O_{2}$ ) → pure. Q2 is where the two-atom worry lives. Both atoms are the same type (oxygen). A compound needs two or more different elements bonded. $O_{2}$ = oxygen + oxygen = still one element. The molecule is just how the element likes to travel.

Figure 2 — what it shows: two molecular sketches side by side. On the left, a single lone violet sphere labelled "Ne — one atom, one kind" (monatomic). On the right, two orange spheres joined by a bond labelled "O₂ — two atoms, same kind." A caption bar underneath reads "SAME kind of atom ⇒ still an ELEMENT." The extra insight over the text: it makes visible that "two spheres" alone never means compound — what would make a compound is two spheres of different colours bonded, which is drawn faded-out and crossed through as the "NOT here" case.

Answer: both are elements (cell A: monatomic; cell B: diatomic molecule).

Verify: neon molar composition = 100% Ne; $O_{2}$ molar composition = 100% O. A compound would require a second element with more than $0%$ . Both fail that, so both are elements. ✓

Ex 2 — A compound with hidden ingredients (cell C) · uses Q1, Q2

Forecast: pure or mixture? Element or compound?

Q1 (ratio fixed?) Every water molecule is $H_{2} O$ — exactly 2 H to 1 O. You cannot buy "extra-hydrogen water." Why this step? Fixed ratio ⇒ pure substance, not a mixture.
Q2 (chemical change able to decompose it?) Yes — electrolysis is a chemical change: it breaks the H–O bonds inside the molecule, producing $H_{2}$ and $O_{2}$ (substances not present before). Why this step? Being splittable by a chemical change (not by boiling or filtering) is the signature of a compound.
Explain the property change. Why this step? This is the conceptual heart. When H and O bond, their electrons rearrange into new molecular orbitals — a genuinely new substance with its own properties. The ingredients' properties are gone, not averaged.

Figure 3 — what it shows: one bent $H_{2} O$ molecule on the left (one navy central O sphere with two small magenta H spheres, bonds drawn), a lightning-bolt arrow labelled "electrolysis = CHEMICAL change (breaks inside-molecule bonds)" pointing right, and on the right two separate $H_{2}$ molecules plus one $O_{2}$ molecule. Beneath, a contrast strip: "boiling / filtering = physical: molecule survives, no split" crossed through. The added insight: it visually separates bond-breaking (chemical, splits the compound) from particle-shuffling (physical, would not), which is exactly the Q2 distinction defined at the top.

Answer: water is a compound (cell C).

Verify: the decomposition $2 H_{2} O \to 2 H_{2} + O_{2}$ conserves atoms — left side: 4 H, 2 O; right side: 4 H, 2 O. Balanced ⇒ consistent with a fixed-formula compound. ✓

Ex 3 — Homogeneous mixture, variable ratio (cell D) · uses Q1, Q3

Forecast: guess the box before step 1.

Q1 (fixed ratio?) No — you chose 5 g vs 20 g. Why this step? Freedom to pick the ratio ⇒ mixture, immediately ruling out element and compound.
Q3 (uniform everywhere?) Yes — dissolved salt is spread at the ionic scale; any dropper-full has the same saltiness. Why this step? Q3 is the homogeneous/heterogeneous fork for mixtures. Uniform ⇒ homogeneous mixture (a solution).
Confirm variability didn't break uniformity. Why this step? Students confuse "variable" (Q1) with "non-uniform" (Q3). A solution can be any concentration you like (variable) yet still be perfectly even within one jar (uniform). Two different ideas.

Figure 4 — what it shows: a single jar of saltwater with a magnifying-glass inset. Inside the inset, magenta "+" spheres (Na⁺ ions) and violet "−" spheres (Cl⁻ ions) are scattered evenly among orange water dots — no clumps, no layers. Two small dropper samples taken from top and bottom are drawn with equal numbers of ions, labelled "same ratio everywhere ⇒ UNIFORM." The added insight: it shows why dissolving at the ionic scale looks the same in every sample — the reason Q3 answers "homogeneous," which pure text cannot picture.

Answer: both jars are homogeneous mixtures (cell D) — just at different concentrations. See Solutions and concentration units for how we pin the number down.

Verify: mass fraction of salt, jar 1: $5/ (5 + 100) = 0.0476 = 4.76%$ . Jar 2: $20/ (20 + 100) = 0.1667 = 16.67%$ . Different numbers ⇒ variable composition ⇒ mixture, confirmed. ✓

Ex 4 — Heterogeneous: you can see the parts (cell E) · uses Q1, Q3

Forecast: homogeneous or heterogeneous? Faster or slower than 1 mm/s?

Q1 then Q3 first. Ratio is your choice (mixture, from Q1); and you can see grains distinct from water (non-uniform, so Q3 says heterogeneous). Why this step? The framework classifies it before any physics ⇒ heterogeneous mixture locked in.
Quantify "settles" with Stokes' Law. Why this tool and not another? Stokes' Law is the one formula that tells us whether particles stay suspended (would look homogeneous) or sink (visibly heterogeneous). It answers "will thermal jiggling or gravity win?" $v = \frac{2 r ^{2} ( ρ _{p} - ρ _{f} ) g}{9 η}$ Every symbol earned: $v$ = sinking speed (m/s); $r$ = grain radius; $ρ_{p} - ρ_{f}$ = density difference, grain minus water; $g = 9.8$ m/s² (gravity); $η$ = water's viscosity (its "thickness"), about $1 0^{- 3}$ Pa·s. The symbol $\approx$ below just means "approximately equals."
Plug numbers. $r = 50 μ m = 5 \times 1 0^{- 5}$ m, $ρ_{p} - ρ_{f} = 1650 - 1000 = 650$ kg/m³ (sand vs water). $v = \frac{2 ( 5 \times 1 0 ^{- 5} ) ^{2} ( 650 ) ( 9.8 )}{9 ( 1 0 ^{- 3} )} \approx 3.5 \times 1 0^{- 3} m/s$

Answer: heterogeneous (cell E), sinking at about $3.5$ mm/s — visibly fast, so layers form in seconds. Compare a 1 μm particle (parent note) that settles $1 0^{5}$ times slower and stays cloudy for hours.

Figure 5 — what it shows: a beaker of water with orange sand grains piled at the bottom (the settled layer) and one magenta grain mid-fall with a downward arrow labelled "v ≈ 3.5 mm/s." The right half writes out Stokes' Law with the three plugged-in numbers ( $r$ , density difference, viscosity) and the punchline "big $v$ ⇒ visible layers ⇒ HETEROGENEOUS." The new insight it adds over the text: it makes visible why a large settling speed is exactly what turns a shaken mixture into a layered, heterogeneous one.

Verify: the VERIFY block computes $v = \frac{2 ( 5 \times 1 0 ^{- 5} ) ^{2} ( 650 ) ( 9.8 )}{9 \times 1 0 ^{- 3}}$ and checks it is about $3.5 \times 1 0^{- 3}$ m/s. ✓

Ex 5 — Degenerate & limiting cases (cells F, G) · uses Q1

Forecast: these two "edge" cases catch almost everyone.

(a) Count the substances. A "mixture" requires two or more distinct substances. Distilled water is one compound. Why this step? The degenerate case (one ingredient) makes Q1 answer "fixed" — it collapses "mixture" back to pure compound, not a mixture.
(b) Is one molecule a real second component? Two different substances are present (water + sugar), so it is a mixture — however tiny the amount. Why this step? The limiting case (amount of sugar shrinking toward zero) shows the definition counts types, not how much. Even nearly-zero sugar means 2 types ⇒ mixture.
Reconcile with fixed-ratio. Why this step? One-molecule-of-sugar water still has variable composition (add a second molecule and the ratio changes), so Q1 says "variable" ⇒ mixture, and it never becomes a compound.

Figure 6 — what it shows: two panels sharing a "count the types" number line. Left panel (a): a beaker of pure water, all orange dots, big label "types = 1" with "1 is NOT ≥ 2 ⇒ PURE." Right panel (b): a vast ocean of orange water dots with a single lone magenta sugar dot circled and arrowed, label "types = 2" with "2 IS ≥ 2 ⇒ MIXTURE." The added insight: it pictures that the definition is a count of kinds, not an amount — one lonely molecule flips the count from 1 to 2 even though the quantity is almost nothing.

Answer: (a) pure compound (cell F — the one-component degenerate case). (b) homogeneous mixture, no matter how dilute (cell G).

Verify: for (a) the count is 1, and "1 ≥ 2" is false ⇒ not a mixture. For (b) the count is 2, and "2 ≥ 2" is true ⇒ mixture. ✓

Ex 6 — The separation test confirms the box (cell H) · uses Q1, Q2, Q3

Forecast: which is the mixture, which is pure?

Q1 for Sample X (composition fixed?) — boiling left a residue, so a physical method separated it into two substances. Why this step? Physical separation working means the composition was not forced ⇒ Q1 answers "variable" ⇒ mixture. Q2 is not reached (Q2 is only for pure substances).
Q3 for Sample X (uniform?) — before boiling it looked clear and even throughout. Why this step? For a mixture we go to Q3; uniform ⇒ homogeneous mixture (saltwater). See Separation techniques in chemistry.
Q1 then Q2 for Sample Y. Why this step? No physical method (boiling) splits Y — it returns identical ⇒ Q1 answers "fixed" ⇒ pure substance. Now Q2: splitting it further would need electrolysis, a chemical change breaking inside-molecule bonds ⇒ Q2 answers "yes" ⇒ pure compound (water). Q3 is not reached (Q3 is only for mixtures).

Answer: X = homogeneous mixture (cell H — the separation test answers Q1: yes, separable). Y = pure compound (physical test: no split; Q2: chemical change needed).

Verify: logic — physically separable ⇒ mixture. X separable = true ⇒ mixture true. Y separable = false ⇒ mixture false. ✓

Ex 7 — Real-world word problem (cell I) · uses Q1, Q3

Forecast: one box for the whole thing? Or does milk get its own?

Scan the smoothie: Q1 then Q3. Ingredients in any ratio (mixture, Q1); strawberry seeds and pulp chunks are visibly separate regions. Why this step? Q3 (uniform?) fails at eye level ⇒ heterogeneous mixture.
Zoom into the milk alone: Q3 at the micron scale. Milk = fat droplets (about $1 μ m$ ) suspended in water — a colloid. Why this step? It looks uniform to the naked eye but scatters light (the Tyndall effect); under a microscope it is non-uniform ⇒ Q3 says heterogeneous (a borderline case, resolved by choosing the right scale).
State the rule you used. Why this step? Classification depends on the scale of observation you're asked about. The whole smoothie: obviously heterogeneous. Milk on its own: heterogeneous colloid.

Answer: smoothie = heterogeneous mixture; milk = heterogeneous colloid (cell I — messy real language, two nested answers).

Verify: visible distinct regions present ⇒ Q3 gives heterogeneous = true for both. Consistent. ✓

Ex 8 — Exam twist: same element, different answers (cell J) · uses Q1, Q2, Q3

Forecast: are these the same category?

$O_{3}$ : Q1 then Q2. Q1: fixed formula (always 3 O atoms) ⇒ pure. Q2: only one type of element, no chemical change makes it a simpler substance (splitting $O_{3}$ just gives more oxygen) ⇒ pure element (an allotrope of oxygen). Why this step? The twist: $O_{2}$ and $O_{3}$ are different substances but both pure elements — many atoms, all oxygen.
Air: Q1. $N_{2}$ about 78%, $O_{2}$ about 21%, plus $Ar$ , $CO_{2}$ — the ratio varies (humidity, altitude). Why this step? Variable ⇒ mixture, so we go to Q3, not Q2.
Air: Q3. Gases mingle at the molecular level, uniform. → homogeneous mixture.

Answer: $O_{3}$ = pure element; air = homogeneous mixture (cell J — "oxygen" language, two totally different boxes).

Verify: $O_{3}$ distinct-element count = 1 ⇒ element. Air distinct-substance count is 2 or more ⇒ mixture. ✓

Ex 9 — Quantitative Law of Definite Proportions (cell K) · uses Q1 with numbers

Forecast: guess Lab B's hydrogen mass.

Find Lab A's mass fraction of hydrogen. $1.008/9.00 = 0.1120$ . Why this step? A fixed fraction is the numerical signature of a compound (Law of Definite Proportions). A mixture would give a different fraction per sample.
Apply the same fraction to Lab B. Why this step? If (and only if) it's a compound, the fraction is a constant of nature. Lab B hydrogen $= 0.1120 \times 18.00 = 2.016$ g.
Cross-check the H:O mass ratio. Oxygen in Lab A $= 9.00 - 1.008 = 7.992$ g. Ratio H:O $= 1.008 : 7.992 = 1 : 7.93$ . Why this step? This matches the parent note's $1 : 7.94$ — the tiny gap is rounding. A constant ratio ⇒ Q1 answer is "fixed" ⇒ compound confirmed.

Answer: Lab B releases 2.016 g of hydrogen; the constant $11.20%$ fraction proves water is a compound (cell K). This is the arithmetic backbone of Stoichiometry and the mole concept.

Verify: the VERIFY block checks $1.008/9.00 = 0.112$ , then $0.112 \times 18.00 = 2.016$ , and the ratio about $7.93$ . ✓

Ex 10 — Boundary case: an alloy is a solid mixture (cell L) · uses Q1, Q3

Forecast: solids feel "fixed" — is brass a compound?

Q1 (fixed ratio?) No — "brass" spans a range of Cu:Zn (musical-instrument brass differs from cartridge brass). Why this step? Variable ratio ⇒ mixture, even though it's a hard solid.
Q3 (uniform at atomic scale?) Zn atoms sit randomly on Cu lattice sites, evenly throughout. Why this step? Uniform ⇒ homogeneous mixture — a solid solution, same box as saltwater (cell D), not sand-water (cell E).
Contrast with a true copper compound. Why this step? $CuO$ (copper oxide) is a compound: fixed 1:1 Cu:O, new properties. Brass has no fixed formula — that's the whole difference. Bonding ideas live in Chemical bonding basics.

Answer: brass = homogeneous mixture (solid solution, cell L). Solids can be mixtures too.

Verify: two different valid Cu fractions (say 0.63 and 0.70) both count as "brass" ⇒ variable ⇒ mixture. The statement $0.63 \neq = 0.70$ is true. ✓

Recall

Recall The three questions

Q1 asks ::: Is the composition fixed? (fixed → pure substance → Q2; variable → mixture → Q3).

Recall Chemical vs physical change

Q2 asks whether a chemical change splits the substance — meaning ::: breaking the bonds inside molecules (boiling/filtering are physical and do not count).

Recall Diatomic ≠ compound

Is $O_{2}$ a compound? ::: No — both atoms are the same element, so it is a pure element.

Recall Separation fingerprint

If a sample splits into components by physical means (boiling, filtering), it must be a ::: mixture (compounds need a chemical change — that is Q1 answered in the lab).

Recall Alloy

Brass (Cu + Zn, variable ratio, uniform) is classified as a ::: homogeneous mixture (solid solution).

Recall Definite proportions number

The H:O mass ratio in every water sample is about ::: 1 : 7.94.