Physical and Chemical Properties and Changes

Every substance carries a set of traits that describe it. It also has behaviors that transform it. Chemists sort these traits into physical properties and chemical properties. They sort the transformations into physical changes and chemical changes. This distinction is one of the very first filters a chemist applies. It applies when looking at any sample of matter. This holds whether it is a chunk of Elements and Compounds or a mixture encountered in the lab. Getting this sorting right matters. It determines what kind of question you are even allowed to ask. If you want to know a substance's melting point, you are asking a physical question. If you want to know whether it will rust, burn, or neutralize an acid, you are asking a chemical question. The same logic separates intensive properties, which do not depend on how much material you have, from extensive properties, which do. This article builds that vocabulary carefully, with numeric examples. This lets it be applied without hesitation later. That includes choosing how to separate mixtures, a topic explored under Separation Techniques.

Key takeaways

A physical property can be measured without changing the chemical identity of a substance (density, melting point, color, hardness).
A chemical property describes how a substance reacts to form a different substance (flammability, reactivity with acid, corrosiveness).
A physical change alters form or state but keeps the same molecules (melting ice, dissolving sugar, crushing chalk).
A chemical change breaks and forms bonds, producing new substances with new properties (rusting, burning, digestion).
Intensive properties (density, temperature, concentration) stay constant regardless of sample size; extensive properties (mass, volume, total energy) scale with the amount of matter.
Key idea: identity of the substance is the dividing line — physical properties/changes never alter it, chemical properties/changes always do.

The core insight

Ask one question to classify anything: after this property is measured, or after this change happens, is the substance still made of the same molecules? Yes → physical. No → chemical.

The physical picture

Consider a silver spoon. Its shininess, its density (10.49 g/cm³), and the temperature at which it melts (961.8 °C) are all physical properties — you can measure every one of them and still have pure silver metal afterward, atom for atom the same Ag atoms bonded the same way. Now suppose the spoon reacts with hydrogen sulfide gas in the air and slowly grows a dull black film. That film is silver sulfide, Ag₂S, a genuinely different substance with different composition and different properties. This is a chemical change, and the fact that silver can tarnish in the presence of sulfur compounds is a chemical property of silver. The same logic applies at the particle level. In a physical change, the collection of molecules or ions is reshuffled in position, spacing, or motion — solid to liquid, dissolved to dispersed, whole to shattered — but each particle retains its own atoms in the same bonding arrangement. In a chemical change, existing bonds break and new bonds form, so the particles themselves become different chemical species, often detectable by a color change, gas bubble, temperature change, precipitate, or light emission. A second, orthogonal distinction runs alongside this one: does a property depend on how much substance you have? Mass and volume obviously do — double the sample, double the mass. But density, the ratio of mass to volume, does not; a 5 g chip of silver and a 500 g bar of the same silver both have density 10.49 g/cm³. Properties that scale with amount are extensive; properties that stay fixed per unit of substance are intensive. Intensive properties are what identify a pure substance, because they don't change when you simply have more or less of it — an idea already implicit in the way melting points and densities are tabulated for Elements and Compounds without reference to sample size.

Figure

Sorting matter's traits: physical vs. chemical, intensive vs. extensive

The core relations

There is no single algebraic formula that "is" this topic — it is a classification scheme — but two quantitative relations make the intensive/extensive distinction precise and testable.

\rho = \frac{m}{V}

where:

$\rho$ = density, an intensive property (units g/mL or g/cm³)
$m$ = mass of the sample, an extensive property (g)
$V$ = volume of the sample, an extensive property (mL or cm³)

\rho_{\text{sample 1}} = \rho_{\text{sample 2}} \quad \text{for any two samples of the same pure substance at the same conditions}

where:

this equality is the operational test for "intensive": if scaling $m \to k m$ forces $V \to kV$ , the ratio $\rho$ is unchanged for any scale factor $k>0$
it fails to hold across different substances or across a chemical change, because a chemical change alters composition and therefore alters $\rho$ itself

Derivation AWhy density is intensive while mass and volume are extensive

Goal. Show algebraically that density is independent of sample size, using a general scale factor $k$ , and confirm this against actual numbers rather than assuming it.

Step 1. Start from a reference sample of a pure liquid with mass $m_0$ and volume $V_0$ , so its density is

\rho_0 = \frac{m_0}{V_0}.

Step 2. Take a second sample of the same substance at the same temperature and pressure, but $k$ times larger, so $m = k m_0$ and $V = k V_0$ (volume scales with mass at fixed density, since more material occupies proportionally more space).

Step 3. Compute the density of the larger sample:

\rho = \frac{m}{V} = \frac{k m_0}{k V_0}.

Step 4. Cancel the common factor $k$ :

\rho = \frac{k m_0}{k V_0} = \frac{m_0}{V_0} = \rho_0.

Step 5. Because $k$ cancelled completely, $\rho$ does not depend on how large the sample is — this is exactly the definition of an intensive property. Mass and volume, by contrast, are each multiplied by $k$ and are therefore extensive.

Step 6 (numeric check). Take ethanol, $\rho_0 = 0.789\ \text{g/mL}$ , with a reference sample $V_0 = 15.0$ mL, so $m_0 = 0.789 \times 15.0 = 11.835$ g. Scale by $k = 7.40$ (an arbitrary, non-round factor):

V = 7.40 \times 15.0 = 111.0\ \text{mL}, \qquad m = 7.40 \times 11.835 = 87.579\ \text{g}.

Recomputing density:

\rho = \frac{87.579\ \text{g}}{111.0\ \text{mL}} = 0.7890\ \text{g/mL}.

This matches $\rho_0 = 0.789$ g/mL, confirming the cancellation is not just symbolic — it holds for an arbitrary, asymmetric scale factor, not merely for doubling or halving.

Worked examples

Example 1 — Classifying properties and computing an intensive quantity. A student is given an irregular 37.6 g chunk of an unknown gray metal. By water displacement, it raises the water level in a graduated cylinder from 50.0 mL to 53.44 mL. The student also observes: (i) the metal is a solid at room temperature, (ii) it conducts electricity, (iii) it reacts vigorously with dilute hydrochloric acid, releasing bubbles of gas. Step 1 — classify each observation. Solid at room temperature: physical property (state of matter, no new substance involved). Conducts electricity: physical property (electrons move without the metal changing composition). Reacts with HCl releasing gas: chemical property (a new substance, hydrogen gas, plus a metal chloride, is produced — bonds are broken and reformed). Step 2 — find the intensive property (density) from the two extensive properties (mass and volume). Volume displaced:

V = 53.44\ \text{mL} - 50.0\ \text{mL} = 3.44\ \text{mL}.

Density:

\rho = \frac{m}{V} = \frac{37.6\ \text{g}}{3.44\ \text{mL}} = 10.93\ \text{g/mL}.

Step 3 — interpret. This value is intensive: if the student had used a 75.2 g chunk of the same metal instead (exactly double the mass), the displaced volume would double to 6.88 mL, and

\rho = \frac{75.2}{6.88} = 10.93\ \text{g/mL}

again — unchanged. The computed density (close to that of pure silver, 10.49 g/mL, or to some silver alloys) is a physical property that helps identify the metal, while the mass 37.6 g and volume 3.44 mL are extensive and specific to this one sample only.

Example 2 — Distinguishing a physical change from a chemical change quantitatively. A 22.50 g sample of solid naphthalene (mothball material, C₁₀H₈) is heated. At 80.2 °C it melts completely into a clear liquid; the liquid is then reweighed and found to be 22.50 g. Separately, a 22.50 g sample of magnesium ribbon is ignited in air and burns with a bright white flame, leaving behind 37.31 g of white powder (magnesium oxide, MgO, plus a trace of magnesium nitride). Step 1 — naphthalene melting. Mass before: 22.50 g. Mass after: 22.50 g. No mass change, no new substance — this is a physical change (a phase transition, solid → liquid), consistent with naphthalene's melting point being a physical property. Step 2 — magnesium burning. Mass of solid product, 37.31 g, is greater than the mass of magnesium metal alone, 22.50 g. This increase happens because oxygen from the air has combined chemically with the magnesium:

2\,\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\,\text{MgO}(s).

Check the mass added:

37.31\ \text{g} - 22.50\ \text{g} = 14.81\ \text{g of oxygen incorporated}.

Using molar masses ( $M_{\text{Mg}} = 24.31$ g/mol, $M_{\text{O}} = 16.00$ g/mol), 22.50 g Mg is

n_{\text{Mg}} = \frac{22.50}{24.31} = 0.9256\ \text{mol Mg},

which by the 2:1 stoichiometry above would combine with

n_{\text{O}_2} = \frac{0.9256}{2} = 0.4628\ \text{mol O}_2 \; \Rightarrow\; m_{\text{O}_2} = 0.4628 \times 32.00 = 14.81\ \text{g},

matching the observed mass gain almost exactly. This confirms a chemical change: a new substance (MgO) formed, elements recombined, and total mass was conserved only once the reacting oxygen is included — an early preview of the mass-balance reasoning used throughout stoichiometric calculations.

Limiting cases & common traps

Situation	Correct classification	Why the trap is tempting
Dissolving salt in water	Physical change	Looks "irreversible" at a glance, but boiling off the water recovers the same NaCl
Burning a candle	Chemical change	The wax "disappears," but it has become CO₂ and H₂O, not vanished
Bending a metal wire	Physical change	Permanent shape change, but atoms retain the same bonding — no new substance
Milk turning sour	Chemical change	No smoke or flame, yet bacteria produce lactic acid, a new compound
Density of a sample	Intensive property	Depends on both mass and volume, but the ratio cancels sample size
Total volume of a gas in a container	Extensive property	Easy to mistake for a "fixed" property, but it scales directly with amount of gas

The chemist behind it

The physical/chemical distinction as a formal teaching tool is inseparable from the broader reform of chemical thinking led by Antoine Lavoisier in the 1770s and 1780s. Before Lavoisier, many "chemical" transformations were poorly distinguished from ordinary physical events, partly because mass was not tracked carefully across a reaction. Lavoisier's insistence on weighing reactants and products on precise balances — famously in his studies of combustion and calcination of metals — established that chemical changes conserve total mass even while transforming substances into new ones with new properties. This is exactly the reasoning used above to confirm that burning magnesium is a chemical change: the solid gained mass because it captured atmospheric oxygen, not because matter appeared from nothing.

Origins · 1 minAntoine-Laurent de Lavoisier and how this was discovered

Antoine-Laurent de Lavoisier

1743–1794 · Founder of modern quantitative chemistry, established the law of conservation of mass through careful weighing of reactants and products in combustion and respiration experiments, helping to distinguish genuine chemical transformation from mere physical rearrangement

Ask the tutor: If dissolving is physical and burning is chemical, which category does dissolving a metal in acid (with visible bubbling) belong to, and why?

Is boiling water a physical or chemical change?

Boiling is a physical change. Liquid water molecules (H₂O) simply move farther apart and enter the gas phase; no bonds within the H₂O molecule are broken. Cooling the vapor condenses it back into liquid water with identical chemical identity, which is the hallmark test for a physical change.

Why is temperature usually treated as intensive rather than extensive?

Temperature measures the average kinetic energy per particle, not a total amount of anything. Splitting a beaker of 80 °C water into two smaller beakers still leaves both portions at 80 °C — the value does not add up or scale with sample size, so it behaves as an intensive property, unlike total thermal energy (heat content), which is extensive.

Can a property be both physical and dependent on chemical composition?

Yes. Physical properties (like boiling point, density, or color) are strongly influenced by what the substance is chemically, but measuring them does not itself change that substance. The property's value depends on chemical identity; the act of measuring it is what keeps it classified as physical.

How does this distinction connect to separating mixtures?

Physical changes are reversible using physical methods, which is exactly why properties like boiling point, solubility, and density are exploited to separate mixtures without altering the chemical identity of the components — a strategy developed further under Separation Techniques.

Is dissolving always a physical change, even for something like an acid tablet fizzing in water?

Not always — the key is whether new chemical species form. Dissolving sugar in water is physical (sucrose molecules just disperse). An effervescent tablet fizzing in water is chemical, because an acid and a carbonate react to release CO₂ gas, forming new substances rather than merely dispersing the original ones.