Partial Molar Quantities: The General Concept

Pour 50 mL of ethanol into 50 mL of water. The final volume is not 100 mL. It is closer to 96 mL. Something has "disappeared." Understanding why requires a new kind of quantity. That quantity is the partial molar property. It measures how an extensive property of a mixture responds to small changes. These properties include volume, enthalpy, and Gibbs energy. Specifically, it tracks the response when you add a sliver of one component. Temperature, pressure, and every other amount stay fixed. This idea underlies mixing, solvation, and eventually the chemical potential itself. This article builds the definition from scratch. You will see why extensive properties fail to add simply in mixtures. Then you will derive the partial molar volume from first principles. Finally, you will see two fully worked numerical examples. Common traps and a compact exam cheatsheet follow as well.

Key takeaways

A partial molar quantity is the derivative of an extensive property with respect to moles of one component, at constant $T$ , $P$ , and all other $n_j$ .
Partial molar volumes explain why volumes of mixing are not additive — mixing ethanol and water shrinks the total volume.
Partial molar quantities are intensive: they depend on composition, $T$ , and $P$ , not on the total size of the sample.
The total extensive property is the mole-weighted sum of partial molar quantities (Euler's theorem for homogeneous functions).
The Gibbs–Duhem equation links how partial molar quantities of different components must change together.
Key idea: a partial molar property tells you the "local" contribution of one substance to a mixture's extensive property, and that contribution generally differs from the substance's own molar property in the pure state.

The core insight

A partial molar quantity, $\overline{X}_i = (\partial X/\partial n_i)_{T,P,n_{j\neq i}}$ , is how much an extensive property $X$ of a mixture changes per mole of component $i$ added, at fixed temperature, pressure, and fixed amounts of everything else — and because molecules experience different environments in a mixture than in the pure substance, $\overline{X}_i$ is generally not equal to the pure-substance molar property $X_{m,i}^*$ .

The physical picture

Extensive properties — volume, internal energy, enthalpy, entropy, Gibbs energy — scale with the size of the system. Double the amount of a pure substance at fixed $T$ and $P$ , and every extensive property doubles. This is the defining behavior of pure, homogeneous samples, and it lets us define ordinary molar quantities like $V_m = V/n$ . Mixtures behave differently. When water and ethanol mix, hydrogen bonding rearranges: water molecules can pack more efficiently around ethanol's ethyl group than they do around each other in pure water. The result is that adding one more mole of ethanol to a large volume of water-ethanol mixture changes the total volume by an amount that depends on the composition already present — not by the fixed increment $V_m^*(\text{ethanol}) = 58.4\ \text{cm}^3\,\text{mol}^{-1}$ you would predict from the pure liquid. This composition-dependent, per-mole response is the partial molar volume, $\overline{V}_i$ . It is defined as a partial derivative at constant $T$ , $P$ , and constant amount of every other component, so that only the "local packing environment" effect is captured — no temperature or pressure change is smuggled into the number, and no other component is secretly being added or removed. The same logic applies to any extensive state function: enthalpy gives partial molar enthalpies (linked to heats of solution), entropy gives partial molar entropies, and Gibbs energy gives the single most important partial molar quantity in chemistry — the chemical potential, $\mu_i = \overline{G}_i$ , which governs phase equilibrium and reaction spontaneity in mixtures.

Figure

Volume is not additive when liquids mix: partial molar volume as a local slope

The core relations

\overline{X}_i = \left(\frac{\partial X}{\partial n_i}\right)_{T,P,n_{j\neq i}}

where: $X$ is any extensive state function (V, H, S, G, U), and the derivative is taken holding $T$ , $P$ , and all other mole numbers $n_j$ fixed.

X = \sum_i n_i \overline{X}_i

where: the total extensive property equals the mole-weighted sum of the partial molar quantities, a consequence of $X$ being a first-order homogeneous function of the $n_i$ at fixed $T,P$ .

\sum_i n_i\, d\overline{X}_i = 0 \quad (\text{at constant } T, P)

where: this is the Gibbs–Duhem equation — the partial molar quantities of different components cannot vary independently with composition.

dX = \overline{X}_1\,dn_1 + \overline{X}_2\,dn_2 + \cdots \quad (\text{at constant } T, P)

where: the total differential of $X$ at fixed $T,P$ is built entirely from the partial molar quantities and the differential mole numbers.

Derivation AFrom the definition to Euler's theorem: X = Σ nᵢ X̄ᵢ

Goal. Show that the intuitive additive formula, $X = \sum_i n_i \overline{X}_i$ , is not an assumption but a mathematical consequence of the definition of $\overline{X}_i$ together with the extensivity of $X$ .

Step 1 — extensivity as a scaling law. At fixed $T$ and $P$ , if we take a mixture and scale every amount by the same factor $\lambda$ , the whole sample is just $\lambda$ times as much material at the same intensive state (same $T$ , $P$ , same mole fractions). Extensive properties must scale the same way:

X(T, P, \lambda n_1, \lambda n_2, \ldots) = \lambda\, X(T, P, n_1, n_2, \ldots)

This says $X$ , viewed as a function of the $n_i$ alone (at fixed $T,P$ ), is a homogeneous function of degree 1.

Step 2 — differentiate the scaling law with respect to λ. Differentiate both sides with respect to $\lambda$ , treating $T$ and $P$ as constants:

\sum_i n_i \left(\frac{\partial X}{\partial (\lambda n_i)}\right)_{T,P} = X(T,P,n_1,n_2,\ldots)

using the chain rule on the left-hand side, where each term is $n_i$ times the derivative of $X$ with respect to its $i$ -th argument.

Step 3 — evaluate at λ = 1. Setting $\lambda = 1$ makes $\lambda n_i \to n_i$ , and each derivative becomes exactly the partial molar quantity defined earlier:

\left(\frac{\partial X}{\partial (\lambda n_i)}\right)_{T,P}\Bigg|_{\lambda=1} = \left(\frac{\partial X}{\partial n_i}\right)_{T,P,n_{j\neq i}} = \overline{X}_i

This is exactly Euler's theorem for homogeneous functions of degree 1, applied to a thermodynamic extensive property.

Step 4 — assemble the result. Substituting Step 3 into Step 2 at $\lambda=1$ gives the compact additive law:

X = \sum_i n_i\, \overline{X}_i

Step 5 — derive Gibbs–Duhem as a check of consistency. Take the total differential of $X$ two ways. From $X=\sum_i n_i\overline{X}_i$ directly:

dX = \sum_i \overline{X}_i\, dn_i + \sum_i n_i\, d\overline{X}_i

But from the definition of $\overline{X}_i$ at constant $T,P$ , the differential of $X$ is also:

dX = \sum_i \overline{X}_i\, dn_i

Subtracting these two expressions for $dX$ forces the extra term to vanish:

\boxed{\sum_i n_i\, d\overline{X}_i = 0 \quad (\text{constant } T,P)}

This boxed Gibbs–Duhem relation is the key constraint: partial molar quantities cannot be assigned arbitrarily as functions of composition. If you know how $\overline{X}_1$ varies with composition, the variation of $\overline{X}_2$ is not free — it is fixed by this equation.

Worked examples

Example 1 — Foundational: reading a partial molar volume from data. A chemist prepares a series of NaCl(aq) solutions at 298 K and 1 bar and measures the total volume as a function of moles of NaCl added to a fixed 1000 g of water. Near a molality of 1.00 mol/kg, the data fit locally to

V(n_{\text{NaCl}}) = V_0 + (16.6\ \text{cm}^3\text{mol}^{-1})\,n_{\text{NaCl}} + (1.2\ \text{cm}^3\text{mol}^{-2}\text{kg})\,n_{\text{NaCl}}^2

with $n_{\text{NaCl}}$ in mol per kg of water and $V_0$ the volume of the pure 1000 g of water (1002.96 cm³ at 298 K, since $\rho_{\text{water}} = 0.99704\ \text{g cm}^{-3}$ ). Step 1. The partial molar volume of NaCl is the derivative of $V$ with respect to $n_{\text{NaCl}}$ at fixed $T$ , $P$ , and fixed water:

\overline{V}_{\text{NaCl}} = \left(\frac{\partial V}{\partial n_{\text{NaCl}}}\right)_{T,P,n_{\text{water}}} = 16.6 + 2(1.2)\,n_{\text{NaCl}}

Step 2. Evaluate at $n_{\text{NaCl}} = 1.00\ \text{mol/kg}$ :

\overline{V}_{\text{NaCl}} = 16.6 + 2(1.2)(1.00) = 19.0\ \text{cm}^3\,\text{mol}^{-1}

Step 3 — sanity check. The molar volume of solid NaCl is about 27.0 cm³/mol (from density 2.165 g/cm³ and $M = 58.44$ g/mol). The partial molar volume in solution, 19.0 cm³/mol, is smaller. This is physically sensible: dissolved Na⁺ and Cl⁻ ions electrostrict the surrounding water, pulling water molecules in tighter and shrinking the effective volume occupied per mole of salt. It can even be negative for some electrolytes at high dilution — a signature that the partial molar quantity is not simply "borrowed" from the pure solid.

Example 2 — Applied: reconstructing total volume of mixing for ethanol–water. At 298 K, in a mixture that is $x_{\text{EtOH}} = 0.40$ (mole fraction ethanol), experiment gives partial molar volumes

\overline{V}_{\text{EtOH}} = 55.0\ \text{cm}^3\,\text{mol}^{-1}, \qquad \overline{V}_{\text{water}} = 17.5\ \text{cm}^3\,\text{mol}^{-1}

(compare pure molar volumes $V_{m,\text{EtOH}}^* = 58.4\ \text{cm}^3\,\text{mol}^{-1}$ and $V_{m,\text{water}}^* = 18.1\ \text{cm}^3\,\text{mol}^{-1}$ — both partial molar volumes are smaller than the pure values, reflecting tighter packing on mixing). Step 1. Take 1.000 mol of total mixture. Then $n_{\text{EtOH}} = 0.400$ mol and $n_{\text{water}} = 0.600$ mol. Step 2. Apply the additive law from the derivation, $V = \sum_i n_i \overline{V}_i$ :

V = (0.400\ \text{mol})(55.0\ \text{cm}^3\text{mol}^{-1}) + (0.600\ \text{mol})(17.5\ \text{cm}^3\text{mol}^{-1})

V = 22.0\ \text{cm}^3 + 10.5\ \text{cm}^3 = 32.5\ \text{cm}^3

Step 3. Compare to the volume if there were no mixing effect at all — i.e., if the liquids simply added their pure molar volumes:

V_{\text{ideal}} = (0.400)(58.4) + (0.600)(18.1) = 23.36 + 10.86 = 34.2\ \text{cm}^3

Step 4 — the volume of mixing. The difference is the (molar) volume of mixing for this 1 mol sample:

\Delta V_{\text{mix}} = V - V_{\text{ideal}} = 32.5 - 34.2 = -1.7\ \text{cm}^3

A negative volume of mixing — the mixture occupies about 1.7 cm³ less than the naive sum, consistent with the everyday observation that mixing ethanol and water causes a slight volume contraction. Both numbers are of the right order of magnitude (a few percent of the total), matching known ethanol–water contraction data.

Limiting cases & common traps

Situation	What happens to $\overline{X}_i$	Why
Pure substance limit ( $x_i \to 1$ )	$\overline{X}_i \to X_{m,i}^*$ , the ordinary molar property	No other component is present to alter the local environment
Ideal mixture (no interaction change)	$\overline{X}_i = X_{m,i}^*$ at all compositions (for $V$ , $H$ , $U$ )	Molecules do not "notice" the difference between like and unlike neighbors
Infinite dilution ( $x_i \to 0$ )	$\overline{X}_i \to \overline{X}_i^{\infty}$ , a finite but distinct limiting value	Component $i$ is fully surrounded by solvent, not by itself
Real, non-ideal mixture	$\overline{X}_i$ varies smoothly with composition	Molecular environment changes continuously with composition

The chemist behind it

The formal machinery of partial molar quantities — and the recognition that chemical potential is the partial molar Gibbs energy — was built out of the extraordinarily systematic thermodynamic framework developed by Josiah Willard Gibbs in the 1870s. Gibbs treated multicomponent, multiphase equilibrium with a rigor that took decades for chemists to fully absorb; his 300-page memoir "On the Equilibrium of Heterogeneous Substances" contains, in embryonic form, essentially the entire apparatus of partial molar properties, chemical potential, and phase rule reasoning used today. Gilbert N. Lewis later popularized and operationalized these ideas for working chemists, introducing fugacity and activity as practical tools built directly on the partial molar Gibbs energy.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and defined the chemical potential as a partial molar quantity

Ask the tutor: How do we measure a partial molar volume experimentally, and what does the "apparent molar volume" method actually plot?

Is a partial molar quantity the same as an ordinary molar quantity?

No. An ordinary molar quantity, like $V_m = V/n$ , is defined for a pure substance and is just total property divided by total moles. A partial molar quantity is defined for one component within a mixture and is a derivative — the marginal change in the mixture's property per mole of that component added, holding the amounts of everything else fixed. The two coincide only in the pure-substance limit or in an ideal mixture.

Can a partial molar quantity be negative even though the property itself, like volume, is always positive?

Yes. A striking example is MgSO₄ in dilute aqueous solution, where the partial molar volume of the salt is negative near infinite dilution. This does not mean "negative volume" exists physically; it means adding a small amount of salt to water causes such strong electrostriction — water molecules packing tightly around the ions — that the total solution volume actually decreases slightly as salt is added, even though the total volume itself stays positive.

Why do we hold both temperature and pressure constant in the definition, rather than temperature and volume?

Chemistry is overwhelmingly done at constant pressure (open beakers, constant atmospheric pressure), so $T$ and $P$ are the natural, experimentally controlled variables. Holding $T,P$ fixed also makes the resulting partial molar Gibbs energy identical to the chemical potential, which governs equilibrium and spontaneity at constant $T,P$ — the conditions of most laboratory and biological chemistry.

How is this concept used later, for chemical potential and phase equilibria?

The chemical potential $\mu_i$ is defined precisely as the partial molar Gibbs energy, $\mu_i = (\partial G/\partial n_i)_{T,P,n_{j\neq i}}$ . Everything proven here for a general property $X$ — the additive law $X=\sum_i n_i\overline{X}_i$ and the Gibbs–Duhem constraint — applies directly to $G$ and $\mu_i$ . This is exactly the machinery used to derive phase equilibrium conditions, Raoult's law, and colligative properties.

Do partial molar quantities depend on which component you call "solvent" versus "solute"?

No — the definition treats all components symmetrically. $\overline{X}_i$ is defined the same way for every component: differentiate $X$ with respect to $n_i$ , holding $T$ , $P$ , and all other $n_j$ fixed. "Solvent" and "solute" are just conventional labels chosen by whichever component is present in large excess; the mathematics does not care about the label.