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A solute is a component of a chemical solution distinguished from its solvent, typically present in a smaller proportion. There can be several solutes. A concentration must name both what is counted and the denominator: molar concentration uses amount of substance per final chemical solution volume; a solute-to-solvent mass ratio has another denominator.^[ref]

“Added sodium chloride” and “dissolved sodium cations” describe different levels of the same inventory. Dissociation of an ionic compound changes the grouping of the particles; it preserves their chemical elements. A dissolved ion remains an ion, and dissolved sucrose remains molecular unless a separate chemical reaction occurs. Thus the name of the weighed solute need not be the name of every dissolved chemical species.

A chemical solution is a homogeneous mixture in which designated solutes are distributed in a solvent. Representative equal-volume samples have the same composition after mixing. Individual molecules and ions remain distinguishable. The solute/solvent labels describe their roles in this mixture.^[ref]

A dissolved solute is dispersed as molecules, chemical atoms, or ions, rather than as the larger solid particles of a particle suspension. For example, sucrose retains its molecular formula during dissolution in water; a chemical reaction is not required. Dissolved sodium chloride supplies sodium cations and chloride anions, surrounded by water molecules through solvation. Dissolved does not mean chemically destroyed or converted into water.^[ref]

This makes a useful connection between microscopic identity and laboratory sampling. If a chemical solution is well mixed, the amount of substance in a measured portion follows from its concentration and volume. A scoop from a poorly mixed particle suspension need not represent the whole vessel. Visual clarity alone cannot establish composition or exclude a fine dispersion; the quantitative claim needs representative sampling.

The denominator of molar concentration is the final volume of the chemical solution, including the dissolved solute. It is not the volume of solvent initially poured in. A volumetric flask fixes the final total volume; weighing the solute supplies its mass, and a specified molar mass converts that mass to amount of substance.^[ref]

Consider an original illustrative account, not reported measurements. Suppose $2.922\mathrm{g}$ of pure dry sodium chloride, with stipulated molar mass $58.44\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, is completely dissolved and quantitatively transferred into a chemical solution whose final volume is $250.0\mathrm{mL}$. Assume no losses, no additional chemical reactions, and complete dissociation of an ionic compound. ThenHere analytical concentration counts the sodium chloride inventory, not intact dissolved $\mathrm{NaCl}$ molecules. Under these assumptions, the amount of substance of sodium cations and of chloride anions is separately $0.05000\mathrm{mol}$; each has molar concentration $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$. Their combined ionic amount of substance is $0.1000\mathrm{mol}$, but this does not double either the salt inventory or the mass. Each sodium cation still contains a sodium atomic nucleus, and each chloride anion a chlorine atomic nucleus. Counting chemical elements survives dissociation of an ionic compound.

The mass ledger is different from the volume ledger. If the final chemical solution had an illustrative net mass of $252.922\mathrm{g}$, the water contribution would be $252.922-2.922=250.000\mathrm{g}$, assuming nothing entered or escaped except those two components. The ratio $2.922/252.922\approx 0.011553$ expresses the solute share of total mass; the separate mass density would be $252.922/250.0=1.011688\mathrm{g}\mathrm{m}{\mathrm{L}}^{-1}$ before rounding. These stipulated quantities are mutually consistent bookkeeping inputs, not a measured mass density for saltwater. Neither establishes that $250.0\mathrm{mL}$ of water was added.

More generally, with full transfer and conserved solute, $n=m/M$ and $c=m/(MV)$. No additive volume or ideal chemical solution assumption is needed when $V$ is the measured final volume. If only a fraction $f$ of the weighed solute reaches the flask, the actual result is $c=fm/(MV)$. This short calculation explains why an undissolved residue or an incomplete rinse creates a systematic deficit even when the final calibration mark is read correctly.

Specific heat capacity is heat capacity per unit mass:Its units are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and the constraint remains part of the definition. In particular, specific heat capacity at constant volume and specific heat capacity at constant pressure are different responses. For homogeneous material, doubling the mass doubles its heat capacity while leaving its specific heat capacity unchanged. ^[ref]

As an illustrative material model, give a fixed-volume body mass $m=2\mathrm{kg}$ and constant specific heat capacity at constant volume $c_V=400\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no phase transition and no other work, increasing its uniform temperature by $10\mathrm{K}$ requiresThese stipulated values describe no measured substance. The right panel of the diagram in heat capacity shows the straight line $Q=m{c}_V\Delta T$: a temperature increase of $5\mathrm{K}$ takes $4000\mathrm{J}$ of heat. If the specific heat capacity at constant volume depends on temperature, use $Q=m{\int }_{T_1}^{T_2}{c}_V(T)dT$ instead. The corresponding fixed-pressure calculation uses specific heat capacity at constant pressure and gives $Q=\Delta H$ under the assumptions in heat capacity.

Choosing between specific heat capacity, molar heat capacity and volumetric heat capacity is choosing whether to compare equal mass, equal amount of substance or equal volume. A small sample with large specific heat capacity can still have less total heat capacity than a much larger sample.

A spectator ion cancels from the chosen net ionic equation because its net inventory is unchanged. This label does not make it physically irrelevant: dissolved spectator ions contribute to ionic strength, which affects activity coefficients. Their concentrations can also change during dilution even though their amount of substance remains fixed.

Stoichiometry relates the quantities consumed and formed in a chemical reaction. Its coefficients are ratios of entity counts, or equivalently of amount of substance, rather than ratios of masses. For the water chemical equation, consuming two molecular hydrogen molecules per molecular oxygen molecule gives the same $2:1$ ratio as consuming two moles of molecular hydrogen per mole of molecular oxygen. Multiplying every entity count by the same factor preserves the ratio. See Flowers et al., Chemistry 2e, reaction stoichiometry.

Coefficients and subscripts do different jobs. The coefficient in $2{\mathrm{H}}_2\mathrm{O}$ counts two water molecules; the subscript says that each contains two hydrogen chemical atoms. Changing coefficients scales quantities. Changing a subscript changes the specified chemical substance.

This distinction catches a common counting error: one mole of water molecules contains two moles of hydrogen chemical atoms and one mole of oxygen chemical atoms. Always specify what is counted.

Stoichiometry records reactant and product multiplicities in a chemical reaction or a chemical reaction network. The change caused by one reaction is its stoichiometric vector; multiplying by the reaction rate gives its contribution to the concentration equations. For two identical reactants the rate-constant convention absorbs any combinatorial factor.

A strong electrolyte yields essentially all ions from its dissolved substance in the specified solvent and conditions. Strength does not specify concentration.

For dissolved sodium chloride in dilute water, one dissolved formula unit supplies one sodium cation and one chloride anion. For hydrogen chloride in water, ions instead arise by proton transfer to the solvent, giving hydronium and chloride. Thus separating existing ions by dissociation of an ionic compound and producing ions from neutral molecules by ionization can both give a strong electrolyte.^[ref]

The denominator in “essentially all” is the dissolved inventory. Solubility asks how much enters the chemical solution; electrolyte strength asks what happens to the part that enters. A sparingly soluble ionic compound can therefore have an essentially fully dissociated dissolved portion. Undissolved solid does not count as neutral dissolved chemical species, and low electrical conductivity alone does not establish a weak electrolyte.^[ref]

The diagram's weak electrolyte has the smaller converted fraction but more ions in the same volume: ten charged symbols versus four. Even at equal total ion concentrations, different ion drift velocities can give different electrical conductivities. Strength is useful for predicting dissolved chemical species in chemical equations; concentration and transport properties are additionally needed to predict electric current.

An inspected motivating experiment is Harvard Natural Sciences Lecture Demonstrations' Conductivity of Solutions by Harvard Natural Sciences Lecture Demonstrations, Setting it Up. Separate labeled beakers hold aqueous sodium chloride, sucrose, acetic acid and hydrochloric acid samples specified at $0.10\mathrm{mol}{\mathrm{L}}^{-1}$, with about $150\mathrm{mL}$ in $250\mathrm{mL}$ beakers; distilled water supplies a blank. Metal electrodes enclosed in tubing connect to a lamp. The readout is lamp brightness; the page reports that an ionic chemical solution lights the lamp. A separate deionized-water rinse station limits carryover between samples.^[ref]

For a controlled interpretation, keep electrode spacing and immersed area fixed, compare at the same temperature, and recheck the blank after rinsing. These are additional controls: the page supplies no numerical electrode geometry, temperature or calibration. Brightness supports the presence of mobile ions; it cannot alone measure the converted fraction because concentration, ion mobility and circuit response also matter. Its classification of insoluble compounds as weak should not replace the dissolved-inventory distinction above.^[ref]

An equilibrium thermodynamic state specifies sufficient independent macroscopic properties, together with material identity and constraints, to determine the other equilibrium properties. For a uniform fixed-composition ideal gas, its amount of substance, temperature and volume determine its pressure through the ideal gas law. These properties specify a state, without specifying a thermodynamic process that prepared it.^[ref]

A titrant is the reagent added in measured quantities during a titration; its known concentration and delivered volume determine the added amount of substance.

Titrant standardization determines the actual titrant concentration by reaction with a known inventory, often a primary standard. It is distinct from calibrating the delivered volume or a glass electrode's response. For a known monoprotic acid inventory $n_s$ and a suitable blank-corrected endpoint,An original illustrative standard of $0.980\mathrm{mmol}$ requiring $9.85\mathrm{mL}$, with matched blank $0.05\mathrm{mL}$, gives $c_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. The measured quantities here are stipulated examples. Independent solution preparation and volumetric calibration tests the pipette, burette, and volumetric flask; titrant standardization tests the reacting inventory per delivered volume.

The useful chain is known mass, known composition, known amount of substance, and then a measured titre through a specified stoichiometry. If each standard unit consumes $r$ units of titrant, the general inventory equation is $c_{B}V=r{n}_s$, where $V$ is the corrected titre. The one-to-one formula above has $r=1$. A titration endpoint supplies an observable stopping rule; whether it represents the equivalence point accurately remains a separate question. Standardization endpoint and blank bias makes the direction of the resulting error explicit.

A titration curve plots a measured or calculated property against added titrant volume or amount of substance. In an acid-base titration, plotting pH exposes gradual changes and a steep endpoint region. A potentiometric titration uses an electrode electric potential difference; a calibrated glass electrode and reference electrode provide operational pH. Finite additions bracket a transition rather than measuring a continuous curve. A chemical equilibrium fit can estimate the equivalence point; identifying the steepest sampled interval alone has a resolution set by the dose spacing and instrument response.

A trypsin colorimetric assay follows enzyme activity through absorbance at $405\mathrm{nm}$. Trypsin cleaves an amide bond in the synthetic arginine derivative BAPNA, releasing p-Nitroaniline.^[ref]

The inspected Sigma procedure prepares fresh trypsin in cold $1\mathrm{mM}$ hydrochloric acid and fully dissolves $1mg/ml$ BAPNA. Its buffer solution stock contains $200\mathrm{mM}$ triethanolamine and $20\mathrm{mM}$ calcium chloride, at pH $7.8$. Three $3.00\mathrm{ml}$ cuvettes compare trypsin, trypsin plus the enzyme inhibitor aprotinin, and a blank replacing trypsin with its acid solvent. Sodium chloride solution replaces aprotinin in the uninhibited control. A thermostatted instrument for spectrophotometry holds $25{}^{\circ }\mathrm{C}$ and a $1\mathrm{cm}$ light path. After equilibration, adding BAPNA starts approximately five minutes of absorbance recording. The uninhibited sample checks activity; the blank checks background; the inhibited sample tests suppression.^[ref]

Let $s_U$, $s_I$ and $s_B$ denote the measured absorbance slopes per minute over the same linear time interval. Blank subtraction gives $s_U-s_B$ and $s_I-s_B$. If p-Nitroaniline accounts for the changing absorbance, the Beer-Lambert law givesHere $c$ is product molar concentration and $l$ is the geometric light path through the chemical solution. Sigma uses $\epsilon =9.96\mathrm{m}{\mathrm{M}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$.^[ref]

Choose original illustrative slopes $s_U=0.1036$, $s_I=0.0538$ and $s_B=0.0040\mathrm{mi}{\mathrm{n}}^{-1}$. The corrected uninhibited slope is $0.0996\mathrm{mi}{\mathrm{n}}^{-1}$, giving $dc/dt=0.0100\mathrm{mM}\mathrm{mi}{\mathrm{n}}^{-1}=10.0\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$ at $l=1\mathrm{cm}$. In $3.00\mathrm{ml}$, the product amount of substance increases at $0.0300\mu \mathrm{mol}\mathrm{mi}{\mathrm{n}}^{-1}$. The corrected inhibited slope is $0.0498\mathrm{mi}{\mathrm{n}}^{-1}$, half as large. The relative suppression is thereforeThe background cancels in the numerator, but remains essential in the denominator. Using uncorrected slopes would instead give about $48.07\%$. Differentiating $A_{405}=\epsilon lc$ with fixed $\epsilon$ and $l$ explains the rate conversion; dimensional analysis checks it independently, since $\mathrm{mi}{\mathrm{n}}^{-1}/(\mathrm{m}{\mathrm{M}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}\mathrm{cm})=\mathrm{mM}\mathrm{mi}{\mathrm{n}}^{-1}$.

The inference is a reaction rate on BAPNA under the stated conditions, rather than a direct count of trypsin molecules or proof of every cut in a protein. Cloudiness, other absorbing solutes, or different product protonation can compromise the Beer-Lambert law interpretation. A matched blank and a linear interval are therefore part of the measurement's meaning. A coloured enzyme inhibitor may require its own matched blank; a changed substrate or pH requires a fresh calibration. The constructed constant slopes above describe only that interval, not indefinite product accumulation.

Suppose a synthetic polymer population contains three chains of molar mass $10000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ for every one chain of molar mass $40000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Multiplying both chain counts by any common factor leaves the averages unchanged. Substitution into the existing number-average molar mass and weight-average molar mass definitions givesThe larger chains are only one quarter of the chain count but supply $40000/70000=4/7$ of the mass. Thus a weighted mean that emphasizes mass gives a larger result than one that weights each molecule equally. Total mass divided by total chain amount of substance gives $M_n$, while $M_w/N_{\mathrm{A}}$ is not the mean mass of a randomly chosen chain. Inferring chain counts from size-exclusion chromatography still requires the existing SEC detector response and SEC molar mass calibration assumptions.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

The Van der Waals equation is a simple real gas equation of state that can also describe a liquid-gas critical point. With particle count $N$ fixed, write the volume per particle as $v=V/N$:or equivalently $(p+a{N}^2/V^2)(V-Nb)=N{k}_{B}T$. Here $p$ is pressure, $T$ is absolute temperature, and $k_B$ is the Boltzmann constant. In SI, $v$ and $b$ have units ${\mathrm{m}}^3$ per particle, $a$ has units $\mathrm{Pa}{\mathrm{m}}^6=\mathrm{J}{\mathrm{m}}^3$, and $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$; the particle count is dimensionless. The pole at $v=b$ excludes smaller volume values from this model.

The $b$ correction represents excluded volume; the attraction term lowers pressure at fixed temperature and particle number density $1/v$. This is a mean-field theory, replacing detailed intermolecular structure by two parameters. At $v\gg b$ with $a/v^2\ll k_{B}T/v$, it approaches the ideal gas law $pv=k_{B}T$. It supplies a useful qualitative route from a dilute gas to phase coexistence, without a claim of quantitative accuracy for every fluid.^[ref]

Do not mix this particle convention with the molar volume convention elsewhere in this article. With amount of substance $n_{\mathrm{mol}}=N/N_A$, Avogadro constant $N_A$, and molar gas constant $R=N_A{k}_B$, one has $v_m=N_{A}v$, $b_m=N_{A}b$, and $a_m=N_A^{2}a$, so $p=RT/(v_m-b_m)-a_m/v_m^2$. The molar parameters have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, respectively.

Johannes Diderik van der Waals introduced this equation of state in 1873, connecting molecular attraction with departures from the dilute ideal gas law. Its historical significance is the unified treatment of gases and liquids, rather than a universally exact fit.^[ref]

The molar volume $v_m=V/n_{\mathrm{mol}}$ is the volume per amount of substance, not the total container volume or the volume per particle. In the molar convention of the Van der Waals equation, take constant positive parametersThenor, for any fixed amount of substance,Both terms in the first equation are pressures. The total excluded volume is $n_{\mathrm{mol}}{b}_m$: using $V-b_m$ would subtract quantities with different units. Changing the sample size at fixed molar volume and temperature changes $V$ and $n_{\mathrm{mol}}$ proportionally, while leaving pressure unchanged. This is a useful unit and extensivity check before fitting a real gas.

For the molar Van der Waals critical point, differentiate with respect to molar volume at fixed temperature:Setting both to zero gives $R{T}_c=2a_m(v_{m,c}-b_m{)}^2/v_{m,c}^3$ and $R{T}_c=3a_m(v_{m,c}-b_m{)}^3/v_{m,c}^4$. Equating them and dividing by the nonzero factors yields $2v_{m,c}=3(v_{m,c}-b_m)$, soHere $v_{m,c}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T_c$ has units $\mathrm{K}$, and $p_c$ has units $\mathrm{Pa}$. This is the molar version of the particle derivation below: $v_{m,c}=N_A{v}_c$, with the same physical $T_c$ and $p_c$. If one writes $v_c=3b$, $T_c=8a/(27Rb)$ and $p_c=a/(27b^2)$ without subscripts, then all of $v_c,b,a$ in those formulas must be molar quantities.

Now define the dimensionless reduced variables explicitly as $x=v_m/v_{m,c}$, $t=T/T_c$ and $q=p/p_c$. Substituting the three critical values cancels $a_m,b_m,R$ and givesBecause $v_m/v_{m,c}=v/v_c$, these are exactly the reduced variables in the Van der Waals isotherm discussion. The Van der Waals law of corresponding states is therefore independent of whether one starts with particle or molar units. It predicts a critical ratio $p_c{v}_{m,c}/(R{T}_c)=3/8$, providing a third check on any proposed parameter fit.

An ideal gas law measurement obtains amount of substance without knowing the chemical formula in advance:Here $p$ is vapor pressure, $V$ is its occupied volume, $T$ is absolute temperature, $R$ is the molar gas constant, and $m$ is the mass of vapor occupying that volume at those conditions.

Santa Monica College's teaching experiment uses a $125\mathrm{mL}$ Erlenmeyer flask with a pinhole foil cap, a $600\mathrm{mL}$ water bath on a hot plate, a thermometer, and a barometer. Excess volatile liquid displaces air. A dissolved dye loses its colour and deposits when the liquid evaporates; three more minutes of heating follow. Cooling condenses retained vapor. Subtracting the dry assembly mass from the cooled assembly mass supplies $m$. The flask volume comes from brim-full water mass and mass density at its measured temperature. Controls include a dry exterior, excluding bath water, and three trials.^[ref]

The inference assumes vapor pressure equals atmospheric pressure, vapor and bath reach thermal equilibrium, the retained vapor mass is recovered on cooling, and the vapor obeys the ideal gas law. Residual air, thermal expansion of the flask, and vapor temperature below bath temperature are explicit error questions in the teaching experiment. Repeating trials assesses repeatability; it does not remove a shared bias.^[ref]

Volume magnetic susceptibility is the dimensionless ratio of longitudinal magnetization to internal magnetic field intensity in a linear regime through the origin. Positive response gives paramagnetism. Always distinguish magnetization $M=\mathcal{M}/V$ from total magnetic moment $\mathcal{M}$: dividing by volume changes the normalization. Mass-normalized and mole-normalized values likewise require the mass density or the volume per amount of substance before comparison with ${\chi }_v$.

In the existing Pauli paramagnetism formula, $M$ denotes total magnetic moment, and $g(E_F)$ is the total spin-summed density of states. Relabel that response asHere ${\mu }_B$ is the Bohr magneton. If its $B$ is a prescribed vacuum/applied magnetic flux density $B_0={\mu }_0{H}_{\mathrm{app}}$, the dimensionless apparent slope is ${\mu }_0{\chi }_B^{\mathrm{vol}}$. It agrees approximately with intrinsic ${\chi }_v$ only when the demagnetizing field correction and magnetic feedback are negligible. The existing Curie's law formula also uses a total-magnetic moment, $B$-derivative convention; its $N$ counts moments rather than giving a number density.

The volume divided by the positive number of particles $N$. The Van der Waals equation uses this convention for $v$, whereas specific volume divides by mass and molar volume divides by the amount of substance. With the Avogadro constant $N_A$, the molar volume is $v_m=N_{A}v$.

Choose approximate atomic weights $A_{\mathrm{r}}(\mathrm{H})=1.008$ and $A_{\mathrm{r}}(\mathrm{O})=15.999$, and use $M_{\mathrm{u}}\approx 1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The chemical formula ${\mathrm{H}}_2\mathrm{O}$ then gives water molar massThese are deliberately chosen rounded inputs, not exact atomic weights or a universal value for every water sample. Different isotopic abundances, notably those in heavy water, change the mean molar mass; chemical formula alone does not specify them. Adding atomic weights also neglects the tiny mass change associated with chemical bond energies, immaterial at this rounding precision.

For an illustrative pure water sample of mass $4.50375\mathrm{g}$, this convention givesThe count refers to water molecules. Their hydrogen chemical atom count is approximately $2N=3.01107038\times 10^{23}$, and their oxygen chemical atom count is approximately $N$. The mean mass per water molecule is $M/N_{\mathrm{A}}\approx 2.99146\times 10^{-26}\mathrm{kg}$. Extra displayed digits expose the arithmetic; they do not imply that the sample composition or mass was measured that precisely.

The same conversion makes stoichiometry a mass calculation. For the chemical equation $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, complete conversion of $0.250000\mathrm{mol}$ of molecular hydrogen and $0.125000\mathrm{mol}$ of molecular oxygen gives $0.250000\mathrm{mol}$ of water. With the chosen approximate atomic weights, the reactant masses are $0.50400\mathrm{g}$ and $3.99975\mathrm{g}$, summing to $4.50375\mathrm{g}$. The $2:1:2$ ratio belongs to molecule counts and amount of substance, not to masses; this bookkeeping neglects the much smaller reaction energy contribution to mass.