Why is ‘amount of substance’ so poorly understood? The mysterious Avogadro constant is the culprit!

Abstract

The base quantity ‘amount of substance’ is poorly understood and the name and symbol usually avoided. This is because of its formal interpretation as the number of entities multiplied by the reciprocal of the mysterious Avogadro constant, N _A. If X signifies the kind of entities involved, the number of entities in a sample, N(X), is easily comprehended, and if m _av(X) is the sample-average entity mass, the total mass, m(X) = N(X)m _av(X)—an aggregate of N(X) average entity masses—is also conceptually straightforward. However, the corresponding amount of substance, n(X) = N(X)(1/N _A)—an aggregate of N(X) ‘reciprocal Avogadro constants’—is incomprehensible unless some physical meaning can be attached to 1/N _A. By contrast, the base unit, mole, is thought of by chemists as an aggregate of a particular number of entities: mol = ${\mathcal N}_{\rm{Avo}}$ ent, where ${\mathcal N}_{\rm{Avo}}$ is the Avogadro number (equal to g/Da) and ent represents one entity. It makes sense, therefore, to interpret amount of substance as an aggregate of a general number of entities: n(X) = N(X) ent—an easily grasped concept. A ‘reciprocal Avogadro constant’ is thus seen to actually be exactly one entity. One mole then corresponds to setting N(X) = ${\mathcal N}_{\rm{Avo}}$, for which the total mass is the relative entity mass in grams—conforming to the original mole concept.

Appendix: Resolution of the ‘(1 + κ)’ problem

With the redefinition of the SI base units scheduled for 2018, the relationship between the number of entities, N(X), and the substance mass expressed in (redefined) grams will not change from that shown in Eq. (12). In dimensionless form, it remains as:

$$ N\left( {\text{X}} \right) /{\mathcal N}_{\rm{Avo}} = m\left( {\text{X}} \right)/\left[ {M_{\text{r}} \left( {\text{X}} \right)\;{\text{g}}} \right] $$

(19)

where ${\mathcal N}_{\rm{Avo}}$ is the (inexactly known) Avogadro number, ${\mathcal N}_{\rm{Avo}}$ = g/m _u, where m _u = m _a(¹²C)/12. However, the relationship between N(X) and the amount of substance, n(X), expressed in (redefined) moles will change from that shown in Eq. (15). Instead, it becomes:

$$ N\left( {\text{X}} \right)/N^{*} = n\left( {\text{X}} \right)/{\text{mol}} $$

(20)

where $N^{*}$ is a fixed exact integer chosen to be as close as possible to ${\mathcal N}_{\rm{Avo}}$ at the time of adoption of the new units. If the new unit definitions were to be adopted today, $N^{*}$ would be set equal to exactly 6.022 140 8568 × 10²³ (the value is adjusted within the uncertainty in ${\mathcal N}_{\rm{Avo}}$ to be an integer multiple of 12). Equations (19) and (20) give:

$$ N\left( {\text{X}} \right) /{\mathcal N}_{\rm{Avo}} = m\left( {\text{X}} \right)/\left[ {M_{\text{r}} \left( {\text{X}} \right)\;{\text{g}}} \right] = (1 \, + \kappa )\left[ {n\left( {\text{X}} \right)/{\text{mol}}} \right] $$

(21)

where (1 + κ) is the ‘molar mass correction factor’, (1 + κ) = $N^{*}$/${\mathcal N}_{\rm{Avo}}$ = $N^{*}$ m _u/g, i.e.:

$$ (1 \, + \kappa ) \, = \, \left( {1000\,N^{*}/12}\right)m_{\text{a}} \left( {^{12} {\text{C}}} \right)/{\text{kg }} = \, [5.018\;451\;714 \times 10^{25} m_{\text{a}} \left( {^{12} {\text{C}}} \right)]/{\text{kg}} $$

(22)

In relating substance mass to amount of substance directly, the architects of the redefined units [17] recommend grouping (1 + κ) with g mol⁻¹:

$$ m\left( {\text{X}} \right)/n\left( {\text{X}} \right) \, = M\left( {\text{X}} \right) \, = \, \left[ {m_{\text{av}} \left( {\text{X}} \right)/m_{\text{u}} } \right] \, [(1 \, + \kappa )\;{\text{g}}\;{\text{mol}}^{ - 1} ] = [ {m_{\text{av}} \left( {\text{X}} \right)/m_{\text{u}} }]\, M_{\text{u}} $$

(23)

where M _u is the (inexactly known) ‘molar mass constant’, M _u = (1 + κ) g mol⁻¹.

However, chemists will continue to work with fixed exact units, grams and moles, not the inexactly known M _u. So it makes much more sense to group the (1 + κ) factor with the term m _av(X)/m _u in equation (23), giving:

$$ M\left( {\text{X}} \right) \, = m_{\text{av}} \left( {\text{X}} \right)/[m_{\text{u}} /(1 \, + \kappa )]{\text{ g}}\;{\text{mol}}^{ - 1} $$

(24)

where the denominator is:

$$ m_{\text{u}} /(1 \, + \kappa ) = {\text{g}}/N^{*} = (1/6.022\;140\;8568 \times 10^{26} )\,{\text{kg}}$$

(25)

an atomic-scale mass exactly related to the (redefined) kilogram.

I have previously recommended that the dalton should be redefined this way, for use in cataloguing (sample-average) entity masses used in stoichiometry—that have relatively low precision due to uncertainties in the isotopic composition and the (usually ignored) mass equivalent of binding energy—while retaining the unified atomic mass unit, u = m _u = m _a(¹²C)/12, for cataloguing individual nuclidic masses (to very high precision) [9]. If we redefine the dalton as:

$$ {\text{Da}} = (1/6.022\;140\;8568 \times 10^{26} )\;{\text{kg}},{\text{exactly}} $$

(26)

Eq. (24) becomes:

$$ M\left( {\text{X}} \right) \, = m_{\text{av}} \left( {\text{X}} \right)/{\text{Da}}\;{\text{g}}\;{\text{mol}}^{ - 1} = M_{\text{r}} \left( {\text{X}} \right)\;{\text{g}}\;{\text{mol}}^{ - 1} $$

(27)

where M _r(X) is the (sample-average) relative entity mass, m _av(X)/Da, referred to the exact dalton. This has the added advantage of fixing the value of the Avogadro number:

$$ {\mathcal N}_{\rm{Avo}} = {\text{g}}/{\text{Da}} = N^{*} = 6.022\;140\;8568 \times 10^{23} ,{\text{exactly}} $$

(28)

so that Eq. (15) can be written:

$$ N\left( {\text{X}} \right)/N^{*} = m\left( {\text{X}} \right)/\left[ {M_{\text{r}} \left( {\text{X}} \right)\,{\text{g}}} \right] = n\left( {\text{X}} \right)/{\text{mol}} $$

(29)

with no explicit or implicit uncertainty factors. The exact dalton also means that the redefined mole can be written as:

$$ {\text{mol}} = N^{*} \;{\text{ent}} = \left( {{\text{g}}/{\text{Da}}} \right)\;{\text{ent}} = {\mathcal N}_{\rm{Avo}} \;{\text{ent}} $$

(30)

an aggregate of an (exact) Avogadro number of entities, as it is thought of by chemists. It also means that we retain the exact relationships between the dalton per entity and the corresponding macroscopic units:

$$ {\text{Da}}\;{\text{ent}}^{ - 1} \equiv{\text{g}}\;{\text{mol}}^{ - 1} \equiv{\text{kg}}\;{\text{kmol}}^{ - 1} $$

(31)

so that the amount-specific mass of a substance, M(X) = m(X)/n(X), can be written directly as:

$$ \begin{aligned} M\left( {\text{X}} \right) & = \, \left[ {N\left( {\text{X}} \right)m_{\text{av}} \left( {\text{X}} \right)} \right]/\left[ {N\left( {\text{X}} \right)\;{\text{ent}}} \right] = m_{\text{av}} \left( {\text{X}} \right)/{\text{ent}} = \left[ {m_{\text{av}} \left( {\text{X}} \right)/{\text{Da}}} \right]\;{\text{Da}}\;{\text{ent}}^{ - 1} \\ & = M_{\text{r}} \left( {\text{X}} \right)\,{\text{Da}}\;{\text{ent}}^{ - 1} = M_{\text{r}} \left( {\text{X}} \right)\;{\text{g}}\;{\text{mol}}^{ - 1} = M_{\text{r}} \left( {\text{X}} \right)\;{\text{kg}}\;{\text{kmol}}^{ - 1} \\ \end{aligned} $$

(32)