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In a recent interesting post, (Is 1 ppb equal to 1 μg/kg?) it was pointed out that IUPAC advises to abandon parts per million, parts per billion quantities and instead it suggests to employ micromole/ mole for ppm and nanomole/mole for ppb. An example is given on page 98 of the Green Book "The volume fraction of helium is 20 ppm".

May be I am missing a point or I see a problem here with the moles in the denominator of these suggested replacement definitions.

a) If we analyze Zr in a rock and find 5 mg Zr in a 20 gram rock, how would one apply the mole definition because one cannot define the moles of a rock. We have to stick to the classical definition:

ppm =10^6 * (mass of analyte / mass of sample).

b) How can one express ppm of Ar in air, and how should we define the moles of air? If we wish to generalize this case, what if our solvent system consists of several components.

The ambiguity I feel in micromole/mole definition is the mole in the denominator. Are we referring to the moles of the solvent or moles of solvent plus solute.

If this replacement is analogous to molal concentration, where the denominator is clearly definied to be kg of solvent not the kg of solvent plus kg of solute, then IUPAC should clearly define it.

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    $\begingroup$ Who said it's all about moles? Have you actually seen other examples from that table you are quoting (I only listed one to illustrate the replacement for ppb)? The idea is not to convert everything to the amount of substance, but to use mg/g, μg/g, pg/g etc. instead when appropriate! With this in mind: a. Use mass fraction (mg/g) instead of ppm; b Use volume fraction (μmol/mol) instead of ppm. It's all in the table you are quoting, really. $\endgroup$
    – andselisk ♦
    Apr 16, 2019 at 3:02
  • $\begingroup$ Yes, I have a copy of the Green book. They do mention above the table that mg/g and other similar quantities can be formulated which are used by analytical chemists. (a) I am still not sure how to calculate "The volume fraction of helium is 20 ppm" in terms of (μmol/mol). (b) They don't specify "moles" of "what" in the denominator? An example would have helped. $\endgroup$
    – AChem
    Apr 16, 2019 at 3:18
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    $\begingroup$ They talk about "volume fraction of helium". It means that it is 20 ul per liter of sampled gas. They are not talking about moles. $\endgroup$
    – user32223
    Apr 16, 2019 at 3:50
  • $\begingroup$ You are right, this "It means that it is 20 ul per liter of sampled gas." is a classical definition. Have a look at the Table on pg 98 of the Green Book, they write μmol/mol as "Replacement". My point is that IUPAC should have provided better examples. $\endgroup$
    – AChem
    Apr 16, 2019 at 3:54

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There are not this kind of problems with the IUPAC recommendation. The only point in it is to specify the parts of what we refer to, nothing more. In such a sense it seems clearer than expression as molality or molarity.

Take an amount of atmosphere at given conditions. The concentration of Ar or whatever component can surely be expressed in terms of mole number in a million or whatever number of moles that amount contains.

As for the rock analysis case in the question, note that IUPAC umol/mol is an example, it doesn't mean it should be preferred to mg/kg (or ul/l etc.). The measured amount remains the most convenient both practically and mathematically.

I add that following recommendations would - perhaps - also help clear another practical con of unspecified ppm and percentage concentration values. Often people forget that 1 l --> 1 kg of solution is acceptable only with very dilute solutions and when the solvent is water and go around telling and even publishing % concentration in various solvent so that the reader must assume a wrong procedure in order to follow up. Explicitly saying mg/kg instead of ppm should evidence that when mg starts to be hundred, than one has to switch concentration units or being more careful when reporting.

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  • $\begingroup$ "The concentration of Ar or whatever component can surely be expressed in terms of mole number in a million or whatever number of moles that amount contains." calculating the mole of Ar is easy, but how would be calculate moles of "air" which is the solvent for Ar. Can we work this example out? $\endgroup$
    – AChem
    Apr 16, 2019 at 12:20
  • $\begingroup$ Yes, for a start I would just consider an ideal gas. Note that this is indipendent of the unit you choose, using ppm would be very much the same (at end it is!). The iupac recommendation just tell you not to use ppm, or, if you really want, specify ppm of what. You can keep part in volume, if you want to skip the average molecular mass of that air. It will be anyway how many molecules of Argon are in a million molecules of air. $\endgroup$
    – Alchimista
    Apr 16, 2019 at 12:28
  • $\begingroup$ @M. Farooq ↑ isn't conceptually any different than thinking of partial Pressure, tough I perfectly understand now your concern. There is not, strictly, a mole of air, if not "engineeringly" speaking. $\endgroup$
    – Alchimista
    Apr 16, 2019 at 12:42
  • $\begingroup$ Thanks. The micromole/mol definition is similar to molal concentration not molar concentration because the denominator does not include the mass of the substance in the numerator. For example if I add 30 g NaCl to 1 kg of water, it is possible to determine molal concentration but impossible to determine its molarity (because we don't know how much volume changed after adding 30 g NaCl to 1 kg of water). $\endgroup$
    – AChem
    Apr 16, 2019 at 12:46
  • $\begingroup$ This is why I dislike in general operational dictated concentration values for which density is required. But at the end, this is pretty much used when we refer to washing/neutralizing/working up solution for which precision is the last concern. However, the umol/mol way of expressing concentration does include the stuff dissolved, as ppm does: 1 ppm = 1 part in 1 million total parts. Again, this is usually neglected when part <<< parts, which in turns lead to rounding 1 ppm = 1 mg/ml for very diluted solutions in water (d = 1). I hope this clarify. @M. Farooq $\endgroup$
    – Alchimista
    Apr 16, 2019 at 13:10

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