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I am starting to study physics in detail and as I read about physical quantities, I was puzzled why mol (amount of substance) is taken as a physical quantity.

A physical quantity is any quantity which we can measure and has a unit associated with it. But a mol represents the amount of substance by telling the number of particles (atoms, molecules, ions, etc.) present. So it is a pure number and numbers are dimensionless. So mol should not be considered a physical quantity.

Also, fundamental physical quantities should be independent of each other. I am wondering whether mass and mol are independent. This is so as they surely affect each other as we can evidently see while calculating the number of moles and using the mass of that sample for calculation.

So how is the mol a fundamental physical quantity and independent of mass?


marked as duplicate by sammy gerbil, ZeroTheHero, David Hammen, Jon Custer, Yashas Mar 24 '17 at 3:46

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  • $\begingroup$ "This is so as the surely affect each other as we can evidently see while calculating number of moles and using mass of that sample for calculation." Would you mind making this example a bit clearer? It sounds like you say that mole and mass can be substituted for each other in some equations - that is not the case. $\endgroup$ – Steeven Mar 23 '17 at 14:06
  • $\begingroup$ I am in high school and in chemistry while calculating number of mole present in x grams of carbon (suppose ) , then we divide x by molar mass of carbon and get the answer. $\endgroup$ – Abhinav Dhawan Mar 23 '17 at 14:09
  • $\begingroup$ I thought this may show a relation between the two... $\endgroup$ – Abhinav Dhawan Mar 23 '17 at 14:10
  • $\begingroup$ The mass density of a material lets you transform between mass and volume for a material, but we recognize these as separate measurements nonetheless. It's the same for molar mass of a material $\endgroup$ – Jim Mar 23 '17 at 14:14
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    $\begingroup$ @AbhinavDhawan There is a relation between the two, yes. The relation is: $$m=nM$$ where $m$ is mass (kg), $n$ amount of substance (moles) and $M$ molar mass (kg/mole). There is also a relation between mass and time, two other fundamental properties, in expressions like Newton's law: $F=m \frac{dv}{dt}$. As you can see, a relation between two things says nothing about them being fundamental or not. You can always invent a number with unit to multiply on to one fundamental property, which converts it into the other - that doesn't say anything special about the two. $\endgroup$ – Steeven Mar 23 '17 at 14:23

The mole definitely isn't a fundamental physical quantity. It's just a shorthand for Avogadro's number, to make really big numbers more tractable. It's purely there for convenience, there's nothing fundamentally physically significant about it.

  • $\begingroup$ Yours, the 'right' answer. 1 mol = 6.023 X 10^23 'things' be it molecules or widgets. A convenience. The cool thing, when you talk about a mol of (ideal) gas at STP it's always ~22.4 liters regardless of what gas or gas mixture you have! $\endgroup$ – docscience Mar 23 '17 at 17:23
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    $\begingroup$ A big problem with his question is he started by referring to it as just a physical quantity. That is quite easy to accept. In the second last paragraph he changes it to "fundamental physical quantity" which implies something entirely different. We can consider it a physical quantity, because by definition it is the quantity of something physical. It isn't fundamental though, it seems like a mistake that OP is not differentiating between fundamental physical quantities and non-fundamental physical quantities. $\endgroup$ – JMac Mar 23 '17 at 17:35
  • $\begingroup$ @JMac - I even think he means "base quantity", not "fundamental quantity". $\endgroup$ – Vera K Jul 25 '17 at 21:11
  • $\begingroup$ @tparker If it's just for convenience, then why is it part of the SI? I thought the SI was about defining units for fundamental physical dimensions. $\endgroup$ – Joffrey Aug 14 '17 at 15:34
  • $\begingroup$ @Joffrey Nope, many of the SI "fundamental base quantities" can be expressed in terms of each other. $\endgroup$ – tparker Aug 14 '17 at 17:37

Mols are a units of quantity. Technically, you can have a two cars, or a mole of cars, two forks, or a mole of forks, two baby rabbits, or a mole of baby rabbits. But since one mole is such a large number, it is only really useful for things that you have lot of, like molecules. In that case, though, it is very useful, since saying one mole is lot faster than enunciating 602 sextillion, or $6.022140857\times10^{23}$.

It is very important to know how many molecules of a particular type there is (for instance) in a beaker. If you have two highly reactive molecules in a beaker, it's probably not too dangerous: these two molecules will only destroy two of the floor's molecules were the beaker dropped. However, if you have a mole of these dangerous molecules, the floor might start to complain.

Mass and mole are completely different things: a mole of cars will weigh more than a mole of H2 molecules.

  • $\begingroup$ I completely agree with you and your answer really helped me . But still the thing that puzzled me is that moles are number only.Like a dozen or a score , mole also indicated a number (which is Avogadro's number). So then why is it called a physical quantity? Is it only door is relative significance ? Or is it something else ? $\endgroup$ – Abhinav Dhawan Mar 23 '17 at 15:22
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    $\begingroup$ I feel like a big problem is that people are confusing mole and the Avogadro constant. As far as I can tell a mole is a unit of measure which has a very specific quantity(which makes it equivalent to the Avogadro constant in value). Mole itself isn't the quantity, mole is the unit of measure, the quantity of which is Avogadro's constant. $\endgroup$ – JMac Mar 23 '17 at 17:00
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    $\begingroup$ It's considered a physical quantity for the same reason as "radians" are considered a measure of angle and "steradians" are considered a measure of solid angle. They too are unitless. I've got a few answers around here on Physics.SE pointing out that the unit systems exist for one reason and one reason alone: to help us understand physics. radians, moles, and steradians are given special status because we have found, over the years, that giving them that special status makes it easier to understand physics and its equations. $\endgroup$ – Cort Ammon Mar 23 '17 at 18:50
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    $\begingroup$ @CortAmmon I it helpful to talk about radians, moles, dozens, steradians, and the like as dimensionless units (not unitless). $\endgroup$ – Chris Chudzicki Mar 23 '17 at 19:21
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    $\begingroup$ @mdavezzc Why pick rabbits when you could have saida mole of moles $\endgroup$ – AHusain Mar 23 '17 at 21:47

Is it a fundamental number in nature? It's (currently) a number resulting from atomic structure (fundamentally defined by the masses of quarks, Planck's constant and the way quantum mechanics works) and our definition of the gram, which is based on the international Kg prototype. Avogadro's constant is currently defined by experiment, and therefore has no absolute "right" number, just an agreed working definition.

This is a messy way to define things though, and there are many arguing that the Kg should be defined in terms of a particular element and Avogadro's number, which would put it on a more "fundamental" level in my book. (See https://en.wikipedia.org/wiki/Kilogram#Avogadro_project )

This would mean "fixing" Avogadro's constant by simply picking a number, then defining the Kg in terms of this, in the same way the second was "fixed" in terms of the so-many-oscillations of a particular frequency of light, rather than being a 60th of a 60th of a 24th of one rotation of Earth (a messy, variable number).

  • $\begingroup$ Note that the Avogadro number is simply the conversion factor between the two units of mass "gram" and "unified atomic mass unit". It has nothing to do with the mass of quarks. $\endgroup$ – Vera K Jul 25 '17 at 21:24
  • $\begingroup$ Note that is it "kg", not "Kg". $\endgroup$ – Vera K Jul 25 '17 at 21:25

True, a mole is a measure of quantity, i.e. it's dimensionless. But that doesn't prevent us from treating it as a physical quantity!

The fact is, units can be treated the same way as the numbers (or symbols) that they apply to - multiplied, divided, reduced, replaced with equivalent expressions.

The special feature of dimensionless units is that they can turn from number to unit and back at will. E.g. when transposing units, you can always replace the $k$ prefix with $10^3$, $M$ with $10^6$ etc and vice versa.


$$ 20\times10^{23} \approx 3,3\,mol\\ 5\,mol \cdot 5\,g/mol = 25\,g\\ N_A=6,022\times10^{23}\,mol^{-1}=1\\ R=kN_A=1,380\times10^{-23}\,J/K \cdot 6,022\times10^{23}\,mol^{-1}=8,314\,\frac{J}{mol\cdot K} $$ "$mol$" is effectively just a multiplier, so $R$ is actually $k$ in different units!

Since $mol$ is dimensionless, you can legitimately (in the mathematical sense) introduce it anywhere in any power. But that likely won't make physical sense since there's no such thing as a "square number of atoms" - so once introduced in an appropriate place, it should rather be treated as a dimension from that moment on to keep everyone's sanity intact :)

So, a mole is often considered a physical quantity because it's convenient to treat is as such - this results in more comprehensible numbers in the numerical part of expression when dealing with practical amounts of substances. Besides, it points out that the number that "$mol$" is used with is not just any number but a number of atoms - a plain number wouldn't carry this meaning.

One other dimensionless entity commonly treated as a physical quantity is the decibel.

  • $\begingroup$ Ivan, your terminology is a bit confusing. I think what you wanted to say is: "the unit mole is used for convenience, it results in more comprehensible numbers in numerical values of quantities". $\endgroup$ – Vera K Jul 25 '17 at 21:18
  • $\begingroup$ @VeraK I also showed how units can be treated the same way as letters and numbers in a formula. This demonstrates how "mol" makes perfect mathematical sense - it can be introduced via equivalence transformations. $\endgroup$ – ivan_pozdeev Jul 26 '17 at 1:07

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