The source of my question is due the nature of a problem which I found misleading in the way how it was presented.

The problem is as follows:

The molecular weight of hydrogen ($\ce{H2}$), helium ($\ce{He}$), nitrogen ($\ce{N2}$) and oxygen ($\ce{O2}$) are $2$, $4$, $28$ and $32$, respectively. If $N$ is the number of molecules of $\ce{H2}$ per mole. Then find the number of molecules per mole of $\ce{He}$, $\ce{N2}$ and $\ce{O2}$ respectively.

The alternatives given are as follows:

$\begin{array}{ll} 1.&\textrm{2N, 14N, 16N}\\ 2.&\frac{N}{2},\,\frac{N}{12},\,\frac{N}{16}\\ 3.&\textrm{N, N, N}\\ 4.&\textrm{4N, 28N, 32N}\\ 5.&\textrm{N, 7N, 8N}\\ \end{array}$

This problem in particular has left me confused on what should be the answer? as in my opinion is not very clear or precise what the author is exactly intending to mean. First of there isn't any indication regarding the weight of any of the gases.

From the way how this question is stated (as giving references of the molar masses of each gas) I think its implying that in all of them there is one mole, if such is the case then there's one mole of helium, one mole of nitrogen and one mole of oxygen, and therefore there are $N$, $N$ and $N$ in all cases.

But the other interpretation which I thought was:

$\textrm{N= number of molecules}$


Therefore for: helium


for nitrogen:


for oxygen:


and it will be:

$2N$, $14N$, $16N$.

Respectively. But what will be the answer for this question?. I am still doubtful due the ambiguity. Will it be $N$, $N$ and $N$ or $2N$, $14N$ and $16N$?. What sort of interpretation should be made here?.

  • 2
    $\begingroup$ Actually none of the answers are correct. The IUPAC definition of a molecule notes that a molecule must have two or more atoms. Thus the He atom is not a molecule. The correct answer would be: 0,N,N. $\endgroup$ – MaxW Feb 25 '20 at 15:03
  • $\begingroup$ @MaxW That's exactly what I was thinking but I suspect that the author "forced" beyond the boundaries of such definition to include He. But I agree with your interpretation. $\endgroup$ – Chris Steinbeck Bell Feb 26 '20 at 10:21

Option $(3) N, N, N$ is correct.

According to IUPAC, the definition of one mole is as follows:

The mole, symbol $mol$, is the SI unit of amount of substance. One mole contains exactly $6.02214076 × 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$, when expressed in $mol^{−1}$, and is called the Avogadro number.

Thus, irrespective of the type of gas, 1 mole of its molecules will have the same number of molecules i.e. $N$.

As far as your second interpretation goes, I can't quite understand it. Here are my thoughts.

  • You assigned $N=2 gmol^{-1}$ which seems dimensionally incorrect. Number of molecules would be a dimensionless quantity.

  • Judging by your interpretation, I think you're trying to apply unitary method of some sorts (like if 2 grams of a gas is 1 mole of it, then 4 grams of it must equal 2 moles for all types of gases). But this is incorrect as $2g$ of not every gas equals 1 mole of it. You should go by its definition.

  • $\begingroup$ The source of confusion was what moles were for each gas? Because whoever proposed this question to my understanding did not provided the mass of the gases from which to find the moles. So I assumed that they were intending to say that there was $\textrm{1 mol}$ of each gas. From it, there can be concluded that there was $N$ molecules of each gas which of course is the Avogadro's number. $\endgroup$ – Chris Steinbeck Bell Feb 26 '20 at 10:38
  • $\begingroup$ Because of that "missing" grams, I attempted to use the unitary method was to bypass that ambiguity by assuming that the molar masses could be interpreted as a "floating" proportion hence the method. $\endgroup$ – Chris Steinbeck Bell Feb 26 '20 at 10:40
  • $\begingroup$ The question is referring to moles of molecules of gas. Now you might be thinking about Helium - because the term 'moles of molecules' won't be valid for Helium because it is one atom. However, the definition for 'molecule' in the Merriam Webster dictionary is: "the smallest particle of a substance that retains all the properties of the substance and is composed of one or more atoms." A molecule has to also exist freely and stably, which Helium does. So we would call He(g) a monoatomic molecule. Now the question seems valid, I guess. $\endgroup$ – Vijay Bharadwaj Feb 26 '20 at 10:51
  • $\begingroup$ I haven't had concrete proof for this statement, calling He(g) a molecule, except the formal definition of a molecule, but I and my professor (just asked him about this) have personally always treated 'He' as such. $\endgroup$ – Vijay Bharadwaj Feb 26 '20 at 10:52
  • $\begingroup$ Okay, interesting. The IUPAC goldbook defines a molecule to be of more than one atom without equality. Seems like there is a conflict of knowledge. In that case, the question must refer to the amount of smallest entity of the standard form of each gas, be it molecule or atom. That does call for more information to be provided in the question, you're right. $\endgroup$ – Vijay Bharadwaj Feb 26 '20 at 10:59

There is a fixed number in this universe called the Avogadro's constant.

One mole of a substance always contains this fixed number of entities (in this case: molecules). That is the very definition of "amount of substance" in moles.

Your answer would be $\mathrm {N, N, N}$, without ambiguity.

P.S. I believe you should consider going through the Concept of moles and Avogadro's number. Here's a decent page to get you started.

  • $\begingroup$ I am well aware of the meaning of mole. But in this given context there was no specification of a mass from which to "find" the number of moles. They were only giving the molar mass that's it, and from there I believed that the intended meaning was that there were $\textrm{1 mol}$ of each gas. That was the source of confusion or ambiguity. $\endgroup$ – Chris Steinbeck Bell Feb 26 '20 at 10:26
  • $\begingroup$ "and from there I believed that the intended meaning was that there were 1 mol of each gas." isn't this true? $\endgroup$ – William R. Ebenezer Feb 26 '20 at 12:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.