Mole Concept

This article is more about the SI unit of amount of substance mole, rather than the unpleasant animal. However, as we shall discover, the mole is better described as a certain number of something, rather than a certain amount.

Let's say that, for example, we are dealing with molecules of hydrogen and oxygen.

As we know, the molecular mass of 1 hydrogen molecule is $2\text{ u}$. There is certainly a number $n$ of hydrogen molecules that would make up 2 grams. Again, the molecular mass of 1 oxygen molecule is $32\text{ u}$. There is certainly a number $m$ of oxygen molecules that would make up 32 grams. We claim that $m=n$. Why?

Because $n \times 2\text{ u} = 2\text{ g}$ implies $n=1\text{ g/u}$, and similarly we have $m = 1\text{ g/u}$, which shows that $m=n$.

This is not just true for a couple of examples, but for all kinds of species. In fact, this ratio, i.e. the number of unified mass units in a gram, is called the Avogadro constant $(N_0)$. An Avogadro's number of anything is called a mole of the same.

So, not just chemicals! We could count hairs in moles, and probably someday, our earth's population will reach a mole!

1 mole of a substance is defined to be the amount of the substance containing as many atoms, molecules, ions, electrons or other elementary entities as there are carbon atoms in 12 grams of $^{12}\ce{C}.$

This number is known as the Avogadro constant $($abbreviated $N_0$ or $N_A).$ It has been experimentally determined that

$N_A = 6.022140 \times 10^{23}.$

So, technically, the earth's population would only be $10^{-14}\text{ mole}$.

The Avogadro hypothesis states the following:

At constant pressure and temperature, equal volumes of gases contain equal numbers of molecules.

(It is possible to derive the Avogadro hypothesis from the foundations of kinetic theory, but we omit that derivation here.)

Interestingly, this tells us that there is a fixed volume that a mole of any gas occupies. This value turns out to be

$V_m = 22.41\text{ L/mol}$

at $0^\circ\text{C}$ and $1\text{ atm}.$ This is called the STP (standard temperature and pressure) condition.

The charge carried by 1 mol of electrons is called 1 Faraday, which is

$F = N_{\rm A} e \approx 96485\text{ C/mol}.$

The gas constant is the thermal energy carried by a mole of molecules per unit temperature, which is

$R = k_{\rm B} N_{\rm A} \approx 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\approx0.0821\text{ L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.$

The gas constant divided by the Avogadro constant is called the Boltzmann constant. It was named after Ludwig Boltzmann.

The essence of mole concept is that it relates the following three ideas:

1. an Avogadro number of anything
2. an amount of substance which is equal to the atomic mass numerically in grams
3. $22.4$ liters of a gas at STP.

The author recommends that you actually understand this equivalence as opposed to memorizing a bunch of formulae.

Calculate the number of molecules and atoms in $24\text{ g}$ of ozone.

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Since the molecular mass of ozone is $3 \times 16=48,$ the number of moles of ozone molecules in 24 grams is $24\div48=0.5.$ Note that we just calculated the moles of ozone molecules present, not the moles of oxygen atoms.

Hence the number of ozone molecules is $0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}.$

Since there are three atoms in each molecule, the total number of atoms would be $9.033 \times 10^{23}.$ $_\square$

Calculate the number of moles of $\ce{O3}$ in one molecule of $\ce{O3}$. Take Avogadro's number as $6 × 10^{23}$.

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By Avogadro's law, the number of molecules in $1$ mole of $\ce{O3}$ is $6 × 10^{23}$. So, the number of moles in $1$ molecule of $\ce{O3}$ is $\dfrac {1}{6 × 10^{23}}= 1.67×10^{-24}$. $_\square$

Calculate the amount of lead which can be obtained from $6.85\text{ kg}$ of red lead $\big(\ce{Pb_3O_4}\big).$ $\ \ [\ce{Pb} = 207; \ce{O} = 16]$

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We have

$\ce {Pb_3O_4 -> 3Pb}.$

Gram molecular mass of $\ce{Pb_3O_4}:$ $3(207) + 4(16) = 685\text{ (g)}$
Gram molecular mass of $\ce{3Pb}:$ $3(207) = 621\text{ (g)}$

Now, $685\text{ g}$ of $\ce{Pb_3O_4}$ yields $621\text{ g}$ of lead metal.

So, $6.85\text{ kg}$ of $\ce{Pb_3O_4}$ yields $\dfrac {621×6.85}{685} = 6.21\text{ (kg)}$. $_\square$

The chemical formula of a compound which contains the actual number of atoms of each element is called its molecular formula.

The chemical formula of a compound which contains the simplest ratio of atoms of each element is called empirical formula.

The molecular formula of glucose is $\ce{C_6H_12O_6}$ while its empirical formula is $\ce{C H2 O}$.

• Concentration Terminology - Molarity, Molality, and Normality