Concept of Mole: From Atomic Scale to Large Scale

PREVIOUS: What is amu?

Using the concept of amu from the previous post, and a bit of creative thinking, we try to devise a technique to count the number of atoms on large scale.

Let’s consider hydrogen gas. One hydrogen atom measures $1 \, amu = 1.66 \times 10^{-24} gm$, same as mass of a proton. Now we ask: If we took $1 \, gm$ of hydrogen atom, exactly how many atoms would be in it? This is a straightforward calculation:
$$\begin{array}{l} \mbox{Number of hydrogen atoms in 1 gm} &=& \frac{\mbox{1 gm}}{\mbox{mass of one hydrogen atom in grams}} \\ & = & \frac{1 gm}{1.66 \times 10^{-24} gm} \\ & = & 6.023 \times 10^{23} \end{array}$$ Note that this is number of atoms and has no units.

Consider the case of helium gas, $^{4}_{2}He$. An atom of helium has 4 nucleons; therefore, its atomic mass is 4 amu. The mass of $6.023 \times 10^{23}$ helium atoms will be
$$\begin{array}{l} 6.023 \times 10^{23} \times 4 \, amu &=& 6.023 \times 10^{23} \times (4 \times 1.66 \times 10^{-24} \, gm) \\ &=& 4 \, gm \end{array}$$

Consider oxygen, $^{16}_{8}O$. An atom of oxygen has 16 nucleons (8 protons and 8 neutrons); therefore, its atomic mass is 16 amu. The mass of $6.023 \times 10^{23}$ oxygen atoms will be $16 \times 6.023 \times 10^{23} \times 1.66 \times 10^{-24} \, gm = 16 \, gm$.

Repeat the calculation for sodium, $^{23}_{11}Na$, which has 23 nucleons (11 protons and 12 neutrons), and you will find that $6.023 \times 10^{23}$ atoms of sodium amount to $23 \, gm$.

The number $6.023 \times 10^{23}$, therefore, has special importance in science. It is referred to as Avogadro's number. It is also known as a mole. One mole of an element has $6.023 \times 10^{23}$ number of atoms. In general, an element $^{A}_{Z}X$ that has mass number $A$, $1 \, mole$ of atoms of that element will have mass $A \, gm$. This realization provides us with a powerful tool for estimating the number of atoms of an element in a given mass.

For instance, say we have been given $20 \, gm$ of sodium $^{23}_{11}Na$ to work with and we want to find out exactly how many sodium atoms there are in $20 \, gm$. We know that there are $1 \, mole$ atoms in $23 \, gm$ of sodium. Therefore, in $20 \, gm$ of sodium there must be $20/23 = 0.8695 \, moles$ atoms, i.e., $0.8695 \times 6.023 \times 10^{23} = 5.237 \times 10^{23}$ atoms.

SUMMARY:

• $1 \, amu = 1.66 \times 10^{-24} \, gm = \mbox{mass of one proton (or neutron)} = \mbox{mass of one hydrogen atom}$
  
  
• $\displaystyle 1 \, mole = \frac{\mbox{1 gm}}{\mbox{mass of one nucleon (or mass of one hydrogen atom) in grams}} =\frac{1 \, gm}{1.66 \times 10^{-24} \, gm}= 6.023 \times 10^{23}$
  
  
• $1 \, mole \, \mbox{of element} \, ^{A}_{Z}X \, \mbox{has mass of} \, A \, gm$.