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:

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