Question

Gold occurs as a face centred cube and it has a density of 19.30 kgdm$^{-3}$. Calculate the atomic radius of gold. (Molar mass of Au = 197)

Answer 1

Hint : In the crystal lattice, the terms are interrelated to each other. With the help of density, we can calculate the volume of the unit cell. We can calculate the atomic radius for a face centred cubic unit cell.

Complete answer :
Firstly, we know the density of gold atoms is 19.3 kgdm$^{-3}$, and the molar mass of Au is 197 g/ mol.
As we know, in the fcc unit cell, there will be 8 atoms of Au at 8 corners, and 6 atoms at 6 face centres of the Au atom.
Thus, the number of atoms in the Au unit cell will be 4, i.e. $\dfrac{1}{8}$ $\times$ 8 + $\dfrac{1}{2}$ $\times$ 6 = 4
We can calculate the mass of 1 Au atom, Avogadro constant = 6.022 $\times$ 10$^{23}$
So, mass of one Au atom = $\dfrac{197}{6.022 \times 10^{23}}$ = 3.271 $\times$ 10$^{-22}$ g
Then, Mass of 4 Au atoms will be 4 $\times$ 3.271 $\times$ 10$^{-22}$ g = 1.308 $\times$ 10$^{-22}$ g
Now, we have the mass of a unit cell i.e. 1.308 $\times$ 10$^{-22}$ g = 1.308 $\times$ 10$^{-24}$ kg
Let’s calculate the volume of a unit cell, as we have the mass of a unit cell, and density too. Thus, Density =$\dfrac{Mass}{volume}$, let volume of unit cell be a$^{3}$.
Now, a$^{3}$ = 6.77$\times$ 10$^{-26}$ dm$^{3}$ = 6.77$\times$ 10$^{-23}$ cm$^{3}$
Then, a = 4.077 $\times$ 10$^{-8}$ cm
If r is the radius of Au atom, we know for the fcc unit cell, r =$\dfrac{a}{2\sqrt{2}}$,
Substituting the value of a, we get r = 1.442 $\times$ 10$^{-8}$ = 144.2pm.
So, in the last, we can conclude that the atomic radius of Au (gold) atom = 144.2pm

Note : The calculation in this question is a bit difficult. There is no need for confusion; just solve it down step by step. Firstly, calculate the number of atoms, then the mass of a unit cell, further proceeding with the volume of a unit cell. Then, the volume of a unit cell will lead towards the atomic radius of a unit atom.