Question

Gold occurs as a face centered cube and it has a density of $19.30kgd{m^3}$. Calculate the atomic radius of gold. (Molar mass of Au = $197gmo{l^{ - 1}}$).

Answer 1

Hint: We have to calculate the numbers of atoms present in the unit cell of FCC atom and then, we have to calculate the mass of the unit cell of FCC type using molar mass of gold and Avogadro number. From the density and mass of the unit cell, we have to calculate the volume of the unit cell. From the volume of the unit cell, we can calculate the edge of the unit cell. The radius of the unit cell is calculated using the edge of the unit cell.

Formula used: We can calculate the atomic radius using the formula,
$r = \dfrac{a}{{\sqrt 8 }}$
Where,
r is radius of the FCC type
a is the edge of the unit cell

Complete step by step answer:
Given data contains,
Gold is a face centered cube
Density of gold is $19.30kgd{m^3}$.
Molar mass of gold is $197gmo{l^{ - 1}}$.
First, we have to calculate the number of atoms present in the unit cell of FCC.
Number of atoms$ = \dfrac{1}{8} \times 8 + 6 \times \dfrac{1}{2}$
Number of atoms = $4$atoms
The number of atoms present in the unit cell of FCC is 4atoms.
Next, we will calculate the mass of the unit cell of FCC using the molar mass of gold and Avogadro number.
The value of Avogadro number is $6.022 \times {10^{23}}$.
Mass of unit cell$ = 4 \times \dfrac{{197}}{{6.022 \times {{10}^{23}}}}$
Mass of unit cell$ = 1.308 \times {10^{ - 22}}g$
Mass of unit cell$ = 1.308 \times {10^{ - 24}}kg$
The mass of the unit cell is $1.308 \times {10^{ - 24}}kg$.
We will now calculate the volume of the unit cell using the density of the unit cell.
We know the formula of density as,
Density of the unit cell$ = \dfrac{{{\text{Mass of unit cell}}}}{{{\text{Volume of unit cell}}}}$
Let us substitute the values of the mass and density of the unit cell
$19.3$=$\dfrac{{1.308 \times {{10}^{ - 24}}}}{{{a^3}}}$
$ \Rightarrow {a^3} = 6.77 \times {10^{ - 26}}\,d{m^3}$
${a^3} = 6.77 \times {10^{ - 23}}c{m^3}$
We shall take the cube root of the volume of the unit cell
${a^3} = 6.77 \times {10^{ - 23}} = 67.77 \times {10^{ - 24}}$
$ \Rightarrow a = \sqrt[3]{{67.77 \times {{10}^{ - 24}}}}cm$
$a = 4.077 \times {10^{ - 8}}cm$
The volume of the unit cell that is the edge length of the unit cell is $4.077 \times {10^{ - 8}}cm$.
We shall calculate the radius of the gold atom using the formula,
$r = \dfrac{a}{{\sqrt 8 }}$
We know that $\sqrt 8 $ can also be written as $2\sqrt 2 $, therefore, the formula becomes,
$r = \dfrac{a}{{2\sqrt 2 }}$
Let us substitute the values of a to calculate the radius of the gold atom
$r = \dfrac{{4.077 \times {{10}^{ - 8}}cm}}{{2\sqrt 2 }}$
On simplifying we get,
$ \Rightarrow r = 1.442 \times {10^{ - 8}}cm$
In terms of picometer, the radius becomes $144.2pm$.
The radius of the unit cell is $144.2pm$ (or)$1.442 \times {10^{ - 8}}cm$.

Note:
An example of a compound that has face-centered cubic lattice is sodium chloride. Lithium fluoride, lithium chloride, Sodium fluoride, potassium fluoride, potassium chloride etc are examples of compounds that contain face centered cubic structures. Simple cubic cell, and body centered cubic cell are the other two crystal structures.