Question

Copper crystallizes with face centered unit cells. If the radius of the copper atom is 127.8 pm. Calculate the density of the copper metal (atomic mass of Cu = 63.55 u).

Answer 1

Hint: FCC means face centered cubic packing. FCC has six faces and eight corners. There is a relationship between edge length of the unit cell and density of unit cell in FCC and it is as follows.
\[d=\dfrac{Z\times M}{{{a}^{3}}\times {{N}_{A}}}\]
Here d = density of the atom
Z = number of the atoms in the unit cell
M = Molecular weight of atomic mass of the atom
a = edge length of the unit cell
${{N}_{A}}$ = Avogadro number

Complete step by step answer:
> In the question it is given that Copper crystallizes with a face centered unit cell and given the radius of the copper atom and we have to calculate the density of the copper metal.
> First we have to calculate the edge length of the unit cell after that we have to calculate the density of the copper atom.
> The edge length of the unit cell can be calculated as follows for FCC.
\[a=\dfrac{4r}{\sqrt{2}}\to (1)\]
Here, r = atomic radius = 127.8 pm = $1.278\times {{10}^{-8}}cm$
a = edge length of the unit cell.

> Substitute all the known values in the above formula (1) to calculate the edge length of the unit cell in FCC and it is as follows.
\[\begin{align}
 & a=\dfrac{4r}{\sqrt{2}} \\
 & a=2\sqrt{2}r \\
 & a=2\sqrt{2}\times (1.278\times {{10}^{-8}})cm \\
\end{align}\]

> Now substitute the edge length of the unit cell in the below formula to get the radius of the copper metal.
\[d=\dfrac{Z\times M}{{{a}^{3}}\times {{N}_{A}}}\to (2)\]
Here d = density of the atom
Z = number of the atoms in the unit cell = 4 atoms
M = Molecular weight of atomic mass of the atom = 63.55
a = edge length of the unit cell
${{N}_{A}}$ = Avogadro number = $6.023\times {{10}^{23}}$

> Substitute all the known values in the above formula (2) to get the density of the copper in FCC and it is as follows.
\[\begin{align}
& d =\dfrac{Z\times M}{{{a}^{3}}\times {{N}_{A}}} \\
& d =\dfrac{4\times 63.55}{{{[2\sqrt{2}\times (1.278\times {{10}^{-8}})]}^{3}}\times 6.023\times {{10}^{23}}} \\
 & d=\dfrac{254.2}{4.723\times {{10}^{-23}}\times 6.023\times {{10}^{23}}} \\
 & d = 8.94g{{m}^{-3}} \\
\end{align}\]

Therefore, the density of the copper metal is 8.94 $g{{m}^{-3}}$ .

Note: We cannot calculate the density of the metal by using the radius of the metal directly. First we have to calculate the edge length of the unit cell from the radius of the atom and later substitute the edge length in the density formula to get the density of the copper metal atom.