Question

Copper crystallizes with a face centred cubic unit cell. If the radius of copper atoms is $127.8pm$ , calculate the density of copper metal?
(Atomic mass of $Cu = 63.55u$ and Avogadro’s number ${N_A} = 6.02 \times {10^{23}}mo{l^{ - 1}}$ ).

Answer 1

Hint: To answer this question the formula of density has to be known. We know the formula for density is mass upon volume. But for a face centred cubic cell we need to find the mass and volume of a unit cell to know the volume.

Complete step by step answer:
The smallest group of atoms which has the overall symmetry of a crystal and from which the entire lattice can be built up by repetition in three dimensions is termed as Unit cell. It can be seen as a three dimensional structure containing one or more atoms.
Density of a unit cell is given as the ratio of mass and volume of the unit cell. The mass of the unit cell is equal to the product of the number of atoms in a unit cell and the mass of each atom in the unit cell. Let’s first see the derivation of the formula of density for a cubic unit cell for better understanding.
We know that the basic formula of density is mass upon volume.
$\Rightarrow d = \dfrac{m}{V}$
Mass of unit cell = number of atoms in unit cell $ \times $ mass of each atom
$ = z \times m$
Mass of an atom can be given with the help of Avogadro number and molar mass as,
$ = \dfrac{{MolarMass}}{{Avogadro'sNumber}}$
$ = \dfrac{M}{{{N_A}}}$
Volume of unit cell, $V = {a^3}$
Therefore, Density is mass upon volume,
$\Rightarrow Density = \dfrac{m}{V} = \dfrac{{z \times m}}{{{a^3}}} = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}$
So, with the knowledge of the number of atoms in a unit cell, edge length and molar mass we can determine the density of a unit cell. Now, we know the formula and we have to substitute values in the formula. So, as per values given in question,
Number of atoms in unit cell, $Z = 4$
This is because in a face centred unit cell the number of atoms in a unit cell is equal to four.
Radius, $r = 127.8pm$
For face centered cubic,
$r = \dfrac{a}{{2\sqrt 2 }}$
Here $a$ is the side of the cell.
$\Rightarrow 127.8 \times 2\sqrt 2 = a$
$\Rightarrow a = 361.42pm$
$\Rightarrow a = 361.41 \times {10^{ - 10}}cm$
Now, Atomic mass of cell is $63.55g/mol$
Now, substituting the values in the formula to find density we get,
\[\Rightarrow Density = \dfrac{{Z \times M}}{{{a^3} \times {N_A}}}\]
\[ = \dfrac{{4 \times 63.55}}{{{{(361.41 \times {{10}^{ - 10}})}^3} \times 6.022 \times {{10}^{23}}}}\]
\[ = 8.9 \times {10^{ - 14}}g/c{m^3}\]

Note:
Face centred cubic is an arrangement of atoms in crystals in which the atomic centres are disposed in space in such a way that one atom is located at each of the corners of the cube and one at the centre of each face.