12.93P - Chemistry: Molecular Nature Of Matter And Change

Question

An element crystallizes in a face-centered cubic lattice and has a density of $\mathbf{1}\mathbf{.}\mathbf{45}\mathbf{ }\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$. The edge of its unit cell is 4.52108 cm.

(a) How many atoms are in each unit cell?

(b) What is the volume of a unit cell?

(c) What is the mass of a unit cell?

(d) Calculate an approximate atomic mass for the element.

Step-by-Step Solution

Answer

1. Four atoms in each unit cell
2. Volume of unit cell is $\mathbf{92}\mathbf{.}\mathbf{41161}\mathbf{ }{\mathbf{cm}}^{\mathbf{3}}$.
3. Mass of unit cell is $\mathbf{16}\mathbf{.}\mathbf{45796}\mathbf{ }\mathbf{g}$.
4. Atomic mass for element is $\mathbf{4}\mathbf{.}\mathbf{11449}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{ }\mathbf{g}\mathbf{ }$

Step 1: Face-centered cubic unit cell

The face-centered cubic unit cell consists of 8 atoms in the corner, thus one atom contributing 1/8 to the unit cell. Along with that, atoms are present in the center of the six respective faces of the cell. Hereby, each atom contributes $\frac{1}{2}$ to the unit cell.

Thus, each unit cell consists of $8\times \left(\frac{1}{8}\right)+6\times \left(\frac{1}{2}\right)=4$ , i.e., four atoms in the unit cell.

Step 2: Volume of the unit cell

The unit cell being a cube have volume of ${\left(\mathrm{edge}\right)}^3$ thus ${\left(4.52108 \mathrm{cm}\right)}^3$, calculated to be $\mathbf{92}\mathbf{.}\mathbf{41161}\mathbf{ }{\mathbf{cm}}^{\mathbf{3}}$.

Step 3: Mass of the unit cell

As the face-centered cubic unit cell contributes for 4 atoms, the mass of the cubic unit cell will be four times the mass of the atom.

The face- centered cubic unit cell has the following pictorial representation:

Thus, considering the diagonal of face, $\sqrt{2}$ times of the edge i.e.,$\sqrt{2}\times 4.52108\mathrm{cm}=6.39280\mathrm{cm}$ .

Now, as each diagonal contributes for two times the diameter (2r) of the atom, thus

$2\times 2\mathrm{r}=6.39280\mathrm{cm}$

or, data-custom-editor="chemistry" $\mathrm{r}=1.59280\mathrm{cm}$

Thus, the radius of the atom is 1.59820 cm.

 

 

 Now, the mass of the atom is calculated as,

$$\begin{gathered} \mathrm{Mass}=\mathrm{Density}\times \mathrm{Volume} \\ \mathrm{Mass}=1.45\times \frac{4}{3}\times \frac{22}{7}\times {\left(1.59820\right)}^3=24.78162 \mathrm{g}/\mathrm{mol} \\ \mathrm{Atomic} \mathrm{mass}\left(\mathrm{g}\right)=4.11449\times {10}^{-23} \end{gathered}$$

As the face-centered cubic unit cell contributes for 4 atoms and the density of unit cell is already calculated, the mass of a single atom will be:

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{unit} \mathrm{cell}=\mathrm{Mass} \mathrm{of} \mathrm{atom}\times 4 \\ \mathrm{Mass} \mathrm{of} \mathrm{unit} \mathrm{cell}=16.45796\times {10}^{-23} \mathrm{g} \end{gathered}$$

Step-4: Calculation of atomic mass of element

The atomic mass of the element is $4.11449\times {10}^{-23} \mathrm{g}$