Question

Mn crystallizes in the same type of cubic unit cell as Cu. Assuming that the radius of Mn is 5.6% larger than the radius of Cu and the density of Cu is 8.96g/cm³, calculate the density of Mn.

Short answer

The density of Mn is equal to$6.65g/c{m}^3$

Step-by-step solution

Determine the mass and volume of copper

The total numberof atoms per unit cell copper is four atoms.

Massof copper:

$$\begin{gathered} mass of Copper=\frac{4}{6.022\times {10}^{22}mo{l}^{-1}}\times 63.55g/mol \\ =4.22\times {10}^{-22}g \end{gathered}$$

$$\begin{gathered} Volume of copper=\frac{mass of copper}{density} \\ =\frac{4.22\times {10}^{-22}g}{8.96g/c{m}^3} \\ =4.71\times {10}^{-23}c{m}^3 \end{gathered}$$

Determine the edge length and radius of copper

$$\begin{gathered} Edge length=\sqrt[3]{Volume} \\ =\sqrt[3]{4.71\times {10}^{-23}c{m}^3} \\ =361pm \end{gathered}$$

$$\begin{gathered} Radius of Cu=\frac{edge length}{\sqrt{8}} \\ =\frac{361pm}{\sqrt{8}} \\ =127.6pm \end{gathered}$$

Determine the edge length and volume of Mn

It is given that the radius of Mn is 5.6% larger than the radius of Cu.

So, the radius of Mn is $127.6pm+0.56\times 127.6pm=134.7pm$

$$\begin{gathered} edge length=\sqrt{8}\text{radius} \\ =\sqrt{8}\times 134.7 \\ =3.8\times {10}^{-8}cm \end{gathered}$$

$$\begin{gathered} Volume=({\text{edge length)}}^3 \\ ={\left(3.8\times {10}^{-8}cm\right)}^3 \\ =5.49\times {10}^{-23}c{m}^3 \end{gathered}$$

Determine the mass and density of manganese Mn.

Mass of Mn:

$$\begin{gathered} Mn=\frac{4}{6.022\times {10}^{22}mo{l}^{-1}}\times 54.94g/mol \\ =3.65\times {10}^{-22}g \end{gathered}$$

Density of Mn

$$\begin{gathered} Mn=\frac{3.65\times {10}^{-22}g}{5.49\times {10}^{-23}c{m}^3} \\ =6.65g/c{m}^3 \end{gathered}$$

Hence the densityof Mn is 6.65g/cm³.