Chapter 9, Problem 56E

A metallic solid with atoms in a face-centered cubic unit cell with an edge length of 392 pm has a density of 21.45 g/cm³. Calculate the atomic mass and the atomic radius of the metal. Identify the metal.

Interpretation Introduction

Interpretation:

• Atomic radius of the metal has to be calculated.
• Atomic mass of the metal has to be calculated.
• The metal has to be identified.

Concept introduction:

            In crystalline solids, the components are packed in regular pattern and neatly stacked. The components are imagined as spheres and closely packed. This phenomenon is called “close packing” in crystals. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell.

           In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

           Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

                     $\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{6}\text{×}\frac{\text{1}}{\text{2}}\text{atoms}\text{in}\text{faces}\text{=}\text{1}\text{+}\text{3}\text{=}\text{4}\text{atoms}\\ \\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge}\text{length of unit cell}\\ \text{R}\text{=}\text{atomic}\text{radius}\text{.}\\ \end{array}$

Answer to Problem 56E

Answer

Atomic radius of the metal is 1.4 $\stackrel{\text{°}}{\text{A}}.$

Atomic mass of the metal is 194 u

The metal is identified as “Platinum”.

Explanation of Solution

Explanation

Calculate the atomic radius of metal.

$\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}\text{392}\text{pm}\text{=}\text{392}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\end{array}$

$\begin{array}{l}\\ \text{R}\text{=}\frac{\text{a}}{\text{2}\sqrt{\text{2}}}\\ \text{=}\frac{\text{392}\text{×}{\text{10}}^{\text{-12}}\text{m}}{\text{2}\text{×}\text{1}\text{.414}}\\ \text{=}\text{138}\text{.6}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{R}\text{=}\text{138}\text{.6}\text{pm}\end{array}$

           The edge length value for the FCC unit cell is given. The edge length value is related to the radius of the atom by the equation $\text{a}\text{=}\text{2R}\sqrt{\text{2}}$. The given value for ‘a’ is substituted and the equation is rearranged to obtain atomic radius ‘R’ of the metal.

Calculate the mass of unit cell of metal.

$\begin{array}{l}\text{given data: density = 21}{\text{.45 g/cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{mass}\text{=}\text{volume}\text{×}\text{density}\\ \text{=}\text{6}\text{.024}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{×}\text{21}\text{.45}{\text{g/cm}}^{\text{3}}\\ \text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{129}\text{.21}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

       Density of the unit cell is given. The mass of unit cell is calculated using the equation $\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\text{.}\text{The values are substituted and the equation is rearranged to obtain the}\\ \text{ mass value}\text{.}\end{array}$

Calculate the atomic mass of metal.

$\begin{array}{l}\text{Average mass of one metal atom}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{4}}\\ \text{=}\frac{\text{129}\text{.21}\text{×}{\text{10}}^{\text{-23}}}{\text{4}}\\ \text{=}\text{32}\text{.3}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\begin{array}{l}\text{Average mass of one metal atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{metal}}{\text{Avogadro}\text{number}}\\ \text{atomic}\text{mass}\text{of}\text{metal}\text{=}\text{Average mass of one metal atom}\text{×}\text{Avogadro}\text{number}\end{array}$

                                                      $\begin{array}{l}\text{=}\text{32}\text{.3}\text{×}{\text{10}}^{\text{-23}}\text{×}\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\\ \text{=}\text{194}\text{.5}\text{u}\\ \end{array}$

               In FCC unit cell of a metal, each unit cell has 4 metal atoms. So each metal atom has an average mass of 1/4^th of the mass of unit cell. The atomic mass of a metal atom corresponds to the Avogadro number of the average mass of a metal atom.

Identify the metal corresponding to the atomic mass calculated.

             The identified metal is “Platinum”.

             Platinum is the metal that possibly agrees best with the calculated atomic mass.

Conclusion

Conclusion

• Atomic radius of the metal is calculated.
• Atomic mass of the metal is calculated.
• The metal is identified.