Chapter 9, Problem 55E

You are given a small bar of an unknown metal X. You find the density of the metal to be 10.5 g/cm³. An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.09 Å (1 Å = 10⁻¹⁰ m). Identify X.

Interpretation Introduction

Interpretation:

            A metal has to be identified given its density and edge length of FCC unit cell.

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 55E

Answer

           The metal is identified as “Silver”.

Explanation of Solution

Explanation

Calculate the volume of unit cell of unknown metal.

$\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}\text{4}\text{.09}\text{×}{\text{10}}^{\text{-10}}\text{m}\\ \text{volume,}{\text{a}}^{\text{3}}\text{=}{\text{(4}}^{\text{.09}}\\ \text{=}\text{6}\text{.84}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

             Density of the titanium is given and mass of the titanium is calculated in the previous step. Substituting these two values in the equation $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$ and rearranging it, volume of the unit cell is obtained.

Calculate the mass of unit cell of unknown metal.

$\begin{array}{l}\text{known}\text{data}\text{: density = 10}{\text{.5 g/cm}}^{\text{3}}\\ \text{volume}=\text{6}\text{.84}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{mass}=\text{volume}\text{×}\text{density}\\ \text{=}\text{6}\text{.84}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{×}10.5{\text{g/cm}}^{\text{3}}\\ \text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}71.82\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

             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 unknown metal.

$\begin{array}{l}\text{Average mass of one metal atom}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{4}}\\ \text{=}\frac{71.82\text{×}{\text{10}}^{\text{-23}}}{\text{4}}\\ \text{=}17.95\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \\ \text{atomic}\text{mass}\text{of}\text{metal}\text{=}\text{Average mass of one metal atom}\text{×}\text{Avogadro}\text{number}\\ \text{=}17.95\text{×}{\text{10}}^{\text{-23}}\text{×}\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\\ \text{=}108.09\text{u}\\ \end{array}$

            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 “Silver”.

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

Conclusion

Conclusion

           Density and edge length of FCC unit cell of a metal is given and it is identified as “Silver”.