Chapter 10, Problem 148IP

A metal burns in air at 600°c under high pressure to form an oxide with formula MO₂. This compound is 23.72% oxygen by mass. The distance between the centers of touching atoms in a cubic closest packed crystal of this metal is 269.0 pm. What is this metal? What is its density?

Interpretation Introduction

Interpretation:

             The metal has to be identified and its density has to be calculated given its percentage mass and distance between the ionic centers of the compound of that element.

Concept introduction:

             In packing of atoms in a crystal structure, the atoms are imagined as spheres. 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,

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

          In FCC unit cell the atoms are assumed to touch along the face diagonal of the cube, so the face diagonal $=\sqrt{2}l$ $\text{=}\text{4r}$. The distance between the atomic/ionic centers is $2\text{r}$ where

 ‘ $\text{r}$’ is the atomic radius.

Answer to Problem 148IP

• The metal is identified as Rhodium.

• The density of Rhodium is calculated as 12.4 g/cm³.

Explanation of Solution

Identify the element using given data.

Given data:  The element burns in air and forms ${\text{MO}}_{\text{2}}$ oxide. It has 23.72 % oxygen by mass.

Assume 100.00g ${\text{MO}}_{\text{2}}$ produced.

                        $\begin{array}{l}{\text{MO}}_{\text{2}}\text{oxide}\text{has}\text{23}\text{.72}\text{g}\text{of}\text{oxygen}\text{.}\\ \text{No}\text{.}\text{of}\text{moles}\text{=}\frac{\text{weight}}{\text{Molecular}\text{weight}}\\ \text{No}\text{.of}\text{moles}\text{of}\text{oxygen}\text{in}\text{23}\text{.72}\text{g}\text{of}\text{Oxygen}\text{is,}\\ \frac{\text{23}\text{.72}\text{g}\text{of}\text{Oxygen}}{\text{16}\text{.00}\text{g}}\text{=}\text{1}\text{.482}\text{mol}\text{of}\text{Oxygen}\\ \text{no}\text{.of moles}\text{of}\text{one}\text{oxygen}\text{atom}\text{=}\frac{\text{1}\text{.482}}{\text{2}}\text{=}\text{0}\text{.741}\text{mol}\\ \text{mass}\text{of}\text{metal}\text{=}\text{100}\text{g}\text{-}\text{mass}\text{of}\text{oxygen}\\ \text{=}\text{100-}\text{23}\text{.72}\text{=}\text{76}\text{.28}\text{g}\\ \text{Weight}\text{of}\text{the}\text{metal}\text{=}\frac{\text{weight}\text{of}\text{the}\text{metal}\text{formed}}{\text{no}\text{.of}\text{moles}}\\ \text{=}\frac{\text{76}\text{.28}}{\text{0}\text{.741}\text{mol}}\text{=}\text{102}\text{.94}\text{g/mol}\\ \\ \end{array}$

       The obtained value of weight is consistent with the atomic weight of Rhodium.

           The atomic weight of the metal is calculated using given data to identify the metal. Since the ${\text{MO}}_{\text{2}}$ oxide has 23.72% of Oxygen by mass, the remaining part must be that of metal. By using this concept, the number of moles of oxygen formed in the product has been determined. Using the number of moles of Oxygen, mass of the metal is determined and it is found to be in consistent with that of Rhodium. Hence the metal is identified as Rhodium.

Determine the volume of FCC unit cell.

$\text{volume}\text{of}\text{FCC}\text{unit}\text{cell}\text{=}{\text{l}}^{\text{3}}$

$\text{given}\text{data:}\text{r}\text{=}\text{269 pm}\text{=}\text{269}\text{×}{\text{10}}^{\text{-12}}\text{m}$

                $\begin{array}{l}\text{volume}\text{of}\text{FCC}\text{unit}\text{cell}\text{=}{\text{l}}^{\text{3}}\\ \\ \text{given}\text{data:}\text{r}\text{=}\text{269 pm}\text{=}\text{269}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \\ \text{face}\text{diagonal}\text{=}\sqrt{\text{2}}\text{l}\text{=}\text{4r}\\ \text{since}\text{the distance between ionic centers is 2r,}\\ \text{face}\text{diagonal}\text{=}\sqrt{\text{2}}\text{l}\text{=}\text{2r}\text{=}\text{2×}\text{269}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{=}\text{538}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{l}\text{=}\frac{\text{538}\text{×}{\text{10}}^{\text{-12}}\text{m}}{\sqrt{\text{2}}}\text{=}\text{380}\text{.5}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ {\text{l}}^{\text{3}}\text{=}{\left(\text{380}\text{.5}\text{×}{\text{10}}^{\text{-12}}\text{m}\right)}^{\text{3}}\\ \text{=}\text{5}\text{.51}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\\ \end{array}$

         The distance between ionic centers is given and the volume is calculated by determining the edge length of the unit cell.

Determine the mass of the FCC unit cell.

              $\begin{array}{l}\text{Average mass of one Rh atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}Rh}{\text{Avogadro}\text{number}}\\ \text{=}\frac{102.94\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}17.09\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

               Each unit cell has 4 Rhodium atoms. Therefore,

                $\begin{array}{l}\text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{4}\text{×}\text{average}\text{mass}\text{of}\text{one}\text{Rh}\text{atom}\\ \text{=}\text{4}\text{×}\text{17}\text{.09}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{68}\text{.4}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

      Each unit cell contains 4 Rhodium atoms. Therefore four times the average mass of one Rhodium atom gives mass of a unit cell.

Determine the density of metal.

                    $\begin{array}{l}\text{known}\text{data}\text{:}\text{mass}\text{=}\text{68}\text{.4}\text{g}\\ \text{volume}\text{=}\text{5}\text{.51}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\\ \\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{\text{68}\text{.4}\text{×}{\text{10}}^{\text{-23}}\text{g}}{\text{5}\text{.51}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}}\text{=}\text{12}\text{.4}{\text{g/cm}}^{\text{3}}\end{array}$

                 Mass and volume of the unit cell is calculated in the previous steps. By substituting the values in the formula $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$, the density of the Rhodium is determined.

Conclusion

The metal is identified as Rhodium and its density is calculated