Chapter 18, Problem 18.93QA

Interpretation Introduction

To find:

a) Sketch the unit cell of $Re{O}_3$.

b) Calculate the density of $Re{O}_3$.

c) Calculate the percent empty space in a unit cell of $Re{O}_3$.

Answer to Problem 18.93QA

Solution:

a)

b) $density= 5.25\frac{g}{c{m}^3}$

c) $empty space=79\%$

Explanation of Solution

1) Concept:

a) Unit cell of $Re{O}_3$: It consists of a cube with rhenium atoms at corners and an oxygen atom on each of the twelve edges. $Re$ atoms are shown by blue circles, and the $O$ atom is shown by a red circle. Therefore, this unit cell is a combination of a simple cubic cell of Rhenium with a simple cubic cell of oxygen.

b) $Re and O$ atoms touch along the edge of the unit cell. There are two $Re$ and one $O$ atoms on an edge length. From the atomic radius of both $Re and O$ atoms, we can get the edge length. Using edge length, we can calculate the volume of a unit cell. From molar mass, Avogadro’s number, and number of atoms per unit cell, we can get the mass of a unit cell. Using mass and volume of a unit cell, we can calculate its density.

c) The packing efficiency is calculated using volumes of one $Re$ sphere and three $O$ spheres and volume of unit cell. Empty space available can be calculated using the total volume of a unit cell and the packing efficiency.

2) Formula:

i) $volume=l^3$    

ii) $density=\frac{mass}{volume}$

iii) $V of sphere=\frac{4}{3}\pi r^3$

iv) $packing efficiency= =\frac{V of spheres}{V of a unit cell}\times 100\%$

3) Given:

i) $atomic radius of Re=137 pm$      

ii) $atomic radius of O=73 pm$      

4) Calculations:

b) $Re and O$ atoms touch along the edge of the unit cell. There are two $Re and one O$ atoms on an edge length. So, the edge length of a unit cell is

$l=\left(2\times radius of Re\right)+diameter of O$

$l=\left(2\times 137 pm\right)+\left(2\times 73 pm\right)=274 pm+146 pm=420 pm$

Volume of a unit cell:$V=l^3={\left(420 pm\times \frac{1 m}{1\times {10}^{12}pm}\times \frac{100 cm}{1 m}\right)}^3=7.4088\times {10}^{-23}c{m}^3$

Molar mass of $Re{O}_3$:

$=(1 mol Re\times atomic mass of Re)+\left(3 mol O\times atomic mass of O\right)$

$=\left(1 mol\times 186.21\frac{g}{mol}\right)+\left(3 mol O\times 16.00\frac{g}{mol}\right)=234.21\frac{g}{mol}$

Mass of a unit cell of $Re{O}_3$= mass of 1 molecule of $Re{O}_3$

$=234.21\frac{g}{mol}\times \frac{1 mol}{6.022\times {10}^{23}molecules}\times 1 molecule=3.889\times {10}^{-22} g$

$density=\frac{mass}{volume}=\frac{3.889\times {10}^{-22} g}{7.4088\times {10}^{-23}c{m}^3}=5.25\frac{g}{c{m}^3}$

Volume of $Re atom$:

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\times 3.14\times {\left(137\times \frac{1 m}{1\times {10}^{12}m}\times \frac{100 cm}{1 m}\right)}^3=1.076\times {10}^{-23}c{m}^3$

Volume of $O atom$:

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\times 3.14\times {\left(73\times \frac{1 m}{1\times {10}^{12}m}\times \frac{100 cm}{1 m}\right)}^3=1.6287\times {10}^{-24}c{m}^3$

So total volume of a $Re{O}_3=\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }Re+\left(3\times \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }O\right)$

$=1.076\times {10}^{-23}c{m}^3+\left(3\times 1.6287\times {10}^{-24}c{m}^3\right)$

$=1.565\times 10^{-23}c{m}^3$

So,$Packing effeciency=\frac{V of Re{O}_3}{V of a unit cell}\times 100\%=\frac{1.565\times 10^{-23}c{m}^3}{7.41\times {10}^{-23}c{m}^3}\times 100\%=21\%$

This is the space occupied by $Re{O}_3$ in a unit cell. If the total space in a unit cell is 100%, then the empty space available in a unit cell

$=100\%-21\%=79\%$

Conclusion:

From the diagram of unit cell of $Re{O}_3$, we got the number of atoms per unit cell and its edge length. From this, we calculated volume, density, and the empty space available in a unit cell.