Chapter 12.6, Problem 3.4ACP

Interpretation Introduction

Interpretation:

The percentage of the space occupied by the tin atoms in tetragonal and cubic crystal lattice has to be calculated.

Concept introduction:

The volume of the atom is given below,

$\text{V}\text{=}{\text{(a)}}^{\text{3}}$

The density of the unit cell is calculated as follows,

$\frac{\text{Mass}}{\text{volume}}\text{ = density}$

$\begin{array}{l}\text{Mass = number of moles × molar mass}\hfill \\ \text{Number of moles = }\frac{\text{Number of atoms per unit cell}}{\text{Avogadro number}}\hfill \end{array}$

The Volume of the one atom is given below

$\text{ V = }\frac{\text{4}}{\text{3}}{\text{Πr}}^{\text{3}}$

Answer to Problem 3.4ACP

• The percentage of the space occupied by the tin atoms in tetragonal is = $\text{43}\text{.5}\text{\%}$
• The percentage of the space occupied by the tin atoms in cubic crystal lattice is =$\text{34}\text{.4}\text{\%}$

Explanation of Solution

Number of atoms per unit is given below,

It has 8 atoms in the corner, 6 atom in the face, 4 atoms in the body.

$\begin{array}{l}\text{Number of atoms per unit}\text{=(8}\text{corner}\text{atoms}\text{×}\text{1/8}\text{+}\text{6}\text{face}\text{atom}\text{×}\text{1/2}\text{+4}\text{body}\text{atoms}\text{×}\text{1)}\\ \text{Number of atoms per unit}\text{=}\text{8}\end{array}$$\begin{array}{l}\text{density =}\text{5}\text{.769}{\text{g/cm}}^{\text{3}}\\ \text{molar}\text{mass}\text{of}\text{tin}\text{=}\text{118}\text{.710 g/mol}\end{array}$

Z = number of atoms per unit.

The Avogadro number ${\text{N}}_{\text{A}}=6.02\times {10}^{23}$

$\begin{array}{l}\text{Mass}\text{of}\text{the}\text{unit}\text{cell}\text{=}\frac{\text{8 × 118}\text{.710}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}}\\ \text{Mass}\text{of}\text{the}\text{unit}\text{cell}\text{=}1.5767\times {\text{10}}^{\text{-21}}g\end{array}$

$\begin{array}{l}\text{Volume}\text{=}\frac{\text{Mass}}{\text{Density}}\\ \text{Volume}\text{=}\frac{\text{1}\text{.5767}{\text{×10}}^{\text{-21}}\text{g}}{\text{5}\text{.769}{\text{g/cm}}^{\text{3}}}\\ \text{Volume}\text{=}\text{2}\text{.733}{\text{×10}}^{\text{-22}}{\text{cm}}^{\text{3}}\\ \text{Volume}\text{=}\text{2}\text{.733}{\text{×10}}^{\text{8}}{\text{pm}}^{\text{3}}\end{array}$

Volume of the one atom is given below

$\text{ V = }\frac{\text{4}}{\text{3}}{\text{Πr}}^{\text{3}}$

Therefore, Volume of the eight atoms is given below,

$\begin{array}{l}\text{ V = 8×}\frac{\text{4}}{\text{3}}{\text{Πr}}^{\text{3}}\\ \text{where,}\\ \text{r=141pm}\\ \text{V = 8×}\frac{\text{4}}{\text{3}}{\text{Π(141 pm)}}^{\text{3}}\\ \text{V = 9}{\text{.397×10}}^{\text{7}}{\text{ pm}}^{\text{3}}\end{array}$

The percentage of space occupied is

$\begin{array}{c}\text{The percentage of space occupied is}\text{=}\frac{\text{9}{\text{.397×10}}^{\text{7}}{\text{ pm}}^{\text{3}}}{\text{2}\text{.733}{\text{×10}}^{\text{8}}{\text{pm}}^{\text{3}}}\text{×100}\\ \text{=}\text{0}\text{.344×100}\\ \text{=}\text{34}\text{.4}\text{\%}\end{array}$

The percentage of space occupied is $=34.4\%$

Cube with a two side of white tin is $\text{583 pm}$, and one side with $\text{318 pm}$ the conversion of pm to cm is $\text{5}{\text{.83 × 10}}^{\text{-8}}\text{cm}$ and $\text{3}{\text{.18 × 10}}^{\text{-8}}\text{cm}$

$\begin{array}{l}\text{V}\text{=}{\text{(a)}}^{\text{3}}\\ \text{Cube with a two side of white tin is}\text{583 pm, and one side }\text{318}\text{pm}\\ \text{V=}{(5.83\times {10}^{-8}\text{cm})}^2\text{(3}\text{.18}\text{×}{\text{10}}^{\text{-8}}\text{cm)}\\ \text{V=}\text{1}\text{.081}\text{×}{\text{10}}^{\text{-22}}\text{cm}\end{array}$

$\begin{array}{l}\text{Molar mass of tin is 118}\text{.7 g}\\ \text{Volume of the unit cell =}\text{}\text{1}\text{.081}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\end{array}$

Therefore,

Volume of the one atom is given below

$\text{ V = }\frac{\text{4}}{\text{3}}{\text{Πr}}^{\text{3}}$

There are four atoms in the unit cell, Therefore, Volume of the four atoms is given below,

$\begin{array}{l}\text{ V = 4×}\frac{\text{4}}{\text{3}}{\text{Πr}}^{\text{3}}\\ \text{where,}\\ \text{r=141pm}\\ \text{V = 4×}\frac{\text{4}}{\text{3}}{\text{Π(141 pm)}}^{\text{3}}\\ \text{V = 4}{\text{.699×10}}^{\text{7}}{\text{ pm}}^{\text{3}}\end{array}$

The percentage of space occupied is

$\begin{array}{c}\text{The percentage of space occupied is}\text{=}\frac{\text{4}{\text{.699×10}}^{\text{7}}{\text{ pm}}^{\text{3}}}{\text{1}\text{.081}{\text{×10}}^{\text{8}}{\text{pm}}^{\text{3}}}\text{×100}\\ \text{=}\text{0}\text{.435×100}\\ \text{=}\text{43}\text{.5}\text{\%}\end{array}$

  The percentage of space occupied is $=43.5\%$

Conclusion

The percentage of the space occupied by the tin atoms in tetragonal and cubic crystal lattice was calculated.