Chapter 11, Problem 146AP

Interpretation Introduction

Interpretation:

The density of a hypothetical ionic compound for given radii and masses of cations and anions, is to be determined.

Concept introduction:

Ionic compounds contain cations and anions that are arranged in a crystal lattice.

A crystal lattice is made of small repeating unit cells.

A unit cell is of primitive or centered type.

Each atom or ion in a unit cell is shared by the adjacent cells. An atom or ions at corners is shared by eight unit cells.

The edge length of a body-centered cubic cell is given by the relation as follows:

$a\text{= }\frac{\text{2}{R}_{\text{cation}}\text{+2}{R}_{\text{anion}}}{\sqrt{\text{3}}}$

Here, $a$

is the edge length, $R_{\text{cation}}$

is the radius of cation, and $R_{\text{anion}}$ is the radius of the anion.

The volume of the unit cell $\left(V\right)$ is given by the relation as follows:

$V=a^{\text{3}}$;

The density $\left(D\right)$ of the unit cell in terms of mass and volume is as:

$D=\frac{m}{V}$;

Here, $m$

is the mass.

Answer to Problem 146AP

Solution: $\text{21}\text{.88}\text{g}/{\text{cm}}^{\text{3}}$

Explanation of Solution

Given information:

Radii of the anion $=1\text{50 pm}$; Radii of cation $=\text{92 pm}$; Mass of the anion $=\text{98 amu}$; and Mass of the cation $\text{=192 amu}$.

The figure is as shown below:

The given unit cell is body-centered, in which there is one cation at the center, and eight anions at the corners, shared by eight unit cells.

The contribution of 8 anions at the corners to one unit cell is:

$\frac{1}{8}\times 8=1$

In the unit cell, one anion and one cation are present.

The mass of an anion is $\text{98 amu}$

and that of a cation is $\text{192 amu}$.

The relation between $\text{amu}$

and $\text{g}$

is as:

$\text{1 g}=\text{6}{\text{.022×10}}^{\text{23}}\text{amu}$

Convert the mass to grams as follows:

$\begin{array}{c}\text{Mass of anion}=\text{98 }\cancel{\text{amu}}\times \frac{\text{1g}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\cancel{\text{amu}}}\\ =16.27\times {\text{10}}^{-\text{23}}\text{ g}\end{array}$

Similarly,

$\begin{array}{c}\text{Mass of cation}=\text{192}\cancel{\text{amu}}\times \frac{\text{1g}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\cancel{\text{amu}}}\\ =31.88\times {\text{10}}^{-\text{23}}\text{ g}\end{array}$

Thus, the mass of the unit cell will be:

$\begin{array}{c}\text{Mass}=\left(16.27\times {\text{10}}^{-\text{23}}\text{g}\right)+\left(31.88\times {\text{10}}^{-\text{23}}\text{g}\right)\\ =48.15\times {\text{10}}^{-\text{23}}\text{ g}\end{array}$

The formula to calculate the edge length is as:

$a\text{= }\frac{\text{2}{R}_{\text{cation}}\text{+2}{R}_{\text{anion}}}{\sqrt{\text{3}}}$

Substitute $92\text{ pm}$

for $R_{\text{cation}}$

and $\text{150 pm}$

for $R_{\text{anion}}$

in the above expression as:

$\begin{array}{c}a=\frac{\text{2}\left(\text{92 pm}\right)\text{+2}\left(\text{150 pm}\right)}{\sqrt{\text{3}}}\\ =\frac{\text{484 pm}}{\text{1}\text{.73}}\\ =\text{280 pm}\end{array}$

Convert the edge length to $\text{cm}$

as follows:

$\begin{array}{c}a=\text{280}\cancel{\text{pm}}\text{×}\frac{{\text{1×10}}^{\text{-12}}\cancel{\text{m}}}{\text{1}\cancel{\text{pm}}}\text{×}\frac{\text{1cm}}{{\text{1×10}}^{\text{-2}}\cancel{\text{m}}}\\ ={\text{280×10}}^{-\text{10}}\text{cm}\\ =\text{2}{\text{.80×10}}^{-\text{8}}\text{cm}\end{array}$

The expression to calculate volume is as:

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

Substitute $2.80\times {10}^{-10}\text{ cm}$

in the above expression as:

$\begin{array}{c}V={\left(2.80\times {10}^{-8}\text{cm}\right)}^3\\ =\text{21}{\text{.95×10}}^{-\text{24}}{\text{ cm}}^{\text{3}}\\ =2.2{\text{×10}}^{-\text{23}}{\text{ cm}}^{\text{3}}\end{array}$

Now, finally calculate density as:

$D=\frac{m}{V}$

Substitute $48.15\times {10}^{-23}\text{ g}$

for $m$ and $2.2\times {10}^{-23}{\text{ cm}}^3$

for $V$

in the above expression as:

$\begin{array}{c}D=\frac{48.15\times {\text{10}}^{-\text{23}}\text{ g}}{2.2{\text{×10}}^{-\text{23}}{\text{ cm}}^{\text{3}}}\\ =\text{21}\text{.88}\text{ g}/{\text{cm}}^{\text{3}}\end{array}$

Conclusion

The density of the hypothetical ionic compound is $\text{21}\text{.88}\text{g}/{\text{cm}}^{\text{3}}$.