Chapter 11, Problem 47QP

Barium metal crystallizes in a body-centered cubic lattice (the Ba atoms are at the lattice points only). The unit cell edge length is 502 pm, and the density of the metal is $\text{3}{\text{.50 g/cm}}^{\text{3}}$. Using this information, calculate Avogadro's number. [Hint: First calculate the volume (in ${\text{cm}}^{\text{3}}$) occupied by 1 mole of Ba atoms in the unit cells. Next calculate the volume (in ${\text{cm}}^{\text{3}}$) occupied by one Ba atom in the unit cell. Assume that 68 percent of the unit cell is occupied by Ba atoms.]

Interpretation Introduction

Interpretation:

The value of Avogadro’s number in a unit cell is to be determined.

Concept introduction:

The relationship between mass $\left(m\right)$, volume $\left(V\right)$, and density $\left(d\right)$ of a solution is given as:

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

Here, $m$ is the mass, and $V$ is the volume of the unit cell.

The volume of a unit cell is given as follows:

$V=a^3$

Here, $a$ is the edge length of the unit cell.

Convert picometer to meter, $1\text{ pm}$ is equal to $1\times {10}^{-12}\text{m}$, conversion factor is as:

$\frac{1\times {10}^{-12}\text{m}}{1\text{ pm}}$

Convert centimeter to meter, $1\text{ cm}$ is equal to $1\times {10}^{-2}\text{m}$, conversion factor is as: $\frac{1\text{ cm}}{1\times {10}^{-2}\text{m}}$

Answer to Problem 47QP

Solution: $6.20\times {10}^{23}\text{atoms}/\text{mol}$

Explanation of Solution

Given information: The unit cell of metallic barium has a body centred cubic cell lattice.

The edge length of the unit cell is $\text{502 pm}$.

The density of metallic barium is $\text{3}\text{.50 g}/{\text{cm}}^3$

Convert the unit cell length from pm to cm as follows:

$\left(502\cancel{\text{pm}}\right)\left(\frac{1\times {10}^{-12}\cancel{\text{m}}}{1\cancel{\text{pm}}}\right)\left(\frac{1\text{ cm}}{1\times {10}^{-2}\cancel{\text{m}}}\right)=5.02\times {10}^{-8}\text{ cm}$

First, calculate the volume of unit cell, that is, as follows:

$V=a^3$

$\begin{array}{c}V={\left(5.02\times {10}^{-8}\text{ cm }\right)}^3\\ =126.5\times {10}^{-24}{\text{cm}}^3\end{array}$

Calculate the mass of the unit cell.

$m=d\times V$

$\begin{array}{c}m=3.50\text{ g}/\cancel{{\text{cm}}^3}\times 126.5\times {10}^{-24}\cancel{{\text{cm}}^3}\\ =442.75\times {10}^{-24}\text{g}\end{array}$

Now, calculate the volume of one mole of barium atom.

$\frac{\text{Volume}}{\text{mass of Ba}}\to \frac{\text{Volume}}{\text{mol of Ba}}$

The mass of barium is $\text{137}\text{.3 amu}$.

The volume of one mole of barium atom is as follows:

$\left(\frac{1{\text{ cm}}^3}{3.50\text{ g Ba}}\right)\left(\frac{137.3\text{ g Ba}}{\text{ 1 mol Ba}}\right)=\frac{39.23{\text{cm}}^3}{\text{1 mol Ba}}$

Volume occupied by two barium atoms in body centered cubic unit is less than the actual volume of body centered cubic unit cell because some of the space is empty due to the 68.0 percent packing efficiency this unit cell.

Calculate the number of barium atoms as follows:

$\begin{array}{c}\left(\frac{\text{Number of Ba atoms}}{{\text{cm}}^3}\right)\left(\frac{{\text{cm}}^3}{1\text{mol}\text{Ba}}\right)\text{=}\frac{\text{Number of Ba atoms}}{\text{1 mol Ba}}\\ \left(\frac{2\text{ Ba atoms}}{126.5\times {10}^{-24}\cancel{{\text{cm}}^3}}\right)\left(\frac{39.3\cancel{{\text{cm}}^3}}{1\text{mol}\text{Ba}}\right)\text{=}\frac{\text{Number of Ba atoms}}{\text{1 mol Ba}}\\ \text{A}=6.20\times {10}^{23}\text{atoms}/\text{mol}\end{array}$

Conclusion

The value of Avogadro’s number is $6.20\times {10}^{23}\text{atoms}/\text{mol}$.