Chapter 11, Problem 11.146QP

Iron crystallizes in a body-centered cubic lattice. The cell length as determined by X-ray diffraction is 286.7 pm. Given that the density of iron is 7.874 g/cm³, calculate Avogadro’s number.

Interpretation Introduction

Interpretation:

The Avogadro’s number of Iron atom in its body centered cubic lattice has to be calculated.

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 forms two types of lattices – body – centered lattice and face – centered lattice.

In body-centered cubic unit cell, each of the six corners is occupied by every single atom. Center of the cube is occupied by one atom.  Each atom in the corner is shared by eight unit cells and a single atom in the center of the cube remains unshared.  Thus the number of atoms per unit cell in BCC unit cell is,

$\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{1}\text{atom}\text{at}\text{the}\text{center}\text{=}\text{1}\text{+}\text{1}\text{=}\text{2}\text{atoms}\\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\frac{\text{4R}}{\sqrt{\text{3}}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\text{.}\\ \end{array}$

Answer to Problem 11.146QP

The Avogadro’s number of Barium atom in its body centered cubic lattice is $6.01\times 10^{23}atoms/mol.$

Explanation of Solution

One mole of Iron has mass of $\text{55}\text{.84 g}$. Density of Iron is given. Volume of one mole of Iron is obtained by the formula

$\text{volume}\text{=}\frac{\text{mass}}{\text{density}}$.

$\begin{array}{l}\text{given}\text{data:}\text{density}\text{=}7.\text{874}{\text{g/cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{volume of 1 mole of Iron}\text{=}\frac{\text{mass}\text{of 1 mole of Iron}}{\text{density}}\\ \text{=}\frac{55.84\text{g}}{\text{7}{\text{.874 g/cm}}^{\text{3}}}\text{=}7.09{\text{cm}}^{\text{3}}\end{array}$

$\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}286.7\text{pm}\text{=}286.7\text{×}{\text{10}}^{\text{-12}}\text{m}\text{=}5.\text{02}\text{×}{\text{10}}^{\text{-10}}\text{m}\\ \text{volume,}{\text{a}}^{\text{3}}\text{=}{\text{(286}\text{.7}\text{×}{\text{10}}^{\text{-12}}\text{m)}}^{\text{3}}\\ \text{=}2.36\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\\ =2.36\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

Edge length of the cubic unit cell is given. The cubic value of edge length gives the volume of the unit cell.  In each cell of unit cell of body centered cubic lattice, 2 Iron atoms are occupied.

The Avogadro’s number of Iron atom in one mole of Iron is,

$\begin{array}{l}\text{Number of Fe atoms in}\text{1}\text{mole of Iron =}\frac{\text{volume}\text{of}\text{1}\text{mole}\text{of}\text{Fe}}{\text{volume}\text{of}\text{1}\text{Fe}\text{in}\text{unit}\text{cell}}\\ \text{=}\frac{\text{7}\text{.09}{\text{cm}}^{\text{3}}}{\frac{\text{2}{\text{.36 ×10}}^{\text{-23}}{\text{cm}}^{\text{3}}}{\text{2}}}\text{=6}\text{.01}{\text{×10}}^{\text{23}}\text{atoms/mol}\end{array}$

Dividing the volume of one mole Iron atoms by volume of one mole of Iron atoms in its unit cell gives the Avogadro’s number of Iron atoms in one mole of Iron.

Conclusion

The Avogadro’s number of Iron atom in its body centered cubic lattice was calculated as $6.01\times 10^{23}atoms/mol.$