Chapter 12, Problem 12.73QP

(a)

Interpretation Introduction

Interpretation: In Buckminsterfullerene, ${\text{C}}_{60}$ molecules are given to form a cubic closest-packed array of spheres with a unit cell edge length of $1410\text{pm}$. The density of crystalline ${\text{C}}_{60}$ and the radius of the ${\text{C}}_{60}$ molecule is to be calculated.

Concept introduction: The density of crystalline ${\text{C}}_{60}$ is calculated by formula,

$d=\frac{Z\times M}{a^3\times {\text{N}}_{\text{A}}}$

The relationship between atomic radius and edge length of cubic is,

$a=2r$

To determine: The density of ${\text{C}}_{60}$ molecule.

Answer to Problem 12.73QP

Solution

The density of ${\text{C}}_{60}$ crystalline is $\underset{\_}{1.707g/c{m}^3}$.

Explanation of Solution

Explanation

Given

The edge length of unit cell is $1410\text{pm}$.

The smallest unit of a crystal is known as a unit cell which repeats itself in different directions to generate the complete crystal lattice.

The cubic closest-packed has eight atoms at the corners of the cube.

Therefore, only $\frac{1}{8}$ of each corner atom are inside each unit cell.

The total corner atoms inside unit cell is,

$\frac{1}{8}\times 8=1\text{atom}\text{per}\text{cubic}\text{cell}$

The cubic closest-packed contains one center atom and has six atoms at the centers of each face of the cube. Two unit cells are shared by the one face atom.

The total face atoms inside unit cell is,

$\frac{1}{2}\times 6=3$

Therefore, number of atoms per unit cell in ccp lattice is $4$.

The molar mass of ${\text{C}}_{60}$ molecule is $720.66\text{g}/\text{mol}$.

The density of crystalline ${\text{C}}_{60}$ is calculated by formula,

$d=\frac{Z\times M}{a^3\times {\text{N}}_{\text{A}}}$

Where,

• $d$ is the density of unit cell.
• $Z$ is the number of atoms per unit cell.
• $a$ is the edge length.
• ${\text{N}}_{\text{A}}$ is the Avogadro constant.
• $M$ is the molar mass.

Conversion of $1\text{pm}$ into $\text{cm}$ is done as,

$1\text{pm}={10}^{-10}\text{cm}$

Therefore, conversion of $1410\text{pm}$ into $\text{cm}$ is,

$1410\text{pm}=1410\times {10}^{-10}\text{cm}$

Substitute the values of $Z$, $a$, $M$, and ${\text{N}}_{\text{A}}$ in the above formula.

$\begin{array}{c}d=\frac{Z\times M}{a^3\times {\text{N}}_{\text{A}}}\\ =\frac{4\times 720.66}{{\left(1410\times {10}^{-10}\right)}^3\times 6.023\times {10}^{23}}\\ =\underset{\_}{1.707g/c{m}^3}\end{array}$

Therefore, the density of ${\text{C}}_{60}$ crystalline is $\underset{\_}{1.707g/c{m}^3}$.

(b)

Interpretation Introduction

To determine: The radius of the ${\text{C}}_{60}$ molecule.

Answer to Problem 12.73QP

Solution

The radius of the ${\text{C}}_{60}$ molecule is $\underset{\_}{11.75pm}$

Explanation of Solution

Explanation

Given

The edge length of ${\text{C}}_{60}$ molecules is $1410\text{pm}$.

Therefore, edge length of one molecule is

$\begin{array}{c}1\text{molecule}=\frac{1410}{60}\\ =23.5\text{pm}\end{array}$

The relationship between atomic radius and edge length of cubic is,

$a=2r$

Where,

• $a$ is the edge length
• $r$ is the radius.

Substitute the value $a$ in above formula.

$\begin{array}{c}a=2r\\ r=\frac{a}{2}\\ =\frac{23.5}{2}\\ =\underset{\_}{11.75pm} \end{array}$

Thus, the radius of the ${\text{C}}_{60}$ molecule is $\underset{\_}{11.75pm}$.

Conclusion

1. a. The density of ${\text{C}}_{60}$ crystalline is $\underset{\_}{1.707g/c{m}^3}$.
2. b. The radius of the ${\text{C}}_{60}$ molecule is $\underset{\_}{11.75pm}$