Chapter 2, Problem 2.19P

To determine

The number of carbon and hydrogen atoms in the unit cell of polyethylene.

Answer to Problem 2.19P

The number of carbon atoms in the unit cell of polyethylene is $4$ and number of carbon atoms is $8$.

Explanation of Solution

Given:

The given figure is,

Figure (1)

The density of crystalline polyethylene ${\rho }_{\text{PE}}$ is $0.996\text{g}/{\text{cm}}^{\text{3}}$.

From above figure,

The lattice parameter $a$ is,

$a=0.741\text{nm}$

The lattice parameter $b$ is,

$b=0.494\text{nm}$

The lattice parameter $c$ is,

$c=0.255\text{nm}$

Formula used:

The equation for density of poly-ethylene is given by,

${\rho }_{\text{PE}}=\frac{n_{\text{C}}\times M_{\text{C}}+n_v\times M_{\text{H}}}{\left(V_{\text{PE}}\right)N_{\text{A}}}$

Here, $n_{\text{C}}$ is the number of carbon atoms, $n_{\text{H}}$ is the number of hydrogen atoms, $M_{\text{C}}$ is the molecular mass of carbon, $M_{\text{H}}$ is the molecular mass of hydrogen, $N_{\text{A}}$ is Avogadro’s number, and $V_{\text{PE}}$ is the volume of polyethylene.

The hydrogen atoms are twice in number as that of carbon atoms.

Substitute $2n_{\text{C}}$ for $n_{\text{H}}$ in above equation.

${\rho }_{\text{PE}}=\frac{n_{\text{C}}\times M_{\text{C}}+2n_{\text{C}}\times M_{\text{H}}}{\left(a\times b\times c\right)N_{\text{A}}}$   ...... (I)

Here, $a$, $b$, $c$ are the lattice parameters.

Calculation:

The molecular mass of carbon is $12.01\times {10}^{-3}\text{kg}/\text{mole}$.

The molecular mass of hydrogen is $1.01\times {10}^{-3}\text{kg}/\text{mole}$.

The value of Avogadro’s number is $6.02\times {10}^{23}\text{atoms}/\text{mole}$.

Substitute $12.01\times {10}^{-3}\text{kg}/\text{mole}$ for $M_{\text{C}}$, $1.01\times {10}^{-3}\text{kg}/\text{mole}$ for $M_{\text{H}}$, $6.02\times {10}^{23}\text{atoms}/\text{mole}$ for $N_{\text{A}}$, $0.996\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\text{PE}}$, $0.741\text{nm}$ for $a$, $0.494\text{nm}$ for $b$ and $0.255\text{nm}$ for $c$ in equation (I).

$\begin{array}{c}0.996\text{g}/{\text{cm}}^{\text{3}}=\left[\frac{n_{\text{C}}\times \left(12.01\times {10}^{-3}\text{kg}/\text{mole}\right)+2n_{\text{C}}\times \left(1.01\times {10}^{-3}\text{kg}/\text{mole}\right)}{\left\{\begin{array}{l}\left(0.741\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ \times \left(0.494\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\\ \times \left(0.255\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\end{array}\right\}\left(6.02\times {10}^{23}\text{atoms}/\text{mole}\right)}\right]\\ =\left[\frac{n_{\text{C}}\times \left(12.01\times {10}^{-3}\text{kg}/\text{mole}\right)+2n_{\text{C}}\times \left(1.01\times {10}^{-3}\text{kg}/\text{mole}\right)}{\left(\begin{array}{l}0.741\times {10}^{-9}\text{m}\times 0.494\times {10}^{-9}\text{m}\\ \times 0.255\times {10}^{-9}\text{m}\end{array}\right)6.02\times {10}^{23}\text{atoms}/\text{mole}}\right]\\ n_{\text{C}}=\frac{0.996\text{g}/{\text{cm}}^{\text{3}}\times 0.562\times {10}^{-4}}{14.01\times {10}^{-3}}\\ \approx 4\text{atoms}\end{array}$

Since $n_{\text{H}}=2n_{\text{C}}$ so,

$\begin{array}{c}{n}_{\text{H}}=2\left(4\text{atoms}\right)\\ =8\text{atoms}\end{array}$

Conclusion:

The number of carbon atoms in the unit cell of polyethylene is $4$ and number of carbon atoms is $8$.