Chapter 16, Problem 16.15P

Interpretation Introduction

(a)

Interpretation:

The weight average molecular weight and degree of polymerization for the given polymer are to be calculated.

Concept Introduction:

The molecular weight of a polymer is not fixed. It has a certain range under which its molecular weight falls. For a linear polymer, degree of polymerization represents the average length of the polymer. Degree of polymerization is defined as:

  $\begin{array}{c}\text{Degree of polymerization}=\frac{\text{average molecular weight of polymer}}{\text{molecular weight of repeating unit}}\\ n=\frac{{\left(\text{MW}\right)}_{\text{av}\text{.}}}{{\left(\text{MW}\right)}_{\text{repeating unit}}}\end{array}$   ....... (1)

Weight average molecular weight $\left({\overline{M}}_w\right)$ of a polymer is calculated by the formula:

  ${\overline{M}}_w={\displaystyle \sum f_i{M}_i}$   ....... (2)

Here, $f_i$ is the weight fraction polymer chains in the size range $i$ and $M_i$ is the mean molecular weight of that size range.

Answer to Problem 16.15P

The weight average molecular weight for the given polymer is $11203.5\text{ g/mol}$.

The degree of polymerization for the given polymer is $211$.

Explanation of Solution

Given information:

A sample of polyacrylonitrile was analyzed and the results are:

  

First calculate the total number of chains as:

  $\begin{array}{c}\text{Total Chains}={\displaystyle \sum _i{N}_i}\\ =10000+18000+17000+15000+9000+4000\\ =73000\end{array}$

Now, calculate the weight for each of the chain category as:

  $\begin{array}{c}{W}_i=N_i\times M_i\\ W_1=N_1\times M_1=10000\times 3000=30000000\\ W_2=N_2\times M_2=18000\times 6000=108000000\\ W_3=N_3\times M_3=17000\times 9000=153000000\\ W_4=N_4\times M_4=15000\times 12000=180000000\\ W_5=N_5\times M_5=9000\times 15000=135000000\\ W_6=N_6\times M_6=4000\times 18000=72000000\end{array}$

Calculate the total weight of all the chains as:

  $\begin{array}{c}W={\displaystyle \sum _i{W}_i}\\ =30000000+108000000+153000000+180000000+135000000+72000000\\ =678000000\end{array}$

Calculate the mass fraction for each of the size range as:

  $\begin{array}{c}{f}_i=W_i/W\\ f_1=W_1/W=30000000/678000000=0.04425\\ f_2=W_2/W=108000000/678000000=0.15929\\ f_3=W_3/W=153000000/678000000=0.22566\\ f_4=W_4/W=180000000/678000000=0.26549\\ f_5=W_5/W=135000000/678000000=0.19912\\ f_6=W_6/W=72000000/678000000=0.10619\end{array}$

Now, use formula (2) to calculate the weight average molecular weight of the polymer sample as:

  $\begin{array}{c}{\overline{M}}_w={\displaystyle \sum f_i{M}_i}\\ =f_1{M}_1+f_2{M}_2+f_3{M}_3+f_4{M}_4+f_5{M}_5+f_6{M}_6\\ =\left(\begin{array}{l}\left(0.04425\times 3000\right)+\left(0.15929\times 6000\right)+\left(0.22566\times 9000\right)\\ +\left(0.26549\times 12000\right)+\left(0.19912\times 15000\right)+\left(0.10619\times 18000\right)\end{array}\right)\\ =11203.5\text{ g/mol}\end{array}$

The repeating unit of the polymer polyacrylonitrile is $-\left[{\text{C}}_3{\text{H}}_3\text{N}\right]-$. Its molecular weight is:

  $\begin{array}{c}{\left(\text{MW}\right)}_{-\left[{\text{C}}_3{\text{H}}_3\text{N}\right]-}=\left(3\times 12\right)+\left(3\times 1\right)+\left(1\times 14\right)\\ =53\text{ g/mol}\end{array}$

Now, use equation (1) and calculate the degree of polymerization as:

  $\begin{array}{c}{n}_w=\frac{{\overline{M}}_w}{{\left(\text{MW}\right)}_{-\left[{\text{C}}_3{\text{H}}_3\text{N}\right]-}}\\ =\frac{11203.5\text{ g/mol}}{53\text{ g/mol}}\\ \approx 211\end{array}$

Interpretation Introduction

(b)

Interpretation:

The number average molecular weight and degree of polymerization for the given polymer are to be calculated.

Concept Introduction:

The molecular weight of a polymer is not fixed. It has a certain range under which its molecular weight falls. For a linear polymer, degree of polymerization represents the average length of the polymer. Degree of polymerization is defined as:

  $\begin{array}{c}\text{Degree of polymerization}=\frac{\text{average molecular weight of polymer}}{\text{molecular weight of repeating unit}}\\ n=\frac{{\left(\text{MW}\right)}_{\text{av}\text{.}}}{{\left(\text{MW}\right)}_{\text{repeating unit}}}\end{array}$   ....... (1)

Number average molecular weight $\left({\overline{M}}_n\right)$ of a polymer is calculated by the formula:

  ${\overline{M}}_n={\displaystyle \sum x_i{M}_i}$   ....... (3)

Here, $x_i$ is the fraction of total polymer chains in the size range $i$ and $M_i$ is the mean molecular weight of that size range.

Answer to Problem 16.15P

The number average molecular weight for the given polymer is $9287.67\text{ g/mol}$.

The degree of polymerization for the given polymer is $175$.

Explanation of Solution

Given information:

A sample of polyacrylonitrile was analyzed and the results are:

  

First calculate the total number of chains as:

  $\begin{array}{c}\text{Total Chains}\left(N\right)={\displaystyle \sum _i{N}_i}\\ =10000+18000+17000+15000+9000+4000\\ =73000\end{array}$

Calculate the fraction of chain in each of the size range as:

  $\begin{array}{c}{x}_i=N_i/N\\ x_1=N_1/N=10000/73000=0.13699\\ x_2=N_2/N=18000/73000=0.24658\\ x_3=N_3/N=17000/73000=0.23288\\ x_4=N_4/N=15000/73000=0.20548\\ x_5=N_5/N=9000/73000=0.12329\\ x_6=N_6/N=4000/73000=0.05479\end{array}$

Now, use formula (3) to calculate the number average molecular weight of the polymer sample as:

  $\begin{array}{c}{\overline{M}}_n={\displaystyle \sum x_i{M}_i}\\ =x_1{M}_1+x_2{M}_2+x_3{M}_3+x_4{M}_4+x_5{M}_5+x_6{M}_6\\ =\left(\begin{array}{l}\left(0.13699\times 3000\right)+\left(0.24658\times 6000\right)+\left(0.23288\times 9000\right)\\ +\left(0.20548\times 12000\right)+\left(0.12329\times 15000\right)+\left(0.05479\times 18000\right)\end{array}\right)\\ =9287.67\text{ g/mol}\end{array}$

The repeating unit of the polymer polyacrylonitrile is $-\left[{\text{C}}_3{\text{H}}_3\text{N}\right]-$. Its molecular weight is:

  $\begin{array}{c}{\left(\text{MW}\right)}_{-\left[{\text{C}}_3{\text{H}}_3\text{N}\right]-}=\left(3\times 12\right)+\left(3\times 1\right)+\left(1\times 14\right)\\ =53\text{ g/mol}\end{array}$

Now, use equation (1) and calculate the degree of polymerization as:

  $\begin{array}{c}{n}_n=\frac{{\overline{M}}_n}{{\left(\text{MW}\right)}_{-\left[{\text{C}}_3{\text{H}}_3\text{N}\right]-}}\\ =\frac{9287.67\text{ g/mol}}{53\text{ g/mol}}\\ \approx 175\end{array}$