Chapter 10.14, Problem 81AAP

To determine

The amount of weight percent sulfur.

Answer to Problem 81AAP

The amount of weight percent sulfur is $\underset{\_}{9.83\text{\%}}$.

Explanation of Solution

Show the chemical structure bonding of polyacrylonitrile and polybutadiene as:

Express the weight per cent of sulfur.

    $\text{W}{\%}_{\text{S}}=\frac{M_{\text{S}}}{100\text{g}+M_{\text{S}}}\times 100\%$     (I)

Here, the molar mass of the sulfur is $M_{\text{S}}$.

Calculate the mass of sulfur at 22%.

    $\text{Mass}\text{of}{\text{S}}_{@22\%}=\left(0.22\right)\left(N_{\text{B}}\right)\left(M_{\text{S}}\right)$  (II)

Here, the number of mole in butadiene is $N_{\text{B}}$.

Calculate the copolymer total mass.

    $\text{Copolymer}\text{total}\text{mass}=\left[\begin{array}{l}\left(5\text{mole}\text{polybutadlene}\times M_{\text{polybutadlene}}\right)\\ +\left(1\text{mole}\text{polyacrylonitrile}\times M_{\text{polyacrylonitrile}}\right)\end{array}\right]$ (III)

Here, the molar mass of the polybutadlene is $M_{\text{polybutadlene}}$ and the molar mass of the $\text{polyacrylonitrile}$ is $M_{\text{polyacrylonitrile}}$.

Conclusion:

The average molecular mass of the $\text{polyacrylonitrile}$ is $53\text{g}/\text{mol}$.

The average molecular mass of the polybutadlene is $54\text{g}/\text{mol}$.

The average molecular mass of the sulfur is $32\text{g}/\text{mol}$.

Substitute $54\text{g}/\text{mol}$ for $M_{\text{polybutadlene}}$ and $53\text{g}/\text{mol}$ for $M_{\text{polyacrylonitrile}}$ in the Equation (III).

    $\begin{array}{c}\text{Copolymer}\text{total}\text{mass}=\left[\begin{array}{l}\left(5\text{mole}\text{polybutadlene}\times 54\text{g}/\text{mol}\right)\\ +\left(1\text{mole}\text{polystyrene}\times 53\text{g}/\text{mol}\right)\end{array}\right]\\ =\left[270\text{g}+53\text{g}\right]\\ =323\text{g}\end{array}$

Calculate the fraction of butadlene.

    $\begin{array}{c}\text{Fraction}\text{of}\text{butadlene}=\frac{270\text{g}}{323\text{g}}\\ =0.836\end{array}$

Calculate the number of moles of butadlene using 200 kg of copolymer.

    $\begin{array}{c}\text{mole}\text{butadlene}=\frac{\text{200}\text{}\text{kg}\times \text{mass}\text{of}\text{butadiene}}{M_{\text{polystyrene}}}\\ =\frac{\text{200}\text{}\text{kg}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)\times \left(0.836\right)}{54\text{g}/\text{mol}}\\ =\frac{167200\text{g}}{54\text{g}/\text{mol}}\\ =3096.3\text{mol}\end{array}$

Substitute 3096.3 mol for mole butadlene and $32\text{g}/\text{mol}$ for $M_{\text{S}}$ in Equation (II).

    $\begin{array}{c}\text{Mass}\text{of}{\text{S}}_{@22\%}=\left(0.20\right)\left(3069.3\text{mol}\right)\left(32\text{g}/\text{mol}\right)\\ =21,798\text{g}\end{array}$

Substitute 21,798 g for $\text{Mass}\text{of}{\text{S}}_{@22\%}$ in Equation (I).

    $\begin{array}{c}\text{W}{\%}_{\text{S}}=\frac{21,798\text{g}}{200\text{kg}+21,798\text{g}}\times 100\%\\ =\frac{21,798\text{g}}{\text{200}\text{}\text{kg}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)+21,798\text{g}}\times 100\%\\ =0.0000005\times 100\%\\ =9.83\%\end{array}$

Thus, the amount of weight percent sulfur is $\underset{\_}{9.83\text{\%}}$.