Chapter 10.14, Problem 79AAP

To determine

The amount of weight percent sulfur.

Answer to Problem 79AAP

The amount of weight percent sulfur is $\underset{\_}{8.49\text{\%}}$.

Explanation of Solution

Show the chemical structure bonding of polybutadiene and polystyrene 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 20%.

    $\text{Mass}\text{of}{\text{S}}_{@20\%}=\left(0.20\right)\left(N_{\text{B}}\right)\left(M_{\text{S}}\right)$  (II)

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

Calculate the copolymer total mass.

    $\text{Copolymer}\text{total}\text{mass}=\left[\begin{array}{l}\left(7\text{mole}\text{polybutadlene}\times M_{\text{polybutadlene}}\right)\\ +\left(1\text{mole}\text{polystyrene}\times M_{\text{polystyrene}}\right)\end{array}\right]$ (III)

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

Conclusion:

The average molecular mass of the polylstyrene is $105\text{g}/\text{mol}$.

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 $105\text{g}/\text{mol}$ for $M_{\text{polybutadlene}}$ and $54\text{g}/\text{mol}$ for $M_{\text{polystyrene}}$ in the Equation (III).

    $\begin{array}{c}\text{Copolymer}\text{total}\text{mass}=\left[\begin{array}{l}\left(7\text{mole}\text{polybutadlene}\times 105\text{g}/\text{mol}\right)\\ +\left(1\text{mole}\text{polystyrene}\times 54\text{g}/\text{mol}\right)\end{array}\right]\\ =\left[378\text{g}+105\text{g}\right]\\ =483\text{g}\end{array}$

Calculate the fraction of butadlene.

    $\begin{array}{c}\text{Fraction}\text{of}\text{butadlene}=\frac{378\text{g}}{483\text{g}}\\ =0.783\end{array}$

Calculate the number of moles of butadlene using 100 g of copolymer.

    $\begin{array}{c}\text{mole}\text{butadlene}=\frac{\text{100}\text{}\text{g}\times \text{mass}\text{of}\text{butadiene}}{M_{\text{polystyrene}}}\\ =\frac{\text{100}\text{g}\times \left(0.783\right)}{54\text{g}/\text{mol}}\\ =\frac{78.3\text{g}}{54\text{g}/\text{mol}}\\ =1.45\text{g}\end{array}$

Substitute 1.45 g for mole butadlene and $32\text{g}/\text{mol}$ for $M_{\text{S}}$ in Equation (II).

    $\begin{array}{c}\text{Mass}\text{of}{\text{S}}_{@20\%}=\left(0.20\right)\left(1.45\text{mol}\right)\left(32\text{g}/\text{mol}\right)\\ =9.28\text{g}\end{array}$

Substitute 9.28 g for $\text{Mass}\text{of}{\text{S}}_{@20\%}$ in Equation (I).

    $\begin{array}{c}\text{W}{\%}_{\text{S}}=\frac{9.28\text{g}}{100\text{g}+9.28\text{g}}\times 100\%\\ =\frac{9.28\text{g}}{109.28\text{g}}\times 100\%\\ =0.084919\times 100\%\\ =8.49\%\end{array}$

Thus, the amount of weight percent sulfur is $\underset{\_}{8.49\text{\%}}$.