Question

A rubber contains 70 wt% butadiene (molar mass = 54 g/mol), 20 wt% isoprene (molar mass = 68 g/mol), 5 wt% sulfur and 5 wt% carbon black. Assume that all the sulfur is present in crosslinks. If each sulfide crosslink contains an average of two sulfur atoms, the percentage of possible crosslinks that are joined by vulcanization is ............%. {(Round off to one decimal place)}

Hint:

To find the number of crosslinks, divide the moles of sulfur by 2. Then, compare that to the total number of chain pairs to get the percentage of chains that are crosslinked.

Answer 1

We are given the following weight percentages:
- Butadiene: \( 70 \, {g} \), \( M = 54 \, {g/mol} \)
- Isoprene: \( 20 \, {g} \), \( M = 68 \, {g/mol} \)
- Sulfur: \( 5 \, {g} \)
- Carbon black: \( 5 \, {g} \) (irrelevant for calculation)
Let’s calculate moles of monomers: \[ {Moles of butadiene} = \frac{70}{54} \approx 1.296 \, {mol} \] \[ {Moles of isoprene} = \frac{20}{68} \approx 0.294 \, {mol} \] \[ {Total moles of monomer units} = 1.296 + 0.294 = 1.59 \, {mol} \] Now calculate moles of sulfur: \[ {Moles of sulfur} = \frac{5}{32} \approx 0.15625 \, {mol} \] Each crosslink contains 2 sulfur atoms, so: \[ {No. of crosslinks} = \frac{0.15625}{2} = 0.0781 \, {mol} \] Percentage of monomer units involved in crosslinking: \[ {Percentage} = \left( \frac{0.0781}{1.59} \right) \times 100 \approx 4.91% \] However, this is the percentage of crosslinks relative to monomer units. To get the percentage of possible crosslinks that are joined, we need to consider pairs of monomer units (since each crosslink connects two chains): Total possible crosslinks (max) = \( \frac{1.59}{2} = 0.795 \) So final percentage: \[ {Percentage joined} = \left( \frac{0.0781}{0.795} \right) \times 100 \approx 9.8% \]