Question

A butadiene-styrene is made by polymerizing 2 monomers of styrene with 6 monomers of butadiene. If 22% of the crosslink sites are to be bonded with sulfur, what wt.% sulfur is required?

Answer 1

Co-polymer has 2 moles of styrene for every 6 moles of buta-diene

6 moles of Polybutadiene X 54g/mol = 324 g

2 moles of polystrene X 105 g/mol = 210 g

Copolymer total mass = 534 g

now, as only the butadiene can participate in the cross-linking with sulfur. We thus need the mass fraction of butadiene: 324/534 =0.606

moles of Butadiene in 100 g of Copolymer = 100 X 0.606/54 g/mol = 1.12

Mass of S required for 22 % cross linking = 0.22 x 1.12 x 32 g/mol = 7.88 g

wt % = (7.88/ 100+7.88) = 7.30%