A sample of natural rubber (200.0 g) is vulcanized, with the complete consumption of 4.8 g of sulfur. Natural rubber is a polymer of isoprene (C5H8). Four sulfur atoms are used in each crosslink connection. What percent of the isoprene units will be crosslinked

1.3% many crosslinks as isoprene units,  

Explanation:

Given:

mass pf natural rubber= 200.0g

mass of sulphur = 4.8g

molar mass of sulphur =32g/mol

molar mass of isoprene = C5H8=( 12x5) +(1x8)= 68g/mol

Solution: we first find no of moles present in each  using

no of moles =

Isoprene: 200.0g x [1mole / 68g] = 2.94moles.

Sulfur: 4.8g x [1mole / 32g] x [1 mole crosslinks / 4 moles S] = 0.0375 moles crosslinks.

to find % crosslinked units, we have  

0.0375 / 2.94 = 1.3% as many crosslinks as isoprene units,  

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