Question

A rubber contains 60% butadiene, 30% isoprene,5% sulfur,5% carbon black. If each sulfide cross-link contains an average of two sulfur atoms,what fraction of possible cross-links are joined by vulcanization?(assume that all the sulfur is present in cross-links).​

Answer 1

Answer: Fraction of possible cross-links are joined by vulcanization = 10.1 %

Explanation:

Chemical formula for Butadiene = $C_4H_6$

∴ Molar mass of Butadiene = $C_4H_6$ = 12 X 4 + 1 X 6 = 48+6 = 54 g/mol

Chemical formula for Isoprene = $C_5H_8$

∴ Molar mass of Isoprene = $C_5H_8$ = 12 X 5 + 1 X 8 = 60 + 8 = 68 g/mol

Molar mass of sulphur = 32 g/mol

One sulphur atom will be required for cross linking

∴ % Fraction of cross links = $\frac{Fraction of Sulphur}{Fraction of butadiene + Fraction of Isoprene}$  X 100

= $\frac{\frac{5}{32} }{\frac{60}{54} + \frac{30}{68} }$   X 100

= 0.101 X 100

= 10.1 %

∴ Fraction of possible cross-links are joined by vulcanization = 10.1 %

Answer 2

Given :

Percentage of butadiene = 60%

Percentage of isoprene = 30%

Percentage of sulphur = 5%

Percentage of carbon black = 5%

To Find :

The fraction of possible links joined by sulphur = ?

Solution :

• For every cross link one atom of suplur will be required

• Molecular mass of butadiene(C₄H₆) = 4(12) + 6(1) = 54g/mol

       Molecular mass of isoprene (C₅H₈) = 5(12) + 8(1) = 68 g/mol

       Molecular mass of sulphur = 32g/mol

• Fraction of cross links of sulphur is

          $Fraction\: of \:cross links = \frac{Fraction\:of\:sulphur}{Fraction\:of\:isoprene + Fraction\:of\:butadiene}$

                                               $=\frac{\frac{5}{32} }{\frac{30}{68} +\frac{50}{64} }$

                                               $=0.101\times 100$

                                               $=10.1\%$

The fraction of cross links in suplur is 10.1%.