Question

10 mL of 0.59M 1,6-diaminohexane and 10 mL of 0.29M Sebacoyl chloride are used to make nylon 6.10. The final product weight was 1.36g.

Calculate the percent yield? Use the MW of one repeating monomer for the weight of the product? Please show all work.

Answer 1

First finding the limiting reactant.

Limiting reactant means which reactant present in low amount.

The number of moles of 1,6-diaminohexane:

10 mL × ( 1 L / 1000 mL ) ×( 0.59 mol / L)

= 5.9×10^-3 mol

The number of moles sebacoyl chloride:

10 mL × (1 L / 1000 mL) × ( 0.29 mol / L)

= 2.9×10^-3 mol

Hence limiting reactant is sebacoyl chloride.

Theoretical yield of nylon 6.10.

Molar mass of nylon 6.10 is 282.43 g / mol ( repeating unit molar mass)

2.9×10^-3 mol ×( 1mol Nylon 6.10 / 1 mol S.C )× ( 282.43 g Nylon 6.10 / a mol nylon 6.10)

= 2.9×10^-3 × 282.43 g

= 0.819 g

Percent yield =( Actual yield / theoretical yield ) × 100%

Percent yield= (1.36 / 0.819 )×100%

Percent yield = 166%

I have checked 3 times but getting the answer 166%.

This the correct procedure to solve the percent yield problems.