Question

Starting diol weight: 1.2g isolated and purified product: .1g MW of diol: 146.23 g/mol MW of...

Starting diol weight: 1.2g

isolated and purified product: .1g

MW of diol: 146.23 g/mol

MW of product: 144.2 g/mol

(3 pts.) Using the molecular weights of the starting diol and the product, and the measured weights of the starting diol and the isolated and purified product, calculate the yield of the reaction. Show all equations and include all UNITS.

Homework Answers

Answer #1

Number of moles of diol (reactant) = Mass/Molar mass = 1.2/146.23 = 0.008206 moles

Number of moles of product (product) = Mass/Molar mass = 1/144.2 = 0.006934 moles

Assuming one mole of diol gives one mole of purified product, hence in case of 100% yield, the number of moles of product formed must be 0.008206  moles

Percent Yield = Actual Yield/Theoritical Yield * 100

=> 0.006934/0.008206 * 100

=> 84.49%

Hence the percent yield of the reaction will be 84.49%

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