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.
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
Hence the percent yield of the reaction will be 84.49%
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