Calculate the percent yield for a student who records the following observations

Question

Calculate the percent yield for a student who records the following observations (clipped to relevant portions). (excerpts from observations) mass 10 mL Erlenmeyer 16.432 g 1000 microliter pipet adjusted to 0.5 mL, dispensed 2-ethyl-1,3-hexanediol into flask, mass 17.047 g... added 3.0 mL of 6% aq. NaOCl dropwise... ether evaporated to give 0.571 g of product. after looking at the IR it was determined to be product 2 (see page 542). (excerpts complete) What is the percent yield, round to one decimal place?

Explanation / Answer

weight of 2-ethyl-1,3-hexanediol = 17.047 - 16.432 = 0.615 g

1 mole of diol would give 1 mole of hydroxyketone

moles of diol = 0.615/146.23 = 4.206 mmol

Theoretical yield of product = 4.206 x 144.23 = 606.63 mg (100% yield)

actual yield = 0.571 g

Thus,

percent yield = (571/606.63) x 100 = 94.1 %

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