Question

Suppose that in the synthesis of isoamyl acetate, a student began with 6.033 grams of acetic acid and 3.849 grams of isoamyl alcohol. If the mass of isoamyl acetate obtained were 4.912, what was the percent yield?

Answer 1

calculate the no of mole sof each

Acetic acid moles = 6.033 / 60.02    = 0.1 moles

no of moles of isoamyl alcohol = 3.849 / 88.15 = 0.043 moles

that means limiting agent is isoamyl alcohol

we should consider isoamyl alcohol as a starting material so theritically we sjould get 0.043 moles of isoamyl acetate

no of moles of isoamyl acetate obtained = 4.912 / 130.18 = 0.037 moles

% of yield = no of moles of product / no of moles of reactant   x 100   = 0.037 / 0.043   = 86.04 %