Question

Given the following atomic weights, that the density of acetic acid is 1.05, and that 9.8 g of isoamyl alcohol, 7.4 mL of acetic acid, and 0.5 mL of sulfuric acid make 3.7 g of isoamyl acetate, what is the % yield of isoamyl acetate from its limiting reagent? Give at least two significant figures. C = 12, H = 1, O =16

Answer 1

mass = density X volume

mass of acetic acid = 1.05 X 7.4 = 7.77 gm

molar mass of acetic acid = 60.05 gm /mole then 7.77 gm acetic acid = 7.77 / 60.05 = 0.1294 mylole

molar mass of isoamyl alcohol = 88.148 gm/mole then 9.8 gm of isoamyl alcohol = 9.8 / 88.148 = 0.1112 mole

According to reaction acetic acid and isoamyl alcohol react in equimolar proportion therefore to react with 0.1111766 mole of isoamyl alcohol required acetic acid = 0.1112 mole but acetic acid given 0.1294 mole thererfore acetic acid is excess reactant and isoamyl alcohol is limiting reactant.

according to reaction 1 mole isoamyl alcohol prodece 1 mole isoamyl alcohol then 0.1111766 mole of isoamyl alcohol produce 0.1111766 mole of isoamyl acetate.

molar mass of isoamyl acetate = 130.19 gm/mole then 0.1111766 mole of isoamyl acetate = 0.1111766 X 130.19 = 14.47 gm

14.47 gm of isoamyl acetate is theoratical yield

% yield = 100 X actual yield / theoratical yield

% yield of isoamyl acetate = 100 X 3.7 / 14.47 = 25.57 %

% yield of isoamyl acetate = 25.57 %