Question

Given the following atomic weights, that the density of acetic acid is 1.05, and that 5.3 g of isoamyl alcohol, 7.6 mL of acetic acid, and 0.5 mL of sulfuric acid make 3.8 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 volume

mass of acetic acid = 1.05 7.6 = 7.98 gm

molar mass of acitic acid = 60.05 gm/mol then 7.98 gm of acetic acid = 7.98/60.05 = 0.132889 mole

molar mass of isoamyl alcohol = 88.148 gm/mol then 5.3 gm of isoamyl = 5.3/88.148 = 0.060126 mole

in the reaction 1 mole of acetic acid react with 1 mole of isoamyl alcohol therefore for 0.132889 mole of acetic acid require 0.132889 mole of isoamyl but isoamyl alcohol given only 0.060126 mole thus, isoamyl alcohol is limiting reactant react completly.

0.060126 mole of isoamyl alcohol produce 0.060126 mole of isoamyl acetate

molar mass of isoamyl acetate = 130.19 gm/mol then 0.060126 mole of isoamyl acetate = 130.19 0.060126 = 7.83 gm

7.83 gm of isoamyl acetate is theoritical yield

7.83 gm = 100% then 3.8 gm =100 3.8/7.83 = 48.53%

% yield = 48.53 %