Question

If 4.2 mL of isoamyl alcohol was reacted with 3.5 mL of acetyl chloride and 2.7 grams of isoamyl acetate was recovered, what is the % yield? Include calculations for density conversions the limiting reagent, theoretical yield and %yield.

Answer 1

First,

identify the theoreticla amount of isoamyl acetate

so

mass of Isoamyl alcohol

D = 0.810 g/mL

mass = D*V = 0.81*4.2 = 3.402 g of Isoamyl Alcohol

mol = mass/MW = 3.402/88.148 = 0.03859 mol of Isamyl alcohol

then

mass of Acetyl chloride = D*V

D = 1.1 g/mL

mass = D*V = 1.1*3.5 = 3.85 g of acetly chloride

now

mol = mass/MW = 3.85/78.49 = 0.049

ratio is 1:1

so

0.03859 : 0.049

0.03859 mol will be formed

0.03859 mol of isoamyl acetate

mass = mol*MW = (0.03859 )*130.19 = 5.0240 g should form at 100%

so

%yield = real / theoretical * 100%

% yield = 2.7/5.0240*100 = 53.742 %