Question

What is percent yield fro your synthseis of octyl acetate (acetic anhydride and 1-octanol)? V of ctyl acetate is 1.53 mL.

3 mL of acetic anhydride and 2 mL 1-octanol. One drop of concentrated sulfuric acid.

Answer 1

Scheme for the given synthesis can be written as

mol wt of n-octanol = 130.23 g/mol

density of n-octanol = 0.83 g/mL

volume used for reaction = 2.0 mL

mass of n-octanol = 0.83 x 2.0 g = 1.66 g

moles of n-octanol = 1.66 g/(130.23 g/mol) = 0.01275 mol = 12.75 mmol

mol wt of acetic anhydride = 102.09 g/mol

density of acetic anhydride = 1.08 g/mL

volume used for reaction = 3.0 mL

mass of acetic anhydride = 1.08 x 3.0 g = 3.24 g

moles of acetic anhydride = 3.24 g/(102.09 g/mol) = 31.80 mmol

Here e see that moles of acetic anhydride are greater than those of n-octanol

Therefore, n-octanol is limiting reagent.

mol wt of octyl acetate = 172.27 g/mol; density = 0.879 g/mL

expected mass of octyl acetate = expected moles x 172.27 = 12.75 x 172.27 = 2.20 g

but obtained volume = 1.53 mL

obtained mass = 1.53 x 0.879 = 1.33 g

yield = (observed mass x 100)/expected mass = 1.33 x 100/2.22 = 60.45 %