Question

Given the following atomic weights, that the density of acetic acid is 1.05, and that 7.0 g of isoamyl alcohol, 6.6 mL of acetic acid, and 0.5 mL of sulfuric acid make 3.2 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

Formula of acetic acid = CH3COOH or C2H4O2, so, M.wt =24+4+32= 60 g/mol

Isoamyl alcohol = (CH3)2CH2CH2CH2OH, or C5H12O, so, M.wt =(5*12+12+16) g/mol = 88 g/mol.

6.6 mL of acetic acid = 6.6*1.05 g=6.93 g = 6.93/60 mols = 0.1155 mols

7.0 g of isoamyl alcohol = 7.0/88 mols = 0.0795 mols.

The balanced reaction (H2SO4 is the catalyst in this esterification reaction) is -

CH3COOH + (CH3)2CH2CH2CH2OH à CH3COOCH2CH2CH2(CH3)2 + H2O

M. wt. of isoamyl acetate = (60+88)-18 = 120 g/mol

So, 3.2 g of isoamyl acetate = 3.2/120 mols =0.0266 mols

Based on the balanced reaction, the limiting reagent is isoamyl alcohol, stoichiometric ratio is 1:1.

So, the theoretical yield = 0.0795 mols of isoamyl acetate.

But the actual yield is 0.0266 mols.

So, the percentage yield = 100*0.0266/0.0795 % = 33.46%