Question

How do I find the theoretical yield for ethyl acetate? Started with 5 mL of an unknown alcohol and added approximately 7 mL of glacial acetic acid, then added 1 mL of concentrated sulfuric acid. We then refluxed the reaction for 65 minutes.

Answer 1

Esterification takes place in the following way where concentrated H2SO4 is aciting as catalyst

R-OH + R'COOH    R'COOR + H2O

(Alcohol) (Carboxylic acid) ( Ester ) ( water)

Accroding to you, your product is ethyl acetate (ester) but you have taken an unknown alcohol and acetic acid.

It means that unknown alcohol must be ethyl alcohol

C2H5OH + CH3COOH    CH3COOC2H5 + H2O

Ethyl alcohol Acetic acid Ethyl acetate

MW of C2H5OH is 40.068, MW of CH3COOH is 60.05 & MW of CH3COOC2H5 is 88.11

5 mL of C2H5OH (density 0.789 g/mL) = ( 0.789 x 5 )g = 3.945 g = 3.945/ 40.068 = 0.09845 mole

7 mL of  CH3COOH (density 1.05 g/mL) = (1.05 x 7 )g = 7.35 g = 7.35 / 60.05 = 0.12239 mole

So. here, C2H5OH is the limiting reagent which will control the amount of ethyl acetate formed in the reaction.

Now,  0.0984 mole of C2H5OH will produce  0.09845 mole of ethyl acetate i.e (  0.0984 x 88.11) = 8.675 g ethyl acetate

Density of ethyl acetate is 0.902 g/mL

So the theoritical yield of ethyl acetate is (8.675 / 0.902 ) = 9.61 mL (ANSWER)