Question

question in first picture

Actual ester yield (calculations in your lab notebook must be present, re-type it in the report). Theoretical yield (calculations in your lab notebook must be present, re-type it in the report). Percent yield (calculations in your lab notebook must be present, re-type it in the report). in loto in table format) drawstructure of your

information located below

mass of empty vial (979.2mg) then mass with crude (1769.1mg)

mass of empty vial (1769.1mg) then yield with pure ester product(18535.6mg)

Answer 1

Mass of empty vial = 17745.1 mg

Mass of vial + ester = 18535.6 mg.

Mass or ester = (mass of vial + ester) – (mass of vial)

= (18535.6 mg) – (17745.1 mg)

= 790.5 mg

= (790.5 mg)*(1 g)/(1000 mg)

= 0.7905 g.

Therefore, actual yield of the ester = 0.7905 g (ans).

Volume of propionic acid taken = 1.5 mL.

Density of propionic acid = 0.99 g/mL.

Mass of propionic acid taken = (volume of acid)*(density of acid)

= (1.5 mL)*(0.99 g)

= 1.485 g.

Molar mass of propionic acid = 74.08 g/mol.

Mol(s) propionic acid corresponding to 1.485 g = (1.485 g)/(74.08 g/mol) = 0.0200 mol.

The word equation for the reaction between propionic acid and the unknown alcohol is given as

Propionic acid + alcohol ---------> Propionate ester + H₂O

As per the stoichiometric equation,

1 mol propionic acid = 1 mol propionate ester.

The identity of the alcohol and hence, the ester is required to answer the remaining question. That’s why you have the NMR data. You have to elucidate the structure with the help of NMR. Please provide the identity of the ester.