Question

Calculate percent yield of acetanilidae if you started with 2 mL of aniline and 2 mL of acetic anhydride and obtained only 1.5 grams of acetanilidae. Show detailed work indicating what is the limiting reagent and theoretical tield, etc.

Answer 1

Hi,

Percent yield = actual yield in mols divided by theoretical yield in mols, X 100

Actual yield is what you get from an experiment.
Theoretical yield is the molar mass the stoichiometry problem makes.

Now

C6H5NH2 + (CH3CO)2O ? C6H5NHCOCH3 + CH3COOH

1 mole aniline reacts with 1 mole acetic anhydride to form 1 mole
acetanilide and 1 mole acetic acid

Therefore, Your theoretical yield is 1 mol.

Molar mass


Acetic anhydride,
102.1 g/mol

Aniline
93.1 g / mol

Acetanilide
135. 2 g/mol

Given

Starting volume of Aniline and acetic anhydride = 2mL

Now amount of aniline weighed = density of aniline x volume of anilline taken = 1.02 g/mL x 2 mL = 2.04g

Similarly amount of acetic anhydride weighed = 1.08 g/mL X 2mL = 2.16 g

2.04 g C5H6NH2 (1mol/93.13g)(2.16 C6H5NHCOCH3/1 C5H6NH2)(137.17g/1mol) = g C6H5NHCOCH3 made
: 2.04*137.17/93.13*2.16 = 0.671 g C6H5NHCOCH3
0.671g (1mol/137.17g) = 0.0048 mol

Actual/Theoretical x 100 ===> 0.0048 mol/1mol = 0.0048 x 100 = 0.48% yield.