After re-weighing your crude 4-bromoacetanilide (214.06 g/mol) this week, you obtain 161 mg of crude product. After recrystallization, you isolate 116 mg of pure product. If you started with 125 mg of acetanilide (135.17 g/mol), calculate the percent yield of 4-bromoacetanilide (214.06 g/mol) for this reaction. Assume an excess of bromine. Enter your answer as digits only, no percent sign, using the proper number of significant figures.

Percent yield of 4-bromo Acetanilide from the reaction is 0.586

Explanation:

The equation of reaction is

Acetanilide + Bromine > 4-bromo Acetanilide.

That is,

C₆H₅NHCOCH₃ + Br₂ > (4-Br)C₆H₄NHCOCH₃

1 mole of Acetanilide gives 1 mole of 4-bromo Acetanilide

We need to to acquire the number of reacted Acetanilide. (the entire Acetanilide reacts, since Bromine is in excess)

Number of moles = mass of Acetanilide/Molar mass of acetanilide

Mass of reacted Acetanilide = 125mg = 0.125g, molar mass of Acetanilide = 135.17 g/mol

Number of moles of reacted Acetanilide = 0.125/135.17 = 0.000925 moles

From the stoichiometric balance of the reaction,

1 mole of Acetanilide gives 1 mole of 4-bromo Acetanilide

0.000925 mole of Acetanilide will give 0.000925 mole of 4-bromo Acetanilide

Theoretical mass of 4-Bromo Acetanilide = Stoichiometric number of moles × Molar mass = 0.000925 × 214.06 = 0.198g = 198mg.

Percentage yield = (Actual yield/Theoretical yield) = 116/198 = 0.586.

Hope this helps!!

it's filtration, but this is an assumption so my apologies if i got this

може ли да зареждате повече информация, моля