Question

Calculations SHOW all work for any credit-SF and UN count:

1) 1.08 g antracene to mmol

a. 0.52 g maleic anhydride to mmol

b. Amount of product IN GRAMS produced from 1.08 g of anthracene

c. Amount of product formed IN GRAMS produced from 0.52 g maleic anhydride

d. * Based on the calculations from a and b, which reagent is considered the limiting reagent? Which reagent is in excess?

e. Convert the theoretical yield (amount of product produced by the limiting reagent) into mmol

f. Calculate the GRAMS left over of the excess reagent.

g. If you isolated 0.99 grams of pure product, what would the percent yield be for your experiment?

Answer 1

Given values:

The values given in the question are as follows.

1. Amount of anthracene = 1.08g
2. Amount of maleic anhydride = 0.52g
3. Amount of pure product = 0.99g

Solution:

Diels Alder reaction between one mole of anthracene (molecular weight = 178.23g/mol) and one mole maleic anhydride (molecular weight = 98.06g/mol) forms one mole of 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (molecular weight = 276.29g/mol).

1. 1.08g anthracene to mmol:

mmol of anthracene = (1.08g / 178.23g/mol) x (1000mmol / 1mol)

=  0.00606 mol x (1000mmol / 1mol) = 6.06mmol

Hence the mmol of anthracene in the reaction is 6.06mmol

a. 0.52g maleic anhydride to mmol:

mmol of maleic anhydride = (0.52g / 98.06g/mol) x (1000mmol / 1mol)

=  0.00530 mol x (1000mmol / 1mol) = 5.30mmol

Hence the mmol of maleic anhydride in the reaction is 5.30mmol

b. Amount of product IN GRAMS produced from 1.08 g of anthracene:

Amount of product = (mmol of anthracene) x (1 mol / 1000mmol) x molecular weight of product

= 6.06mmol  x (1 mol / 1000mmol) x 276.29g/mol

= 1.6743g = 1.67g

Hence 1.67g of product produced from 1.08 g of anthracene

c. Amount of product IN GRAMS produced from 0.52 g of maleic anhydride:

Amount of product = (mmol of maleic anhydride) x (1 mol / 1000mmol) x molecular weight of product

= 5.30mmol  x (1 mol / 1000mmol) x 276.29g/mol

= 1.4643g = 1.46g

Hence 1.46g of product produced from 0.52 g of maleic anhydride

d. 6.06mmol of anthracene and 5.30mmol of maleic anhydride are used in the reaction. Hence maleic anhydride is the limiting reagent in the reaction and anthracene is used in excess.

e. Convert the theoretical yield (amount of product produced by the limiting reagent) into mmol:

We know that the limiting reagent is maleic anhydride and Theoretical yield of product produced will be 1.46g (Refer c for calculation). Hence the mmol of product is equal to the mmol of maleic anhydride. That is mmol of product is 5.30mmol

f. Grams left over of the excess reagent:

Amount of anthracene used in the reaction is 5.30mmol due to availability of maleic anhydride.

Excess anthracene = 1.08g - [(5.30mmol x 178.23g/mol) (1mol / 1000mmol)

= 1.08g - 0.9446g = 0.1354g = 0.14g

Hence 0.14g of anthracene left over.

g. Percent yield of pure product:

Percent yield of pure product = (Weight of isolated product / Theoretical yield based on maleic anhydride) x 100

= (0.99g / 1.46g) x 100

= 0.678 x 100

= 67.8%

Hence the yield of pure product based on maleic anhydride is 67.8%.