Question

QUESTION 7

1. Calculate the % yield of the following reaction. You have isolated 1.5 g of benzil from the reaction of 1.95 g of benzoin and 1.0 g of ammonium nitrate.

   		75.5%
   		76.9%
   		77.6%
   		78.1%

Answer 1

Question 7

Option 3 is correct

Benzil is prepared by the oxidation of Benzoin in presence of an oxidizing agent such as Copper (II) acetate in catalytic amount. Ammonium nitrate is used for the regeneration of Cu(II) catalyst once Cu(II) is reduced to Cu (I) while oxidizing Benzoin. Thus the ammonium nitrate do not take part in the reaction directly with Benzoin.

So in this reaction Benzoin is the sole reactant and Benzil is the sole product.

So 212 g Benzoin = 1 mole Benzoin

so, 1.95 g Benzoin = (1.95 / 212) =   9.2 x 10^-3 mole Benzoin

So, from 9.2 x 10^-3 mole Benzoin if we had got 9.2 x 10^-3 mole Benzil we would say 100% yield was achieved.

Now, 1 mole Benzil = 210 g Benzil

So, 9.2 x 10^-3 mole Benzil = 9.2 x 10^-3 x 210 = 1.932 g Benzil

So for 100% yield theoretical yield = 1.932 g

But we got 1.5 g Benzil as actual yield

Now, % yield = (actual yield / theoretical yield) x 100%

So, % yield = (1.5 g / 1.932) x 100% = 77.6 %

Therefore % yield = 77.6 %