Question

Find the Percentage yield of [1,3,5-C6H3(CH3)3]Mo(CO)3 and discussion of factors responsible for the low yield?

Procedure: 0.5 g (~ 1.9 mmol) of Mo(CO)6 was added with 5mL (~36mmol) of mesitylene.  The apparatus is assembled in the hood using a straight reflux condenser. The Mo(CO)6 is volatile when heated and will sublime into the condenser during the reaction. Pass nitrogen over the reaction mixture continuously during the reaction. After 25 min, stop reflux When the solution has cooled to room temperature, add 8 mL of hexane to complete the precipitation. Suction filter the solution and rinse the yellow product (that is contaminated with black metallic molybdenum), with 5 mL of hexane. Purify the crude product by dissolving it in a minimum of CH2Cl2(no more than ~ 5 mL). Filter your solution and then add ~15 mL of hexane to the filtrate liquid to precipitate the product. Suction filter off the yellow [1,3,5-C6H3(CH3)3]Mo(CO)3 precipitate, wash twice with 4 mL of hexane, and allow the product to dry.

Initial mass 0.5 g of Mo(CO)6, product at the end 0.05g

Answer 1

Mo(CO)6     --------------> [1,3,5-C6H3(CH3)3]Mo(CO)3

264 g                                  300 g

0.5 g                                    ?

? =  ( 0.5 g/ 264 g ) x 300 g of product

    = 0.568 g

This is theoretical yield of product = 0.568 g

But, given that actual yield of the product = 0.05g

Hence,

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

               = (0.05 g/ 0.568 g) x 100

               = 8.8 %

% yield of [1,3,5-C6H3(CH3)3]Mo(CO)3 = 8.8 %