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...

3.9 Ratings (609 Votes)

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

264 gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 300 g

0.5 gÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ?

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

Â Â Â = 0.568 g

This is theoretical yield of product = 0.568 g

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

Hence,

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (0.05 g/ 0.568 g) x 100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 8.8 %

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

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