Determine the yield of your crude product (in grams and in %) reagent 1: m-toluoyl chloride...

Determine the yield of your crude product (in grams and in %)

reagent 1: m-toluoyl chloride MW: 154.60 g Density: 1.173 Quantity 0.90g

reagent 2: diethylamine: MW: 73.14g Density: 0.707 Quantity: 3.50mL

product: MW: 191.27g Density: 0.996 Quantity: 0.251 g

Solutions

Expert Solution

1 mol of m-toluoyl chloride = 154.60 g

0.90 g of m-toluoyl chloride = 0.90/154.60 = 0.0058 mol

3.50 mL of diethylamine corresponds to 0.707x3.50=2.47g (m=dxv)

2.47 g of diethylamine = 2.47/73.14 = 0.0338 mol

mol equivalent of diethylamine = 0.0338/0.0058 = 5.83

since the limiting reagent is m-toluoyl chloride, which is 0.0058 mol, the product should also be 0.0058 mol according to the stoichiometry of the reaction.

0.0058 mol of product in grams = 0.0058 x 191.27 = 1.109 g

so the theoretical yield is 1.109 g

the given practical yiled is 0.251 g

% practical yield = practical yield/theoretical yield x100

=0.251/1.109 x 100

= 0.2263 x100

= 22.63

hence the pracctical yield is 22.63%

queen_honey_blossom answered 1 year ago

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