Question

1. Balance the reaction equation below. Note: consider the equilibrium shown on p. 47 of your lab manual. MOBI pyridinium tribromide E-stilbene meso-stibene dibromide 2. Complete the following table, using your actual measured amounts. Greyed-out cells do not need to be filled. bp MW vol. mass (C) (g/mol) (mL) (mg) mmol equiv E-stilbene 180.25 12.10.28% acetic acid (glacial) 118 Iml pyridinium tribromide 319.82 10.312 meso-stilbene dibromide 236- 237 340.05 0.051 3. Mass of product:_ 0.019 4. mp of product: 290

Answer 1

The theoretical yield is obtained by first identifying the limiting reactant, this can be done by calculating the number of equivalents of each reactant. The reaction is 1:1, 1 mol of pyridinium tribromide reacts with 1 mol of stilbene.

As each mol of pyridinium tribromide produces one reactive unit (bromine) , each mol of pyridinium tribromide produces 1 equivalent, as you're using mmol the easiest units will be mili-equivalents (meq):

$equiv=(0.312mmol)(\frac{1meq}{1mmol})=0.312meq$

The same is done for stilbene, each mol of stilbene needs one reactive unit so each mol equals to one equivalent:

$equiv=(0.288mmol)(\frac{1meq}{1mmol})=0.288meq$

The limiting reactant has to be stilbene because there are less equivalents than for pyridinium tribromide so, this will the reagent used to obtain the theoretical yield (this reactant must react completely) with a unit factor:

$yield_{theo}=0.288mmol_{stilbene}(\frac{1mmol_{dibromide}}{1mmol_{stilbene}})(\frac{340.05mg_{tribromide}}{1mmol_{tribromide}})=97.934mg_{tribromide}\equiv 0.0979g_{tribromide}$

So, the percent yield that you got is obtained as:

$\%yield=\frac{experimental}{theoretical}*100=\frac{0.0190g}{0.0979g}*100=19.41\%$