Percent Yield

Question

Aspirin, C₉H₈O₄ (180.2 g/mol) is prepared by the following reaction:

 

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + C₂H₂O₂

 

If you need to prepare 45 g of aspirin, and the yield for the reaction is 85%, how many grams of C₇H₆O₃ (138.1 g/mol) must you use?

Answer 1

You want 45 g of aspirin. Howmany moles is that.

45 g x 1 mole/180.2 g = 0.2497 moles

BUT, your yield is only 85%, so you actually need to make more. How many moles do you THEORETICALLY need.  That woud be 0.2497/0.85 = 0.2938 moles

From the balanced eq. you see 1 mol C7H6O3 makes 1 mole aspirin. So you need 0.2938 moles. 

Mass needed = 0.2938 mol x 138.1g/mol = 40.57 g = 41 grams. 

Answer 2

To begin you need to find the theoretical yield from the reaction. Using the equation for percent yield you can find it by setting up the equation like this:

 

85= (45 g/theoretical yield)*100

0.85*(theoretical yield)= 45 g

theoretical yield= 53 g

 

Now we can use stoichiometry to find the amount of C₇H₆O₃ needed:

 

53 g aspirin * 1 mol/180.2 g= 0.29 mol aspirin

0.29 mol aspirin * 1 mol aspirin/1 mol C₇H₆O_{3 *} 138.1 g/ 1 mol= 41 g C₇H₆O₃

 

41 grams of C₇H₆O₃ is needed to yield 45 g of aspirin.

Answer 3

Since the yield is 85%, we need to start with 45g/.85 = 52.9 g aspirin

 

(52.9 g C₉H₈O₄)(1 mole C₉H₈O₄ / 180 g C₉H₈O₄)(1 mole C₇H₆O₃ / 1 mole C₉H₈O₄)(138.1 g C₇H₆O₃ / 1 mole C₇H₆O₃)

 

= 40.59