Theoretical yield of Acetaminophen Reaction?

Question

I really need help figuring out the theoretic yield for the acetaminophen lab in organic chemistry. I haven't done this in a while and completely forgot how to do this.

The starting materials are 0.400 g of p-aminophen molecular weight=109.1, 1.2 ml water, and 0.450 ml acetic anhydride molecular weight = 102.1

Please explain how do figure this out, any help is appreciated!

Answer 1

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Properties of compounds:



p-aminophen (4-aminophenol)

C6H7NO

MW = 109.126 g/mol



acetic anhydride

C4H6O3

MW = 102.089 g/mol

Density = 1.082 g/mL (wikipedia)



acetaminophen

C8H9NO2

MW = 151.163 g/mol



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The equation for this reaction is:

(copy and past the equation in google,

and you'll come up with the same equation.)



p-aminophen + acetic anhydride ---> acetaminophen + acetic acid

C6H7NO + C4H6O3 ---> C8H9NO2 + C2H4O2



This is a limiting reagent problem.

When there are competing reactants/reagents,

one of them only allows the reaction to go so far.



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reagent p-aminophen:



mol acetaminophen from p-aminophen =



(0.400 g of C6H7NO)

x (1mol C6H7NO / 109.126 g C6H7NO)

x (1mol acetaminophen / 1mol C6H7NO)



= 0.0036655 mol acetaminophen



-----------------------------------------------------------

reagent acetic anhydride:



mol acetaminophen from acetic anhydride =



0.450 mL C4H6O3

x (1.082 g C4H6O3/ 1 mL C4H6O3)

x (1 mol C4H6O3 /102.089 g C4H6O3)

x (1 mol acetaminophen / 1mol C4H6O3)



= 0.0047694 mol acetaminophen



-----------------------------------------------------------

p-aminophen limits the amount

(moles) of acetaminophen produced,

and is the limiting reagent.



-----------------------------------------------------------

The theoretical yield can be found from the

0.0036655 mol acetaminophen:



0.0036655 mol C8H9NO2

x (151.163 g C8H9NO2 / 1 mol C8H9NO2)

= 0.5540861 g



= 0.554 g acetaminophen (theoretical)



(at 3 sig. figs.)

Answer 2

Acetaminophen Density

Answer 3

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Lisa's answer is not entirely correct. What she said is correct, but the theoretical yield she gave you is only relevant if you started with 1 mole of Na2CO3, but you didn't. You started with 15 mL. The density of Na2CO3 = 1.55g/mL (found this from an online reference) so: 15 mL X 1.55g/mL = 23.25g of Na2CO3 Molar mass of Na2CO3 = 2 (Molar mass of Na) + Molar Mass of Carbon + 3(Molar mass of Oxygen) 2 (22.99) + 12.01 + 3 (16.00) = 105.99 g/mol Na2CO3 23.25g x 1mol Na2CO3 / 105.99g = 0.2194 moles Na2CO3 That's what you actually started with, so since Na2CO3 stands in a 1:1 mole ratio with SrCO3, the theoretical yield would be 0.2194 moles SrCO3. Since your actual yield is in grams, we'll convert the theoretical yield into grams as well. Molar mass SrCO3 = Molar mass of Sr + Molar mass of C + 3 (Molar mass of O) 87.62 + 12.01 + 3(16.00) = 147.63 g/mol. So to convert the 0.2194 moles SrCO3: 0.2194 moles x 147.63g / 1mol = 32.39g If your theoretical yield is 32.39g and your actual yield is .44g then you can calculate your % yield. actual yield / theoretical yield x 100% = % yield, so: .44 / 32.39 x 100 = 1.36% This means you only obtained 1.36% of what you could have theoretically obtained. That's a pretty horrific number. Are you sure you reported your actual yield correctly? EDIT #2: Ok, let's calculate the theoretical yield based on Strontium chloride. Molarity of SrCO3 solution = .3 mol/L so 35mL solution = 0.105 moles. So, you're right! SrCl2 is the limiting reagant. In this case: theoretical yield = 1.55g so % yield = 28% Still a very low yield. Are you sure the filter paper was 1.02g and not 0.02? In that case, your precipitate would weigh 1.44g which is a much more conceivable result. EDIT: Also...Lisa, two things: 1) SrCO3 does not weigh 147.63 moles. Moles are not a measurement of mass or weight, they're a measurement of quantity. SrCO3 weighs 147.63 amu OR 147.63 g/mol. 2) Only losers use Bing. I would have expected more from a top contributor. ;D

Answer 4

Well, what do you expect when a legend gets such a poor decision that goes against him! After all there wasn't even a great appeal? I think Dravid saw how preposterous the decision was & more so shocked to see the umpire raise his finger in utmost confidence. That's what made him explode. The umpiring in this IPL has been nothing short of mediocre especially for Indian umpires

Answer 5

theoretical yield acetaminophen reaction

Answer 6

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RE:

Theoretical yield of Acetaminophen Reaction?

I really need help figuring out the theoretic yield for the acetaminophen lab in organic chemistry. I haven't done this in a while and completely forgot how to do this.



The starting materials are 0.400 g of p-aminophen molecular weight=109.1, 1.2 ml water, and 0.450 ml acetic anhydride...

Answer 7

here are the usual steps for problems of this type.



1) write a balanced equation

2) convert everything to MOLES

3) determine limiting reagent

4) convert moles limiting reagent to moles of other species via coefficients of the balanced equation

5) convert moles back to mass.. this is THEORETICAL mass aka THEORETICAL YIELD.

6) % yield = actual measured mass / theoretical mass x 100%



you were asked to complete the first 5 steps. Can you do it?



here are some hints.



step 1. self explanitary.



step 2.

0.400 g x ( 1 mole / 109.1 g) = ? moles P-aminophen

1.2 mL H2O x (1 g / mL) x (1 mole / 18.0 g) = ? moles H2O

0.450 mL x (??? g / mL) x (1 mole / 102.1 g) = ? moles acetic anhydride

look up the density of acetic anhydride. (see the reference)



step 3... compare mole ratios calculated in step 2 to that of the balanced equation



step 4... from the balanced equation X moles limiting reagent ----> Y moles product.



so ? moles limiting reagent from step 2 times ( Y / X ) = moles product



step 5... moles product x ( ?? g / mole acetaminophen) = ??? g acetaminophen...

Answer 8

Acetaminophen Molecular Weight

Answer 9

Chemical Formula For Acetaminophen