Theoretical Yield and Percentage Yield of Organic Molecules

[fishes]

Homework Statement

1. A student investigated several esters, and made octyl acetate, an ester with the fragrance of oranges. The student used the acid derivative, acetyl chloride (as you saw decanedioyl dichloride used to make 6,10 nylon) and 1-octanol. Using the following information, calculate the theoretical yield for the student. (Hint: you need to calculate the molar masses and to identify the limiting reagent).

Structural Equation

Quantities used:

acetyl chloride: 0.654 g

1-octanol: 1.401 g.

Calculate theoretical yield. Enter your answer with three significant figures in grams.

2.
A student began the aspirin synthesis with 2.083 g salicylic acid and excess acetic anhydride. The student's actual weighed yield was 2.534 g. Calculate the % yield. Give your answer as % with 3 significant figures (that is 1 decimal place in your %age). Do not include % symbol.

Homework Equations

Percentage Yield = Actual yield/theoretical yield

The Attempt at a Solution

1. I figured out the equation: C8H18O + CH3COCl ----> C10H20O2 + HCL (1:1 Ratio)
Figuring out the moles for 1-octanol and acetyl chloride gives me acetyl chloride as the limiting reagent with 0.00833moles. Multiplying this with the molar mass of octyl acetate gives me 1.44g. Is this right?

2. Once again we have a 1:1 ratio. We are given that salicylic acid is the limiting reagent. 2.083/138.13 = 0.0150 moles theoretically of aspirin. 0.0150mol x molar mass aspirin (180.17g/mol) = 2.7169g.
2.534/2.7169 = 93.3%.
Thanks for the help, I think my answers are right but I normally get something simple wrong.

 

[Borek]

Your images don't show for us.

Yes, acetyl chloride is a limiting reagent, yes 1.44 g of octyl acetate looks OK.

 

[fishes]

Sorry about that, think I fixed it now

 

[Borek]

Second image has nothing to do with the reaction from the question.

 

[fishes]

That's embarrassing! Sorry about that.

 

[Borek]

Other than that logic looks good.

 

[Jacob Mierendorff]

THANKYOU SO MUCH! I am studying for an exam at the moment and this helped me so much!

Cheers fam :)