What is the Theoretical and Percent Yield in Cholesterol Nonanoate Synthesis?

[~Sam~]

Homework Statement

A reaction of nonanoyl chloride with cholesterol to form cholesterol nonanoate, pyridine is also used in the reaction.

1.If a student starts with 0.2543 g of cholesterol, 1.4 mL of pyridine, and 0.14 g of nonanoyl chloride, what would be the theoretical yield of cholesteryl nonanoate?

2.At the end of the experiment, the student isolates 0.2254 g of cholesteryl nonanoate. What is the percent yield for this student’s synthesis?

Homework Equations

C9H17ClO + C27H46O -----(C5H5N) -----> C36H62O2 + C5H6NCl- (I believe that is the balance equation, any confirmation?

We are also given:
nonanoyl chloride M.W. = 176.7 g/mol
cholesterol M.W. = 386.6 g/mol
cholesteryl nonanoate M.W. = 527.2 g/mol

The Attempt at a Solution

I might have the solution but I'm not sure. First I believe that they all react with a ratio of 1mol, Using the balanced equation which says that ideally 1 mole of each of the three react together, it shows that C27H46O is the limiting substance as there it least of this. Therefore 1 mole cholesterol makes 1 mole of cholesteryl nonanoate, so 386g of cholesterol makes 526 g of cholesteryl nonanoate, so 0.2543g of cholersterol makes (526/386) x 0.2543 = 0.3465 g of cholesteryl nonanoate, which is the theoretical yield.

For 2) is it just % yield = (0.2254/theoretical yield) x 100

 

[Borek]

Approach seems to be OK, but I have just skimmed. I haven't checked numbers.

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[Blueshift5]

% = (0.2254/0.3465) x 100% = 65.1%

I would confirm that your calculations are correct. The balanced equation you have provided is correct and the ratio of 1 mole of each reactant is accurate. Your reasoning for cholesterol being the limiting substance is also correct.

For the first part, the theoretical yield would indeed be (0.2543 g cholesterol)(527.2 g cholesteryl nonanoate / 386.6 g cholesterol) = 0.3466 g cholesteryl nonanoate. So your calculation is very close to the theoretical yield.

For the second part, your calculation for percent yield is also correct. The student's actual yield is 0.2254 g, so the percent yield would be (0.2254 g / 0.3466 g) x 100% = 65.1%. This means that the student's experiment yielded 65.1% of the expected amount of cholesteryl nonanoate.

In addition, it would be helpful to mention the possible sources of error in this experiment. For example, the student may have lost some product during the isolation process or there could have been some side reactions that affected the yield. it is important to always consider and address potential sources of error in any experiment.