[skeme]

Can anyone walk me through how i would calculate the theoretical yield and percentage yeild of the following

C6H5COCl   + C6H5NH2 -----> C13H11NO + HCl

[Dan]

Theoretical yield = 100% yield

To calculate yields you must know the mass or molar quantity of your starting material (limiting reagent).

You also need to show that you have attempted the question.

[skeme]

Benzoyl chloride

4.4g x 1 mole x 100 = 3.14 mole
           140g

Aniline

2.6g x 1 mole x 100 = 2.80 mole
            93g

The limiting reagent is the aniline

here's where i get confused

[Dan]

what is the maximum molar amount of the amide product you could possibly get?

[skeme]

2.80?

[Dan]

yes, now work out the mass of that yield, and you have your theoretical yield.

Now your percentage yield is the percentage of that mass of product that you obtain after your experiment.

[skeme]

Benzoyl chloride

4.4g x 1 mole x 100 = 3.14 mole
           140g

Aniline

2.6g x 1 mole x 100 = 2.80 mole
            93g

theoretical yield
Benzanilide

5.23 x 1 mole x 100 = 2.65 mole
           197g

2.65/2.80 * 100 = 95% theoretical yield?

[Dan]

Benzoyl chloride

4.4g x 1 mole x 100 = 3.14 mole
           140g

Aniline

2.6g x 1 mole x 100 = 2.80 mole
            93g

theoretical yield
Benzanilide

5.23 x 1 mole x 100 = 2.65 mole
           197g

2.65/2.80 * 100 = 95% theoretical yield?

Hang on a minute, why are you multiplying by 100 in the highligted areas? This is incorrect.

(mass, g)/(rel molecular weight, g mol^-1) = (amount, mol)

If your mass of benzanilide was 5.23g, then your yield is indeed 95%, the factors of 100 cancelled, so your yield still has the correct value, but your values for number of moles are all out by a factor of 100.

[skeme]

so my theoretical yield is 95% (the 5.23g was my crude product)

after recrystillation i had 1.23g

so would it be 1.23/197 = 0.0624 mol

0.0624/0.0279 x 100 = 22% for my percentage yeild?

[bnstria]

so my theoretical yield is 95% (the 5.23g was my crude product)

after recrystillation i had 1.23g

so would it be 1.23/197 = 0.0624 mol

0.0624/0.0279 x 100 = 22% for my percentage yeild?

Your theoretical yeild needs to be quoted as a mass not a percentage. as then it is just a percent yeild.

Try to follow this through...

Moles of Benzoyl chloride = (4.4 g/140.52) = 0.031312268 moles    --- try to keep you decimal places while you work these things out.  If you begin to round too early then your final answer may be out and you won't get full marks.

Moles of aniline = (2.6 g/93.126) = 0.027919163   - so this is your limiting reagent - do you understand why the mass is divided by the molar mass?  If not then just ask.

So aniline is your liming reagent, but you already know this anyway.  Therefore the largest number of moles of product you can get is the same as your limiting reagent which is therefore 0.027919163 moles.

Then you need to get back to grams for your theoretical yield.

Multiply the number of moles by the molar mass of your product.

like this 0.027919163 x 197.228 = 5.50644 g  <-- this is your theoretical yeild.

When quoting your percent yeild you should not use your crude yeild but your recrystallised yield which in your case is 1.23g

so 1.23/5.506 x 100 = ?  this should be your percent yeild

Hope you can follow this through, let me know if you get stuck.

[bnstria]

Oops, I should mention that you don't have to write all of the number down each time, just as long as you perpetuate it through your calculations. i.e. you don't have to write 0.027919163 mol in your book but as long as you keep it in your calculator.

[Custos]

Moles of Benzoyl chloride = (4.4 g/140.52) = 0.031312268 moles    --- try to keep you decimal places while you work these things out.  If you begin to round too early then your final answer may be out and you won't get full marks.

I don't agree with that. If I saw a student write "0.031312268 moles" I would assume he/she had limited understanding of significant figures and error calculations. Obviously if the benzoyl chloride has been weighed out as 4.4 grams the degree of accuracy in the number of moles calculation is, at best, 3 significant figures. Why not get into the practice of using the right number of significant figures? How often do students follow through these calculations and end up with a yield of 78.653% or some other nonsense... because they are afraid to discard an insignificant trailing digit that the calculator said was correct? Sorry, just a pet peeve.