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Topic: Can someone double check the % yield here?  (Read 2226 times)

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Offline yourdeath01

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Can someone double check the % yield here?
« on: November 06, 2019, 10:41:34 PM »
Reaction 1:


I have this reaction: https://imgur.com/MJBdOif

Flowchart of experiment: https://imgur.com/Yb6O7nz

I know p-nitroaniline we use 10 mmol of it in the diazotization step and I know we use 10 mmol of the 2-napthol and since everything is 1:1 therefore I can multiply 10 mmol x product (para red which has 293.28 mw) to get a theoretical yield of 0.01 mols x 293.28 = 2.9328 theoretical yield of para red. Is the theoretical yield correct here? I did not consider the reagents Hcl and NaNO2, is that okay?

Now what we did here is that we split our product where half of it was coupled with 2-napthol and measured and weighted for a % yield and the other half was coupled with 2-napthol again but it was actually used for the dye process and not measured and weighted. Therefore, my final mass of the product was 0.634 g therefore 0.634/2.932 = 21.62% yield, is this % yield correct? Do I write down 21.62% and while it is very low, I can mention that half of it was used for the dye process. But even still if it wasn't cut in half it would be a final yield of 1.268 for a yield of 43.24 % which is still very low so what you guys think I should put down? I feel like since my final mass was 0.634 g therefore final yield was 21.62% regardless of what was split or not!

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Reaction 2: https://imgur.com/lB9yVCN

This time p-toluidine 10 mmol reacted with phenol 10 mmol to produce the azo dye product. The MW of final product is 212.245 mw therefore 0.01 mol x 212.245 g/mol = 2.122 theoretical yield, is this correct?

My final product was 1.2879 g which is divided by 2.122 g = 60.69% yield correct? Again it was split in 2 so my yield could have probably been higher. But your thoughts guys?

Offline chenbeier

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Re: Can someone double check the % yield here?
« Reply #1 on: November 07, 2019, 01:34:32 AM »
The theoretical yields are ok, but the measured ones are rubbish. The mistake is that you don't know the weight of the half of the product. Did you split to 5 mmol or what was the number. You need the value of the real half.

Offline anonymous10012

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Re: Can someone double check the % yield here?
« Reply #2 on: November 07, 2019, 11:22:04 AM »
I assume you did not split it in half by exact weight. You have to weigh it before you do anything else with the product.  :(

AP/TAP x 100 = % Yield

Offline chenbeier

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Re: Can someone double check the % yield here?
« Reply #3 on: November 07, 2019, 02:53:53 PM »
Exactly, that is the point.

Offline yourdeath01

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Re: Can someone double check the % yield here?
« Reply #4 on: November 07, 2019, 03:08:00 PM »
Hey guys thanks for replies just want to to clear something as you mentioned:

Now what we did here is that the lab is that it tells us to split our diazo salt componenet in half as best as possible (we do not measure this) basically one half of the diazo componenet will be coupled with one half of 2-napthol and measured and weighted for a % yield and the other half of the diazo componenet will be coupled with the other half of 2-napthol but will not be weighted but actually used for the dye process. Therefore, my final mass of the product (half of diazo + half of coupling component) was measured to be 0.634 g therefore 0.634/2.932 = 21.62% yield, is this % yield correct? Do I write down 21.62% and while it is very low, I can mention that half of it was used for the dye process? If the % yield above is correct what you think caused me to have such a low yield? Probably the transfer of products from one container to the other caused this I would think.

Offline anonymous10012

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Re: Can someone double check the % yield here?
« Reply #5 on: November 08, 2019, 10:21:28 AM »
I think I see what you did here. If you theoretically had 10 mmol of diazonium salt then you split it in half to proceed with the next step in the synthesis, you only reacted (roughly) 5 mmol of your salt with about 5 mmol of 2-napthol. It is likely that the first step to form the diazonium salt did not give you 100 percent yield, but I guess the teacher isn't asking for your percent yield of diazonium salt. You were supposed to have weighed the portion of diazonium salt (about 5 mmol) and the 2-napthol that you reacted with it. The reactant that you had more moles of (diazonium salt or 2-napthol) is the limiting reagent. You can only make as much para red as you have limiting reagent. I would try to work through it with a classmate and compare yields. If you are still lost, maybe it is a strange experiment that your teacher designed, and we are not communicating effectively about it.

I hope this helps. Bye.

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