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Topic: Organic Chem Lab - % Yield, mmmol, limiting reagents  (Read 6345 times)

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Offline nozo

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Organic Chem Lab - % Yield, mmmol, limiting reagents
« on: April 29, 2007, 08:18:32 PM »
Hi all, I am doing my % yields and mmols, but somewhat confused on how to do the calculations taking into consideration the limiting reagents.

For instance, on my reagent table, mass of salicylic acid (limiting reagent) = 1.00 g (MWT = 138); acetylsalicylic acid (product) = 1.30 g (MWT = 180)

On my actual experiment, my salicylic acid = 1.10 g and my product came to 1.26 g

Normally, I would calculate my % yield to be 1.26/1.30 = 97% yield
And my mmol = 1.26/180 = 7 mmol

Taking into account the limiting reagent, how do I solve for my % yield and mmol?

Thank you so much~
nozomi

Offline Wisemanleo

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Re: Organic Chem Lab - % Yield, mmmol, limiting reagents
« Reply #1 on: April 30, 2007, 12:21:30 AM »
Hey nozomi~

To take into account your limiting reagent, the big hint is that the moles of your limiting reagent will equal the moles of your product.

Try to work with that hint, and let me know if you're still stuck  :)
Scientia est potentia.

Offline english

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Re: Organic Chem Lab - % Yield, mmmol, limiting reagents
« Reply #2 on: April 30, 2007, 12:30:45 AM »
Number one rule:

Convert everything to moles first.  Then follow the initial problem.

Offline nozo

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Re: Organic Chem Lab - % Yield, mmmol, limiting reagents
« Reply #3 on: April 30, 2007, 01:31:37 AM »
Ok, so I would have .007 mols of product ==> .007 mols of my limiting reagent
.007 mol x 138 g/mol = .966 g

Then 1.10 g of my actual limiting reagent / .966 g = 113 % yield (is it possible to have >100% yield??)

Erm.. not sure if I did it correctly  :P

Offline english

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Re: Organic Chem Lab - % Yield, mmmol, limiting reagents
« Reply #4 on: April 30, 2007, 02:07:58 AM »
Ok, so I would have .007 mols of product ==> .007 mols of my limiting reagent
.007 mol x 138 g/mol = .966 g

Then 1.10 g of my actual limiting reagent / .966 g = 113 % yield (is it possible to have >100% yield??)

Erm.. not sure if I did it correctly  :P

No.  Tell me your secret.  113 % yield?   :P

You said you have 1.10 g of salicyclic acid, we'll denote as "SA."  You also said your product, acetylsalicylic acid (ASA) was 1.26 g. 

According to your molecular weights, which I'm too tired to check, this means you have 0.0079710 moles of SA (not 0.007), your limiting reagent.  I have not rounded this number yet.  Do this last in any calculation.

So if this is how many moles of limiting reagent you have, this must mean you have 0.0079710 moles of ASA as well, your product.  You can convert this to grams.

A percent yield calculation is simply   actual amount  x 100
                                               theoretical amount

You just calculated your theoretical amount of ASA.  Your actual was 1.26 g.  You have a percent now, which is might I say quite good, 87.8 % .

Offline nozo

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Re: Organic Chem Lab - % Yield, mmmol, limiting reagents
« Reply #5 on: April 30, 2007, 08:49:00 AM »
Thank you so much g_english! I really shouldn't forget what I learned from gen chem  :-[

I think I could've done it this way also instead of below...

.966 g / 1.10 g limiting reagent = 87.8 %

But your way made much more sense... thanks for your *delete me*!

Then 1.10 g of my actual limiting reagent / .966 g = 113 % yield (is it possible to have >100% yield??)

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