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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: sarah9625 on October 16, 2007, 07:12:48 PM

Title: Percent yield in the lab
Post by: sarah9625 on October 16, 2007, 07:12:48 PM
I did a lab on the Synthesis of ter-butyl chloride via SN1 reaction. I have to find the percent yield.
I used 15ml concentrated HCl, 5ml tert-butyl alcohol and then added 40ml saturated sodium bicarb. The end weight of my product was 2.088g. The vial that I weighted it in (clean per-weight) was 14.551g.

I know that the percent yield is my end product / theoretical yeild x 100. I do not know how to get the theoretical yeild.
so far I have:
tert-but chloride as 2.088g/92.5g=0.0226 moles
ter-but alcohol as 5ml x (0.78g/1ml) x (1mole/74.12g)= 0.0526 moles
Conect. HCL as 15ml x (1.69g/ml) x (1mole/36.47g)= 0.695 moles
 I am not sure what to do next
Any help would be wonderful.
Title: Re: Percent yield in the lab
Post by: TheBigF on October 16, 2007, 07:25:52 PM

As you see in your calculations the HCl is in excess (as is the bicarb - to remove unreacted HCl), so the tert-butyl alcohol is the limiting reagent.

For a 100% reaction you would expect to use get 0.0526 moles of product from 0.0526 moles of starting material.

Your yield is therefore (0.0226 moles/0.0526 moles) X 100% = 43%
Title: Re: Percent yield in the lab
Post by: sarah9625 on October 16, 2007, 07:52:46 PM
THANK YOU SOOOOOO MUCH, here is a snack:)
Title: Re: Percent yield in the lab
Post by: Borek on October 17, 2007, 03:23:20 AM
For a 100% reaction you would expect to use get 0.0526 moles of product from 0.0526 moles of starting material.

While this is true in this particular case it is hardly correct approach - you should start with reaction equation to know what is molar ratio of reagent to product. You are probably aware of the fact it is 1:1 in this case, but your answer may suggest to others that it is 1:1 always.
Title: Re: Percent yield in the lab
Post by: TheBigF on October 18, 2007, 02:39:32 PM
My answer is, as you suggest, specific for this question (or all others involving a 1:1 stoichiometry). But then I was answering a specific question - not a general one.

Of course, in general, you need to bear in mind the stoichiometry of the reaction.