[nj_bartel]

The Question: Calculate the amount of isopentyl acetate that should be present in the reaction mixture at equilibrium, based on the quantities of starting materials you used and a value of 4.2 for the equilibrium constant.  (Use the quadratic equation; because volumes cancel out, moles can be used in place of molar concentrations.)

My Attempt (has to be incorrect, but here it is anyway): I took my original expected yield (which was calculated assuming the reaction was irreversible and there was 100% yield) and multiplied it by (4.2/5.2), as I assumed an equlibrium constant of 4.2 means there are 4.2 moles of products for every 1 mole of reactants at equilibrium.  I came up with this:

19.53 grams expected yield * 4.2/5.2 = 15.77 grams true expected yield
This must be the incorrect method of going about this however, as it says to use the quadratic equation.  What forumula should I be using for this?

Thanks.

[macman104]

What is your equilibrium expression?  Is the reaction a 1:1?  Is it merely A + B ---> C?

[nj_bartel]

Isopentyl Alcohol + Acetic Acid  Isopentyl Acetate

So 4.2 = [Isopentyl Acetate] / {[Isopentyl Alcohol][Acetic Acid]}

And I'm stuck!

[macman104]

So, now do an ICE table, use your initial moles, and then -x, and +x on the product side, and solve!

[nj_bartel]

Well, I solved the quadratic for x / (0.150 - x)² = 4.2

and came up with 0.492 and 0.0457.

The first is impossibly large, so I have to take the second one, which leaves me with (0.0457 moles)(130.2 grams/mole) = 5.95 g Isopentyl acetate.

That result works out to me having about 200% yield.

My crude product mixture after reflux was distilled, and essentially the entire mixture distilled from 130 - 133 ^oC.  In the reaction mixture was the isopentyl acetate product, isopentyl alcohol and excess acetic acid reactants, and phosphoric acid catalyst.  How could that mixture of substances have all distilled in such a narrow range?

[nj_bartel]

I thought maybe I was supposed to use the actual moles of acetic acid, not the equimolar amount, so I plugged that in and came up with an x value of 0.015 and a percent yield of almost 600% 

Edit: In case this needs to be known, the alcohol is the limiting reactant, so acetic acid was added in excess to increase yield via LeChatlier's.

[macman104]

Can you provide the starting mole amounts for your starting materials.  I mean, unless I'm missing something, you should just take

Isopentyl Alcohol + Acetic Acid  Isopentyl Acetate

  starting moles     starting moles                   0
     -x                     -x                               +x
starting moles - x     starting moles - x         +x


So we have

4.2 = x/((moles_{isopentyl alcohol} - x)*(moles_{acetic acid} - x))

I don't think you would do it any other way...

[nj_bartel]

0.2967 mol anhydrous acetic acid

0.150 mol isopentyl alcohol

1 mL 85% phosphoric acid (catalyst)


You set it up the same way I did.  It works out to x = mol isopentyl acetate = 0.0158

I recovered 11.6 grams of product via simple distillation. (10 - 12 was the norm for the class)

(0.0158 mol isopentyl acetate)*(130.2 grams / mol) = 2.06 grams isopentyl acetate product

(11.6 / 2.06)*100% = 563% Yield.

Which goes back to what I was saying.  I collected distillate from 120 Celsius to (distillate actually didn't even begin dropping until 130 Celsius) 133 Celsius.  During this range of 3 degrees Celsius, nearly all of the product mixture distilled through the condenser.  The error was not that I heated too quickly, because the distillation took approximately 8 minutes if I recall correctly, from first drop into collection vial to last.  So I was wondering how that mixture of different substances could have all distilled in essentially the same temperature band (this was not under vacuum, so 3 degrees Celsius is a pretty narrow range).

Thanks.

[macman104]

Hmm...well it (wikipedia that is) says isopentyl alcohol boils at 132 C, and that isopentyl acetate boils at 142 C.

Actually, I don't think it matters what the limiting reagent is.  The starting concentration/moles of acetic acid are independent of what your stoichiometry was.  It also doesn't even sound like you would have gotten the isopentyl acetate, because that boiling was not high enough...

Do you disagree with the following picture setup?  I didn't catch it before, but I think you don't take 0.150 for each of them...

[nj_bartel]

And I could definitely accept that reasoning if it weren't for the fact that isopentyl acetate, and isopentyl acetate alone, smells distinctly like bananas, which my reaction container did, big time.

The text had us collecting distillate from 136 - 143 Celsius.  However, before we began our distillations, the TA's told us that we were to begin collecting distillate at 120 Celsius onward because if we waited until 136 celsius, there would be nothing left for us to collect.  I can't really put together what exactly was in my distilled product mixture based on the information I have.  I know I have product, due to the smell, but there also have to be a plethora of impurities in there as well.  The boiling point range for the distillate was also so tight, yet the whole crude product mixture boiled off as one.

=/

Thanks for the help, at least I know it's not my algebra =P

[macman104]

Posting just so it's obvious that I completely changed what I was writing above, and it took longer than I thought, so I should have posted after nj.

[nj_bartel]

That is what I did in my second attempt, except I forgot to multipy the .044505 by the 4.2, which changes the answer tremendously and makes my experimental findings much more reasonable.

Math kills me, thanks a bunch.

(edit: it still doesn't really explain why the acetate would be boiling lower than it should be, but I'm comfortable letting that one go  )

[macman104]

That is what I did in my second attempt, except I forgot to multipy the .044505 by the 4.2, which changes the answer tremendously and makes my experimental findings much more reasonable.

Math kills me, thanks a bunch.

Haha, as you've said in other threads, lol.

(edit: it still doesn't really explain why the acetate would be boiling lower than it should be, but I'm comfortable letting that one go  )

Yea, that is still kinda weird.  The only thing I can think of is an azeotrope forming between them.

[leaftye]

Can you provide the starting mole amounts for your starting materials.  I mean, unless I'm missing something, you should just take

Isopentyl Alcohol + Acetic Acid  Isopentyl Acetate

  starting moles     starting moles                   0
     -x                     -x                               +x
starting moles - x     starting moles - x         +x


So we have

4.2 = x/((moles_{isopentyl alcohol} - x)*(moles_{acetic acid} - x))

I don't think you would do it any other way...

The problem I have with this setup is that it doesn't account for water.  Water doesn't normally go in an ICE table, but in this case it is a product.  Also, the lab book (Lehman, exp 5) says the theoretical yield should be 67% with equimolar concentrations.  I can't get 67% unless I put water in the ICE table.  Here's how I'd change your math:

Isopentyl Alcohol + Acetic Acid  Isopentyl Acetate

  starting moles     starting moles                   0
     -x                     -x                               +x
starting moles - x     starting moles - x         +x


So we have

4.2 = x²/((moles_{isopentyl alcohol} - x)*(moles_{acetic acid} - x))

4.2 = x² - 1.87614x + 0.186921
x² - 1.87614x + 0.186921 = 0

x = 0.1272
x = 0.4590

Theoretical yield = 84.8%