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Topic: Percent yield - a challenge  (Read 9133 times)

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Offline pfnm

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Percent yield - a challenge
« on: May 04, 2010, 06:27:58 AM »
From the lab, I'm finding some particular challenges:

10ml of pure ethylendiamine (8.99g based on density) in 25mL water + 5mL of 10M HCl, + 8.03g of CoSO4.7H2O dissolved in 25mL, then air is passed through.

12(en) + 4HCl(aq) + 4CoSO4.7H2O(aq) + O2(g) ->   4[Co(en)3]ClSO4(aq) + 30H2O(l)

(en) = C2H8N2

This means CoSO4.7H2O is limiting, and the product 4[Co(en)3]ClSO4(aq) formed is 0.02857 mol - 10.59g.

I'm asked to find the percent yield, but the product was never measured. Do I assume the theoretical yield=actual yield, so 100% yield? Do I write ('percent yield not measured'?)

This affects the next reaction:

We added Barium L-tartrate to the solution, to precipitate BaSO4, and slowly crystallise (+)[Co(en)3](+)tartrateCl.5H2O.

That equation I think, is:

[Co(en)3]ClSO4(aq) + Ba(L-tart)(aq) + 5H2O -> BaSO4(s) + [Co(en)3]tart.Cl.5H2O(s)

So if my equations are correct, how am I supposed to get the percent yield of the
final product [Co(en)3]tart.Cl.5H2O(s)? Is the amount of reagent [Co(en)3]ClSO4(aq) presumed to be equal to it's theoretical yield? (ie 10.59g)?

I've been at this for many many hours, any help would be much appreciated,

Thanks
« Last Edit: May 04, 2010, 06:41:02 AM by pfnm »

Offline Grundalizer

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Re: Percent yield - a challenge
« Reply #1 on: May 04, 2010, 06:11:01 PM »
Did YOU not measure it or were you told not to measure it?

Offline pfnm

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Re: Percent yield - a challenge
« Reply #2 on: May 04, 2010, 07:46:18 PM »
No one measured it, how could we measure it, it was in solution?

Quote
"(a) Preparation of [Co(en)3]3+

Prepare in a filter flask a solution containing of pure ethylenediamine (10.0mL) in water (25mL). After cooling the solution in an ice bath, add concentrated HCl (5mL), CoSO4.7H2O (8.0g), dissolved in cold water (25mL) and then add activated charcoal (2g). Place the solution in a labelled Buchner flask and give it to the technical officer (air will be passed through the solution for 3.5 hours.

At your next practical session filter the solution by gravity to remove the charcoal. Add 1M HCl ethylenediamine as required to the solution until its pH is 7.0-7.5. Heat the mixture in an evaporating dish on a steam bath. Cool the solution to room temperature.

(b) Resolution of [Co(en)3]3+

To the above [Co(en)3]3+ filtrate solution add barium L-tartrate (8.5g). After heating the mixture on a steam bath for 30 mins with vigorous stirring, filter off the precipitated BaSO4 and wash it with 10mL of hot water. Evaporate the solution on the hot plate to 30mL and allow the crystals of (+)[Co(en)3](+)tartrateCl.5H2O to precipitate over a week.

Filter off the crystals by suction, wash, air dry."
Thats from the lab book.


Furthermore in part (d) of the prac we are to convert the (-)[Co(en)3](+)tartrate.5H2O remaining in solution, into
(-)[Co(en)3]I3.H2O.

Again percent yield here is predicated on how much was in the filtrate, right? Which was not measured, which we were not told to measure, which we could not measure, which we did not measure.

« Last Edit: May 04, 2010, 07:59:32 PM by pfnm »

Offline pfnm

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Re: Percent yield - a challenge
« Reply #3 on: May 04, 2010, 07:53:52 PM »
The BEST I can come up with is this:

12(en) + 4HCl(aq) + 4CoSO4.7H2O(aq) + O2(g) ->   4[Co(en)3]ClSO4(aq) + 30H2O(l)

CoSO4.7H2O limits (I measured 8.032g for 0.028573 mol), so 0.028573 mol of 4[Co(en)3]ClSO4(aq) = 10.593g
 (I just wrote in the lab 'since actual yield wasn't measured, yield is presumed to be 100%'.)

Equation for resolution into diastereomers:

2[Co(en)3]ClSO4 + 2Ba(L-Tart) + 10H2O---> 2BaSO4 +  (+)[Co(en)3](+)tart.Cl.5H2O(s) + (-)[Co(en)3](+)tart.Cl.5H2O

0.028573 mol/2=0.014287                                        


And so from then on, using that 0.014287 molar amount to determine, firstly, how much (+)[Co(en)3](+)tart.Cl.5H2O was resolved, and secondly, how much [Co(en)3](+)tart.Cl.5H2O remains in solution (ie 0.014287 mol in both cases).

the lab book says, once the (+)[Co(en)3](+)tart.Cl.5H2O has precipitated, that the (-) isomer appears as                 (-)[Co(en)3](+)tartrate.5H2O (without the Cl bonded to the outer coordination sphere).

Is it a reasonable jump to make, the presumption of 100 percent yield, with no other possible (or at least obvious, to me) way to proceed in determining amounts?

Offline AWK

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Re: Percent yield - a challenge
« Reply #4 on: May 05, 2010, 01:52:51 AM »
Quote
Filter off the crystals by suction, wash, air dry."

then weight! Weighting the final product usually follows each synthesis.
AWK

Offline Borek

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Re: Percent yield - a challenge
« Reply #5 on: May 05, 2010, 03:19:53 AM »
I'm asked to find the percent yield, but the product was never measured.

Quote
I've been at this for many many hours, any help would be much appreciated,

And you may waste zillion of additional hours without a result. If you have not MEASURED amount of the product, you can't calculate percent yield.
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