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Topic: Yield/Purity of Alum  (Read 14762 times)

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Offline ILoveISO

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Yield/Purity of Alum
« on: November 11, 2009, 10:44:29 PM »
Calculate the theoretical yield of your product in grams. How do I find theoretical yield? What info do I need?

Offline BetaAmyloid

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Re: Yield/Purity of Alum
« Reply #1 on: November 11, 2009, 10:55:42 PM »
   Actual Yield (in grams)      x   100%   =   Percent Yield
Theoretical Yield (in grams)
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Offline Entrigued

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Re: Yield/Purity of Alum
« Reply #2 on: November 11, 2009, 11:24:29 PM »
Theoretical yield is just the values you calculate on paper, without the experimental flaws considered. Therefore, the theoretical yield of a particular product should be calculated stoichiometrically, which is just a fancy way of saying; using the quantitative relationship between reactants and products to calculate the amount of product or reactant. For instance using the number of moles relationship between compounds as a conversion factor to calculate a unknown, such as volume. 

Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #3 on: November 11, 2009, 11:54:59 PM »
I started with 1.15g of alum now the equation I get is

Al2(SO4)3 + K2SO4 + 24H2O --> 2KAI(SO4)2 * 12H2O

How would I find the theoretical yield for my product?

Offline Borek

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Re: Yield/Purity of Alum
« Reply #4 on: November 12, 2009, 03:18:35 AM »
You started with 1.15g of alum? So how come alum is the product?
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Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #5 on: November 12, 2009, 02:47:06 PM »
started with 1.15g of aluminum foil lol im supposed to assume the starting material is pure aluminum so any idea how to find theoretical yield?

Offline sjb

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Re: Yield/Purity of Alum
« Reply #6 on: November 12, 2009, 02:56:52 PM »
started with 1.15g of aluminum foil lol im supposed to assume the starting material is pure aluminum so any idea how to find theoretical yield?

Well, how did you convert the aluminium to its sulfate salt as required by your proposed balanced equation?

Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #7 on: November 12, 2009, 03:10:39 PM »
I put the alum foil pieces into a beaker and heated it up then cooled it down in a ice beaker and collected the crystals with a vacuum filtration

Offline sjb

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Re: Yield/Purity of Alum
« Reply #8 on: November 12, 2009, 03:12:58 PM »
Presumably you added something else to the foil - I have just finished reheating some leftovers for my tea covered in aluminium foil and not much seemed to happen to the foil, certainly not formation of the sulfate....

Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #9 on: November 12, 2009, 03:30:40 PM »
I added potassium hydroxide, that was what was heated then I added sulphuric acid to dissolve it

Offline Borek

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Re: Yield/Purity of Alum
« Reply #10 on: November 12, 2009, 03:50:17 PM »
Look at all formulae. How many moles of alum are produced from 1 mole of aluminum foil?
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Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #11 on: November 12, 2009, 03:57:41 PM »
So do I use 1.15g divide by mass of aluminum? to get the moles of that then use the mass I get in the end which was 12g and divide that by the mass too?

Offline Borek

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Re: Yield/Purity of Alum
« Reply #12 on: November 12, 2009, 04:30:31 PM »
1.15g divided by mass of aluminium - which happens to be 1.15 - is 1. I guess you mean molar mass, but you better assume you are surrounded by idiots that are not able to read your mind and need you to expilcitely state such things.

And I am not able to guess what is 12g and why. Molar mass of carbon? And by what mass do you want to divide that mass that was the mass that made it complete mess?
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Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #13 on: November 12, 2009, 07:14:37 PM »
12g is the amount of product I ended up with after I burnt and dissolved the foil....

Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #14 on: November 12, 2009, 07:16:11 PM »
I wanted to use 1.15g divide by molar mass of aluminum and then use that and divide it by the mass of my end product which came out to be 12g is that how I would find theoretical yield?

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