November 13, 2019, 04:20:44 PM
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Topic: Yield/Purity of Alum  (Read 14668 times)

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Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #15 on: November 12, 2009, 09:33:28 PM »
I get 3.2g as theoretical... less than my experimental... is this correct?

When I go to find %yield it's 12g / 3.2g x100% and that's over 100 what did I do wrong?

Offline Borek

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Re: Yield/Purity of Alum
« Reply #16 on: November 13, 2009, 02:36:04 AM »
Neither 12 nor 3.2 seems to be correct. Show how you got them.
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Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #17 on: November 13, 2009, 08:21:01 PM »
well 12.15g is the mass I weighed for my product here's how I got 3.2g theoretical

1.115g starting aluminum foil x 2/1 (mole ratio) xmol/342.2g = 0.006755 mols x 474.4g/mol = 3.2g of KAI(SO4)2*12H20

Equation I used is

Al2(SO4)3 + K2SO4 + 24H20 --> 2KAI(SO4)2*12H2O

Offline Borek

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Re: Yield/Purity of Alum
« Reply #18 on: November 14, 2009, 04:55:13 AM »
1.115g starting aluminum foil x 2/1 (mole ratio) xmol/342.2g = 0.006755 mols x 474.4g/mol = 3.2g of KAI(SO4)2*12H20

Watch your units - this conversion is wrong.
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Offline ILoveISO

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Re: Yield/Purity of Alum
« Reply #19 on: November 14, 2009, 09:10:21 PM »
How is it wrong? I went from grams of aluminum to moles of it... then to grams of 2KAI o.o what did I do wrong?

Offline Borek

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Re: Yield/Purity of Alum
« Reply #20 on: November 15, 2009, 04:39:33 AM »
I went from grams of aluminum to moles of it...

No.
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Offline Entrigued

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Re: Yield/Purity of Alum
« Reply #21 on: November 18, 2009, 11:27:14 AM »
I went from grams of aluminum to moles of it...

No.
when I have time I'll look into this problem but a fact is that your theoretical yield should be higher than your experimental yield, since the theoretical yield does not take experimental errors into consideration. ok. Continue to tackle the problem, with perseverance you can do it. 

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