Chemistry HELP!!!???

Question

Ok. Im having a VERY hard time figuring out the "theoretical yield of Alum." We took .997 g of aluminum scrap = .0370 mol. We ended up with 12.771 g of Alum (KAl(SO4)2 *12 H2O) which I **think** is .0269 mol (12.771g/474.2=.0269) So I took .997g * (.0370/.997)*(12.771/.0269)=17.6 However, if I take 12.771/17.6 my % yield is like 72% which CANT be right. I think Im messing up the formula and Dont know where.... Do any chemistry EXPERTS out there know what the heck Im doing wrong??? Please help me out. Thanks

PS: The reaction is well behaved and a percent yield of near 100% should be expected. That is why 73% cant be right. Please let me know. Thanks

Answer 1

A very rough check of your math indicates to me that you arrived at the correct figure, approx 72% . But you may want to recalculate to get a more accurate figure...given the problem's figures, you should report your answer to three significant figures, eg 72.3%.

When your math is correct, why question a very reasonable answer?...Apparently, your 'scrap' was a 72% Al alloy.
But if it IS an alloy, other metals present may be interfering...would need details of your analytical procedure to determine this.

One other possibility is that since some of the .12H2O is loosely held, you may have over-dried the product...to check that, re-wet it and dry it mildly, eg over silica gel.

Lots of 'figures' in above...your avatar has me thinking that figures figure...:-))

Answer 2

Umm...I'm not quite sure what you did with that calculation, but here's my work.

Your actual yield (multiply by 27 to convert to grams)
.12.771 g * ( 1 mol KAl(SO4)2 *12 H2O/ 474.2 g) * (1 mol Al/ 1 mol KAl(SO4)2 *12 H2O)= .0269 mol Al

Your theoretical yield is given by the sounds of it: .0370 mol Al

Your percent yield:
.0269 mol/.0370 mol * 100% = 73%

Either you're mistaken about the content of the aluminum scrap, you misread the periodic table (I don't have one with me, so I assumed you had the right molecular weight), or 73 is right. Why couldn't it be? You're not over 100%.

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EDIT: Are you sure the scrap contains 100% aluminum and that you calculated the molecular masses correctly? If so, then you must have messed up during the experiment. It happens.

Answer 3

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Answer 4

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Answer 5

I did the calculation another way (using the percentage of Al in Alum) and got the same number you have: 72.9%.

Did you cool the stuff?