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Author: Subject: Molecular Weight Determination
RedsAreRaw
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[*] posted on 6-1-2017 at 08:48
Molecular Weight Determination


Ran across a question when working through my Morrison and Boyd Organic book that went: The hormone insulin contains 3.4% sulfur. (a) What is the minimum molecular weight of insulin? (b) The actual molecular weight is 5734; how many sulfur atoms are probably present per molecule?

I honestly just took a stab in the dark by dividing 3.4/32.06 to get about 10% and divided this into the remaining 96.6% to get a ratio of about 1 to 10 for C,H, and O. So I just guess C10H10O10S to get an answer of 322.23 for (a). and then divided that into 5734 to get 18 for S.

The answer was actually 942 for (a) and 6 for (b).

How do you go about solving this question?

Thanks
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Texium (zts16)
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[*] posted on 6-1-2017 at 09:27


First you do 32.06/.034 which gives you 942. That would be if 1 molecule of sulfur represented 3.4% of the mass of the molecule.

Then you simply divide 5734 by 942 to get about 6. Or you can get the same answer by doing 5734 x .034, and then that answer /32.06, to give approximately 6.




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aga
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[*] posted on 6-1-2017 at 11:49


If it's any consolation, i stared at that for a while, and was a bit lost.

Overthinking things tends to be a recurring problem.

Now the problem has been worked through, it seems so obvious.

Thanks zts16 !




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[*] posted on 9-1-2017 at 06:45


Quote: Originally posted by aga  

Now the problem has been worked through, it seems so obvious.

That's exactly what I thought as well.

Thanks zts16 for the quick reply!
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