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F.21

Posted: Thu Oct 01, 2015 9:16 pm
by Nivi Ahlawat 3I
Hi, I have a question about number 21 in the F section of the Fundamental Problems.

Q: In 1978, scientists extracted a compound with antitumor and antiviral properties from marine animals in the Caribbean Sea. A sample of the compound didemnin-A of mass 1.78 mg was analyzed and found to have the following composition: 1.11 mg C, 0.148 mg H, 0.159 mg N, and 0.363 mg O. The molar mass of didemnin-A was found to be 942 g/mol. What is the molecular formula of didemnin-A?

A: The answer key says C49H78N6O12, but I got C48H78N6O12. I found the number of moles for each element and then found the ratios: 8.143 for C, 13 for H, 1 for N, and 2 for O. After dividing 942 g/mol (the molar mass of didemnin) by 155.197 g/mol (the empirical mass) and got 6.0697. I multiplied 8 by 6 to get C48. Am I supposed to multiply 8.143 by 6.0697 to get C49? Thank you.

Re: F.21

Posted: Thu Oct 01, 2015 9:30 pm
by Chem_Mod
It turns out that 8.143 is the correct ratio for C when you do not round until the end.
Then the "empirical mass" is 156.716 (in quotes, because you can't have a compound with 8.143 carbons)
942 divided by 156.716 = 6.01
multiplying 8.143 by 6 gives 48.86 which indicates 49 carbons

(Technically, the empirical formula is the SAME as the final molecular formula because the ratios are irreducible)

Re: F.21

Posted: Thu Apr 12, 2018 10:39 pm
by Cindy Nguyen 1L
I understand how the answer is so, but if the problem asked to find the empirical formula of the compound didemnin-A, how can the ratio for Carbon be 8.143 when the number of atoms should/is always a whole number?

Re: F.21

Posted: Sun Apr 15, 2018 3:50 pm
by Anna De Schutter - 1A
I believe that if the problem had asked to find the empirical formula, we would have also found C49H78N6O12 because as said by Dr. Lavelle just above, the empirical formula and molecular formula are the same in this problem because the ratios of the molecular formula (C49H78N6O12) are irreducible. Moreover, the molar mass of C49H78N6O12 is 942g/mol (which is what we asked for in this problem) so the empirical formula is the same as the molecular formula.

I hope this helps! :)
Anna De Schutter - section 1A

Re: F.21

Posted: Sun Apr 15, 2018 5:09 pm
by Anthony Mercado 1K
As an inquiry into discovering the molecular formula of didemnin-A, in the primary steps towards conversion would one have to convert mg units to g towards the end or beginning of calculations? * Would these affect the number of sig figs due to constant conversions?

Re: F.21

Posted: Sun Apr 15, 2018 6:17 pm
by Chem_Mod
When multiplying by a conversion, you are mutliplying by an exact number. An exact number has infinite sig figs, so it does not affect the number of significant figures you get as a result.