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Topic: Molar Mass Determination  (Read 6964 times)

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Offline OrganicSynthesis

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Molar Mass Determination
« on: April 13, 2010, 02:05:14 AM »
Quote
13. Flutamide (Eulexin) is an important organic compound containing
three fluorine atoms in each molecule. It is used in the treatment of
prostate cancer. An analytical chemist extracted flutamide from a
commercial tablet weighing 203.21 mg leaving a residue (containing
non-medicinal ingredients) that weighed 128.23 mg. Elemental analysis
of the extracted flutamide revealed the presence of 15.47 mg of
fluorine. What is the molar mass (in g mol–1) of flutamide?

A. 232.8
B. 254.5
C. 276.2
D. 286.9
E. 303.1

I feel quite stumped on this question. It is a question from a past competition, and the answer is C.

The progress I have made so far is that the mass of the flutamide is 75.98mg, with 15.47mg of it being fluorine. Otherwise, I can't seem to make the logical next step. Anyone feel like pushing me in the right direction? :P

Offline UG

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Re: Molar Mass Determination
« Reply #1 on: April 13, 2010, 02:15:27 AM »
Quote
I feel quite stumped on this question. It is a question from a past competition, and the answer is C.

The progress I have made so far is that the mass of the flutamide is 75.98mg, with 15.47mg of it being fluorine. Otherwise, I can't seem to make the logical next step. Anyone feel like pushing me in the right direction? :P
Calculate the number of moles of fluorine in 15.47 mg. How many molecules of F are in one flutamide molecule. How does knowing these two pieces of information allow you to work out the number of moles of flutamide and then the molar mass? Have a think.

Offline OrganicSynthesis

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Re: Molar Mass Determination
« Reply #2 on: April 13, 2010, 02:28:53 AM »
Oh! This makes perfect sense now.

I found the number of moles of Fluorine

n = 15.47/(3)(19)

Then I just plugged it into the molar mass formula

M = (3)(19)(75.98)/15.47

The closest answer is C

Thank you very much!

Offline UG

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Re: Molar Mass Determination
« Reply #3 on: April 13, 2010, 02:32:32 AM »
Haha, glad to help you out. I know all about tricky questions  ;D although probably not as much as I should :-\

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