QUESTION F.21 HW

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Daniel Martinez 1k
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Joined: Wed Sep 18, 2019 12:16 am

QUESTION F.21 HW

Postby Daniel Martinez 1k » Wed Oct 02, 2019 10:39 pm

Can someone help me with this question? I've tried it multiple times and can't seem to get the right answer. Thanks in advance!

madijohnson_4A
Posts: 40
Joined: Fri Aug 30, 2019 12:18 am

Re: QUESTION F.21 HW

Postby madijohnson_4A » Wed Oct 02, 2019 10:57 pm

Was F.21 assigned in the homework?

Julie_Reyes1B
Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

Re: QUESTION F.21 HW

Postby Julie_Reyes1B » Wed Oct 02, 2019 11:15 pm

Hi!
Here is the problem for anyone wondering:
"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 0.383 mg O. The molar mass of didemnin-A was found to be 942 g/mol. What is the molecular formula of this compound?"
The first thing to do is convert each of the element's individual masses into moles. This is done by dividing each element by its molar mass. Since the masses are measured in mg, your answer will be in mmol. If you are not comfortable with this, you can always convert each mass to grams before proceeding. Here is C as an example:
1.11 mg C/12.01 g mol-1= 0.0924 mmol.
H=0.147 mmol, N=0.0113 mmol, O=0.0227 mmol
Next, divide each value by the smallest amount of moles, which in this case would be 0.0113 mmol. Once this is done, you will have the following:
C=8.18, H=13.0, N=1.00, O=2.00.
Your goal is to have all of the above values be integers, so in order to do this you must multiply each by 6. This will be the result:
C=49.08 (close enough to round to 49), H=78.0, N=6.00, O=12.00.
This would make the empirical formula: C49H78N6O12.
We aren't done yet. Now, we have to compare the molar mass of the empirical formula to the molar mass to the compound and multiply each subscript by a factor if needed.
The molar mass of the empirical formula is calculated as follows:
(12.01x49 + 1.01x78 + 14.01x6 + 16.00x12)942 g. This matches the molar mass of the unknown compound, meaning the molecular formula would also be:
C49H78N6O12.
Hope this helps!

504939134
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

Re: QUESTION F.21 HW

Postby 504939134 » Wed Oct 02, 2019 11:17 pm

You would first find the number of moles of each element. For example for the moles of Carbon you would divide the mass of the element in the compound which is 1.11mg (remember to convert this to grams) to the molar mass of carbon which is 12.01 g. Mol^-1. You would then find the moles of the rest of the elements, and divide each amount by the smallest amount of moles. Hope this kind of helps!


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