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Topic: Molecular Formula  (Read 10774 times)

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Offline cjtm92

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Molecular Formula
« on: March 13, 2009, 08:29:46 PM »
Okay, so I'm getting my Chemistry homework done right now, and I'm really stuck on a problem, which is:

"Tooth enamel, or hydroxyapatite, has a molar mass of 422 g/mol. Its composition is 28.5 percent calcium, 22 percent phosphorus, 49.3 percent oxygen, and 0.2 percent hydrogen. Caclulate the molecular formula."

I found the empirical formula, which I believe to be right -- C8H11O3N

Then, when I divide the molar mass by the empirical molar mass (979.36 g/mol), I get a 0.42 number, which wouldn't make any sense when I would multiply the subscripts. If you could guide me in the right direction, I would really appreciate it. Thank you!

Offline enahs

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Re: Molecular Formula
« Reply #1 on: March 13, 2009, 08:39:19 PM »
Where did you get calcium or nitrogen?

Offline ARGOS++

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Re: Molecular Formula
« Reply #2 on: March 13, 2009, 08:51:58 PM »

Dear cjtm;

Allow me to ask you where is the phosphorus in your empiric formula?

And from where you get an empirical molar mass of   979.36 g/mole?

And why you need any empirical molar mass, when you have the real molar mass instead?

Have you ever thought about: How many oxygen are 49.3% of 422 g/mole?
(I don’t believe that will be only 3.)

Good Luck!
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Offline cjtm92

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Re: Molecular Formula
« Reply #3 on: March 13, 2009, 08:53:32 PM »
Oh jeeze, I was looking at the wrong problem for some areas. The Empirical Formula I got is Ca7P7030H2. Sorry about that haha, but everything else seems to be right.

Offline ARGOS++

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Re: Molecular Formula
« Reply #4 on: March 13, 2009, 09:11:43 PM »
Dear cjtm;

In this moment please tell what are 422 g/mole, and what is 0.42 inside your question?
Or where is your question at all?

Good Luck!
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Offline cjtm92

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Re: Molecular Formula
« Reply #5 on: March 13, 2009, 09:25:29 PM »
When I divide the Molar Mass, that is given, by the Empirical Molar Mass [(422g/mol / 979.36g/mol)], I get a ratio of 0.43. My ultimate question is, what is the Molecular Formula? I don't know what to do with the 0.43, because it will not give me whole numbers for subscripts. Sorry, for the confusion, and if it's not clear still let me know.

Offline ARGOS++

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Re: Molecular Formula
« Reply #6 on: March 13, 2009, 09:29:17 PM »

Dear cjtm;

How you got the empirical molar mass of   979.36 g/mole?

Good Luck!
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Offline Borek

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Re: Molecular Formula
« Reply #7 on: March 14, 2009, 06:02:23 AM »
When I divide the Molar Mass, that is given, by the Empirical Molar Mass [(422g/mol / 979.36g/mol)], I get a ratio of 0.43.

Why does the empirical mass differ from the given?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline cjtm92

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Re: Molecular Formula
« Reply #8 on: March 14, 2009, 09:19:11 AM »
Because, if you are just told to find the Empirical Formula, it's a given that you are finding the formula of the atoms in the simplest proportion for a given molecule, for Molecular Formula, you are finding the exact number of atoms.

Example: If you found the Empirical Formula for H2O, but you were given the molar mass is 36g/mol, then it would be a 1:2 ratio, making the Molecular Formula H4O2.

But, I think I found the answer to my problem, I would multiply the subscripts by 0.43, so my Molecular Formula would be Ca3P3O13H

Offline ARGOS++

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Re: Molecular Formula
« Reply #9 on: March 14, 2009, 10:00:49 AM »

Dear cjtm;

Your formula is now correct for the given question!

But you could do it next time simpler in case you think:
How many Ca are 28.5% of 422 g/mole? – and so on!
(It gives quite exact numbers, because the correct MW is given!)

Good Luck!
                    ARGOS++

Offline cjtm92

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Re: Molecular Formula
« Reply #10 on: March 14, 2009, 08:14:46 PM »
I'm not sure I fully understand. 0.285 * 422 = 120.27, but how would that help me find how many atoms of each element are in the given compound?

Offline ARGOS++

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Re: Molecular Formula
« Reply #11 on: March 14, 2009, 08:50:24 PM »

Dear cjtm;

Now divide your result (120.27 g/mole) by the molar mass of  Ca and be surprised how exact the number will be!
The same you can do for each other element too!

Good Luck!
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Offline cjtm92

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Re: Molecular Formula
« Reply #12 on: March 14, 2009, 09:03:09 PM »
That's so awesome! I wish our teacher taught us that way... would of helped to check answers. Thanks again so much!

Offline ARGOS++

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Re: Molecular Formula
« Reply #13 on: March 14, 2009, 09:05:54 PM »

Dear cjtm92;

You 're welcome!   ─   Soon again.

Good Luck!
                    ARGOS++

 P.S.: Thank you too!
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