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October 31, 2014

Homework Help: Chemistry

Posted by Helpme on Sunday, October 16, 2011 at 7:34pm.

Hi Im reposting incase my original post doesn't get read. Please see bottom for additional info that I need help getting verified. Many thanks


Hi Please help verify my answer to the following question:
0.500g of pure phosphorous was burned in excess pure oxygen to give a product that has a mass of 1.145 g. What is the empirical formula of the resulting compound? By the use of a mass spectrometer the molecular mass of this compound (phosphorous oxide) was determined to be approximately 285 amu. What is it's molecular formula?


I was able to determine the empiricle formula as being PO3 (P 0.0161446 O 0.0403125)but don't know how to find the molecular mass.

Please verify my empiricle formula too because when I google formula for phosphorous oxide it tells me something different.

Please a step by step explaination for Molecular mass would be helpful as I don't have an example in my text to refer to which is why I posted here.

Best Regards




Chemistry - DrBob222, Saturday, October 15, 2011 at 11:18pm
The empirical formula is P2O5; you should go back over your calculation and find the error. Post your work if you want me to check it.
For the molecular formula, it is done this way.
empirical formula mass P2O5 = 141
molar mass = 285
285/141 = 2.02 which rounds to 2.0 so the molecular formula is two units of the empirical formula or
(P2O5)2 or P4O10.




Chemistry - Helpme, Sunday, October 16, 2011 at 7:32pm
Here is wht I did if you can please tell me where I went wrong would be appreciated.

resulting compound phosphorous oxide mass = 1.145g

# of moles of Phosphorus = 0.500g/30.97g/mol = 0.0161446mol

# of grams of combined phosphorus oxide = 1.145g-0.500g (P) = 0.645g of oxygen

#of moles of oxygen = 0.645g/16.00g/mol = 0.0403125mol

Therefore P 0.0161446 O 0.0403125 which divided into the smallest = P1 and O2.4969 which is then rounded to P1O3???? Please verify all of this against the question and let me know where I'm going wrong.....

Best Regards

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