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
2 answers
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.
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