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

Here is what 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
This step is OK.

# of grams of combined phosphorus oxide = 1.145g-0.500g (P) = 0.645g of oxygen
This step is ok if you relabel it as grams phosphorus oxide - g P = grams oxygen = 0.645g

#of moles of oxygen = 0.645g/16.00g/mol = 0.0403125mol
This step is ok.

Therefore P 0.0161446 O 0.0403125 which divided into the smallest = P1 and O2.4969 which is then rounded to P1O3????This is where you are going wrong. You have P1 O 2.4969 and you should round; however, 2.4969 can't be rounded to 3 or 2. You could round it to 2.5 (and should) but you can see that P1 O 2.5 actually is P2O5 using small whole numbers. The general rule for rounding in elementary classes is to round 1.1 or so to 1 and 1.9 or so to 2 but 1:1.25 should be counted as 4:5 and 1:1.33 should be counted as 3:4. I find the easiest way to see this is to take a number like 1:2.249 and multiply both numbers by 2,3,4,5 sequentially to see if you can get small whole numbers out of it. Multiplying by 2 gives 2:4.98 and we can round that to 2:5. If we had P1 O 1.33 we would try multiplying by 2 to give 2:2.66(which CAN'T be rounded to whole numbers) so then we try multiplying by 3. That gives us 3:3.99 and we can round that to 3:4. Etc. Please verify all of this against the question and let me know where I'm going wrong.....

Best Regards

This is very helpful. I'm sure this will be reviewed in upcoming classes and understanding this information will help me be a bit ahead of the game.

Thank you very much for your help, I'm sure it wasn't an easy feast to type all this information down. Very appreciated.

To verify the empirical formula, we need to determine the ratio of moles of each element in the compound. Let's first calculate the moles of phosphorus and oxygen:

1. Moles of phosphorus:
Mass of phosphorus = 0.500 g
Molar mass of phosphorus = 30.97 g/mol
Moles of phosphorus = 0.500 g / 30.97 g/mol = 0.0161446 mol

2. Moles of oxygen:
Mass of oxygen = 1.145 g - 0.500 g (mass of phosphorus) = 0.645 g
Molar mass of oxygen = 16.00 g/mol
Moles of oxygen = 0.645 g / 16.00 g/mol = 0.0403125 mol

Now, let's compare the moles of phosphorus and oxygen to find the empirical formula ratio.

Moles of phosphorus: 0.0161446 mol
Moles of oxygen: 0.0403125 mol

Dividing both values by the smallest mole value (0.0161446 mol), we get:

Moles of phosphorus = 1 mol (approximately)
Moles of oxygen = 2.4969 mol (approximately)

Rounding these values to the nearest whole number, we have:

Moles of phosphorus = 1
Moles of oxygen = 2

Therefore, the empirical formula is P1O2, which simplifies to PO2.

To calculate the molecular formula, we need to determine the relationship between the empirical formula mass and the molar mass.

Empirical formula mass of PO2:
Mass of phosphorus (P) = 30.97 g/mol
Mass of oxygen (O) = 16.00 g/mol x 2 = 32.00 g/mol
Empirical formula mass = (30.97 g/mol) + (32.00 g/mol) = 62.97 g/mol

Molar mass of the compound = 285 g/mol

Dividing the molar mass by the empirical formula mass:

Molar mass / Empirical formula mass = 285 g/mol / 62.97 g/mol ≈ 4.52

Rounding this value to the nearest whole number, we get 4.

So, the molecular formula is four units of the empirical formula, which is (PO2)4. Simplifying this gives us the molecular formula as P4O8.

Therefore, the empirical formula of the resulting compound is PO2 and the molecular formula is P4O8.