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

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.

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

To verify the empirical formula, we need to calculate the number of moles of each element in the compound.

1. Calculate the number of moles of phosphorus:
- Mass of phosphorus = 0.500 g
- The molar mass of phosphorus (P) is approximately 31.0 g/mol.
- Moles of phosphorus = mass / molar mass = 0.500 g / 31.0 g/mol ≈ 0.0161 mol

2. Calculate the number of moles of oxygen:
- Mass of oxygen = 1.145 g (product mass) - 0.500 g (phosphorus mass) = 0.645 g
- The molar mass of oxygen (O) is approximately 16.0 g/mol.
- Moles of oxygen = mass / molar mass = 0.645 g / 16.0 g/mol ≈ 0.0403 mol

3. Divide the number of moles of each element by the smallest number of moles:
- Phosphorus: 0.0161 mol / 0.0161 mol = 1
- Oxygen: 0.0403 mol / 0.0161 mol ≈ 2.5

4. Multiply the subscripts by 2 to obtain whole numbers:
- Phosphorus: 1 × 2 = 2
- Oxygen: 2.5 × 2 = 5

Therefore, the empirical formula of the compound is P2O5.

To determine the molecular formula, we need to use the molecular mass of the compound. The empirical formula mass of P2O5 can be calculated as follows:

- Molar mass of P2O5 = (2 × molar mass of P) + (5 × molar mass of O)
- Molar mass of P2O5 = (2 × 31.0 g/mol) + (5 × 16.0 g/mol) = 142.0 g/mol

We are given that the molecular mass determined by a mass spectrometer is approximately 285 amu.

Comparing the empirical formula mass (142.0 g/mol) to the molecular mass (285 amu), we can calculate the molecular formula mass:

- Molecular formula mass = empirical formula mass × n
- Molecular formula mass = 142.0 g/mol × n = 285 amu

Solving for n:

- n = 285 amu / 142.0 g/mol ≈ 2

Therefore, the molecular formula of the compound is twice the empirical formula, which is P4O10.

To verify the empirical formula and calculate the molecular formula of the compound, we can follow these steps:

1. Calculate the moles of phosphorous (P) and oxygen (O) in the compound:
- Mass of phosphorous = 0.500 g
- Mass of oxygen = (Mass of the compound) - (Mass of phosphorous) = 1.145 g - 0.500 g = 0.645 g

Moles of phosphorous (P) = (Mass of phosphorous) / (Molar mass of phosphorous)
The molar mass of phosphorous is 31.0 g/mol.

Moles of phosphorous (P) = 0.500 g / 31.0 g/mol ≈ 0.0161 mol

Moles of oxygen (O) = (Mass of oxygen) / (Molar mass of oxygen)
The molar mass of oxygen is 16.0 g/mol.

Moles of oxygen (O) = 0.645 g / 16.0 g/mol ≈ 0.0403 mol

2. Divide the moles of each element by the smallest number of moles to get the simplest whole number ratio:

Divide moles of phosphorous (P) and oxygen (O) by 0.0161 (the smallest number of moles):

Moles of P = 0.0161 mol / 0.0161 mol ≈ 1
Moles of O = 0.0403 mol / 0.0161 mol ≈ 2.5

So, the empirical formula would be P2O5.

To find the molecular mass, we need the empirical formula, which we have already determined as P2O5.

3. Calculate the empirical formula weight:
The empirical formula weight is the sum of the atomic masses of all the atoms in one empirical formula unit.

Molar mass of P = 31.0 g/mol
Molar mass of O = 16.0 g/mol

Empirical formula weight = (2 * Molar mass of P) + (5 * Molar mass of O)
= (2 * 31.0 g/mol) + (5 * 16.0 g/mol)
= 62.0 g/mol + 80.0 g/mol
= 142.0 g/mol

4. Determine the molecular formula:
To find the molecular formula, we need to compare the empirical formula weight (142.0 g/mol) to the given molecular mass (285 amu).

Molecular formula weight = 285 amu

The molecular formula mass must be a whole number multiple of the empirical formula weight:

Whole number multiple = Molecular formula weight / Empirical formula weight
= 285 amu / 142.0 g/mol
≈ 2

So, the molecular formula would be twice the empirical formula: (P2O5)2, which simplifies to P4O10.

Therefore, the empirical formula of the compound is P2O5, and its molecular formula is P4O10.