The question says: when .422 g of phosphorus is burned, .967 g of a white oxide is obtained. Determine the empirical formula of the oxide.

I did the calculations and got that there would be 1 mole Phosphorus and 4.29 mol Oxygen. To get the empirical formula, I would need to get them to whole numbers, right? What would my whole numbers be?

Nevermind, I think I've figured out what I was doing wrong!!

To determine the empirical formula of the oxide, you need to find the simplest whole-number ratio between phosphorus and oxygen atoms in the compound.

Based on the information given, you have 0.422 g of phosphorus and 0.967 g of the white oxide. To find the number of moles of each element, you can use their molar masses.

The molar mass of phosphorus (P) is 30.97 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.

For phosphorus:
0.422 g P × (1 mol P / 30.97 g P) = 0.0136 mol P

For oxygen:
0.967 g oxide - 0.422 g P = 0.545 g O
0.545 g O × (1 mol O / 16.00 g O) = 0.0341 mol O

Now we have the mole ratios between phosphorus and oxygen:
Phosphorus: 0.0136 mol P
Oxygen: 0.0341 mol O

To convert these ratios into whole numbers, you need to divide each value by the smallest value. In this case, the smallest value is 0.0136 mol P.

Dividing both values by 0.0136 mol P:
Phosphorus: 0.0136 mol P / 0.0136 mol P = 1 mol P
Oxygen: 0.0341 mol O / 0.0136 mol P = 2.5 mol O

Now, we have a ratio of 1 phosphorus atom to 2.5 oxygen atoms. To make it a whole number ratio, you can multiply both values by 2:

Phosphorus: 1 mol P × 2 = 2 mol P
Oxygen: 2.5 mol O × 2 = 5 mol O

Therefore, the empirical formula of the white oxide is P2O5, indicating that it contains 2 phosphorus atoms and 5 oxygen atoms.