How would you find the empirical formula for this... a substance was found by analysis to contain 20% by mass calcium and 80% by mass bromine.

Would it be CaBr2?

This is what I did...

20g Ca * 20 mol Ca/801.56g Ca = .499 mol Ca

80g Br * 80 mol Br/6392.32g Br = 1.001 mol Br

1.001/.499 = 2 so CaBr2

I meant 20g Ca not 50. Sorry! Would that be right? I multiplied 40.078g Ca (which is one mole) times 20 to get 801.56g Ca (mass of 20 moles) and I used the same method to find the mass for 80 moles of Bromine.

That will work but it's more complicated than it needs to be.

20% Ca and 80% Br.
You take a 100 g sample which gives you 20 g Ca and 80 g Br.
Then 20g Ca x (1 mole Ca/40.078 g Ca) = 0.499 mole Ca which is the same as

20 g Ca x (20 moles Ca/20*40.078) = 0.499 mole Ca. My point to all of this is you don't need 20 moles Ca and you could have just canceled the 20 moles in the form I wrote it and have a simpler math situation. But your method will work if you want to do it the long way and work with larger numbers. :-)

Yes, your revised calculations are correct. To find the empirical formula, you first need to determine the moles of each element present in the substance.

Using the given information, you correctly calculated the number of moles of calcium (Ca) and bromine (Br). Using the molar mass of calcium, which is approximately 40.078 g/mol, you multiplied it by the mass of calcium (20 g) to get 801.56 g Ca.

Similarly, using the molar mass of bromine, which is approximately 79.904 g/mol, you multiplied it by the mass of bromine (80 g) to get 6392.32 g Br.

Dividing the moles of bromine (1.001 mol Br) by the moles of calcium (0.499 mol Ca), you get a ratio of approximately 2. This means that the empirical formula is CaBr2, indicating that there are two bromine atoms for each calcium atom.

To summarize, based on your calculations, the empirical formula for the substance containing 20% by mass calcium and 80% by mass bromine would indeed be CaBr2.