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. :-)
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.
1 answer
1 answer