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...
50g 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
2 answers
Yes, the answer is CaBr2 but I don't buy the way you obtained that formula. Did you just pull 50 g Ca and 80 g Br out of thin air? The proper way to work it is the way I showed you in your first post.
First we find the moles of each assuming that mixture is having total mass of 100 grams. So mass of Ca = 20 g and Br =80
Moles of Ca = 20 g/40.08 (g/mole)= 0.499 Mole
Moles of Br = 80 g/ 1.001 moles
So now we can find the molar ratio = 0.499/1.001 =1/2
That means Ca1Br2 so the empirical formula would be CaBr2.
This is absolutely right.
Moles of Ca = 20 g/40.08 (g/mole)= 0.499 Mole
Moles of Br = 80 g/ 1.001 moles
So now we can find the molar ratio = 0.499/1.001 =1/2
That means Ca1Br2 so the empirical formula would be CaBr2.
This is absolutely right.
1 answer