General Chemistry I
posted by Emily on .
A sample of a compound containing only Br and O reacts with an excess of H2 to give 6.601 g of HBr and 3.919 g of H2O. What is the empirical formula of this compound?
e. None of the above
I already know the answer is (e), but I am not entirely sure how this answer is reached. Please correct me if I am wrong in any of my calculations:
____ + H2 --> HBr + H2O
(the space is the unknown compound)
6.601 g * (1 mol/80.9119 g) * (1 mol Br/1 mol HBr) = .0815 mol Br
3.919 g * (1 mol/18.0148 g) * (1 mol O/1 mol H2O) = .2175 mol O
This was as far as I got while it still made sense; as I continued I got some outrageous answer like Br19O50. Any help on figuring this out is greatly appreciated!
I have the following although I don't think I've ever seen this compound and I don't know what it is.
BrxOy + H2 ==> HBr + H2O
mols HBr = 0.08159
mols H2O = 0.2177
There is 1 mol Br in 1 mol HBr so we mut have started with 0.08159 mol Br in BrxOy.
I see only 1 mol O in H2O; therefore, we must have started with 0.2177 mol O in BrxOy.
Therefore I wrote the formula of BrxOy as
mols Br = 0.08159
mols O = 0.2177
Divide by the smaller number to get a ratio with 1.00 being the smallest number.
Br = 0.08160/0.08159 = 1.00
O = 0.2177/0.08159 = 2.668
2.668 is too far from 2.0, 2.5 or 3.0 to round; therefore we try find multiples of 1.00 and 2.668.
1.00 x 2 = 2.00
2.668 x 2 = 5.34. Still not whole numbers.
1.00 x 3 = 3.00
2.668 x 3 = 8.004 and we can round this to 3:8
I would write the formula as Br3O8.
Thank you so much, this makes much more sense than what I was doing; I was dividing each amount by the number of moles of O, not Br, which explains why I was getting crazy numbers.