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March 28, 2017

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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?

a. BrO
b. BrO2
c. Br2O
d. BrO3
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!

  • General Chemistry I - ,

    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
    Br(0.08159)O(02177)
    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.

  • General Chemistry I - ,

    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.

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