Asked by sam

A sample of a compound of mercury and bromide with a mass of .389 was found to contain .111g bromine. Its molar mass was found to be 561 g mol-1. What are its empirical and molecular formula. Please detail each step in the process. Thanks in advance!
Answered by DrBob222
Omitting coefficients:
..........Hg + Br ==> HgBr
..........x..0.111....0.389
So Hg must be 0.389-0.111 = 0.278g

moles Hg = 0.278/200.6 =0.00139
moles Br = 0.111/79.9 = 0.00139

So ratio is 1:1 and formula is HgBr.
Check my work.
Answered by Van M
Can you explain, why did you omit coefficients? and what about the molecular formula, i think you only did the empirical.
Answered by DrB
DrBob222 is SOO close... Yes, the ratio is 1:1, but how does the molecular mass of HgBr compare to the molecular mass given in the problem (this will give you the answer)?
Answered by Maria
Molecular formula is Hg2Br2. you divide the molecular mass by the molar mass of the imperical formula
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