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!

Question

DrBob222 · Answer

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.

Van M · Answer

Can you explain, why did you omit coefficients? and what about the molecular formula, i think you only did the empirical.

DrB · Answer

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)?

Maria · Answer

Molecular formula is Hg2Br2. you divide the molecular mass by the molar mass of the imperical formula