posted by sam on .
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!
..........Hg + Br ==> HgBr
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.
Can you explain, why did you omit coefficients? and what about the molecular formula, i think you only did the empirical.
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)?
Molecular formula is Hg2Br2. you divide the molecular mass by the molar mass of the imperical formula