Xenon of mass 5.08 g reacts with fluorine to form 9.49 g of a xenon fluoride. What is the empirical formula of this
compound?
A) XeF6
B) XeF4
C) XeF2
D) XeF
E) Xe2F
I am not familiar with these types of empirical formula problems. If I'm given the percentages of the elements in the compound, I know what to do, but I have no idea what to do here.
Thanks!
6 answers
Oh, and the answer is A. I just don't understand how you get it.
This is very like the percentage problems. A useful way to do the percentage problems is to follow the steps
1. Percentages (make sure they add up to 100)
2. masses (assuming 100 g total so numerically same as percentages)
3. convert mass to moles (by dividing by relative atomic mass)
4. divide by smallest number of moles (which gives ratio of the atoms for empirical formula)
here we have the masses so start at step 3.
1. Percentages (make sure they add up to 100)
2. masses (assuming 100 g total so numerically same as percentages)
3. convert mass to moles (by dividing by relative atomic mass)
4. divide by smallest number of moles (which gives ratio of the atoms for empirical formula)
here we have the masses so start at step 3.
If I don't know the empirical formula, what am I supposed to use for the relative atomic mass here for xenon fluoride to get moles?
Never mind! :) I think I see what I need to do. I simply subtract the initial mass from the mass of the compound to get the mass of the Chlorine and go from there. Correct?
Yes, subtract the two so you have
F Xe
4.41 g 5.08 g
Number of moles?
Divide by smallest.
F Xe
4.41 g 5.08 g
Number of moles?
Divide by smallest.
As the number of moles in the balanced chemical reaction is equal for both xe and the formed compound using weight by gram molecular weight for both calculate molecular weight of the compound.then substract the mass of xenon from the obtained molecular weight. At last we will obtain the molecular weight of all the flourine atoms in the compound.