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!
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.
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.
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.
To determine the empirical formula of the compound, you need to know the masses of the elements present in the compound. In this case, you are given the mass of Xenon (Xe) as 5.08 g and the mass of the xenon fluoride compound as 9.49 g.
To find the empirical formula, you'll need to determine the mole ratio of the elements present.
1. Convert the grams of Xenon to moles:
- Use the molar mass of Xenon (Xe), which is 131.29 g/mol:
moles of Xe = (mass of Xe) / (molar mass of Xe)
= 5.08 g / 131.29 g/mol
≈ 0.0386 mol Xe
2. Convert the grams of the xenon fluoride compound to moles:
- Use the molar mass of the xenon fluoride compound, which is 131.29 g/mol (molar mass of Xe) + 18.9984 g/mol (molar mass of F):
moles of XeF(n) = (mass of XeF(n)) / (molar mass of XeF(n))
= 9.49 g / (131.29 g/mol + 18.9984 g/mol)
≈ 0.0630 mol XeF(n)
3. Find the mole ratio of Xenon to Fluorine:
Divide the number of moles of each element by the smaller number of moles obtained.
The closest ratio is approximately XeF2:
Xe : F ≈ 0.0386 mol : (0.0630 mol / 2) ≈ 1 : 3/2
Multiplying the ratio by 2 gives: Xe : F ≈ 2 : 3
Therefore, the empirical formula of the compound is XeF2, which corresponds to answer choice C.