chemistry
posted by Deanne .
Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid containing only uranium, oxygen and fluorine and 0.970 gram of gas. The gas is 95.0% flourine, and the remainder is hydrogen. From data determine the empirical formula of the gas.
What fraction of the fluorine of the original compound is in the solid and what fraction in the gas after the reaction?
What is the formula of the solid product?
Write a balance equation for the reaction between UF6 and H2O. Assume that the empirical formula of the gas is the true formula

Part 1.
Take a 100 g sample which will give you 95 g F and 1 g H.
Convert to moles.
95/atomic mass F = 95/19 = 5
5/atomic mass H = 5/1 = 5
So the ratio is 1:1 and the formula is HF.
part 2. How much F do you have total (in UF6)? Convert 4.267 g UF6 to moles.
4.267/molar mass UF6. Convert that to moles F (multiply by 6), then convert moles F to grams F (multiply by atomic mass F). I get something like 1.4 but you should do it more accurately. Some of the F will be in the solid compound and some in the HF. How many grams in the HF. HF has a mass of 0.97, it has a 95% F; therefore, g F in HF is 0.97 x 0.95 = 0.9 (again you do it more accurately. Subtract from the 1.4 to determine amount F in the solid (about 0.5 or so). Then % in gas (HF) is (0.9/1.4)*100 = ??
% in solid is 100%  % in HF = ??
part 3.
I don't how you have been taught to do these but I do them, probably a little unconventional, this way.
moles UF6 = 4.267/352.02 = 0.010.012 or so. So 0.012 moles UO_{x}F_{y} will be formed. moles = grams/molar mass. You know moles and grams, solve for molar mass. I get something like 308. Subract 238 for U to leave 70. Subtract 1F and 1 O to leave 701916 = 35. That looks like another O and another F (16+19) would do it so the formula should be UO2F2.
Part 4.
Use the percent you found in part 2 to help answer this, although you really don't need that. If you write
UF4 + H2O ==> UO2F2 + HF it's easy enough to balance as is. The percents, if you wish to use them, tells you that the HF is twice the F that is in UO2F2.