Ap Chem. Help-Stoichiometry?

Question

Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid containing only uranium, oxygen and flourine and 0.970 grams of a gas. The gas is 95.0% flourine, and the remainder is hydrogen.

A. From this data determine the empirical formula of the gas
B. What fraction of the fluorine of the original compound is in the solid, and what fraction in the gas after the reaction.
C. Whta is the formula of the solid product.
D. Write a balance equation for the reaction between UF6 and H20. Assume the empirical formula of the gas is the true formula.

Answer 1

Part A: 95% F 5% H just divide both values by the smallest; it becomes 19:1 which just happens to correspond to the weight ratio of F to H so gas is HF.

Part B: total flourine can be calculated as weight fraction of flourine in compound x weight of compound:

(6 x 19)/[(6 x 19) + 238] x 4.267g = 1.38g

Weight of flourine in the evolved gas is weight of evolved gas x weight percent flourine in the gas:

0.970g x 0.95 = 0.922g

Amount remaining in compound is total flourine - evolved flourine:

1.38g - 0.922g = 0.458g

Fraction of flourine remaining in compound is retained weight/total weight:

0.458/1.38 = 0.332 0r 33.2% in compound

Fraction of evolved flourine is evolved weight/total wt:

0.922/1.38 = 0.668 or 66.8% in the HF

Part C; since 2/3 of the 6 flourines leave the compound after reaction with water, there are 2 flourines remaining behind. From the valence of the uranuim we can see you need 2 oxygens to complete the compound, but if you want to calculate it; here is how it goes.

Determine moles U; total weight of initial compound - weight of contained flourine divided by molecular weight of U:

(4.267 - 1.38)g/238g/mol = 0.0121mol

Determine moles of oxygen which combined; weight of "new" compound - weight of contained uranuim and flourine = weight of oxygen (divide by 16 to get moles oxygen).

[3.730g - (2.887g + 0.458g)]/16g/mol = 0.0241mol

The mole ratio of U to O in "new" compound is 1:2.

So formula is UF2O2

Part D; the balanced equation is:

UF6 + 2 H2O >>> UF2O2 + 4 HF

Answer 2

A. Find the weight of fluorine and hydrogen.
Divide the weight of fluorine by 19 to get g-atoms and the weight of hydrogen by 1 to get g-atoms. Make a ratio:
moles of fluorine/moles of hydrogen in gas = #F in empirical formula.

B. If 4.267 g is pure UF6 (BTW: Don't try this, U is radioactive and the gas is poisonous), you can compute fluorine content in mass: divide UF6 wt by 114/352. You know from part A the mass of fluorine in the gas product. So the rest is in the solid product. So you can calculate relative fractions.

C. Since the hydrogen in the gas came from water, the oxygen in the water must be tied up in the solid material. Since you know the mass of hydrogen in the gas, you can back-calculate the mass of oxygen in the solid:
8 x mass of hydrogen in gas = mass of oxygen in solid.

D. Now you can compute the g-atoms of U, F and O in the product solid from the masses you computed. If we call these GAU, GAF and GAO,
you can find an empirical formula by the ratios"
GAU/1 = GAF/#F=GAO/#O where "#" the number in the formula for fluorine and oxygen.