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 fluorine and 0.970 grams of a gas. The percent composition of the gas is 95.0% fluorine and 5.0% hydrogen.

a)     From this data, determine the empirical formula of the gas

b)     What percent of the fluorine of the original compound is in the solid

c)     What is the empirical formula of the solid product

d)     Write a balanced equation for the reaction between UF₆ and H₂O. Assume that the empirical formulas of the products are the true formulas

Answer 1

a) Finding empirical formula of the gas

	F	H
assuming % as mass	95	5
finding moles
finding simple ratio

The empirical formula of gas is = HF

----------------------------------------------------

b) Molar mass of UF₆ = 238 + 6(19) = 352 g/mol

Finding mass of F present in 4.267 g UF₆=

=1.3818 g of Fluorine

Mass of F present in 4.27 g UF₆ = 1.3818 g

Mass of HF gas is given = 0.970 g

Mass of F present in 0.970 g of HF gas =

=0.9215 g

Mass of F present in HF gas = 0.9215 g

Mass of F present in solid = mass of F in original compound - mass of F in Gas = 1.3818 g - 0.9215 g = 0.4603 g

percent of flourine in solid compred to F in original compound =

=33.3%

Percent of the fluorine of the original compound is in the solid = 33.3%

C) Solid contains U, F and O

Mass of U present in solid sample =

=0.01212 mol of U

Moles of U present in UF₆ = 0.01212 mol

the same moles of U must present in solid

We already found mass of F in solid = 0.4603 g

moles of F =

=0.02423 mol

To find mass of O in solid we can solve as shown below

Given that H₂O is added to UF₆

Law of conservation of mass

UF₆ + H₂O ---> solid + gas

So mass of H₂O = (solid + gas mass)-mass of UF6

Maas of H₂O = (3.730 + 0.970)g - 4.267 g  

= 4.70 - 4.267

= 0.433 g H2O

Mass of H2O = 0.433 g H2O

Mass of O in solid =

=0.02406 mol O

Now we have moles

	U	F	O
	0.01212 mol	0.02423 mol	0.02406 mol
simple ratio

Therefore empirical formula of solid = UO₂F₂

d)

We can write the reaction

UF₆(s) + H₂O(l) ----> UO₂F₂(s) + HF(g)

after balancing the equation we get

UF₆(s) + 2H₂O(l) ----> UO₂F₂(s) + 4HF(g)