Question

A compound containing xenon and fluorine was prepared by shining sunlight on a mixture of Xe (0.526 g) and excess F2 gas. If you isolate 0.678 g of the new compound, what is its empirical formula

Answer 1

The empirical formula of the new compound formed from xenon and fluorine is XeF2, determined by calculating the mole ratio of xenon to fluorine which came out to be approximately 1:2.

Step-by-step explanation:

The student is asking for the empirical formula of a compound formed by xenon (Xe) and fluorine (F) after a reaction where sunlight was shone on the mixture. Given the masses of xenon and the compound formed, we can determine the empirical formula by finding the ratio of moles of xenon to moles of fluorine in the compound.

To find the moles of xenon, we use its molar mass (131.29 g/mol):

• Moles of Xe = 0.526 g / 131.29 g/mol = 0.00401 moles.

Since the mass of the compound is 0.678 g and xenon's mass is 0.526 g, the mass of fluorine in the compound is:

• Mass of F = 0.678 g - 0.526 g = 0.152 g.

Using the molar mass of fluorine (19.00 g/mol):

• Moles of F = 0.152 g / 19.00 g/mol = 0.00800 moles.

The ratio of moles of Xe to moles of F is approximately 1:2, which indicates that the empirical formula of the compound is XeF2.

Answer 2

`XeF_2`

Step-by-step explanation:

Hello!

In this case, according the law of conservation of mass, we can see that, since the total mass of the Xe-F compound is 0.678 g and that of xenon is 0.526 g, we have 0.152 g as the mass of reactants equal the mass of the products. It means that we can compute the moles of each atom in the compound as shown below:

`n_(Xe)=0.526g/131.3g/mol=0.004mol\\\\n_F=0.152g/19.0g/mol=0.008mol`

Now we divide the moles of both reactants by the moles of xenon as those are the fewest ones in order to find their subscripts in the empirical formula:

`Xe=0.004/0.004=1\\\\F=0.008/0.004=2`

Thus, the empirical formula is:

`XeF_2`

Best regards!