Question

A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 16 torr. Hydrogen was added to the bulb until the pressure was 64 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 32 torr. What is the empirical formula of the xenon fluoride in the original sample?

Answer 1

Answer:

XeF₂

Explanation:

Let's call the compound as XeaFb, where a and b are the number of atoms of each element. In the beginning, the partial pressure of XeaFb was 16 torr. By Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of the components, so:

pH₂ = 64 - 16 = 48 torr

The reaction that happened was:

XeaFb + (b/2)H₂ → aXe + bHF

The stoichiometry of the reaction is 1:b/2:a:b. If all the XeaFb reacts:

1 torr of XeaFb -------- b/2 torr of H₂

16 torr -------- x

By a simple direct three rule:

x = 16b/2 = 8b torr of H₂ reacts

1 torr of XeaFb -------- a torr of Xe

16 torr -------- x

x = 16a torr of Xe is formed

Thus, 8b + 16a = 32

b + 2a = 4

The empirical formula is the minimum integer number of the elements in the compound. So, let's suppose a = 1 because fluorine can do only one bond, thus it's unprovable that will be more than one Xe bonded with one F:

b + 2*1 = 4

b = 2

Thus, the empirical formula is XeF₂.