Question

A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. The passage of an electric spark through the mixture produced X_e and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)

Short answer

Using the ideal gas equation, and treating gaseous mixtures as one gas, the compound is found to be XeF₂.

Step-by-step solution

Consolidating information provided

The notations and values to be used in this exercise are:

n(Xe), n(F), n(H)- Total number of moles of xenon, fluorine, and hydrogen, respectively, in the process;

n(H’) - Number of moles of hydrogen that do not react;

V, T- Volume and temperature (in K) of the bulb;

p₁= 18 torr- Pressure of the compound of xenon and fluorine in the bulb at the beginning of the process;

p₂= 72 torr- Pressure of the starting compound and hydrogen combined in the bulb; and

p₃= 36 torr- Pressure of the xenon and remaining hydrogen in the bulb.

Formulating equations

It is given that the starting compound is XeFn.

n is the number of atoms of fluorine per atom of xenon in the compound. That is the ratio in which all involved xenon and fluorine is introduced; therefore

$$\mathrm{n}=\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{X}\mathrm{e})}......(1)$$

All the mixtures are gaseous, and so they can be modelled using the ideal gas equation

pV =n’RT

where p, V, Tare respectively pressure, volume and temperature of n’ moles of molecules of a gas, and R is a universal gas constant.

State that this law holds in the given situations

At the beginning - pressure is p_1,number of moles of molecules is n(XeFn).

After the addition of hydrogen - pressure is p_2,number of moles of molecules is n(XeFn)+n(H).

AfterHFis removed - pressure is p₂, number of moles of molecules is n(XeFn)+n(H’).

In all three situations, the volume is V and temperature T.

So, it can be written that

$$\begin{array}{r}l\mathrm{p}1\mathrm{V}=\mathrm{n}(\mathrm{X}\mathrm{e}\mathrm{F}\mathrm{n})\mathrm{R}\mathrm{T}......(2)\\ {\mathrm{p}}_2\mathrm{V}=\left[\mathrm{n}(\mathrm{X}\mathrm{e}\mathrm{F}\mathrm{n})+\mathrm{n}(\mathrm{H})\right]\mathrm{R}\mathrm{T}......(3)\\ {\mathrm{p}}_3\mathrm{V}=\left[\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}({\mathrm{H}}^{\prime })\right]\mathrm{R}\mathrm{T}.......(4)\end{array}$$

Finding the formula for xenon fluoride

Since there is one atom of xenon per one molecule of

(Xe)$$\mathrm{X}\mathrm{e}\mathrm{F}\mathrm{n}\to \mathrm{n}(\mathrm{X}\mathrm{e}{\mathrm{F}}_{\mathrm{n}})=\mathrm{n}(\mathrm{X}\mathrm{e})$$.

Dividing equations (2)/(3) and (4)/(3) and using n(XeFn) = n(Xe), we have

$$\begin{array}{r}\frac{18}{72}=\frac{\mathrm{n}(\mathrm{X}\mathrm{e})}{\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}(\mathrm{H})}\\ \frac{36}{72}=\frac{\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}\left({\mathrm{H}}^{\prime }\right)}{\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}(\mathrm{H})}\end{array}$$

On further simplification –

$$\begin{array}{r}4\mathrm{n}(\mathrm{X}\mathrm{e})=\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}(\mathrm{H})\\ 2\left(\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}\left({\mathrm{H}}^{\prime }\right)\right)=\mathrm{n}(\mathrm{X}\mathrm{e})+\mathrm{n}(\mathrm{H})\\ 3\mathrm{n}(\mathrm{X}\mathrm{e})=\mathrm{n}(\mathrm{H})......(5)\\ \mathrm{n}(\mathrm{X}\mathrm{e})+2\mathrm{n}\left({\mathrm{H}}^{\prime }\right))=\mathrm{n}(\mathrm{H})\end{array}$$

Subtracting the equations –

$$2\mathrm{n}(\mathrm{X}\mathrm{e})-2\mathrm{n}\left({\mathrm{H}}^{\prime }\right)=0\iff \mathrm{n}(\mathrm{X}\mathrm{e})=\mathrm{n}\left({\mathrm{H}}^{\prime }\right)......(6)$$

Similar to n(XeFn) = n(Xe), the number of fluorine atoms is the same as the number of molecules of HF that have been created and then removed. This is the number of hydrogen atoms used, which is the difference between the hydrogen added to the process – n(H) and the hydrogen remaining after the process – n(H’).

$$\begin{array}{r}\to \mathrm{n}(\mathrm{F})=\mathrm{n}(\mathrm{H}\mathrm{F})\\ =\mathrm{n}(\mathrm{H})-\mathrm{n}\left({\mathrm{H}}^{\prime }\right)\\ \mathrm{n}(\mathrm{F})=\mathrm{n}(\mathrm{H})-\mathrm{n}\left({\mathrm{H}}^{\prime }\right)\end{array}$$

The previous equation, (5) and (6), yield

$$\begin{array}{r}\mathrm{n}(\mathrm{F})=\mathrm{n}(\mathrm{H})-\mathrm{n}\left({\mathrm{H}}^{\prime }\right)\\ =3\mathrm{n}(\mathrm{X}\mathrm{e})-\mathrm{n}(\mathrm{X}\mathrm{e})\\ =2\mathrm{n}(\mathrm{X}\mathrm{e})\end{array}$$

Using identity (1) and the previous equation, we have

$$\begin{array}{r}\mathrm{n}=\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{X}\mathrm{e})}\\ =\frac{2\mathrm{n}(\mathrm{X}\mathrm{e})}{\mathrm{n}(\mathrm{X}\mathrm{e})}\\ =2\end{array}$$

Therefore, the formula is XeF₂.