Question

A given sample of a xenon fluoride compound contains molecules of the type ${\mathbf{X}}_{\mathbf{e}}{\mathbf{F}}_{\mathbf{n}}$,where n issome whole number. Given that$\mathbf{9}\mathbf{.}\mathbf{03}\mathbf{\times }{\mathbf{10}}^{\mathbf{20}}$molecules of${\mathbf{X}}_{\mathbf{e}}{\mathbf{F}}_{\mathbf{n}}$, weigh$\mathbf{0}\mathbf{.}\mathbf{368}\mathbf{}\mathbf{g}$, determine thevalue for n in the formula.

Short answer

The value for n in the formula ${\mathrm{XeF}}_{\mathrm{n}}$is 6 and xenon fluoride compound contains molecules of the type${\mathrm{X}}_{\mathrm{e}}{\mathrm{F}}_6$.

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained in one mole of the substance, and it is equalto the total mass of the component atoms. It may be stated mathematically as follows:

$\mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Weight}\mathrm{in}\mathrm{g}}{\mathrm{Number}\mathrm{of}\mathrm{moles}}$

Determining the Moles of XeFn

We know that, one mole $=6.023\times {10}^{23}$particles

Let’s calculate the number of moles of xenon fluoride present in $9.0\times {10}^{20}$molecules of xenon fluoride.

$\begin{array}{rcl}\mathrm{Moles}\mathrm{of}{\mathrm{X}}_{\mathrm{e}}{\mathrm{F}}_{\mathrm{n}}& =& \left(9.03\times {10}^{20}\mathrm{moleculs}{\mathrm{X}}_{\mathrm{e}}{\mathrm{F}}_{\mathrm{n}}\right)\frac{1\mathrm{mol}{\mathrm{X}}_{\mathrm{e}}{\mathrm{F}}_{\mathrm{n}}}{6.023\times {10}^{23}\mathrm{moleculas}{\mathrm{X}}_{\mathrm{e}}{\mathrm{F}}_{\mathrm{n}}}\\ & =& 1.50\times {10}^{-3}\mathrm{molXeFn}\end{array}$

Determining the Molar mass of XeFn

Now, because a substance's molar mass is equal to the total mass of its constituent atoms, the value of n in may be computed using the molar mass obtained above as follows:

$\begin{array}{rcl}\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{XeFn}& =& \frac{\mathrm{Weight}\mathrm{of}\mathrm{XeFn}}{\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{XeFn}}\\ & =& \frac{0.368\mathrm{g}}{1.49\times {10}^{-3\mathrm{mol}}}\\ & =& 245.33\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\end{array}$

Determining the value of n

$\begin{array}{rcl}MolarmassofXe{F}_n& =& MassofXe+n\left(massofF\right)\\ 245.33g& =& 131.3g+n\left(19.0g\right)\\ n& =& \frac{245.33-131.3}{19}\\ n& =& 6\end{array}$

Therefore the value of n is and the molecular formula of is ${\mathrm{XeF}}_{\mathrm{n}}\mathrm{is}{\mathrm{XeF}}_6$.