# A sample of a compound of xenon and fluorine contains molecules of a single type; XeF_n, where n is a whole number. If 9.03 * 10^20 of these XeF_n, molecules have a mass of 0.311 g, what is the value of n?

Nov 26, 2015

$n = 4$

#### Explanation:

The idea here is that you need to use the number of molecules to determine how many moles you have, then use that and the mass of the sample to figure out the compound's molar mass.

So, you know that one mole of any substance contains exactly $6.022 \cdot {10}^{23}$ molecules of that substance - this is known as Avogadro's number.

You can thus use Avogadro's number to determine how many moles you have in that sample

9.03 * 10^(20)color(red)(cancel(color(black)("molecules XeF"_n))) * overbrace(("1 mole XeF"_n)/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules XeF"_n)))))^(color(blue)("Avogadro's number")) = 1.4995 * 10^(-3)"moles XeF"_n

The molar mass of a substance tells what the exact mass of one mole of that substance is. In your case, the molar mass of the compound will be

$\textcolor{b l u e}{{M}_{\text{M}} = \frac{m}{n}}$

${M}_{\text{M" = "0.311 g"/(1.4995 * 10^(-3)"moles") = "207.4 g/mol}}$

Now, the molar mass of a compound can also be found by adding the molar masses of each atom that's part of that compound's molecule or formula unit.

In your case, you know that one molecule of the compound contains

• $1$ atom of xenon, $\text{Xe}$
• $\textcolor{b l u e}{n}$ atoms of fluorine,$\text{F}$

The molar masses of these two elements are

• $\text{Xe: " "131.293 g/mol}$
• $\text{F: " "18.9984 g/mol}$

This means that you can write

$1 \times {M}_{\text{M Xe" + color(blue)(n) xx M_"M F" = "207.4 g/mol}}$

$1 \times 131.293 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g/mol"))) + color(blue)(n) xx 18.9984 color(red)(cancel(color(black)("g/mol"))) = 207.4 color(red)(cancel(color(black)("g/mol}}}}$

The value of $\textcolor{b l u e}{n}$ will thus be

$\textcolor{b l u e}{n} = \frac{207.4 - 131.293}{18.9984} = 4.006 \approx \textcolor{g r e e n}{4}$

The compound is ${\text{XeF}}_{4} \to$ xenon tetrafluoride