# A compound, XF_5, is 42.81% fluorine by mass. What is the element X? What is the molecular structure of XF_5?

Mar 12, 2016

Element $\text{X}$ is iodine.

#### Explanation:

Notice that the problem provides you with the percent composition of fluorine in this unknown compound ${\text{XF}}_{5}$.

This means that you can use a $\text{100-g}$ sample of this compound to determine how many grams of fluorine it contains, then use fluorine's molar mass to determine how many moles of fluorine it contains.

Once you know that, use the fact that you get $5$ moles of fluorine for every $1$ mole of $\text{X}$ in the compound's molecular formula to find the number of moles of $\text{X}$ in this sample.

So, a $\text{100-g}$ sample of ${\text{XF}}_{5}$ will contain $\text{42.81 g}$ of fluorine and

${m}_{\text{X" = "100 g" - "42.81 g" = "57.19 g X}}$

Use fluorine's molar mass to get the number of moles of fluorine in the sample

42.81 color(red)(cancel(color(black)("g"))) * "1 mole F"/(18.9984color(red)(cancel(color(black)("g")))) = "2.2533 moles F"

Use the $1 : 5$ mole ratio that exists between the two elements in the molecular formula to find the number of moles of $\text{X}$

2.2533 color(red)(cancel(color(black)("moles F"))) * "1 mole X"/(5color(red)(cancel(color(black)("moles F")))) = "0.45066 moles X"

Well, if $\text{57.19 g}$ of $\text{X}$ contain $0.45066$ moles of $\text{X}$, it follows that one mole of $\text{X}$ will contain

1color(red)(cancel(color(black)("mole X"))) * "57.19 g"/(0.45066color(red)(cancel(color(black)("moles X")))) = "126.90 g"

This means that element "X" has a molar mass of ${\text{126.90 g mol}}^{- 1}$.

A quick look in the periodic table will reveal that element $\text{X}$ is iodine, $\text{I}$.

Your unknown compound is iodine pentafluoride, ${\text{IF}}_{5}$. Iodine pentafluoride has an octahedral molecular geometry. The central iodine atom is surrounded by six regions of electron density

• five single bonds to each of the five fluorine atoms
• one lone pair of electrons