# Fluorine is so reactive that it forms compounds with materials inert to other treatments. When 0.455g of platinum is heated in fluorine, 0.721g of a dark red, volatile solid forms. When 0.372g of this red solid reacts with 0.158g of xenon gas, 0.53g of an orange-yellow solid forms. What is the empirical formula of this product, the first to contain a noble gas?

In the first step, fluorine gas is added to platinum metal.  The mass of the product has increased, thus fluorine atoms have added to the platinum metal.  Fluorine gas is F2, so the general reaction can be written below:

Pt + F2 --> PtF

What we need to find out is how many fluorines added to the metal.  The atomic weight of Pt is 195.08 and we started with 0.455 g of it, so let's convert that to moles.

0.455 g * (mole/195.08 g) = 0.0023 moles Pt

Now we know that the added mass to form the product comes entirely from fluorine atoms.  The mass difference is 0.721-0.455=0.266 g.  The atomic weight of fluorine is 19 so we can convert that to moles too.

0.266 g * (mole/19 g) = 0.014 moles F

If we divide the two molar values:

0.014/0.0023 = 6.08

This value is approximately equal to 6, so we have 6 times as many moles of F as we do Pt.  This means the the empirical formula for the product is PtF6, meaning that 3 equivalents of F2 was added.  The balanced equation is shown below:

Pt + 3F2 --> PtF6

Now let's look at the second reaction.  The PtF6 is added to xenon gas (Xe) and the product gains mass.  So some amount of Xe is added to the PtF6.  Lets divide the mass of each by the molecular weights as before to get moles that we can directly compare with each other.

0.372 g PtF6 (mole/309.08 g) = 0.0012 moles PtF6

0.158 g Xe (mole/131.29 g) = 0.0012 moles Xe

So they add together in a 1 to 1 molar fashion.  This is shown in the equation below:

PtF6 + Xe --> XePtF6

The empirical formula of the product is XePtF6.  It is called xenon hexafluoroplatinate and it was first made in 1962 by Neil Bartlett.