Question

What is the empirical formula for perfluoropropane if the compound contains 81% fluorine and 19% carbon by mass?

Answer 1

`C_3F_8`

Explanation:

As always with the problems it is useful to assume an `100*g` mass of compound, and then interrogate the atomic composition...

`"Moles of fluorine"=(81*g)/(19.0*g*mol^-1)=4.26*mol`.

`"Moles of carbon"=(19*g)/(12.011*g*mol^-1)=1.59*mol`.

We then divide thru by the LOWEST molar quantity to get a trial empirical formula of ....`C_((1.59*mol)/(1.59*mol))F_((4.26*mol)/(1.59*mol))=CF_2.68`...

But we know by specification, that the empirical formula is the simplest WHOLE number ratio....and so we mulitply by three to get...`C_3F_8`...`"perfluoropropane"`...this is the old `"Freon 218"`...and you must have twisted some analyst's arm to make the determination.