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

1 Answer
Mar 29, 2018

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.