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

1 Answer
Mar 29, 2018

Answer:

#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.