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

Mar 29, 2018

${C}_{3} {F}_{8}$

#### Explanation:

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

$\text{Moles of fluorine} = \frac{81 \cdot g}{19.0 \cdot g \cdot m o {l}^{-} 1} = 4.26 \cdot m o l$.

$\text{Moles of carbon} = \frac{19 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 1.59 \cdot m o l$.

We then divide thru by the LOWEST molar quantity to get a trial empirical formula of ....${C}_{\frac{1.59 \cdot m o l}{1.59 \cdot m o l}} {F}_{\frac{4.26 \cdot m o l}{1.59 \cdot m o l}} = C {F}_{2.68}$...

But we know by specification, that the empirical formula is the simplest WHOLE number ratio....and so we mulitply by three to get...${C}_{3} {F}_{8}$...$\text{perfluoropropane}$...this is the old $\text{Freon 218}$...and you must have twisted some analyst's arm to make the determination.