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

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

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Answer 1 · 2018-03-29T04:06:29

$C_{3} F_{8}$ $C_3F_8$

Explanation:

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

$Moles of fluorine = \frac{81 \cdot g}{19.0 \cdot g \cdot m o l^{- 1}} = 4.26 \cdot m o l$ $"Moles of fluorine"=(81*g)/(19.0*g*mol^-1)=4.26*mol$ .

$Moles of carbon = \frac{19 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}} = 1.59 \cdot m o l$ $"Moles of carbon"=(19*g)/(12.011*g*mol^-1)=1.59*mol$ .

We then divide thru by the LOWEST molar quantity to get a trial empirical formula of .... $C_{\frac{1.59 \cdot m o l}{1.59 \cdot m o l}} F_{\frac{4.26 \cdot m o l}{1.59 \cdot m o l}} = C F_{2.68}$ $C_((1.59*mol)/(1.59*mol))F_((4.26*mol)/(1.59*mol))=CF_2.68$ ...

But we know by specification, that the empirical formula is the simplest WHOLE number ratio....and so we mulitply by three to get... $C_{3} F_{8}$ $C_3F_8$ ... $perfluoropropane$ $"perfluoropropane"$ ...this is the old $Freon 218$ $"Freon 218"$ ...and you must have twisted some analyst's arm to make the determination.

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

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