Question

A compound with a molar mass of 100.0 g/lol has an elemental composition of 24.0% C, 3.0% H, 16.0% O and 57.0%F. What is the molecular formula of the compound?

Answer 1

If I made no mistake in calculation, the given answer must be correct...(tried my best)



elements   :                        carbon     hydrogen    oxygen       Fluorine

 composition [C]                     24             3                16                57

M r                                          12             1                16                19

(divide C by Mr)                       2               3                 1                  3


(Divide by smallest value)       2                3                  1                  3

(smallest value = 1...so all value remained constant)

Empirical formula : C2H3OF3


if molar mas = 100 g per mole, then

first step calculate Mr. of empirical formula:  [= 100]


Them molecular formula = empirical formula