Question

Determine the empirical and molecular formula of a compound composed of its 18.24 g c, 0.51 g h and 16.91 g flourine has a molar mass 562.0 g/ mol

Answer 1

empirical = C3HF2, molecular = C24H8F16

Explanation:

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                     carbon                   hydrogen                   fluorine

mass             18.24                       0.51                           16.91

no. of            18.24/12                  0.51/1                         16.91/19

moles            = 1.52                     = 0.51                         = 0.89

ratio             1.52/0.51                  0.51/0.51                    0.89/0.51

                    ≈ 3                           = 1                               ≈ 2

empirical formula = C3HF2

(find number of moles by dividing mass by element's mass number. find ratio by dividing number of moles by overall smallest number of moles, in this case hydrogen)

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let the molecular formula be (C3HF2)n

molar mass = 562.0g/mol

*[3(12) + 1 + 2(19)]n = 562.0

n = 562/75

  = 7.49

  ≈ 8

molecular formula = (C3HF2)8

                              = C24H8F16

*substitute C, H, and F with their mass numbers