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3 years ago

7

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

1 answer:

OLga [1]3 years ago

8 0

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)

__________________________________

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

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