Question

A 21.31 gram sample of cobalt is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 41.93 g. Determine the empirical formula of the metal fluoride

Answer 1

Given mass of cobalt , m = 21.31 g

Molar mass of Co is 59 g/mol

Let the balanced chemical reaction is   Co + n F₂ CoF_2n

Molar mass of CoF_2n is = ( 59) + (2n x 19 ) = 59 + 38n          ----(1)

From the balanced equation ,

1 mole of Co produces 1 mole of CoF_2n

                                OR

1x 59 g of Co produces 1 x (59+38n) g of CoF_2n

21.31 g of Co produces M g of CoF_2n

M = ( (59+38n) / (59) ) x 21.31 g of CoF_2n

   But given M = 41.93 g

( (59+38n) / (59) ) x 21.31   = 41.93

Upon solving we get 59+38n = 116

                                            n = 1.5

So the formula of metal fluoride is CoF_2n = CoF_(2x1.5)

                                                       CoF₃

So the emperical formula of metal fluoride is CoF₃