Question

A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. Determine the empirical formula of the metal fluoride.

Answer 1

`CuF_2` the empirical formula of the metal fluoride.

Step-by-step explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

`13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g`

Moles of copper :

`= (8.249 g)/(63.546 g/mol)=0.1298 mol`

Moles of fluorine:

`= (4.931 g)/(18.998 g/mol)=0.2596 mol`

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

`Copper= (0.1298 mol)/(0.1298 mol)=1\\Fluorine = (0.2596 mol)/(0.1298 mol)=2`

The empirical formula of the copper fluoride =
`CuF_2`

`CuF_2` the empirical formula of the metal fluoride.