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

Answer:

$CuF_2$ the empirical formula of the metal fluoride.

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 :

$= \frac{8.249 g}{63.546 g/mol}=0.1298 mol$

Moles of fluorine:

$= \frac{4.931 g}{18.998 g/mol}=0.2596 mol$

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

$Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2$

The empirical formula of the copper fluoride = $CuF_2$

$CuF_2$ the empirical formula of the metal fluoride.