Respuesta :
Answer:
[tex]CuF_2[/tex] 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
[tex]13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g[/tex]
Moles of copper :
[tex]= \frac{8.249 g}{63.546 g/mol}=0.1298 mol[/tex]
Moles of fluorine:
[tex]= \frac{4.931 g}{18.998 g/mol}=0.2596 mol[/tex]
For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.
[tex]Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2[/tex]
The empirical formula of the copper fluoride = [tex]CuF_2[/tex]
[tex]CuF_2[/tex] the empirical formula of the metal fluoride.
