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ryzh [129]
3 years ago
7

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. D

etermine the empirical formula of the metal fluoride.
Chemistry
1 answer:
levacccp [35]3 years ago
6 0

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.

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