Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95e: 93

$XeF_2$

Atomic weights (g/mol): $Xe: 131.29,\ F: 18.998$ Molar mass $F_2$: 38.00 g/mol Mass of Xe: 0.526 g Mass of the product: 0.678 g Mass of fluorine $F_2$: 0.678 - 0.526 = 0.152 g Number of moles: Xe: $0.526/131.29=3.99\times10^{-3}\ mol$ $F_2$: $0.152/38.00=4.00\times10^{-3}\ mol\ F_2 = 8.0\times10^{-3}\ mol\ F$ Normalizing by the smallest amount of moles: Xe: 1.0 F: 2.0 Empirical formula of the product: $XeF_2$