Rubidium forms an ionic compound with silver and iodine
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Assume that you have 100 g of compound. then based on the percentages given, there are 7.42 g Rb, 37.48 g Ag, and 55.10 g I. Convert those to moles.
7.42 g Rb x (1 mole Rb / 85.47 g Rb) = 0.0868 moles Rb
37.48 g Ag x (1 mole Ag / 107.87 g Ag) = 0.347 moles Ag
55.10 g I x (1 mole I / 126.90 g I) = 0.434 moles I. . .divide each of those by the smallest (0.0868).
Rb: 0.0868 / 0.0868 = 1
Ag: 0.347 / 0.0868 = 4.0
I: 0.434 / 0.0868 = 5.0
The empirical formula is RbAg4I5
7.42 g Rb x (1 mole Rb / 85.47 g Rb) = 0.0868 moles Rb
37.48 g Ag x (1 mole Ag / 107.87 g Ag) = 0.347 moles Ag
55.10 g I x (1 mole I / 126.90 g I) = 0.434 moles I. . .divide each of those by the smallest (0.0868).
Rb: 0.0868 / 0.0868 = 1
Ag: 0.347 / 0.0868 = 4.0
I: 0.434 / 0.0868 = 5.0
The empirical formula is RbAg4I5
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