An unknown ion was tested an shown to have a (2-) charge and mass percent composition of 48.2% Cr and 51.8% O.

Question

An unknown ion was tested an shown to have a (2-) charge and mass percent composition of 48.2% Cr and 51.8% O. What is the empirical formula and charge for this ion?

Answer 1

Assume you have 100 g of unknown.

48.2% Cr = 48.2 g Cr

51.8% O = 51.8 g O

moles Cr = 48.2 g x 1 mole/52 g = 0.927 moles Cr

moles O = 51.8 g x 1 mole/16 g = 3.24 moles O

Divide both by 0.927 to get 1 mole Cr and 3.5 moles O

Multiply both by 2 to get 2 moles Cr and 7 moles O

Empirical formula = Cr₂O₇^2-