What is the empirical formula of a compound which consists of 89.14% au and 10.80% o?

In order to find the compound's empirical formula, you must know the smallest whole number ration that is found between its constituent elements.
This means that you can use the molar mass of each element to look for how many moles you can get in that sample.
For Au: 89.14g x 1 mole Au/196.97 g = 0.45256 moles AuFor O: 10.86 g x 1 mole O/ 15.9994 g = 0.67876 moles O
To get the ratio, divide both values by the smallest one.
For Au: 0.45256 moles/ 0.45256 moles = 1For O = 0.67876 moles/ 0.45256 moles = 1.4998 or 1.5
So the Empirical formula is Au1O1.5 or simply Au2O3.

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes