Question

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

Answer 1

Basis: 1 g of the compound

Calculate for the masses of gold (Au) and oxygen (O) in the compound by multiplying the decimal equivalent of the percentages to the mass of the basis.

Au:   m = (0.8914)(1) = 0.8914 g
O:     m = (0.1080)(1) = 0.1080 g

Then, calculate for the number of moles of the elements by dividing their masses with their molar masses which are 196.97 g/mol for gold and 16 g/mol for oxygen.

 
Au:    n = 0.8914 g/ (196.97 g/mol) = 4.5255 x 10^-3 mols
O:      n = (0.1080 g)/ (16 g/mol) = 6.75 x 10^-3 mols

Then, divide the number of mols with the lower number between the two, this is 4.5255 x 10^-3 mols.

 Au = 4.5255 x 10^-3 / 4.5255 x 10^-3 moles = 1
 O = 6.75 x 10^-3 mols/ 4.5255 x 10^-3 moles = 1.5

The equation therefore is AuO3/2. Eliminate the fraction by multiplying the numbers by 2. This gives us the final answer of,
   
    Au2O3

Answer 2

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.