# What is empirical formula of a compound which consists of 89.14% Au and 10.80% of O?

Mar 13, 2016

${\text{Au"_2"O}}_{3}$

#### Explanation:

In order to find a compound's empirical formula you must determine the smallest whole number ratio that exists between its constituent elements.

In this case, your unknown compound is said to contain only gold, $\text{Au}$, and oxygen, $\text{O}$. The compound's percent composition essentially tells you how many grams of each constituent element you get per $\text{100 g}$ of compound.

In this case, a $\text{100-g}$ sample of this unknown compound will contain $\text{89.14 g}$ of gold and $\text{10.86 g}$ of oxygen.

This means that you can use the molar mass of each element to find how many moles you get in that $\text{100-g}$ sample. You will thus have

$\text{For Au: " 89.14 color(red)(cancel(color(black)("g"))) * "1 mole Au"/(196.97color(red)(cancel(color(black)("g")))) = "0.45256 moles Au}$

$\text{For O: " 10.86color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.67876 moles O}$

To find the mole ratio that exists between the two elements, divide both values by the smallest one. This will get you

"For Au:" (0.45256color(red)(cancel(color(black)("moles"))))/(0.45256color(red)(cancel(color(black)("moles")))) = 1

"For O: " (0.67876color(red)(cancel(color(black)("moles"))))/(0.45256color(red)(cancel(color(black)("moles")))) = 1.4998 ~~ 1.5

Now, it's important to remember that you're looking for the smallest whole number ratio here, which means that you're going to have to multiply both values by $2$ to get

("Au"_1"O"_1.5)_2 implies color(green)(|bar(ul(color(white)(a/a)"Au"_2"O"_3color(white)(a/a)|)))

The empirical formula of the compound will be ${\text{Au"_2"O}}_{3}$.