Question

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

Answer 1

`"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, `"Au"`, and oxygen, `"O"`. The compound's percent composition essentially tells you how many grams of each constituent element you get per `"100 g"` of compound.

In this case, a `"100-g"` sample of this unknown compound will contain `"89.14 g"` of gold and `"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 `"100-g"` sample. You will thus have

`"For Au: " 89.14 color(red)(cancel(color(black)("g"))) * "1 mole Au"/(196.97color(red)(cancel(color(black)("g")))) = "0.45256 moles Au"`

`"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 `"Au"_2"O"_3`.