Question

A compound is known to contain only gold and oxygen. A sample of this compound is placed in a clean crucible that has a mass of 10.313 g. The crucible and sample have a mass of 10.517 g. The crucible is heated until the compound decomposes to the elements. The oxygen is lost to the air and the gold remains in the crucible. The mass of the crucible and gold is 10.495 g. What is the empirical formula of this compound?

Answer 1

`Au_2O_3`

Step-by-step explanation:

Hello,

Based on the initial data, the mass of sample is:

`m_(sample)=10.517g-10.313g=0.204g`

The final mass accounts for the present gold grams into the sample, thus:

`m_(Au)=10.495g-10.313g=0.182g`

So the oxygen grams are:

`m_(O)=0.204g-0.182g=0.022g`

Based on this fact, the moles of both gold and oxygen are:

`n_(Au)=0.182g*(1mol)/(197g)=0.000924molAu \\n_(O)=0.022g*(1mol)/(16g)=0.001375molO`

Now, the ratio O:Au allows us to establish the mole relationship between gold and oxygen:

`0.001375/0.000924=1.5`

Thus:

`AuO_(1.5)=Au_2O_3`

So
`Au_2O_3` is the empirical formula for the given compound which is auric oxide.

Best regards.