Question

Question #05d76

Answer 1

The `"empirical formula"` is `As_2O_5`.

Explanation:

As with all these problems, we ASSUME that there are `100*g` of unknown compound.

Given the elemental percentages, thus there are:

`"Moles of oxygen"` `=` `(34.8*g)/(15.999*g*mol^-1)=2.18*mol`.

`"Moles of arsenick"` `=` `(65.2*g)/(74.9*g*mol^-1)=0.871*mol`.

We divide thru by the smallest molar quantity to get, `O:As`, `2.5:1.0`.

But by definition, the `"empirical formula"` is the simplest WHOLE number ratio that defines constituent atoms in a species, so we take this preliminary determination, and DOUBLE it to give WHOLE numbers:

`As_2O_5`. Capisce?