Question

What is the empirical formula of an `"aluminum sulfide"`, for which a mass of `2.03*g` contains a `0.73*g` mass of `"aluminum"`?

Answer 1

We interrogate the empirical formula, and get `Al_2S_3`....

Explanation:

And so we work out moles of aluminum and moles of sulfur.

`"Moles of Al"=(0.73*g)/(26.98*g*mol^-1)=0.0271*mol`

`"Moles of S"=(2.03*g-0.73*g)/(32.06*g*mol^-1)=0.0406*mol`

We divide thru by the smallest molar quantity...........

And get `AlS_(1.498)`. But the `"empirical formula"` is specified to be the `"simplest WHOLE number ratio"`.........and thus we get an empirical formula of `Al_2S_3`.........

How did I know that there were `1.30*g` of sulfur?