Question

A `2.07*g` mass of a sulphide of arsenic contained `1.0*g` of arsenic. What is the formal charge on arsenic, and what is its empirical formula of the compound?

Answer 1

Well, we gots `As^(3+)`

Explanation:

We calculate the empirical formula of the stuff....

`"Moles of sulfur"=(1.07*g)/(32.06*g*mol^-1)=0.0334*mol`

`"Moles of arsenic"=(1.00*g)/(74.9*g*mol^-1)=0.0134*mol`

...we divide thru by the LOWEST molar quantity to get an empirical formula of `As_((0.0134*mol)/(0.0134*mol))S_((0.0334*mol)/(0.0134*mol))-=AsS_(1.5)`...but we require a WHOLE NUMBER ratio, so we double this result to get....

`As_2S_3`...

Now for valence we mean the oxidation state, and since here, sulfur is more electronegative than arsenic, we has `As(+III)`...