Question

1g of ZnS combined with `O_2` to produce 0,8375g of product. What is the empirical formula of the product ?

Answer 1

We ASSUME that the final product is `ZnO(s)`

Explanation:

...the which results from the stoichiometric reaction....

`ZnS(s) + 3/2O_2(g) rarr ZnO(s) + SO_2(g)uarr`

And so we start with `(1.0*g)/(97.47*g*mol^-1)=0.0103*mol` with respect to zinc sulfide. Of which there were `0.0103*molxx65.4*g=0.6710*g` WITH RESPECT TO ZINC.

But we end with `0.8375*g` of zinc oxide...of which necessarily there were...`0.8375*g-0.6710*g=0.1665*g` with respect to oxygen....

And so we take the molar quantities of element to arrive at the empirical formula....

`Zn_((0.6710*g)/(65.4*g*mol^-1))O_((0.1665*g)/(16.0*g*mol^-1))-=Zn_(0.0103)O_0.0104-=ZnO` as we would anticipate....