...the which results from the stoichiometric reaction....
ZnS(s) + 3/2O_2(g) rarr ZnO(s) + SO_2(g)uarrZnS(s)+32O2(g)→ZnO(s)+SO2(g)↑⏐
⏐⏐
And so we start with (1.0*g)/(97.47*g*mol^-1)=0.0103*mol1.0⋅g97.47⋅g⋅mol−1=0.0103⋅mol with respect to zinc sulfide. Of which there were 0.0103*molxx65.4*g=0.6710*g0.0103⋅mol×65.4⋅g=0.6710⋅g WITH RESPECT TO ZINC.
But we end with 0.8375*g0.8375⋅g of zinc oxide...of which necessarily there were...0.8375*g-0.6710*g=0.1665*g0.8375⋅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-=ZnOZn0.6710⋅g65.4⋅g⋅mol−1O0.1665⋅g16.0⋅g⋅mol−1≡Zn0.0103O0.0104≡ZnO as we would anticipate....
