Question

Question #31a4e

Answer 1

`BaF_2`......

Explanation:

As always we assume an `100*g` mass of stuff, and work out the atomic composition......

`"Moles of barium"=(78.3*g)/(137.3*g*mol^-1)=0.570*mol`

`"Moles of fluorine"=(21.7*g)/(19.00*g*mol^-1)=1.140*mol`

You note that I divided thru by the ATOMIC mass.......

And we simply divide thru by the element present in LEAST molar quantity (barium) to get ........

`BaF_2`...........this is an ionic compound that would be formulated as same. If we do this for an organic formula, we need a molecular mass BEFORE we deduce the molecular formula......

`"molecular formula"=nxx"empirical formula"`............