As always we assume an 100*g100⋅g mass of stuff, and work out the atomic composition......
"Moles of barium"=(78.3*g)/(137.3*g*mol^-1)=0.570*molMoles of barium=78.3⋅g137.3⋅g⋅mol−1=0.570⋅mol
"Moles of fluorine"=(21.7*g)/(19.00*g*mol^-1)=1.140*molMoles of fluorine=21.7⋅g19.00⋅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_2BaF2...........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"molecular formula=n×empirical formula............
