Question #31a4e

1 Answer
anor277
Jun 21, 2017

#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"#............

Related questions
See all questions in Empirical and Molecular Formulas
Impact of this question
1205 views around the world
You can reuse this answer
Creative Commons License