# Question #31a4e

Jun 21, 2017

#### Answer:

$B a {F}_{2}$......

#### Explanation:

As always we assume an $100 \cdot g$ mass of stuff, and work out the atomic composition......

$\text{Moles of barium} = \frac{78.3 \cdot g}{137.3 \cdot g \cdot m o {l}^{-} 1} = 0.570 \cdot m o l$

$\text{Moles of fluorine} = \frac{21.7 \cdot g}{19.00 \cdot g \cdot m o {l}^{-} 1} = 1.140 \cdot m o l$

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 ........

$B a {F}_{2}$...........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......

$\text{molecular formula"=nxx"empirical formula}$............