Question #31a4e

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Question #31a4e

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Answer 1 · 2017-06-21T04:29:16

$B a F_{2}$ $BaF_2$ ......

Explanation:

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

$Moles of barium = \frac{78.3 \cdot g}{137.3 \cdot g \cdot m o l^{- 1}} = 0.570 \cdot m o l$ $"Moles of barium"=(78.3*g)/(137.3*g*mol^-1)=0.570*mol$

$Moles of fluorine = \frac{21.7 \cdot g}{19.00 \cdot g \cdot m o l^{- 1}} = 1.140 \cdot m o l$ $"Moles of fluorine"=(21.7*g)/(19.00*g*mol^-1)=1.140*mol$

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 ........

$B a F_{2}$ $BaF_2$ ...........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......

$molecular formula = n \times empirical formula$ $"molecular formula"=nxx"empirical formula"$ ............

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Question #31a4e

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