Question #05d76

1 Answer
anor277
Dec 16, 2016

The #"empirical formula"# is #As_2O_5#.

Explanation:

As with all these problems, we ASSUME that there are #100*g# of unknown compound.

Given the elemental percentages, thus there are:

#"Moles of oxygen"# #=# #(34.8*g)/(15.999*g*mol^-1)=2.18*mol#.

#"Moles of arsenick"# #=# #(65.2*g)/(74.9*g*mol^-1)=0.871*mol#.

We divide thru by the smallest molar quantity to get, #O:As#, #2.5:1.0#.

But by definition, the #"empirical formula"# is the simplest WHOLE number ratio that defines constituent atoms in a species, so we take this preliminary determination, and DOUBLE it to give WHOLE numbers:

#As_2O_5#. Capisce?

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