What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

1 Answer
anor277
Dec 19, 2016

#ZnO_2#

Explanation:

With all these problems, we assume that there are #100*g# of unknown compound, and then we proceed to calculate the empirical formula:

#"Moles of zinc"# #=# #(67.1*g)/(65.4*g*mol^-1)=1.03*mol#

#"Moles of oxygen"# #=# #(32.9*g)/(15.999*g*mol^-1)=2.06*mol#

Given the molar ratios, we divide thru by the lowest molar quantity to get an empirical formula of #ZnO_2#. The examiner has been nasty in that the expectation would have been #ZnO#. However, zinc does form a peroxo species composed of #Zn^(2+)# and #O_2^(2-)# ions.

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