A compound contains 63g Mn and 37g O, what is the empirical formula?

1 Answer
anor277
Sep 19, 2016

#MnO_2#.

Explanation:

We divide each constituent mass thru by the ATOMIC mass of each element:

#"Moles of manganese"# #=# #(63*g)/(54.94*g*mol^-1)# #=# #1.15*mol#.

#"Moles of oxygen"# #=# #(37*g)/(15.99*g*mol^-1)# #=# #2.31*mol#.

We divide thru by the smallest molar quantity, that of the metal, to give an empirical formula of #MnO_2#.

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