What is the empirical formula for the compound consisting of 63% manganese and 37% oxygen by mass?

1 Answer
anor277
Aug 21, 2016

#MnO_2#

Explanation:

As with all empirical formula calculations, a starting mass of #100# #g# is assumed.

There are thus #(63*g)/(54.94*g*mol^-1)# #=# #1.15*mol# #Mn#, and #(37*g)/(15.999*g*mol^-1)# #=# #2.132*mol# with respect to #O#.

Now, we simply divide thru by the LOWEST molar quantity, which here is the metal to give an empirical formula of #MnO_2", manganese(IV) oxide"#.

AS always, the empirical formula is the simplest, whole number ratio defining constituent atoms in a species.

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