Question

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

Answer 1

`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`.