Question

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

Answer 1

`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.