# Question d46e7

Oct 2, 2017

${\text{MCl}}_{3}$

#### Explanation:

The trick here is to realize that the number of moles of electrons needed to reduce $1$ mole of "M"^(?+) cations will give you the net charge of the metal cations.

In your case, you know that $3$ moles of electrons will reduce the "M"^(?+) cations to $1$ mole of $\text{M}$ metal, so you can say that

${\text{M"_ ((l))^(?+) + 3"e"^(-) -> "M}}_{\left(s\right)}$

As you know, in any chemical reaction, charge must be conserved.

Since the metal is reduced to its elemental form, i.e. the charge on the metal will be $0$, you can say that you have

(?+) + 3 * (-) = 0

This implies that

? = 3

Therefore, your unknown chloride contained $3 +$ metal cations. Since chlorine forms $1 -$ anions, i.e. the chloride anion, ${\text{Cl}}^{-}$, you can say that the empirical formula of the chloride will be

["M"]^(3+) + 3["Cl"]^(-) -> "MCl"_3#

This tells you that you need three $1 -$ chloride anions to balance the overall $3 +$ positive charge of a single ${\text{M}}^{3 +}$ cation.