A 1.000 g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous...

We can represent the chemical reaction that occurs by using the chemical equation below:

{eq}\rm MCl_2(aq) + AgNO_3(aq) \to 2AgCl(s) + M(NO_3)_2 {/eq}

By adding excess silver nitrate, we can precipitate all of the chloride ions. The mass of AgCl can be used to determine the mass of the chloride. The first step is to determine the moles of chloride ions present. The molar mass of AgCl is 143.32 grams per mole.

{eq}\rm \require{cancel} 1.286~\cancel{g~AgCl}\times\dfrac{1~\cancel{mol~AgCl}}{143.32~\cancel{g~AgCl}}\times\dfrac{1~mol~Cl^-}{1~\cancel{mol~AgCl}} = 8.972\times10^{-3}~mol~Cl^- {/eq}

Next, we determine the mass of chloride ions that were present in the metal chloride. To do so, we use the molar mass of Cl.

{eq}\rm \require{cancel} 8.972\times10^{-3}~\cancel{mol~Cl^-}\times\dfrac{35.45~g}{1~\cancel{mol~Cl^-}} = 0.318~g~Cl {/eq}

Subtracting the mass of Cl from the mass of the metal chloride, we can establish the mass of the metal M.

{eq}\rm mass~M = 1.000~g - 0.318~g \\ mass~M = 0.682~g {/eq}

The atomic mass of M can be determined by taking the ratio between the mass in grams of M and the moles of M present. The moles of M present can be determined by using the subscripts from the given chemical formula of the metal chloride.

{eq}\rm \require{cancel} 8.972\times10^{-3}~\cancel{mol~Cl^-}\times\dfrac{1~mol~M}{2~\cancel{mol~Cl^-}} = 4.486\times10^{-3}~mol~M {/eq}

Calculating for the molar mass, we get the following value. The molar mass is numerically equivalent to the atomic mass of an element.

{eq}\rm Molar~Mass~M = \dfrac{0.682~g}{4.486\times10^{-3}~mol~M} \\ Molar~Mass~M = 152\frac{g}{mol} \\ \boxed{ \mathbf{ Atomic~Mass~M = 152~amu }} {/eq}