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


A 1.000 g sample of a metal chloride, {eq}MCl_2 {/eq}, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that forms weighs 1.286 g. Calculate the atomic mass of {eq}M {/eq}.

Precipitation Reactions:

A precipitation reaction occurs when a cation and anion are able to meet with one another in an aqueous solution to form an insoluble product known as a precipitate. The key in determining whether a precipitate forms is our knowledge of the solubility rules. For instance, silver halides are insoluble in an aqueous solution and precipitate out of the solution.

Answer and Explanation: 1

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}

Learn more about this topic:

Precipitation Reactions: Predicting Precipitates and Net Ionic Equations


Chapter 10 / Lesson 9

Study precipitate reactions. Determine precipitate solubility according to solubility rules. Learn to write ionic equations by following a molecular reaction.

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