Chapter 4, Problem 137CP

In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A 0.256-g sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it bad a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g.

a. What is the percent composition, by mass, of the compound?

b. Assuming the compound contains one cobalt ion per formula unit, what is the formula?

c. Write balanced equations for the three reactions described.

Interpretation Introduction

Interpretation: The mass percentage of compound along with balanced chemical equation and formula has to be written.

Concept introduction: The mass percent of compound is given by the calculated mass of the compound to the total mass. The mass percent of compound is given by the formula,

$\text{Mass}\text{percent(in\%)=}\frac{\text{Calculated}\text{mass(in}\text{g)}}{\text{Total}\text{mass(in}\text{g)}}$

Answer to Problem 137CP

The mass percentage of $\text{Cobalt}\text{\&}\text{Chlorine}$ are $\text{29}\text{.7\%}\text{and}\text{24}\text{.8\%}$

Explanation of Solution

Given:

Record the given info

Mass of sample containing chlorine = $\text{0}\text{.256}\text{g}$

Mass of sample containing cobalt    = $\text{0}\text{.416}\text{g}$

Mass of silver chloride                     = $\text{0}\text{.308}\text{g}$

Mass of cobalt (III) oxide                 = $\text{0}\text{.145}\text{g}$

The mass of samples containing chlorine and cobalt are recorded with the masses of silver chloride and cobalt (III) oxide as shown above.

To calculate the mass percent of ${\text{Cl}}^{\text{-}}$

Molar mass of Chlorine          = $\text{35}\text{.45}\text{g}$

Molar mass of silver chloride = $\text{143}\text{.4}\text{g}$

Moles of ${\text{Cl}}^{\text{-}}$= $\text{0}\text{.308}\text{g}\text{AgCl×}\frac{\text{35}\text{.45}\text{g}\text{Ag}}{\text{143}\text{.4}\text{g}\text{AgCl}}\text{=0}\text{.0761}\text{g}\text{Cl}$

Therefore, the mass percent of ${\text{Cl}}^{\text{-}}$ is,

Mass percent of ${\text{Cl}}^{\text{-}}$= $\frac{\text{0}\text{.0761}\text{g}}{\text{0}\text{.256}\text{g}}\text{=29}\text{.7\%}$

Mass percent of ${\text{Cl}}^{\text{-}}$= $29.7\%$

The mass percent of ${\text{Cl}}^{\text{-}}$ is calculated by plugging in the values of moles of ${\text{Cl}}^{\text{-}}$ to the total mass of sample. The moles of ${\text{Cl}}^{\text{-}}$ is calculated by plugging in the values of mass of silver chloride and molar masses of chlorine and silver chloride. The mass percent of ${\text{Cl}}^{\text{-}}$ is $29.7\%$

To calculate the mass percent of ${\text{Co}}^{\text{2+}}$

Molar mass of cobalt                   = $\text{117}\text{.86}\text{g}$

Molar mass of cobalt (III) oxide = $\text{165}\text{.86}\text{g}$ we

Moles of ${\text{Co}}^{\text{2+}}$ = $\text{0}\text{.145}\text{g}{\text{Co}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{117}\text{.86g}\text{Co}}{\text{165}\text{.86}\text{g}{\text{Co}}_{\text{2}}{\text{O}}_{\text{3}}}\text{=0}\text{.103}\text{g}\text{Co}$

Therefore, the mass percent of ${\text{Co}}^{\text{2+}}$ is given by,

$\text{Mass}\text{percent}\text{of}{\text{Co}}^{\text{2+}}\text{=}\frac{\text{0}\text{.103}\text{g}}{\text{0}\text{.416}\text{g}}\text{×100=24}\text{.8\%}$

Mass percent of ${\text{Co}}^{\text{2+}}$ is $\text{24}\text{.8\%}$

The mass percent of ${\text{Co}}^{\text{2+}}$ is calculated by plugging in the values of moles of ${\text{Co}}^{\text{2+}}$ to the total mass of sample. The moles of ${\text{Co}}^{\text{2+}}$ is calculated by plugging in the values of mass of silver chloride and molar masses of cobalt and cobalt (III) oxide. The mass percent of cobalt is $\text{24}\text{.8\%}$.

To calculate the mass percent of water

Molar mass of water = $\text{18}\text{.02g}$

$\text{100-(24}\text{.8+27}\text{.90)=45}\text{.5\%}\text{is}\text{water}$

Assume that 100g of compound is,

$\begin{array}{l}\text{45}\text{.5}\text{g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{2}\text{.016}\text{g}\text{H}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\text{=5}\text{.09}\text{g}\text{H}\\ \text{Mass}\text{percent}\text{of}{\text{H}}^{+}\text{=}\frac{\text{5}\text{.09}\text{g}\text{H}}{\text{100}\text{.0}\text{g}}\text{×100=5}\text{.09\%}\end{array}$

$\begin{array}{l}\text{45}\text{.5}\text{g}{\text{H}}_{\text{2}}\text{0×}\frac{\text{16}\text{.00}\text{g}\text{O}}{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}\text{=40}\text{.4}\text{g}\text{O}\\ \text{Mass}\text{percent}\text{of}{\text{O}}^{\text{-}}\text{=}\frac{\text{40}\text{.4}\text{g}\text{O}}{\text{100}\text{g}}\text{×100=}\text{40}\text{.4\%}\end{array}$

The mass percent of hydrogen and oxygen is calculated by plugging in the molar mass of water and molar masses of hydrogen and oxygen to the total mass of the sample. The molar masses of hydrogen and oxygen were found to be $\text{5}\text{.09\%}$ and $\text{40}\text{.4\%}$

Conclusion

The mass percentages of ${\text{Co}}^{\text{2+}}$, ${\text{Cl}}^{\text{-}}$, ${\text{H}}^{+}$ and ${\text{O}}^{\text{-}}$ were calculated using their respective moles and total masses of the samples. The mass percentages of ${\text{Cl}}^{\text{-}}$ and ${\text{Co}}^{\text{2+}}$ were found to be $29.7\%$, $\text{24}\text{.8\%}$, $\text{5}\text{.09\%}$ and $\text{40}\text{.4\%}$ respectively.

Interpretation Introduction

Interpretation: To calculate the empirical formula of the compound

Concept introduction: The representation of simplest positive integer of a atoms in a compound is called as empirical formula.

Answer to Problem 137CP

The formula for the compound is ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$

Explanation of Solution

To calculate the empirical formula of the compound

Out of 100 g of compound, there are

$\begin{array}{l}\text{24}\text{.8}\text{g}\text{Co×}\frac{\text{1}\text{mol}}{\text{58}\text{.93}\text{g}\text{Co}}\text{=0}\text{.421}\text{mol}\\ \text{29}\text{.7}\text{g}\text{Cl×}\frac{\text{1}\text{mol}}{\text{35}\text{.45}\text{g}\text{Cl}}\text{=0}\text{.839}\text{mol}\\ \text{5}\text{.09}\text{g}\text{H×}\frac{\text{1}\text{mol}}{\text{1}\text{.008}\text{g}\text{H}}\text{=5}\text{.05}\text{mol}\\ \text{40}\text{.4}\text{g}\text{O×}\frac{\text{1}\text{mol}}{\text{16}\text{.00}\text{g}\text{O}}\text{=2}\text{.53}\text{mol}\end{array}$

Dividing the moles by the smallest number,

The empirical formula of the compound becomes ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$ where, ${\text{6H}}_{\text{2}}\text{O}$ the water molecules present in the compound

The empirical formula of the compound is calculated by calculating the mole ratio of individual elements divide by the smallest number. The empirical formula of the compound is found to be ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$.

Conclusion

The empirical formula of the compound was calculated by using the mole ratio of individual elements divided by the smallest number. The empirical formula of the compound is found to be ${\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}\text{O}$.

Interpretation Introduction

Interpretation: To write the balanced equation of the precipitation reactions.

Concept introduction:

When two solutions containing soluble salts are mixed together, an insoluble salt so called precipitate is obtained and the reaction is called as precipitation reaction. These precipitation reactions help in the determination of various ions in the solution.’

Answer to Problem 137CP

$\begin{array}{l}{\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2AgNO}}_{\text{3(aq)}}\to {\text{2AgCl}}_{\left(\text{s}\right)}\text{+Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2NaOH}}_{\left(\text{aq}\right)}\to \text{Co}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}{\text{+2NaCl}}_{\left(\text{aq}\right)}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Co}{\left(\text{OH}\right)}_{\text{2}}\to {\text{Co}}_{\text{2}}{\text{O}}_{\text{3}}\\ {\text{4Co(OH)}}_{\text{2(s)}}{\text{+O}}_{\text{2}}{}_{\left(\text{g}\right)}\to {\text{2Co}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

This is redox reaction. Hence, an oxidizing agent is required and the oxidizing agent is ${\text{O}}_{\text{2}}$

Explanation of Solution

To write the balanced equation of the precipitation reactions.

The reaction between cobalt chloride hexahydrate with base such as silver nitrate and sodium hydroxide yields precipitates of silver chloride and cobalt hydroxide with release of water and sodium chloride. The equation for this reaction can be given as,

$\begin{array}{l}{\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+AgNO}}_{\text{3(aq)}}\to {\text{AgCl}}_{\left(\text{s}\right)}\text{+Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+NaOH}}_{\left(\text{aq}\right)}\to \text{Co}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}{\text{+NaCl}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Co}{\left(\text{OH}\right)}_{\text{2}}\to {\text{Co}}_{\text{2}}{\text{O}}_{\text{3}}\\ {\text{Co(OH)}}_{\text{2(s)}}{\text{+O}}_{\text{2}}{}_{\left(\text{g}\right)}\to {\text{Co}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

Cobalt hydroxide oxidizes to cobalt (III) oxide and water.

Two moles of silver nitrate are required to react with cobalt chloride hexahydrate to give 2 moles of silver chloride as precipitate with side products being cobalt nitrate and water. Cobalt nitrate being water soluble remains inside the solution, thus precipitating silver chloride out of the solution.

Two moles of sodium hydroxide are required to react with cobalt chloride hexahydrate to give 2 moles of cobalt hydroxide with sodium chloride and water. Sodium chloride being soluble in water, dissociates as spectator ions thus remaining in the solution and a precipitate of cobalt hydroxide is precipitated out.

Cobalt hydroxide on heating is oxidized to cobalt (III) oxide with water.

The balanced form of these equations can be given as,

$\begin{array}{l}{\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2AgNO}}_{\text{3(aq)}}\to {\text{2AgCl}}_{\left(\text{s}\right)}\text{+Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2NaOH}}_{\left(\text{aq}\right)}\to \text{Co}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}{\text{+2NaCl}}_{\left(\text{aq}\right)}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Co}{\left(\text{OH}\right)}_{\text{2}}\to {\text{Co}}_{\text{2}}{\text{O}}_{\text{3}}\\ {\text{4Co(OH)}}_{\text{2(s)}}{\text{+O}}_{\text{2}}{}_{\left(\text{g}\right)}\to {\text{2Co}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

The reaction of heating cobalt hydroxide is oxidation-reduction reaction, where oxygen is used as oxidizing agent and cobalt (III) oxide is obtained.

Conclusion

The given reactions were found to be precipitation reaction and moles on the reactant and the product were obtained. The reaction of heating cobalt hydroxide is oxidation-reduction reaction, where oxygen is used as oxidizing agent to oxidized cobalt hydroxide to cobalt (III) oxide. The balanced equations are,

$\begin{array}{l}{\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2AgNO}}_{\text{3(aq)}}\to {\text{2AgCl}}_{\left(\text{s}\right)}\text{+Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}\left(\text{aq}\right)}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{CoCl}}_{\text{2}}{\text{.6H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}{\text{+2NaOH}}_{\left(\text{aq}\right)}\to \text{Co}{\left(\text{OH}\right)}_{\text{2}\left(\text{s}\right)}{\text{+2NaCl}}_{\left(\text{aq}\right)}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ \text{Co}{\left(\text{OH}\right)}_{\text{2}}\to {\text{Co}}_{\text{2}}{\text{O}}_{\text{3}}\\ {\text{4Co(OH)}}_{\text{2(s)}}{\text{+O}}_{\text{2}}{}_{\left(\text{g}\right)}\to {\text{2Co}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$