Chapter 4, Problem 126IL

Interpretation Introduction

Interpretation:

The correct formula of iron containing compound has to be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

     $Numberofmole=\frac{Massingram}{Molarmass}$

     $Massingramofthesubs\tan ce=Numberofmole\times Molarmass$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• $\text{Amount}\text{of substance}\text{=}\text{Concentration}\text{of}\text{substance}\text{×volume}\text{of}\text{the}\text{substance}$

Answer to Problem 126IL

${\text{K}}_{\text{3}}{\text{[Fe(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{3}}\text{]}$ is the correct formula

Explanation of Solution

Herein two iron compounds are given

1. (a) $K[Fe{(C_2{O}_4)}_2{(H_{2}O)}_2]$
2. (b) $K_3[Fe{(C_2{O}_4)}_3]$

To find which one is correct, a weighed sample of the compound is dissolved in acid forming oxalic acid.  This acid is titrated with potassium permanganate.  The balanced net ionic equation for the titration is,

  $5H_2{C}_2{O}_{4(aq)}+2Mn{O}_{(aq)}^{-}+6H_3{O}_{(aq)}^{+}\stackrel{}{\to }2M{n}_{(aq)}^{2+}+10C{O}_{2(g)}+14H_2{O}_{(l)}$

Amount of $KMn{O}_4$ in $34.50mLof0.108MKMn{O}_4$ is,

$\text{Amount}{\text{of KMnO}}_{\text{4}}\text{=}\frac{\text{0}\text{.108}\text{mol}}{\text{L}}\text{×}\text{0}\text{.0345}\text{=}\text{0}\text{.003726}\text{mol}{\text{KMnO}}_{\text{4}}$

Molecular mass of $K[Fe{(C_2{O}_4)}_2{(H_{2}O)}_2]$ is $307.0119g/mol$ and the balanced chemical equation for the reaction of $K[Fe{(C_2{O}_4)}_2{(H_{2}O)}_2]$ with $KMn{O}_4$ is,

  $K[Fe{(C_2{O}_4)}_2{(H_{2}O)}_2]+4KMn{O}_4\stackrel{}{\to }K[Fe{(Mn{O}_4)}_4{(H_{2}O)}_2]+2K_2{C}_2{O}_4$

The mass of $K[Fe{(C_2{O}_4)}_2{(H_{2}O)}_2]$ can be calculated from the amount of $KMn{O}_4$ as follows,

  $\text{0}\text{.003726}\text{mol}{\text{KMnO}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}{\text{K[Fe(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{2}}{{\text{(H}}_{\text{2}}\text{O)}}_{\text{2}}\text{]}}{\text{4}\text{mol}{\text{KMnO}}_{\text{4}}}\text{×307}\text{.0119}\text{g/mol}\text{=}\text{0}\text{.286}\text{g}$

Similarly, the mass of $K_3[Fe{(C_2{O}_4)}_3]$ can be determined from the amount of $KMn{O}_4$ as follows,

The balanced chemical equation for the reaction of $K_3[Fe{(C_2{O}_4)}_3]$ with $KMn{O}_4$ is,

$10K_3[Fe{(C_2{O}_4)}_3]+12KMn{O}_4+48H_{2}S{O}_4\to 5F{e}_2{(S{O}_4)}_3+12MnS{O}_4+21K_{2}S{O}_4+60C{O}_2+48H_{2}O$

Molar mass of $K_3[Fe{(C_2{O}_4)}_3]$ is $437.1969g/mol$.  Hence mass of $K_3[Fe{(C_2{O}_4)}_3]$ is,

$\text{0}\text{.003726}\text{mol}{\text{KMnO}}_{\text{4}}\text{×}\frac{\text{10}\text{mole}{\text{K}}_{\text{3}}{\text{[Fe(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{3}}\text{]}}{\text{12}\text{mole}{\text{KMnO}}_{\text{4}}}\text{×}\text{437}\text{.1969}\text{g/mol}\text{=1}\text{.357}\text{g}$

The given mass of iron compound is $1.356g$ and the calculated mass of iron compounds are $\text{0}\text{.286}\text{g}\text{(compound a)}$ and $\text{1}\text{.357}\text{g}\text{(compound b)}$.

The mass value of compound b is closer to the given mass of iron compound.  Hence formula matches with the compound being titrated is ${\text{K}}_{\text{3}}{\text{[Fe(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{3}}\text{]}$

Conclusion

The correct formula for iron containing compound was determined.