Chapter 4, Problem 122IL

Interpretation Introduction

Interpretation:

The weight percent of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given sample 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.
• $\text{Amount}\text{of substance}\text{=}\text{Concentration}\text{of}\text{substance}\text{×volume}\text{of}\text{the}\text{substance}$
• Weight percent or mass percent of elements of a compound is the ratio of weight of element to the weight of whole compound and multiplied with hundred.

Answer to Problem 122IL

The mass percent of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given sample is $\text{51}\text{.41}\text{\%}$ .

Explanation of Solution

The amount of hydrochloric acid in $50.0mLof0.100MHCl$ can be determined as follows,

  $\begin{array}{l}\text{Amount}\text{of HCl}=\text{Concentration}{}_{\text{HCl}}\times \text{volume}{}_{\text{HCl}}\\ =\frac{0.100\text{mol}}{\text{L}}\text{×0}\text{.05}\text{L}\\ \text{=}\text{0}\text{.005}\text{moles }\text{of}\text{HCl}\end{array}$

Therefore the initial amount of $HCl$ is $0.005mol$

Amount of $NaOH$ used in reaction with $\text{HCl}$ is $0.0013mol$

It is because $1000mL$ contains $0.121moles$, then

$11.1mL(usedinreactionwithHCl)$ contains $\frac{0.121}{1000}\times 11.1=0.0013moles$

The balanced equation for the reaction of $HClandNaOH$ is,

  $HCl+NaOH\stackrel{}{\to }NaCl+H_{2}O$

The stoichiometric ratio between the reactant $NaOH$ and $\text{HCl}$ is $1:1$

Therefore $0.0013moles$ of $NaOH$ reacts with $0.0013moles$ of $\text{HCl}$.

So the final amount of $\text{HCl}$ is $0.0013moles.$

Amount of $\text{HCl}$ reacted with $N{H}_3$ is,

$\begin{array}{l}InitialamountofHCl-FinalamountofHCl=\text{0}\text{.005}\text{-}\text{0}\text{.0013}\\ \text{=}\text{0}\text{.0037}\text{moles}\end{array}$

Here to assess the purity of a sample of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$$\text{0}\text{.475}\text{g}$ sample of impure ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ is dissolved in aqueous $KOH$

The balanced equation for this reaction is,

  ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}{\text{+2KOH}}_{\text{(aq)}}\stackrel{}{\to }{\text{2NH}}_{\text{3(aq)}}\text{+}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4((aq)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

From the balanced equation it is clear that 2 moles of $N{H}_3$ are produced from 1 mole of ${(N{H}_4)}_{2}S{O}_4$.

Therefore the amount of pure ${(N{H}_4)}_{2}S{O}_4$ is,

$0.0037moleofN{H}_3\times \frac{1moleof{(N{H}_4)}_{2}S{O}_4}{2moleofN{H}_3}=0.00185molesofpure{(N{H}_4)}_{2}S{O}_4$

Mass of pure ${(N{H}_4)}_{2}S{O}_4$ is,

$0.00185mol\times 132g/mol=0.2442g$ of pure ${(N{H}_4)}_{2}S{O}_4$ in impure sample.

Therefore,

$\begin{array}{l}Weightpercentofpure {\left(N{H}_4\right)}_{2}S{O}_4=\frac{Massofpure{\left(N{H}_4\right)}_{2}S{O}_4}{Massofimpure{\left(N{H}_4\right)}_{2}S{O}_4}\times 100\\ =\frac{0.2442g}{0.475g}\times 100\\ \text{=}\text{51}\text{.41}\text{\%}\end{array}$

Conclusion

The mass percent of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given sample was determined.