Q.14E - Chemistry (OpenStax)

Question

A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?

Step-by-Step Solution

Answer

The original compound has 29.23% of \(C{l^ - }\)and salt RbCl.    

Step 1: calculating moles:

We need to write down the reaction that occur; 

\(MCl + AgN{O_3} \to MN{O_3} + AgCl\)

 

We have simple formula from which we can calculate moles (n) of AgCl from mass:

\(\begin{aligned}n(AgCl) &= \frac{{m(AgCl)}}{{M(AgCl)}}\\n(AgCl) &= \frac{{3.03707g}}{{143.32g/mol}}\\n(AgCl) &= 0.0211mol\end{aligned}\)

 

We can see that if we separate AgCl into ions, we have the same concentration of \(A{g^ + }\)and \(C{l^ - }\):

\(AgCl \to A{g^ + } + C{l^ - }\)

 

Which means moles of AgCl are equal to moles of Cl which is \(0.0211mol\).

Step 2: Calculating the mass of Cl:

Calculating mass of \(C{l^ - }\)

 

\(\begin{aligned}n\left( {C{l^ - }} \right) &= \frac{{m\left( {C{l^ - }} \right)}}{{M\left( {C{l^ - }} \right)}}\\m\left( {C{l^ - }} \right) &= n\left( {C{l^ - }} \right).M\left( {C{l^ - }} \right)\\m\left( {C{l^ - }} \right) &= 0.0211mol.35.45gmo{l^{ - 1}}\\m\left( {C{l^ - }} \right) &= 0.7490g\end{aligned}\)

Step 3: Calculating the percentage of Cl:

Calculate percentage of \(C{l^ - }\)ion in the original compound:

\(\begin{aligned}w\left( {C{l^ - }} \right) &= \frac{{m\left( {C{l^ - }} \right)}}{{{m_{original}}}}.100\\w\left( {C{l^ - }} \right) &= \frac{{0.7490g}}{{2.5624g}}.100\\w\left( {C{l^ - }} \right) &= 29.23\% \end{aligned}\)

 

\(\begin{aligned}m(compound) &= m(metal) + m(C{l^ - })\\m(metal) &= m(compound) - m(C{l^ - })\\m(metal) &= 2.5624g - 0.7490g\\m(metal) &= 1.8134g\end{aligned}\)

 

To find identity of the salt we need to get the atomic mass of \({M^ + }\)ion:

\(MCl \to {M^ + } + C{l^ - }\)

 

Mole of \(C{l^ - }\) are equal to mols of \({M^ + }\) which is 0.0211 mol

\(\begin{aligned}n(metal) &= \frac{{m(metal)}}{{M(metal)}}\\M(metal) &= \frac{{m(metal)}}{{n(metal)}}\\M(metal) &= \frac{{1.8124g}}{{0.0211mol}}\\M(metal) &= 85.943gmo{l^{ - 1}}\end{aligned}\)

 

Hence, The original compound has 29.23% of \(C{l^ - }\)and salt RbCl.