Question

A solid mixture weighing 0.05485 gcontained only ferrous ammonium sulfate and ferrous chloride. The sample was dissolved in $\mathbf{1}{\mathbf{MH}}_{\mathbf{2}}\mathbf{}{\mathbf{SO}}_{\mathbf{4}}$, and the ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$ required 13.39mLof $\mathbf{0}\mathbf{.}\mathbf{01234}{\mathbf{MCe}}^{\mathbf{4}\mathbf{+}}$for complete oxidation to${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{(}{\mathbf{Ce}}^{\mathbf{4}\mathbf{+}}\mathbf{+}{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{\to }{\mathbf{Ce}}^{\mathbf{3}\mathbf{+}}\mathbf{+}{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{)}$. Calculate the weight percent CI ofin the original sample.

Short answer

The weight percent of Cl in sample is 8.17%

Step-by-step solution

Find the sample mass value.

The mass of sample containing${\mathrm{FeSO}}_4\cdot ({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O}$and${\mathrm{FeCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$is 0.05485g

Volume of$0.01234\mathrm{M}{\mathrm{Ce}}^{4+}$required for complete oxidation to${\mathrm{Fe}}^{3+}$

$$\begin{gathered} =13.39mL \\ =0.01339L \end{gathered}$$

Molar mass of$$\begin{gathered} {\mathrm{FeSO}}_4\cdot ({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O} \\ =392.13\mathrm{g}/\mathrm{mol} \end{gathered}$$

Molar mass of${\mathrm{FeCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$

$=234.84\mathrm{g}/\mathrm{mol}$

Let${\mathrm{FeSO}}_4\cdot ({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O}$ denoted as x and${\mathrm{FeCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$ denoted as y

The number of moles in Ce required

$$\begin{gathered} \mathrm{x}+\mathrm{y}=0.01339\mathrm{L}\left(0.01234\frac{{\mathrm{molCe}}^{4+}}{\mathrm{L}}\right) \\ \mathrm{x}+\mathrm{y}=1.652326\times {10}^{-4}\mathrm{molCe} \end{gathered}$$

The mass of the sample.

$392.13\mathrm{x}+234.84\mathrm{y}=0.05485$

Calculate the mass of the Cl

From above equation we have$\mathrm{x}=1.652326\times {10}^{-4}-\mathrm{y}$

Replacing x in the equation$392.13\mathrm{x}+234.84\mathrm{y}=0.05485$ we get

$392.13(1.652326\times {10}^{-4}-\mathrm{y})+234.84\mathrm{y}=0.05485$

Now we need to calculate the value of y,

$=6.32123\times {10}^{-5}\mathrm{mol}$

Calculate the weight percent of Cl

Now, we can solve for the mass of Cl using mole concept.

$$\begin{gathered} \mathrm{mass}\mathrm{of}\mathrm{Cl}=6.32123\times {10}^{-5}{\mathrm{molFeCl}}_2\left(\frac{2\mathrm{mol}\mathrm{Cl}}{1\mathrm{mol}\mathrm{FeCl}}\right)\left(\frac{35.45\mathrm{gCl}}{1\mathrm{molCl}}\right) \\ =4.482\times {10}^{-3}\mathrm{gCl} \end{gathered}$$

$$\begin{gathered} \%\mathrm{w}/\mathrm{w}=\frac{\mathrm{mass}\mathrm{of}\mathrm{substance}\mathrm{of}\mathrm{interest}}{\mathrm{mass}\mathrm{of}\mathrm{sample}}\times 100 \\ =\frac{4.482\times {10}^{-3}\mathrm{gCl}}{0.05485\mathrm{g}\mathrm{sample}}\times 100 \\ =8.17\% \end{gathered}$$

Thus, the weight percent of Cl in sample in 8.17%