A solid mixture weighing 0.05485 gcontained only ferrous ammonium sulphate and ferrous chloride. The sample was dissolved in$\mathbf{1}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$, and the ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$required 13.39 mL oflocalid="1663577431957" $\mathbf{0}\mathbf{.}\mathbf{01234}{\mathbf{MCe}}^{\mathbf{4}\mathbf{+}}$for complete oxidation to localid="1663577453691" ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$. Calculate the weight percent of Cl in the original sample. If you need refreshing, see Section 7-2 for an example of a titration of a mixture.

localid="1663574328546" ${\mathbf{FeSO}}_{\mathbf{4}}\mathbf{\times }{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\times }\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{FeCI}}_{\mathbf{2}}\mathbf{\times }\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Ferrous ammonium sulfate Ferrous chloride

FM 392.13

FM 234.84

• The second-lightest of the halogen elements is chlorine (Cl), which belongs to Group 17 (Group VIIa) of the periodic table.
• Greenish-yellow gas known as chlorine is toxic, corrosive, and irritates the eyes and respiratory system.

It is necessary to compute the weight % of chlorine in the original sample.

The mass percentage or weight percentage of a compound can be computed by multiplying the calculated mass of the substance by 100 to get the total mass of the sample.

The formula Mass percent can be used to compute the mass percent (in percentage) $=\frac{\mathrm{Calculated}\mathrm{mass}\left(\mathrm{ingrams}\right)}{\mathrm{total}\mathrm{mass}\left(\mathrm{in}\mathrm{grams}\right)}\times 100$

Given data

Weight of a ferrous ammonium and ferrous chloride solid mixture =0.05485g Consolidation of ${\mathrm{H}}_2{\mathrm{SO}}_4=1\mathrm{M}$

Concentration and amount of ${\mathrm{Ce}}^{4+}$compelled by ${\mathrm{Fe}}^{2+}=13.39\mathrm{mL}\mathrm{of}0.01234{\mathrm{MCe}}^{4+}$

To calculate, divide the weight percentage of chlorine in the original sample by the weight % of chlorine in the original sample.

$$\begin{gathered} \mathrm{x}={\mathrm{mgofFeSO}}_4\times \left({\mathrm{NH}}_4\right){\mathrm{SO}}_4\times 6{\mathrm{H}}_2\mathrm{O} \\ \left(54.85-\mathrm{x}\right)={\mathrm{mgofFeCI}}_2\times 6{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Let ${\mathrm{mmolofCe}}^{4+}={\mathrm{mmol}}^2{\mathrm{FeSO}}_4{\left({\mathrm{NH}}_4\right)}_{{\mathrm{so}}_4\times 6{\mathrm{H}}_2\mathrm{O}}+{\mathrm{mmolofFeCI}}_2\times 6{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} \left(13.39\mathrm{mL}\right)\left(0.01234\mathrm{M}\right)=\frac{\mathrm{xmq}}{392.13\mathrm{mg}/\mathrm{mmal}}+\frac{\left(54.85-\mathrm{x}\right)}{224.84\mathrm{mg}/\mathrm{mmol}} \\ \mathrm{x}=40.01\mathrm{mgFeSO}{4}_4\left({\mathrm{NH}}_4\right){\mathrm{SO}}_{4}6{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Mass of ${\mathrm{FeCI}}_2\times 6{\mathrm{H}}_2\mathrm{O}=14.84\mathrm{mg}=0.0619\mathrm{mmol}\approx 4.48\mathrm{mgCI}$