A solid mixture weighing 0.5485 gcontained only ferrous ammonium sulfate hexahydrate and ferrous chloride hexahydrate. The sample was dissolved in $\mathbf{1}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ , oxidized to ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$ with ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$, and precipitated with cupferron. The ferric cupferron complex was ignited to produce 0.1678 gof ferric oxide, ${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$(FM 159.69). Calculate the weight percent of Clin the original sample.

The Weight Percentage is simply the ratio of a solute's mass to the mass of a solution multiplied by 100. The Weight Percentageis also referred to as the Mass Percentage.

$Percentbyweight=\frac{gramofsolute}{100gofsolution}$

Here we have a solid mixture which weighs 0.5485 g and contains only ferrous ammonium sulfate hexahydrate and ferrous chloride hexahydrate. We need to calculate the weight percent of Cl in the original sample.

Consider the following:

x = m (ferrous ammonium sulfate hexahydrate)

y = m (ferrous chloride hexahydrate)

x + y = 0.585g

First we will calculate the moles of Fe in ${\mathrm{Fe}}_2{\mathrm{O}}_3$:

$$\begin{gathered} n\left(F{e}_2{O}_3\right)=\frac{m}{M} \\ n\left(F{e}_2{O}_3\right)=\frac{0.167g}{159.69g/mol} \\ n\left(F{e}_2{O}_3\right)=0.0010508mol \\ n\left(Fe\right)=2n\left(F{e}_2{O}_3\right)=0.0021016mol \end{gathered}$$

Use the information from Step 2 and determine the value of y :

$n\left(Fe\right)=n(ferrousammoniumsulfatehexahydrate)+n(ferrouschloridehexahydrate)$$$\begin{gathered} 0.0021016mol=\frac{x}{392.13g/mol}+\frac{y}{234.84g/mol} \\ 0.0021016mol=\frac{(0.5485g-y)}{(392.13g/mol)}+\frac{y}{234.84g/mol} \\ y=0.41146g \end{gathered}$$

Next we will calculate the mass of 1Cl in ferrous chloride hexahydrate which contains 2Cl so the following would be:

$$\begin{gathered} m\left(Cl\right)=2\times \frac{M\left(Cl\right)}{M(ferrouschloridehexahydrate)}\times y \\ m\left(Cl\right)=2\times \frac{35.453g/mol}{234.84g/mol}\times (0.41146g) \\ m\left(Cl\right)=0.12423g \end{gathered}$$

Then we can calculate the weight percent of Cl in in the original sample:

$$\begin{gathered} wt\%=\frac{m\left(Cl\right)}{m\left(sample\right)} \\ wt\%=\frac{0.12423g}{0.5485g} \\ wt\%=0.2265 \end{gathered}$$

Therefore the weight percent of Cl in the original sample is 0.2265%.