Question

A mixture contains only NaCl and Fe(NO₃)₃. A 0.456-g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Fe(OH)₃. The precipitate is filtered, dried, and weighed. Its mass is 0.107 g. Calculate the following.

a. the mass of iron in the sample

b. the mass of Fe(NO₃)₃ in the sample

c. the mass percent of Fe(NO₃)₃ in the sample

Short answer

a. The mass of iron in the sample is 0.0588 g.

b. The mass of Fe(NO₃)₃ in the sample is 0.242 g.

c. The mass percent of Fe(NO₃)₃ in the sample is 53.07%.

Step-by-step solution

Determination of mass of iron in the sample

The precipitation reaction is

${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{}\mathbf{3}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\to }\mathbf{Fe}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\downarrow }$

Moles of Fe(OH)₃ are,

$$\begin{gathered} \mathbf{moles}\mathbf{}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{molar}\mathbf{}\mathbf{mass}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{107}\mathbf{g}}{\mathbf{106}\mathbf{.}\mathbf{85}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{mol} \end{gathered}$$

From the above reaction, it can be concluded that 1 mol of Fe³⁺ produces 1 mol of Fe(OH)₃. So, the mol of Fe³⁺ are 0.001 mol.

The mass of Fe in the sample is:

$$\begin{gathered} \mathbf{mass}\mathbf{=}\mathbf{moles}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{molar}\mathbf{}\mathbf{mass} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{mol}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{55}\mathbf{.}\mathbf{85}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{558}\mathbf{g} \end{gathered}$$

Thus, the mass of iron in the sample is 0.0558 g.

Determination of mass of Fe(NO3)3 in the sample

The chemical reaction for the process is:

$\mathbf{Fe}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{NaOH}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{Fe}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

From the above reaction, it is concluded that 1 mol of Fe(NO₃)₃ produces 1 mol of Fe(OH)₃.

Moles of Fe(OH)₃ are 0.001 mol.

So, 0.001 mol of Fe(OH)₃ can be produced from 0.001 mol of Fe(NO₃)₃.

The mass of Fe(NO₃)₃ in the sample is:

$$\begin{gathered} \mathbf{mass}\mathbf{=}\mathbf{moles}\mathbf{\times }\mathbf{molar}\mathbf{}\mathbf{mass} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{mol}\mathbf{\times }\mathbf{241}\mathbf{.}\mathbf{86}\mathbf{g} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{242}\mathbf{g} \end{gathered}$$

Thus, the mass of Fe(NO₃)₃ in the sample is 0.242 g.

Determination of mass percent of  in the sample

The mass percent of Fe(NO₃)₃ in the sample is:

$$\begin{gathered} \mathbf{mass}\mathbf{\%}\mathbf{=}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{Fe}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{mixture}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{242}\mathbf{g}}{\mathbf{0}\mathbf{.}\mathbf{456}\mathbf{g}}\mathbf{\times }\mathbf{100} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{53}\mathbf{.}\mathbf{07}\mathbf{\%} \end{gathered}$$

Thus, the mass percent of Fe(NO₃)₃ in the sample is 53.07%.