Question

Douglasite is a mineral with the formula $\mathbf{2}\mathbf{KCI}.{\mathbf{FeCl}}_2.\mathbf{2}{\mathbf{H}}_2\mathbf{O}$ . Calculate the mass percent ofdouglasite in a 455.0-mg sample if it took 37.20 mL of a 0.1000 M ${\mathbf{AgNO}}_3$solution to precipitate all the${\mathbf{Cl}}^{-}$ as$\mathbf{AgCl}$ . Assume the douglasite is the only source of chloride ion.

Short answer

The mass percent ofdouglasite in a 455.0-mg sample if it took 37.20 mL of a 0.1000 M${\mathrm{AgNO}}_3$ solution to precipitate all the ${\mathrm{Cl}}^{-}$ as $\mathrm{AgCl}$is $63.74\mathrm{\%}$.

Step-by-step solution

Definition 

The mass percent of an atom in a compound is equal to the ratio of the atom's total mass to the complex's total mass multiplied by 100. It can be stated as follows:

_{$\mathrm{Mass}\mathrm{percent}\mathrm{of}\mathrm{an}\mathrm{atom}=\left(\frac{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{atom}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{substance}}\right)\times 100$}

Determination of moles of  AgNO3

Moles of ${\mathrm{AgNO}}_3=\frac{0.1000\mathrm{mol}{\mathrm{AgNO}}_3}{1\mathrm{L}{\mathrm{AgNO}}_3}\times \frac{1\mathrm{L}{\mathrm{AgNO}}_3}{1000\mathrm{mL}{\mathrm{AgNO}}_3}\times 37.20\mathrm{mL}{\mathrm{AgNO}}_3$

$=3.72\times {10}^{-3}\mathrm{mol}{\mathrm{AgNO}}_3$

Determination of moles of  Cl−

The reaction of precipitation of $\mathrm{AgCl}$:

${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\to \mathrm{AgCl}\left(\mathrm{s}\right)$

The moles of:

$3.72\times {10}^{-3}\mathrm{mol}{\mathrm{AgNO}}_3=3.72\times {10}^{-3}\mathrm{mol}{\mathrm{AgNO}}_3\times \frac{1\mathrm{mol}{\mathrm{Ag}}^{+}}{1\mathrm{mol}{\mathrm{AgNO}}_3}\times \frac{1\mathrm{mol}{\mathrm{Cl}}^{-}}{1\mathrm{mol}{\mathrm{Ag}}^{+}}$

$\text{=}3.72\times {10}^{-3}\mathrm{mol}{\mathrm{Cl}}^{-}$

Determination of molar mass of douglasite

Themolar mass of douglasite:

$=(2(39.10)+4(35.45)+1(55.85)+4(1.008)+2(16.00))\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$

$=311.88\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$

$=3.72\times {10}^{-3}\mathrm{mol}{\mathrm{Cl}}^{-}\times \frac{1\mathrm{mol}\mathrm{douglasite}}{4\mathrm{mol}{\mathrm{Cl}}^{-}}\times \frac{311.88\mathrm{g}\mathrm{douglasite}}{1\mathrm{mol}\mathrm{douglasite}}$

$\text{=0.290}\mathrm{g}\mathrm{douglasite}$

Determination of mass percent of douglasite

Mass percent of douglasite in the sample:

$\begin{array}{c}\mathrm{\%}\mathrm{douglasite}=\frac{\mathrm{mass}\mathrm{of}\mathrm{Douglasite}\mathrm{required}}{\text{Total}\mathrm{mass}\mathrm{of}\mathrm{Douglasite}}\times 100\mathrm{\%}\\ =\frac{\text{0.290}\mathrm{g}}{455\mathrm{mg}}\times \frac{1000\mathrm{mg}}{1.00\mathrm{g}}\times 100\mathrm{\%}\\ =63.74\mathrm{\%}\end{array}$

Therefore, the mass percent of douglasite is $63.74\%$.