Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000-M \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). Assume the douglasite is the only source of chloride ion.

We know that \(37.20\,\mathrm{mL}\) of a \(0.1000\,\mathrm{M}\,\mathrm{AgNO_3}\) solution reacted with the sample. Let's first convert the volume of \(\mathrm{AgNO_3}\) solution to liters for easier calculations: \[ V = 37.20\,\mathrm{mL} \times \frac{1\,\mathrm{L}}{1000\,\mathrm{mL}} = 0.03720\,\mathrm{L} \] Now we can find the moles of \(\mathrm{AgNO_3}\) used: \[ n(\mathrm{AgNO_3}) = M \times V = 0.1000\,\mathrm{M} \times 0.03720\,\mathrm{L} = 0.003720\,\mathrm{mol} \] Since the only source of \(\mathrm{Cl}\) in the sample is douglasite, and \(\mathrm{AgNO_3}\) reacts with chloride ions in a 1:1 ratio, the moles of chloride ions in the sample must be equal to the moles of \(\mathrm{AgNO_3}\): \[ n(\mathrm{Cl}^-) = n(\mathrm{AgNO_3}) = 0.003720\,\mathrm{mol} \]

From the formula of douglasite, \(2\,\mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2\,\mathrm{H}_{2}\mathrm{O}\), we can see that there are \(2\,\mathrm{mol}\) of \(\mathrm{KCl}\) and \(1\,\mathrm{mol}\) of \(\mathrm{FeCl}_{2}\) in one mole of douglasite. This means there are a total of \(2 + 1 = 3\,\mathrm{mol}\) of \(\mathrm{Cl}\) ions in one mole of douglasite. Now, we can calculate the moles of douglasite in the sample, using the moles of \(\mathrm{Cl}\) ions: \[ n(\mathrm{douglasite}) = \frac{n(\mathrm{Cl}^-)}{3} = \frac{0.003720\,\mathrm{mol}}{3} = 0.001240\,\mathrm{mol} \]

To find the mass of douglasite, we first need its molar mass. The molar mass of douglasite can be determined from the molar masses of each element present in its formula: \[ M_{\mathrm{douglasite}} = 2\,M_{\mathrm{KCl}} + M_{\mathrm{FeCl}_{2}} + 2\,M_{\mathrm{H}_{2}\mathrm{O}} = 2(39.10 + 35.45) + (55.85 + 2 \times 35.45) + 2(2 \times 1.01 + 16.00) \] \[ M_{\mathrm{douglasite}} = 2(74.55) + (126.75) + 2(18.02) = 149.10 + 126.75 + 36.04 = 311.89\,\mathrm{g/mol} \] Now we can find the mass of douglasite in the sample: \[ m_{\mathrm{douglasite}} = n(\mathrm{douglasite}) \times M_{\mathrm{douglasite}} = 0.001240\,\mathrm{mol} \times 311.89\,\mathrm{g/mol} = 0.386\,\mathrm{g} \]

Now we can find the mass percent of douglasite in the sample using the mass of douglasite and the given sample mass: \[ \mathrm{Mass \%\: of\: douglasite} = \frac{m_{\mathrm{douglasite}}}{m_{\mathrm{sample}}} \times 100\% = \frac{0.386\,\mathrm{g}}{0.455\,\mathrm{g}} \times 100\% = 84.84\% \] Hence, the mass percent of douglasite in the \(455.0\,\mathrm{mg}\) sample is \(\boxed{84.84\%}\).