When solid \(\mathrm{KClO}_{3}\) is heated strongly, it decomposes to form solid potassium chloride, \(\mathrm{KCl}\), and \(\mathrm{O}_{2}\) gas. \(\mathrm{A}\) \(0.415 \mathrm{g}\) sample of impure \(\mathrm{KClO}_{3}\) is heated strongly and the \(\mathrm{O}_{2}\) gas produced by the decomposition is collected over water. When the wet \(\mathrm{O}_{2}\) gas is cooled back to \(26^{\circ} \mathrm{C}\), the total volume is \(229 \mathrm{mL}\) and the total pressure is 323 Torr. What is the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample? Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 25.22 Torr at \(26^{\circ} \mathrm{C}\).

We need to figure out the number of moles of oxygen gas collected. To do this, we need to correct the total pressure given to the pressure of oxygen by subtracting the vapor pressure of the water. Therefore, the pressure of oxygen, \(P_{O_{2}} = \text{total pressure} - \text{vapor pressure of water} = 323 \text{Torr} - 25.22 \text{Torr} = 297.78 \text{Torr}\). Next, we convert the pressure to atmosphere (1 Torr = 1/760 atm), so \(P_{O_{2}} = 297.78/760 \text{atm}\). Now, we convert the temperature from Celsius to Kelvin scale (K = \(^{\circ}C + 273.15)\), which gives T = 26 + 273.15 = 299.15 K. Given that the volume V = 229 ml = 0.229 L, we can use the ideal gas law equation \(PV = nRT\) (where n is the number of moles we are seeking, R is the ideal gas constant) to calculate the number of moles of oxygen (\(n_{O_{2}})\). Therefore, \(n_{O_{2}}= \frac{P_{O_{2}}V}{RT} = \frac{(297.78/760) * 0.229}{0.0821 * 299.15}\).

From the reaction equation, we know that for every 2 moles of \(O_{2}\) produced, 2 moles of \(\mathrm{KClO}_{3}\) decomposes. Hence, the number of moles of \(\mathrm{KClO}_{3}\) is equal to the number of moles of \(O_{2}\). So, \(n_{\mathrm{KClO}_{3}} = n_{O_{2}}\).

The molar mass of \(\mathrm{KClO}_{3}\) is 39.1 (K) + 35.5 (Cl) + 16*3 (O) = 122.6 g/mol. Thus, the mass of \(\mathrm{KClO}_{3}\) is equal to the molar mass of \(\mathrm{KClO}_{3}\) times the number of moles, and the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample is \(\frac{\text{mass of } \mathrm{KClO_{3}}}{\text{original mass of sample}} \times 100\%\).