The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(\mathrm{s})\): $${\text { Compound }} \frac{\text { Hot }}{\text { \(\mathrm{CuO}(\mathrm{s})\) }} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2} .\) After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a 0.253 -g sample of a compound produced 31.8 \(\mathrm{mL} \mathrm{N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is 23.8 torr.

The Ideal Gas Law is given as \(PV = nRT\). Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin. First, we need to convert the given pressure and temperature to appropriate units for the given gas constant, R. Given: - Dry nitrogen gas volume, V = 31.8 mL = 0.0318 L (converting mL to L) - Gas at 25°C saturated with water vapor – We need to account for the pressure of the water vapor: Total pressure = 726 torr Vapor pressure of water at 25°C = 23.8 torr Pressure of dry nitrogen gas, P = (726 - 23.8) torr = 702.2 torr = \(93.49\ kPa\) (converting torr to kPa) Potential usable value for the gas constant R: \(8.3145\ J\ mol^{-1} K^{-1}\) or \(8.3145\ kPa\ L\ mol^{-1} K^{-1}\). Now, we also need to convert the temperature to Kelvin: Temperature, T = 25°C + 273.15 = 298.15 K. Now, we can use the Ideal Gas Law, \(n = \frac{PV}{RT}\)