An organic compound contains \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) . Combustion of 0.1023 \(\mathrm{g}\) of the compound in excess oxygen yielded 0.2766 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.0991 $\mathrm{g} \mathrm{H}_{2} \mathrm{O} .\( A sample of 0.4831 \)\mathrm{g}$ of the compound was analyzed for nitrogen by the Dumas method (see Exercise 137\() .\) At STP, 27.6 \(\mathrm{mL}\) of dry \(\mathrm{N}_{2}\) was obtained. In a third experiment, the density of the compound as a gas was found to be 4.02 \(\mathrm{g} / \mathrm{L}\) at \(127^{\circ} \mathrm{C}\) and 256 torr. What are the empirical and molecular formulas of the compound?

From the combustion of 0.1023 g of the compound, we get 0.2766 g of CO2 and 0.0991 g of H2O.

Using the molecular mass ratios in CO2 (1 mole of C produces 1 mole of CO2) and H2O (2 moles of H produce 1 mole of H2O). Mass of C in CO2 = 0.2766 g × \(\frac{12.01 g}{44.01 g}\) = 0.0756 g Mass of H in H2O = 0.0991 g × \(\frac{2.02 g}{18.02 g}\) = 0.0111 g

Using the volume of N2 and the ideal gas law, we can find the mass of N in the sample by converting it to moles and then to grams. (0.0276 L × 1 mol/22.4 L) × (28.02 g/mol) = 0.0343 g of N

Find the mass of O in the compound by subtracting the masses of C, H, and N from the initial mass. Mass of O = 0.1023 g - (0.0756 g + 0.0111 g + 0.0343 g) = 0.0113 g. Convert these masses to moles: moles of C = \(\frac{0.0756}{12.01}\) = 0.00630 moles moles of H = \(\frac{0.0111}{1.01}\) = 0.0110 moles moles of N = \(\frac{0.0343}{14.01}\) = 0.00245 moles moles of O = \(\frac{0.0113}{16.00}\) = 0.0007063 moles _empirical formula_: \(CH_{\text{a}}N_{\text{b}}O_{\text{c}}\) Find the lowest whole number ratio of the constituent moles: Divide each mole by the smallest mole: a: 0.00630 moles / 0.0007063 moles = 8.92 b: 0.0110 moles / 0.0007063 moles = 15.6 c: 0.00245 moles / 0.0007063 moles = 3.47 Thus, the empirical formula is \(C_9H_{15}NO_3\).

Calculate the weight of the empirical formula: Empirical formula weight = \(9(12) + 15(1) + 1(14) + 3(16) = 236 \, \text{g/mol}\)

Use the density data to determine the molecular weight of the compound. Given density: 4.02 g/L Temperature: \(127^{\circ}C + 273.15\) K = 400.15 K Pressure: \(\frac{256 \, \text{torr}}{760 \, \text{torr/atm}}\) = 0.3368 atm Now we use the ideal gas law equation: \(n = \frac{PV}{RT}\) Where: n = number of moles P = pressure (0.3368 atm) V = volume (1 L) R = gas constant (0.08206 L atm/mol K) T = temperature (400.15 K) Solving for n, we get: n = \(\frac{0.3368 \times 1}{(0.08206)(400.15)}\) = 0.01024 moles Given that 4.02 g of the gas corresponds to 0.01024 moles, the molecular weight is: Molecular weight = \(\frac{4.02 \, \text{g}}{0.01024 \, \text{moles}}\) = 392.57 g/mol

Calculate the ratio between the molecular weight and the empirical formula weight to find the molecular formula. Ratio = \(\frac{392.57}{236}\) ≈ 1.66 Since the ratio is approximately equal to an integer, we multiply each subscript in the empirical formula by the ratio to find the molecular formula. Molecular formula: \(C_{9\cdot1.66}H_{15\cdot1.66}N_{1\cdot1.66}O_{3\cdot1.66}\) The molecular formula is approximately \(C_{15}H_{25}N_{2}O_{5}\).