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 129 ). At \(\mathrm{STP}, 27.6 \mathrm{mL}\) 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?

Since we are given the masses of CO₂ and H₂O that are produced upon combustion, we can use these values to find the masses of carbon and hydrogen in the original compound. Mass of \(\text{carbon in CO}_2 = \frac{\text{mass of CO}_2 \times \text{mass of C in CO}_2}{\text{molar mass of CO}_2} = \frac{0.2766 \times 12}{44} = 0.07560\) g Mass of \(\text{hydrogen in H}_2\text{O} = \frac{\text{mass of H}_2\text{O} \times \text{mass of H in H}_2\text{O}}{\text{molar mass of H}_2\text{O}} = \frac{0.0991 \times 2}{18} = 0.01101\) g Since the total mass of the compound is given to be \(0.1023\) g, we can now find the mass of nitrogen and oxygen in the compound. Mass of \(\text{N}_2 = \frac{27.6 \ \text{mL} \ \text{N}_2\times\text{molar mass of N}_2}{22.4\ \text{L/mol} \times 1000\ \text{mL/L}}=\frac{27.6\times28}{22.4 \times 1000}=0.03409\ \text{g}\) Therefore, mass of noncarbon and non-hydrogen atoms = total mass of the compound - mass of carbon - mass of hydrogen \(= 0.1023 - 0.07560 - 0.01101 = 0.01569\ \text{g}\) Since these are only nitrogen and oxygen atoms, mass of oxygen = mass of noncarbon and non-hydrogen atoms - mass of nitrogen \(= 0.01569 - 0.03409 = 0.04978\ \text{g}\)

Next, we need to convert the masses of carbon, hydrogen, nitrogen and oxygen into moles: Moles of C \(= \frac{0.07560\ \text{g}}{12\ \text{g/mol}} = 0.00630\ \text{mol}\) Moles of H \(= \frac{0.01101\ \text{g}}{1\ \text{g/mol}} = 0.01101\ \text{mol}\) Moles of N \(= \frac{0.03409\ \text{g}}{14\ \text{g/mol}} = 0.002435\ \text{mol}\) Moles of O \(= \frac{0.04978\ \text{g}}{16\ \text{g/mol}} = 0.003112\ \text{mol}\)

To find the empirical formula, we need to determine the simplest whole-number ratio of moles of C, H, N, and O: Molar ratios: C: \(\frac{0.00630}{0.002435} = 2.586\) H: \(\frac{0.01101}{0.002435} = 4.518\) N: \(\frac{0.002435}{0.002435} = 1\) O: \(\frac{0.003112}{0.002435} = 1.278\) Since all ratios are close to whole numbers, we can round them to get the empirical formula: C₃H₄NO₂

To calculate the empirical formula mass, we add the molar masses of C₃H₄NO₂: Empirical formula mass \(= 3 \times 12 + 4 \times 1 + 14 + 2 \times 16 = 88\ \text{g/mol}\)

Next, we need to use the given density, temperature, and pressure to find the molar mass of the compound: Molar mass \(= \frac{density \times R \times T}{P} = \frac{4.02\ \text{g/L} \times 0.0821\ \text{L atm/K mol} \times (127 + 273)\ \text{K}}{256\ \text{torr} \times \frac{1\ \text{atm}}{760\ \text{torr}}} = 176\ \text{g/mol}\) Now let's find the ratio of molecular formula mass to empirical formula mass: Ratio \(= \frac{176}{88} = 2\) This indicates that the molecular formula is two times the empirical formula, which gives us: Molecular formula: C₆H₈N₂O₄