Anthraquinone contains only carbon, hydrogen, and oxygen. When 4.80 \(\mathrm{mg}\) anthraquinone is burned, 14.2 \(\mathrm{mg} \mathrm{CO}_{2}\) and 1.65 \(\mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) are produced. The freezing point of camphor is lowered by \(22.3^{\circ} \mathrm{C}\) when 1.32 \(\mathrm{g}\) anthraquinone is dissolved in 11.4 \(\mathrm{g}\) camphor. Determine the empirical and molecular formulas of anthraquinone.

Given that 14.2 mg of CO2 and 1.65 mg of H2O are produced when burning 4.80 mg of anthraquinone, we can use the stoichiometry of these reactions to find out the mass of carbon and hydrogen in anthraquinone. For Carbon(C): CO2 produced: 14.2 mg Moles of CO2 produced: \(\frac{14.2}{44.01} = 0.3224\) moles (using the molar mass of CO2: 44.01 g/mol) Moles of Carbon in anthraquinone: 0.3224 moles (1 mole of C in 1 mole of CO2) Mass of Carbon in anthraquinone: \(0.3224 \times 12.01 = 3.870\) mg (using the molar mass of C: 12.01 g/mol) For Hydrogen (H): H2O produced: 1.65 mg Moles of H2O produced: \(\frac{1.65}{18.02} = 0.0916\) moles (using the molar mass of H2O: 18.02 g/mol) Moles of Hydrogen in anthraquinone: \(0.0916 \times 2 = 0.1832\) moles (2 moles of H in 1 mole of H2O) Mass of Hydrogen in anthraquinone: \(0.1832 \times 1.01 = 0.185\) mg (using the molar mass of H: 1.01 g/mol)

Since anthraquinone contains only carbon, hydrogen, and oxygen, we can find the mass of oxygen by subtracting the mass of carbon and hydrogen from the initial mass of anthraquinone: Mass of Oxygen: 4.80 mg - 3.870 mg - 0.185 mg = 0.745 mg

Once we have the mass of carbon, hydrogen and oxygen in anthraquinone, we can find the percent composition by dividing the respective mass by the total mass of the compound (4.80 mg): Percent of Carbon: \(\frac{3.870}{4.80} \times 100 = 80.63\% \) Percent of Hydrogen: \(\frac{0.185}{4.80} \times 100 = 3.854\% \) Percent of Oxygen: \(\frac{0.745}{4.80} \times 100 = 15.52\% \)

We are given that the freezing point of camphor is lowered by 22.3°C when 1.32 g of anthraquinone is dissolved in 11.4 g of camphor. Using the formula: ΔTf = Kf . \( \frac{moles \, of\, solute}{mass \, of\, solvent} \) where ΔTf is the freezing point depression, Kf is the cryoscopic constant for camphor (40.7 K kg/mol) and moles of solute (anthraquinone) can be calculated using the molecular weight (M). We can rewrite the formula to solve for M: M = \(\frac{Kf \times moles \,of\, solute}{ΔTf \times mass \,of\, solvent}\) Plugging in the given values: M = \(\frac{40.7 \times \frac{1.32}{M}}{22.3 \times 11.4}\) Solving for M gives, M = 208.2 g/mol (molecular weight of anthraquinone)

To find the empirical formula, we need to determine the lowest whole number ratio of C, H, and O atoms in the compound using their moles: Moles of Carbon: \(\frac{3.870}{12.01} = 0.3224\) moles Moles of Hydrogen: \(\frac{0.185}{1.01} = 0.1832\) moles Moles of Oxygen: \(\frac{0.745}{16.00} = 0.0466\) moles To find the lowest whole number ratio: \( \frac{0.3224}{0.0466} = 6.92 \approx 7\) \( \frac{0.1832}{0.0466} = 3.93 \approx 4\) \( \frac{0.0466}{0.0466} = 1\) Thus, the empirical formula is C14H8O2. To find the molecular formula, we will compare the molecular weight (M) to the empirical formula weight (EFW): EFW = 14(12.01) + 8(1.01) + 2(16.00) = 208.3 g/mol Since the molecular weight of anthraquinone (208.2 g/mol) is approximately equal to the empirical formula weight, the molecular formula of anthraquinone is the same as its empirical formula: C14H8O2.