Anthraquinone contains only carbon, hydrogen, and oxygen. When \(4.80 \space \mathrm{mg}\) anthraquinone is burned, \(14.2 \space \mathrm{mg}\space \mathrm{CO}_{2}\) and \(1.65 \space \mathrm{mg} \space \mathrm{H}_{2} \mathrm{O}\) 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 g camphor. Determine the empirical and molecular formulas of anthraquinone.

From the given data, when 4.80 mg anthraquinone is burned, 14.2 mg CO2 and 1.65 mg H2O are produced. We can calculate the masses of carbon and hydrogen in anthraquinone as follows: Mass of carbon in CO2 = Mass of CO2 x (Mass of C in CO2 / Mass of CO2) Mass of hydrogen in H2O = Mass of H2O x (Mass of H in H2O / Mass of H2O) The atomic masses of C, H, and O are approximately 12, 1, and 16, respectively. The molecular mass of CO2 = 12 + 2(16) = 44 The molecular mass of H2O = 2(1) + 16 = 18

Now, we can determine the mass of oxygen in anthraquinone as follows: Mass of anthraquinone = Mass of carbon + Mass of hydrogen + Mass of oxygen => Mass of oxygen = Mass of anthraquinone - Mass of carbon - Mass of hydrogen

Now we will calculate the moles of each element: Moles of carbon = Mass of carbon / Atomic mass of carbon Moles of hydrogen = Mass of hydrogen / Atomic mass of hydrogen Moles of oxygen = Mass of oxygen / Atomic mass of oxygen

We will now determine the empirical formula for anthraquinone by finding the lowest whole number ratio of moles of carbon, hydrogen, and oxygen. Divide each mole value by the smallest mole value obtained in step 3. If necessary, multiply the resulting ratios by the smallest whole number to obtain whole number ratios. The empirical formula is written using these whole number ratios as subscripts.

To find the molecular formula of anthraquinone, we will first find the molecular weight using the freezing point depression data provided. We know that: Freezing point depression = Kf x molality x i Where Kf is the molal freezing point depression constant for camphor, molality is the molality of the solution and i is the van't Hoff factor (equal to 1 for molecular solutes). Given the freezing point depression of camphor is 22.3 °C when 1.32 g anthraquinone is dissolved in 11.4 g camphor, we can calculate the molality of the solution. Molality of solution = (Weight of anthraquinone in g) / (Weight of camphor in g) The molal freezing point depression constant for camphor (Kf) is 40.0 ºC/m. Now, we can find the molecular weight of anthraquinone. Molecular weight of anthraquinone = (Freezing point depression / (Kf x molality)) x 1000 Using the molecular weight of anthraquinone and the empirical formula, we can now determine the molecular formula by finding the whole number ratio of the empirical formula mass to the molecular weight. The molecular formula is found by multiplying the empirical formula by this whole number ratio.