Integrative Problems. These problems require the integration of multiple concepts to find the solutions. Helicenes are extended fused polyaromatic hydrocarbons that have a helical or screw-shaped structure. a. \(A 0.1450\) -g sample of solid helicene is combusted in air to give \(0.5063 \mathrm{g} \mathrm{CO}_{2}\). What is the empirical formula of this helicene? b. If a \(0.0938-g\) sample of this helicene is dissolved in \(12.5 \mathrm{g}\) of solvent to give a 0.0175 \(M\) solution, what is the molecular formula of this helicene? c. What is the balanced reaction for the combustion of this helicene?

Given: Mass of helicene = 0.1450 g Mass of CO₂ = 0.5063 g First, find the moles of carbon in CO₂: Molar mass of carbon = 12 g/mol Molar mass of CO₂ = 44 g/mol Moles of carbon in CO₂ = (Moles of CO₂) x (Moles of carbon per mole of CO2) Moles of CO₂ = (0.5063 g) / (44 g/mol) = 0.0115 mol Moles of carbon in helicene = 0.0115 mol Now, find the mass of hydrogen in helicene: Mass of helicene = Mass of carbon + Mass of hydrogen Mass of hydrogen = Mass of helicene - Mass of carbon Mass of hydrogen = 0.1450 g - (0.0115 mol x 12 g/mol) = 0.010 g Next, find the moles of hydrogen in helicene: Molar mass of hydrogen = 1 g/mol Moles of hydrogen in helicene = (0.010 g) / (1 g/mol) = 0.010 mol Finally, find the empirical formula: C:H ratio = (0.0115 mol) / (0.010 mol) ≈ 1.15 Empirical formula = CH (Since the ratio is approximately 1:1)

Given: Mass of helicene = 0.0938 g Volume of solvent = 12.5 g Molarity of the solution = 0.0175 M First, find the moles of helicene in the solution: Moles of helicene = Molarity of the solution x Volume in liters (Note: the volume of the solvent is given in grams, but we need to convert it into liters. For this problem, we will assume that the solvent's density is approximately equal to 1 g/mL, so we can assume that the volume in liters is equal to 0.0125 L.) Moles of helicene = (0.0175 M) x (0.0125 L) = 0.00021875 mol Now, find the molecular weight of helicene: Molecular weight of helicene = Mass of helicene / Moles of helicene Molecular weight of helicene = (0.0938 g) / (0.00021875 mol) ≈ 429 g/mol Finally, find the molecular formula: The molecular weight of CH (our empirical formula) = 12 + 1 = 13 g/mol Number of CH units in the molecular formula = (Molecular weight of helicene) / (Molecular weight of CH) Number of CH units = (429 g/mol) / (13 g/mol) ≈ 33 Molecular formula = C₃₃H₃₃

Combustion reactions involve the reaction between the fuel (in this case, helicene) and oxygen (O₂), producing CO₂ and H₂O. The general form of a combustion reaction is: Fuel + O₂ → CO₂ + H₂O Now, we just need to balance the reaction, taking into account that we have 33 carbon atoms and 33 hydrogen atoms in the molecular formula of helicene. C₃₃H₃₃ + O₂ → 33CO₂ + 33/2 H₂O To balance the oxygen atoms, we need to multiply the number of O₂ molecules on the reactant side by 33/2: C₃₃H₃₃ + 33/2 O₂ → 33CO₂ + 33/2 H₂O The balanced reaction for the combustion of helicene is: C₃₃H₃₃ + 33/2 O₂ → 33CO₂ + 33/2 H₂O