Helicenes are extended fused polyaromatic hydrocarbons that have a helical or screw-shaped structure. a. A \(0.1450-\mathrm{g}\) 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 \(0.0938-\mathrm{g}\) sample of this helicene is dissolved in \(12.5 \mathrm{~g}\) of solvent to give a \(0.0175 \mathrm{M}\) solution, what is the molecular formula of this helicene? c. What is the balanced reaction for the combustion of this helicene?

Given the mass of \(\mathrm{CO_{2}}\) produced by combustion, we can find the moles of \(\mathrm{CO_{2}}\) and thus the moles of carbon in the helicene sample. The molar mass of \(\mathrm{CO_{2}}\) is 44.01 g/mol. Moles of \(\mathrm{CO_{2}}\) = mass of \(\mathrm{CO_{2}}\)/molar mass of \(\mathrm{CO_{2}}\) Moles of \(\mathrm{CO_{2}}\) = \(0.5063 \mathrm{g} \mathrm{CO_{2}}\) / \(44.01 \mathrm{g/mol}\) = 0.01150 mol \(\mathrm{CO_{2}}\) Since there is one carbon atom per \(\mathrm{CO_{2}}\) molecule, the moles of carbon in the helicene sample is also 0.01150 mol.

Once the moles of carbon are known, we can find the moles of hydrogen in the helicene sample. With the mass of the helicene sample, we can find the mass of hydrogen present. The molar mass of hydrogen is 1.01 g/mol. Mass of hydrogen = mass of helicene - mass of carbon Mass of hydrogen = \(0.1450 \mathrm{g}\) - (0.01150 mol carbon × \(12.01\mathrm{g/mol}\)) = 0.00692 g hydrogen Moles of hydrogen = mass of hydrogen / molar mass of hydrogen Moles of hydrogen = 0.00692 g / 1.01 g/mol = 0.00685 mol hydrogen

To find the empirical formula, we will take the ratio of moles of carbon to moles of hydrogen and simplify the ratio. Ratio of carbon to hydrogen: \(\frac{0.01150 \: \mathrm{mol \: carbon}}{0.00685 \: \mathrm{mol \: hydrogen}}\) = 1.678 Since this ratio is close to 2, we can round it to the closest whole number ratio of 2:1. Therefore, the empirical formula is C2H.

Now, we will use the given solution data and molar mass of the empirical formula to find the molecular formula of the helicene. Since the empirical formula is C2H, the molar mass of this formula is 25.03 g/mol (2 × 12.01 g/mol + 1 × 1.01 g/mol). We are given the concentration and mass of a solution of the helicene and solvent. First, we need to find the moles of the helicene in this solution: Moles of helicene in solution = concentration × volume Moles of helicene = \(0.0175\mathrm{M}\) × (12.5 g / 1.00 g/mL) = 0.0002187 mol helicene Now, we can find the molar mass of the molecular formula using the given mass of helicene sample, 0.0938 g: Molar mass of molecular formula = mass of helicene / moles of helicene Molar mass = 0.0938 g / 0.0002187 mol = 429 g/mol To find the molecular formula, we will divide the molar mass of the molecular formula by the molar mass of the empirical formula: \(\frac{429\mathrm{g/mol}}{25.03\mathrm{g/mol}}\) ≈ 17 Since the ratio is close to 17, we multiply the empirical formula (C2H) by 17: Molecular formula = (C2H) × 17 = C34H17

Now that we know the molecular formula, we can write the balanced combustion reaction for the helicene: C34H17 + 51.5 O2 ⟶ 34 CO2 + 17 H2O This is the balanced combustion reaction for this helicene.