Problem 160 - Chemistry An Atoms First Approach

Question

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?

Step-by-Step Solution

Answer

The empirical formula of the helicene is CH. The molecular formula is C₃₃H₃₃. The balanced reaction for the combustion of helicene is C₃₃H₃₃ + 33/2 O₂ → 33CO₂ + 33/2 H₂O.

Step 1: Find the empirical formula based on the mass of CO2 produced

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)

Step 2: Find the molecular formula based on the given molarity of the solution

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

Step 3: Find the balanced reaction for the combustion of helicene

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

Key Concepts

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a molecule. It's different from the empirical formula, which only gives the simplest whole-number ratio of atoms. To determine the molecular formula, we need to find both the empirical formula and the molecular weight of the compound. In the exercise, the empirical formula of helicene was determined to be CH because the calculated carbon to hydrogen ratio was about 1:1. This is just a theoretical ratio. Once we find the empirical formula, the molecular formula is determined using the molar mass. In this problem, the molar mass of helicene was calculated based on the provided molarity of the solution, which led us to find that helicene has a molecular formula of C₃₃H₃₃. This means that the molecule contains 33 carbon atoms and 33 hydrogen atoms.

• Empirical vs. Molecular Formula: Empirical shows the simplest ratio, molecular shows actual counts of atoms.
• Importance: Knowing the molecular formula can give insights into the structure and properties of the compound, especially in complex hydrocarbons like helicene.

Understanding the difference and relationship between these formulas is crucial for analyzing compounds in chemistry.

Combustion Reaction

A combustion reaction is a chemical process where a substance reacts with oxygen to release energy in the form of light and heat. This type of reaction is common for hydrocarbon compounds. In our exercise, helicene undergoes combustion, and we can describe this reaction simply as helicene reacting with oxygen to produce carbon dioxide and water. The generic formula for a combustion reaction is: Fuel + O₂ → CO₂ + H₂O. Here, the fuel is helicene (C₃₃H₃₃). The task was to write a balanced chemical equation for the combustion of this compound. A balanced equation ensures that the number of each type of atom is the same on both sides of the equation. In this case, the balanced combustion equation was established as: C₃₃H₃₃ + \[\frac{33}{2}\] O₂ → 33CO₂ + \[\frac{33}{2}\] H₂O.

• Energy Release: Combustion reactions are exothermic; they release energy.
• Balancing Equations: A necessary step to accurately represent the reaction.
• Reactants and Products: For complete combustion, the products are always CO₂ and H₂O.

Mastering combustion reactions is vital for understanding energy production and chemical transformations.

Fused Polyaromatic Hydrocarbons

Fused polyaromatic hydrocarbons (PAHs) such as helicenes are compounds with multiple aromatic rings fused together. These structures are usually flat and have unique chemical and physical properties, highly characteristic due to their large conjugated systems. Helices, being a type of fused PAH, feature a helical or screw-shaped configuration.

• Aromatic Structure: These structures increase stability through resonance.
• Properties: They are known for being highly stable and often possess unique electronic properties owing to their structure.
• Applications: Fused PAHs are significant in materials science, having applications in areas like organic semiconductors.

Fused PAHs like helicenes have been the subject of intense study due to their interesting structural and electronic characteristics. Grasping these hydrocarbons' basics is essential for those interested in organic chemistry and materials science. Their size and shape affect how they interact with light and other molecules, contributing to their functionality in advanced applications, making them a fascinating topic for chemists and material scientists alike.