Question

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?

Short answer

The empirical formula of helicene is C_5H_8. The molecular formula of helicene is C_30H_48. The balanced combustion reaction for helicene is C_30H_48 + 45O2 → 30CO2 + 24H2O.

Step-by-step solution

Part a: Calculate the moles of CO2 and identify the empirical formula

1. To find the empirical formula, we need to first calculate the moles of carbon dioxide produced. The mass of CO2 is given as \(0.5063 g\). Use the molar mass of CO2 (\(44.01 g/mol\)) to find the moles. Moles of CO2 = mass of CO2 / molar mass of CO2 Moles of CO2 = \(0.5063 g / 44.01 g/mol = 0.01150 mol\) 2. The moles of carbon in the helicene are equal to the moles of CO2 produced, which is \(0.01150 mol\). Calculate the mass of carbon in helicene using the molar mass of carbon (\(12.01 g/mol\)) as follows: Mass of carbon = moles of carbon × molar mass of carbon Mass of carbon = \(0.01150 mol * 12.01 g/mol = 0.1380 g\) 3. The only remaining element in the polyaromatic hydrocarbon is hydrogen. Find the mass of hydrogen in helicene: Mass of hydrogen = mass of helicene - mass of carbon Mass of hydrogen = \(0.1450 g - 0.1380 g = 0.0070 g\) 4. Calculate moles of hydrogen using its molar mass (\(1.008 g/mol\)): Moles of hydrogen = mass of hydrogen / molar mass of hydrogen Moles of hydrogen = \(0.0070 g / 1.008 g/mol = 0.00695 mol\) 5. Find the mole ratio of carbon and hydrogen to find the empirical formula: Mole ratio C : H = Moles of C / Moles of H Mole ratio C : H = \(0.01150 mol / 0.00695 mol = 1.65\) 6. Since the mole ratio is approximately \(1.65\), we can write the empirical formula as C_5H_8.

Part b: Calculate the molecular weight and find the molecular formula

1. Calculate the molecular weight for the empirical formula (C_5H_8) using the molar mass of carbon and hydrogen: Molecular weight = \(5 * 12.01 g/mol + 8 * 1.008 g/mol = 68.04 g/mol\) 2. Calculate the moles of helicene using the given molarity and mass of the solvent: Moles of helicene = Molarity × Volume (in liters) Moles of helicene = \(0.0175 m * 12.5 g / 1 = 0.2188 mol\) 3. Calculate the mass of helicene using the moles of helicene from step 2 and the \(0.0938-g\) sample: Mass of helicene = moles of helicene × molecular weight of helicene Mass of helicene = \(0.2188 mol * molecular weight of helicene = 0.0938 g\) 4. Using the mass of helicene and the molecular weight of the empirical formula, find the molecular weight of the actual helicene: Molecular weight of helicene = mass of helicene / moles of helicene Molecular weight of helicene = \(0.0938 g / 0.2188 mol = 429g/mol\) 5. Divide the molecular weight of helicene by the molecular weight of the empirical formula to find the number of times the empirical formula is present in the molecular formula: \(429 g/mol / 68.04 g/mol = 6.30\) It is approximately 6 times, which gives the molecular formula: C_30H_48.

Part c: Write the balanced combustion reaction

1. Combustion of a hydrocarbon involves the reaction with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). Write the unbalanced reaction for the combustion of helicene using the molecular formula (C_30H_48): C_30H_48 + O2 → CO2 + H2O 2. Balance the atoms in the reaction by adding appropriate coefficients: C_30H_48 + 45O2 → 30CO2 + 24H2O The balanced combustion reaction for the given helicene is: C_30H_48 + 45O2 → 30CO2 + 24H2O.

Key concepts

Combustion Analysis

Combustion analysis is a lab technique used to determine the empirical formula of a compound, particularly organic compounds. The process involves burning a known mass of the substance in excess oxygen and analyzing the resultant carbon dioxide (CO2) and water (H2O). To comprehend combustion analysis, let's consider a sample burning to completion and turning into CO2 and H2O. By measuring the mass of CO2 and H2O produced, students can backtrack to find out the amount of carbon and hydrogen in the original sample. The mass of any residual ash can provide information about other elements present.

These masses are then converted into moles using the mole concept, which then leads to understanding the mole ratio of elements in the sample. This ratio provides the empirical formula—a simplified version of the molecular formula, which doesn't necessarily give information about the actual number of atoms but offers the lowest whole number ratio of elements in the compound.

Mole Concept

The mole concept is central to chemistry and is the bridge between the atomic world and the laboratory world. One mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12, which is approximately \(6.022 \times 10^{23}\) entities and is known as Avogadro's number.

To calculate the moles of a substance, students use the mass of the sample and the molar mass of the substance (the mass of one mole of that substance). In the helicene example, using the molar masses of carbon and hydrogen, students calculated the moles of carbon and hydrogen, setting the stage for identifying the empirical formula.

Stoichiometry

Stoichiometry refers to the quantitative relationships between reactants and products in a chemical reaction. It is based on the balanced chemical equations and the law of conservation of mass. In practice, stoichiometry involves calculations using the mole concept to find out how much of each substance is involved in a reaction.

In the case of helicene, knowing the moles of CO2 produced in the combustion allows one to calculate the moles and thereby the mass of carbon in the helicene. Stoichiometric calculations can also determine the amount of helicene required to produce a certain amount of product, indicating its molecular weight. Accurate stoichiometric calculations are essential for predicting yields in chemical reactions and are a cornerstone of practical chemistry.

Balanced Chemical Reactions

Balanced chemical reactions are the final piece of the puzzle when studying combustion analysis, the mole concept, and stoichiometry. A balanced reaction is one where the number of atoms for each element is the same on both the reactant and the product sides, in accordance with the law of conservation of mass. Balancing the reaction allows for stoichiometric calculations to be performed accurately.

In the example provided, the balanced combustion reaction for helicene displays a ratio that ensures the same number of carbon and hydrogen atoms on both sides of the equation. This step is critical for accurately determining the stoichiometry of the reaction and for calculating precise amounts of reactants and products in real-life applications.