Question

Consider a sample of a hydrocarbon at 0.959 atm and 298 K. Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 K. This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon and name it.

Short answer

The molecular formula of the hydrocarbon is C₃H₈ and its name is propane.

Step-by-step solution

Determine the partial pressures of CO₂ and H₂O in the mixture.

Since the volume of the CO₂ and H₂O mixture is four times the volume of the hydrocarbon, we can calculate the pressure of the hydrocarbon before combustion by using the ideal gas law. Let's denote the initial volume of the hydrocarbon as V and the final volume of the mixture as 4V. From the ideal gas law, we have: P₁V = P₂(4V), where P₁ is the initial pressure of the hydrocarbon (0.959 atm), and P₂ is the final total pressure of the mixture (1.51 atm). Solving for P₂, we get: P₂ = 4 × 0.959 = 3.836 atm Now, the final pressure of the CO₂ and H₂O mixture (1.51 atm) is the sum of their partial pressures, so: Partial pressure of CO₂ + Partial pressure of H₂O = 1.51 atm

Use the ideal gas law to determine the moles of CO₂ and H₂O, and from that, the empirical formula of the hydrocarbon.

We are given the density of the collected gas mixture (1.391 g/L). We can calculate the mass (m) of the gaseous mixture using the density formula and volume of the mixture as: Mass (m) = Density × Volume = 1.391 g/L * 4V Now we can use the mass and the molecular weights of CO₂ and H₂O to determine the moles of CO₂ and H₂O in the mixture: nCO₂ = mCO₂ / MCO₂ nH₂O = mH₂O / MH₂O where nCO₂ and nH₂O are the moles of CO₂ and H₂O, mCO₂ and mH₂O are their respective masses, and MCO₂ and MH₂O are their molecular weights (44.01 g/mol and 18.02 g/mol, respectively). To find the moles of carbon (C) and hydrogen (H) in the hydrocarbon, we can use the ratio of moles of CO₂ and H₂O to moles of C and H: nC = nCO₂ nH = 2 * nH₂O

Determine the molecular formula from the empirical formula.

Using the moles of C and H, we can represent the empirical formula of the hydrocarbon as CxHy. Dividing both sides by the smallest value (either x or y), we will find the simplest whole number ratio of x and y to obtain the molecular formula. Now, we can name the hydrocarbon using the IUPAC system, considering its carbon and hydrogen atoms in the molecular formula.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle to understand the behavior of gases. It combines several gas laws into one formula: \( PV = nRT \), where:

• \( P \) represents pressure,
• \( V \) is volume,
• \( n \) is the number of moles,
• \( R \) is the universal gas constant, and
• \( T \) is temperature in Kelvin.

This formula helps us relate the physical properties of gases. In this exercise, it is used to compare the pressures and volumes before and after combustion. By setting up the equation with the initial conditions and solving for the final conditions, we can understand how the volume of gases changes after a chemical reaction. The Ideal Gas Law is key for finding how much a gas can expand or contract under different conditions.

Molecular Formula Determination

To determine the molecular formula of a hydrocarbon, we use data from experiments or given conditions after a chemical reaction. The first step is to ascertain the empirical formula or the simplest ratio of atoms in the molecule.
From the empirical formula, we can deduce the molecular formula by considering the molecular mass determined experimentally and comparing it with the molar mass of the empirical formula. The molecular formula is usually expressed as \(C_xH_y\), illustrating the number of carbon \(C\) and hydrogen \(H\) atoms.Subsequently, if the molar mass of the compound is known, we compare it with the empirical formula mass. This helps adjust the subscripts \(x\) and \(y\) to reflect the actual number of atoms in a molecule, providing the molecular formula. This requires solving for the totals using the actual molecular weight.

Partial Pressures

The concept of partial pressures is crucial when dealing with gas mixtures. According to Dalton's Law of Partial Pressures, each gas in a mixture contributes its own pressure as if it were the only gas present. It is calculated as the fraction of the total pressure exerted by that particular gas. In this exercise:

• The total pressure of the gaseous mixture of CO₂ (carbon dioxide) and H₂O (water vapor) is given.
• To find the individual partial pressures, we determine how each gas contributes to the total.

We use this information to find out how many moles are present, and ultimately, how many molecules are involved in the chemical reaction. Understanding partial pressures is important for analyzing results in reactions involving multiple gas products.

Empirical Formula

The empirical formula represents the simplest whole number ratio of atoms within a compound. It is derived from the amount of each element present. For example, if a compound contains carbon and hydrogen, we calculate the number of carbon and hydrogen atoms involved per molecule.
In a combustion reaction, a hydrocarbon's empirical formula can be deduced from the amount of CO₂ and H₂O generated. We compare their moles and establish ratios to express the formula as \(C_xH_y\), where \(x\) and \(y\) are whole numbers.
The empirical formula gives us a basic understanding of the composition and ratio but does not convey the exact number of atoms, which is highlighted in the molecular formula. This distinction is essential for accurately naming and understanding the properties of chemical compounds.