Question

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at \(0.959\) atm and \(298 \mathrm{~K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at \(1.51\) atm and \(375 \mathrm{~K}\). This mixture has a density of \(1.391 \mathrm{~g} / \mathrm{L}\) and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.

Short answer

Upon solving the given problem, we find out that the molecular formula of the hydrocarbon is \(C_{2}H_{4}\).

Step-by-step solution

Calculate the moles of hydrocarbon using the initial pressure and temperature

We can start by calculating the number of moles of the hydrocarbon using the ideal gas law: \(PV = nRT\). Given, \(P = 0.959\) atm, \(T = 298\) K, \(R = 0.0821\) L atm/mol K, let's calculate the volume (V) and moles (n) of hydrocarbon gas. The equation becomes, \(n = \frac{PV}{RT}\).

Calculate the volume of CO2 and H2O mixture after combustion

We are given that the volume of the mixture is 4 times the volume of the hydrocarbon. Let \(V_{hydrocarbon} = V\), then \(V_{mixture} = 4V\).

Determine the moles of the mixture of CO2 and H2O

Now, let's determine the number of moles of the mixture of CO2 and H2O, using the density, volume, and molar masses of CO2 and H2O. The given density of the mixture is \(1.391\) g/L. Let \(x\) be the moles of CO2 and \(y\) be the moles of H2O. Then, \[ \frac{44x + 18y}{4V} = 1.391\]

Calculate the moles of carbon and hydrogen in the hydrocarbon

Now, let the hydrocarbon be \(C_{m}H_{n}\). Since all the carbon in the hydrocarbon forms CO2, we can say that \(m\) moles of carbon are present in \(x\) moles of CO2. Therefore, \(x = m\). Similarly, as all the hydrogen in the hydrocarbon forms H2O, we can say that \(n\) moles of hydrogen are present in \(y\) moles of H2O. Therefore, \(y = \frac{n}{2}\).

Determine the empirical formula of the hydrocarbon

Substituting \(x\) and \(y\) in the equation, we have \[ \frac{44m + 18(\frac{n}{2})}{4V} = 1.391\] Upon simplification, we get \[ \frac{44m + 9n}{V} = 1.391 × 4 = 5.564\] We also know that moles of hydrocarbon are equal to the sum of the moles of CO2 and H2O: \[n = m + \frac{n}{2}\] Solving these two equations, we get the values of \(m\) and \(n\).

Calculate the molecular mass of the hydrocarbon

Now that we have the values of \(m\) and \(n\), we can calculate the molecular mass of the hydrocarbon as: Molecular mass = (12 × m) + (1 × n)

Determine the molecular formula of the hydrocarbon

Finally, using the molecular mass and empirical formula from Steps 5 and 6, we can determine the molecular formula of the hydrocarbon.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates pressure (P), volume (V), temperature (T), and the amount of gas in moles (n) through the formula PV = nRT. In this formula, R represents the ideal gas constant, which has a value of 0.0821 L atm/mol K. The law assumes that gases behave ideally, which means that gas particles move randomly, are not attracted to or repelled by one another, and do not occupy any space.

In the exercise, we use the ideal gas law to calculate the number of moles of the hydrocarbon before combustion. By knowing the pressure, temperature, and using the gas constant, we are able to determine V and n. This step is crucial as it sets the stage for the remaining calculations necessary to find the hydrocarbon's molecular formula.

Combustion Analysis

Combustion analysis is a technique used to determine a compound's empirical formula by combusting it and analyzing the resulting products. For a hydrocarbon, when it combusts, it forms carbon dioxide (CO2) and water (H2O). Knowing the products and their amounts is crucial to back-calculate the ratio of carbon and hydrogen in the original compound.

This technique helps us determine the number of moles of carbon and hydrogen, by linking the amount of CO2 to carbon atoms and the amount of H2O to hydrogen atoms. In the given exercise, we examined the ratio of CO2 and H2O to find out the empirical formula of the hydrocarbon. The efficiency of this analysis depends on precise measurements and the understanding that the original elements in the hydrocarbon are entirely converted into the analyzed gases.

Empirical Formula

The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element present in the compound. It is derived from the percent composition or the mass of each element, which can be translated into moles. In the case of the exercise, the empirical formula is deduced by equating the moles of CO2 and H2O to the moles of carbon and hydrogen in the hydrocarbon.

To determine the empirical formula, it is often necessary to solve a set of equations that represent the conservation of mass, since the atoms are neither created nor destroyed during the reaction. The empirical formula may differ from the molecular formula but is always a whole-number multiple of it. It's the first step toward determining the molecular structure of the compound.

Hydrocarbon

A hydrocarbon is an organic compound consisting entirely of hydrogen and carbon atoms. Hydrocarbons are the simplest form of organic compounds and serve as the fundamental building blocks for more complex molecules. They are categorized into different types depending on their bond structures, such as alkanes, alkenes, and alkynes.

In this exercise, we are dealing with an unknown hydrocarbon undergoing combustion. This process is a reaction with oxygen that produces CO2 and H2O, from which we can deduce the original composition of the hydrocarbon. Determining its molecular formula involves finding the number of carbon and hydrogen atoms, which requires a clear understanding of their properties and how they react during combustion.