Question

The density of a gaseous compound was found to be \(0.943 \mathrm{~g} \cdot \mathrm{L}^{-1}\) at \(298 \mathrm{~K}\) and \(53.1 \mathrm{kPa}\). What is the molar mass of the compound?

Short answer

The molar mass of the compound is approximately 44.05 g/mol.

Step-by-step solution

Write down the ideal gas law

The ideal gas law relates the pressure (P), volume (V), temperature (T), and amount in moles (n) of a gas with the equation: PV = nRT, where R is the ideal gas constant.

Calculate moles of the gas using the density

The density given is 0.943 g/L, which means that there is 0.943 g of the gas in every liter. You can calculate the number of moles (n) in this 1 liter by using the density and the molar mass (M, which is what we're trying to find) with the formula: n = mass (m) / Molar mass (M)

Rearrange the ideal gas law to solve for the molar mass

Firstly, we need to express P in Pascals by multiplying the given kPa value by 1000. We rewrite the ideal gas law in terms of molar mass (M): M = mRT / PV. After calculating n from the previous step, we replace it in the ideal gas law and solve for M.

Insert the values and calculate the molar mass

Now insert the known values into the rearranged equation. Take care to use the correct units for R: 8.314 J/(mol K), P in Pascals, T in Kelvins, and mass (m) in grams. Solve for M to find the molar mass of the gas.

Perform the calculation

Using R = 8.314 J/(mol K), T = 298 K, P = 53.1 kPa which is 53100 Pa, and the given density to find the mass m = 0.943 g, we calculate the molar mass M.

Key concepts

Understanding the Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, crucial for understanding how gases behave under different conditions of temperature, pressure, and volume. It is represented by the formula: \begin{align*}PV = nRT\text{or}\P = \frac{nRT}{V}\text{or}\P\cdot V = \frac{mRT}{M}\text{or}\M = \frac{mRT}{P\cdot V}\text{or} = \frac{P\cdot V}{RT}\text{or}\V = \frac{nRT}{P}\text{or}\T = \frac{P\cdot V}{nR}\text{or}\R = \frac{P\cdot V}{nT}\text{or} = \frac{m}{M}\text{or}\m = n\cdot M\text{or}\M = \frac{m}{n}\text{or}\V = \frac{m}{\rho}\text{or}\rho = \frac{m}{V}\text{or}\V = \frac{P\cdot V}{RT}\text{or}\T = \frac{P\cdot V}{nR}\text{or}\R = \frac{P\cdot V}{nT}\text{or} = \frac{m}{M}\text{or}\m = n\cdot M\text{or}\M = \frac{m}{n}\text{or}\V = \frac{m}{\rho}\text{or}\rho = \frac{m}{V}\text{or}\P = \frac{\rho RT}{M}\text{or}\M = \frac{\rho RT}{P}\text{or}\R = \frac{\rho RT}{PM}\text{or}n = \frac{m}{M}\text{or}m = n\cdot M\text{or}M = \frac{m}{n}\text{or}V = \frac{m}{\rho}\text{or} \rho = \frac{m}{V} \text{or}P = \frac{\rho RT}{M} \text{or}M = \frac{\rho RT}{P}\text{or}R = \frac{\rho RT}{PM}\text{where:}\begin{itemize} &P&= Pressure (in Pascals, Pa) &V&= Volume (in Liters, L) &n&= Moles of gas &R&= Ideal gas constant (8.314 J/(mol\cdot K)) &T&= Temperature (in Kelvin, K) &M&= Molar mass (in grams per mole, g/mol) &m&= Mass (in grams, g)\end{itemize}By manipulating the equation, we can isolate and solve for any variable, provided we have the other values. In our textbook exercise, we are challenged to find the molar mass (M) of an unknown gas. The key to utilizing the ideal gas law in solving for molar mass lies in understanding that the amount (n) of gas and its mass (m) are directly related through the molar mass (M), letting us use the density (\(\rho\)) as a link between them. Understanding the principles of pressure conversions is also critical in ensuring proper use of the law, as the pressure must be in Pascals (Pa) to match the ideal gas constant units.

Exploring Gas Density

Gas density (\begin{align*}\rho\end{align*}) is defined as the mass of the gas per unit volume, expressed as grams per liter (g/L). It is an important property used to relate a gas's physical characteristics to the Ideal Gas Law. In the context of our problem, knowing the gas density allows us to directly calculate the mass of the gas within a known volume, which then can be used to find the molar mass (M). The importance of understanding gas density is mostly because it offers a bridge from measurable quantities, such as mass and volume, to molar quantities that play into gas laws and chemical computations.To interpret density in terms of the Ideal Gas Law, we use the formula: \begin{align*}\rho = \frac{m}{V}\text{or}\M = \frac{\rho RT}{P}\text{or}\P = \frac{\rho RT}{M}\text{or}V = \frac{m}{\rho}\text{or}n = \frac{m}{M}\text{or}\rho = \frac{m}{V}\text{or}\V = \frac{m}{\rho}\end{align*}where \(\rho\) is density, \(m\) is mass, and \(V\) is volume. By understanding the relationship between the mass of the gas and its volume via density, students can then relate this to the amount of substance (n) and its molar mass (M), which is essential for calculations involving moles and molecular weight.

The Role of Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. It is founded on the laws of conservation of mass and the concept of the mole. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction, making it a critical tool in laboratory and industrial chemical processes.

In relation to the Ideal Gas Law and gas density, stoichiometry can be used to determine the molar mass of unknown gases, something we're directly addressing in our exercise. By working out the mass-to-mole ratio from the given density and applying the Ideal Gas Law, we can calculate the molar mass. This process essentially reveals the stoichiometric relationships between mass, moles, volume, and gas constants.

For instance, once the density and molar mass are known, stoichiometry can extend this knowledge to determine how much gas would react with other substances or be produced in a chemical reaction. It's vital for students to understand that stoichiometry isn't limited to solid substances and is equally significant when working with gaseous compounds, as shown by our exercise.