Problem 88 - Chemistry The Central Science

Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 \(\mathrm{L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{L} ;\) in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Step-by-Step Solution

Answer

The molar mass of the unknown gas is approximately 87.5 g/mol.

Step 1: Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass, mathematically represented as: \(rate_1 / rate_2 = √(M_2 / M_1)\) Here, \(rate_1\) and \(rate_2\) are the rates of effusion of gas 1 and gas 2, and \(M_1\) and \(M_2\) are the molar masses of gas 1 and gas 2, respectively.

Step 2: Calculate the rates of effusion

We are given the time required for 1.0 L of the unknown gas and \(\mathrm{O}_2\) gas to effuse. We can use these timings to calculate the rates of effusion for both gases. The rate of effusion is given by the volume of gas effused per unit time. For the unknown gas: \(rate_1 = V / t_1 = (1.0 \,L) / (105 \,s)\) For the \(\mathrm{O}_2\) gas: \(rate_2 = V / t_2 = (1.0 \,L) / (31 \,s)\)

Step 3: Apply Graham's Law of Effusion

Now, we will use the calculated rates of effusion and Graham's Law of Effusion to find the molar mass of the unknown gas. \((1.0 \,L / 105 \,s) / (1.0 \,L / 31 \,s) = √(M_2 / M_1)\) Where \(M_1 = 32 \,g/mol\) (molar mass of \(\mathrm{O}_2\)), and \(M_2\) is the molar mass of the unknown gas.

Step 4: Solve for the molar mass of the unknown gas

We can rearrange the equation and solve for \(M_2\), the molar mass of the unknown gas. \(√(M_2 / 32) = (31 / 105)\) Squaring both sides, we get: \(M_2 / 32 = (31 / 105)^2\) Now, solve for \(M_2\): \(M_2 = 32 * (31 / 105)^2\) Calculate the value of \(M_2\): \(M_2 ≈ 87.5 \,g/mol\) The molar mass of the unknown gas is approximately 87.5 g/mol.

Key Concepts

Effusion of Gases

Effusion is the process where gas molecules pass through a tiny opening from a container into a vacuum. It's one of the intriguing aspects of gas behavior, showing how gases move and spread out over time. When it comes to effusion, light gases zip through small holes faster than heavier ones. Imagine two balloons, one filled with helium and the other with carbon dioxide, each with a small puncture. The helium balloon would deflate quicker due to its lighter gas molecules effusing at a higher rate.

Molar Mass Calculation

Molar mass is essentially the weight of one mole of any chemical substance, be it an element or a compound. It's commonly expressed in grams per mole (g/mol) and is calculated by summing up the atomic weights of each atom in a molecule. For instance, water (H2O) has a molar mass of approximately 18 g/mol, as there are two hydrogen atoms (each about 1 g/mol) and one oxygen atom (about 16 g/mol). Knowing the molar mass is pivotal, as it allows chemists to measure out exact amounts of material for reactions, simply by weighing.

Gas Laws

The behavior of gases can be described by several fundamental laws collectively known as the gas laws. Boyle's Law, for example, tells us that pressure and volume are inversely related, whereas Charles's Law states that volume and temperature are directly proportional, given a constant pressure. Avogadro's Law links volume and the amount of gas, assuming constant temperature and pressure. Collectively, these relationships allow us to predict how a gas will behave under different conditions and are encapsulated in the Ideal Gas Law: PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

Chemical Kinetics

Chemical kinetics focuses on the speed at which chemical reactions occur and the factors affecting those speeds. Reaction rates can be influenced by various factors such as temperature, concentration, surface area, and the presence of catalysts. As reactions proceed, the concentrations of reactants decrease while those of products increase, impacting the rate of the reaction. Kinetics is a key part of understanding not just how quickly reactions happen, but also the mechanisms by which they proceed, which is vital for everything from creating new materials to understanding biological processes.