Problem 49 - Chemistry and Chemical Reactivity

Question

A gas whose molar mass you wish to know effuses through an opening at a rate one third as fast as that of helium gas. What is the molar mass of the unknown gas?

Step-by-Step Solution

Answer

The molar mass of the unknown gas is 36 g/mol.

Step 1: Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) and \( r_2 \) are the effusion rates of gas 1 and gas 2, and \( M_1 \) and \( M_2 \) are their respective molar masses.

Step 2: Identify the Known and Unknown Quantities

We know the rate of effusion for helium and the gas being studied (unknown gas). The rate for the unknown gas is one third that of helium (\( \frac{1}{3} \)), and we know the molar mass of helium, \( M_{\text{He}} = 4 \text{ g/mol} \). We need to find the molar mass of the unknown gas, \( M_2 \).

Step 3: Substitute Values into Graham's Law

Using the expression \( \frac{r_{\text{unknown}}}{r_{\text{He}}} = \frac{1}{3} \) and \( M_1 = 4 \text{ g/mol} \), substitute into Graham's Law: \[ \frac{1}{3} = \sqrt{\frac{M_{\text{unknown}}}{4}} \]

Step 4: Solve the Equation

First, square both sides to remove the square root:\[ \left( \frac{1}{3} \right)^2 = \frac{M_{\text{unknown}}}{4} \]\[ \frac{1}{9} = \frac{M_{\text{unknown}}}{4} \]Now, solve for \( M_{\text{unknown}} \) by multiplying both sides by 4:\[ M_{\text{unknown}} = 4 \times \frac{1}{9} \]\[ M_{\text{unknown}} = \frac{4}{9} \times 9 \]\[ M_{\text{unknown}} = 36 \text{ g/mol} \]

Key Concepts

Molar Mass Calculation

The molar mass of a gas is an essential concept to understand when studying gas behavior, as it directly influences various physical properties, like density and effusion rate. To calculate molar mass, one needs to know the mass of a given quantity of substance, typically a mole, and then apply the known values in formulas that connect them with observable properties. In our specific solved problem, molar mass plays a pivotal role when determining the characteristics of an unknown gas through its effusion rate compared to a reference gas like helium. Calculating unknown molar mass involves using formulas derived from concepts like Avogadro's principle and applying them through methods such as dimensional analysis, leading into the application of Graham’s Law of Effusion to find the desired quantity.

Effusion Rate

Effusion is the process where gas particles escape through a small hole into a vacuum. The rate at which this occurs is partially dictated by the particle's velocity and inversely by its square root of the molar mass. Graham's Law of Effusion mathematically establishes this relationship by showing that lighter gases will effuse more rapidly than heavier ones. In the provided exercise, helium is used as a reference gas - known for its low molar mass of 4 g/mol. By using its effusion rate as a benchmark, it is possible to compare it to the unknown gas whose effusion rate is a third that of helium. Using such comparisons, we can utilize Graham’s Law to determine the molar mass of the unknown gas using the known scenarios and apply them to the formula: \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \], making it a significant aspect of the calculation process.

Unknown Gas Identification

Identifying an unknown gas using its effusion rate requires a systematic approach. By comparing it to a known gas, like helium, whose characteristics are already thoroughly documented, it becomes simpler to refine the search. Through Graham's Law, it is possible to derive the unknown’s molar mass using the relationship between effusion rates and molar masses of gases. In the exercise, once the calculated molar mass was determined to be 36 g/mol, it narrows down potential gas candidates fitting this description. This method is a practical technique often used in laboratory environments to rapidly identify unknown gases by contrasting them with known standards. Such a process underscores the importance of understanding gas laws and their applications for effective and quick gas identification.