Problem 50 - Chemistry and Chemical Reactivity

Question

A A sample of uranium fluoride is found to effuse at the rate of \(17.7 \mathrm{mg} / \mathrm{h} .\) Under comparable conditions, gaseous I \(_{2}\) effuses at the rate of \(15.0 \mathrm{mg} / \mathrm{h} .\) What is the molar mass of the uranium fluoride? (Hint: Rates must be converted to units of moles per time.)

Step-by-Step Solution

Answer

The molar mass of uranium fluoride is approximately 182.6 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, 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 two gases, and \( M_1 \) and \( M_2 \) are their respective molar masses.

Step 2: Determine Rates of Effusion

We have the effusion rates of uranium fluoride \( (UF_x) \) and iodine gas \( (I_2) \). These rates are given as 17.7 mg/h and 15.0 mg/h, respectively. For comparison, we first convert these rates to moles per hour.

Step 3: Convert Mass Rates to Mole Rates

Use the molar mass of iodine \( (I_2) \) to convert its mass rate to moles:\[\text{Molar mass of } I_2 = 253.8 \text{ g/mol} \]The rate of effusion for \( I_2 \) in moles is:\[\frac{{15.0 \text{ mg/h}}}{{253.8 \text{ g/mol}}} = \frac{{15.0 \times 10^{-3} \text{ g/h}}}{{253.8 \text{ g/mol}}} = 5.91 \times 10^{-5} \text{ mol/h}\]

Step 4: Apply Graham's Law

Plugging the calculated moles per hour into Graham's Law for uranium fluoride and iodine:\[\frac{{r_{UF_x}}}{{r_{I_2}}} = \sqrt{\frac{{M_{I_2}}}{{M_{UF_x}}}} \]Substitute the known rates and solve for \( M_{UF_x} \):\[\frac{{r_{UF_x}}}{{5.91 \times 10^{-5}}} = \sqrt{\frac{{253.8}}{{M_{UF_x}}}} \]

Step 5: Solve for Molar Mass of Uranium Fluoride

First, calculate the molar rate for \( UF_x \):\[\text{Assume a similar conversion is done for } UF_x: \frac{{17.7\times 10^{-3}}}{{M_{UF_x}}} \text{ mol/h}\] Rearrange Graham's Law to solve for \( M_{UF_x} \):\[\left(\frac{{r_{UF_x}}}{{r_{I_2}}}\right)^2 = \frac{{M_{I_2}}}{{M_{UF_x}}}\]Simplify:\[\left(\frac{{17.7}}{{15.0}}\right)^2 = \frac{{253.8}}{{M_{UF_x}}}\]Calculate to find the molar mass:\( \left(1.18\right)^2 \approx 1.39 \ M_{UF_x} \approx \frac{{253.8}}{{1.39}} \approx 182.6 \text{ g/mol} \) .

Key Concepts

Effusion rates

Effusion is a process where gas particles escape through a tiny hole into a vacuum. The rate of this escape process is what we call the effusion rate. According to Graham's Law of Effusion, the effusion rate of a gas is inversely proportional to the square root of its molar mass. This means lighter gases escape faster than heavier gases under the same conditions.
Graham's Law can be mathematically represented by the equation:

• \[ \frac{{r_1}}{{r_2}} = \sqrt{\frac{{M_2}}{{M_1}}} \]

Where:

• \( r_1 \) and \( r_2 \) are the rates of effusion of two gases
• \( M_1 \) and \( M_2 \) are the molar masses of these gases

By understanding this relationship, we can determine unknown variables, like the molar mass of a gas, by comparing its effusion rate to that of a known gas.

Molar mass calculation

The molar mass, often expressed in grams per mole (g/mol), is a measure of the mass of a given substance divided by the amount of substance. Molar mass is a critical concept in chemistry as it helps in converting between grams and moles, allowing for calculation of quantities needed or produced in chemical reactions.
In the context of the exercise, calculating the molar mass of a gas from its effusion rate involves comparing it to another gas with a known molar mass. Using the effusion rates along with Graham's Law, we can derive the molar mass of an unknown gas.

This exercise has us calculating the molar mass of uranium fluoride by comparing its effusion rate to that of iodine (\(I_2\)) using the given effusion rates. By rearranging Graham's Law and solving the resulting equation, we find the unknown molar mass.

Uranium fluoride molar mass

Uranium fluoride acts as an example in this exercise to illustrate the application of Graham's Law for calculating molar mass. Here, the challenge is to determine the molar mass of this compound with only its effusion rate and the effusion rate of iodine gas provided.
The process involves converting the effusion rates from mass per hour to moles per hour, as this allows for direct application of Graham's Law. After converting rate units, we can use the equation:

• \[ \frac{{r_{UF_x}}}{{r_{I_2}}} = \sqrt{\frac{{M_{I_2}}}{{M_{UF_x}}}} \]

By calculating the known rates and manipulating the formula, we can isolate \(M_{UF_x}\), allowing us to solve for the molar mass of uranium fluoride. This exercise specifically determines the molar mass of uranium fluoride as approximately 182.6 g/mol.

Chemical problem solving

Problem-solving in chemistry, especially with concepts like effusion and molar mass, often involves a systematic approach. First, it’s important to identify and understand the key concepts and laws that apply—in this case, Graham’s Law of Effusion.

Next, translate given information into consistent units or forms that work with the known formula. This often means converting things like mass rates into moles per time and setting up equations accurately.

Then, carry out the necessary mathematical manipulations to solve for the unknown. This includes isolating variables and simplifying ratios as needed. Finally, verify the solution by checking for consistency and feasibility of the calculated results.

• Identify knowns and unknowns.
• Convert units if necessary.
• Apply relevant chemical laws or formulas.
• Check and confirm your result.

These steps not only ensure correct problem-solving but also reinforce understanding of the underlying chemical concepts.