Problem 33 - Chemistry: An Atoms-Focused Approach

Question

A flask of ammonia is connected to a flask of an unknown acid HX by a \(1.00 \mathrm{m}\) glass tube. As the two gases diffuse down the tube, a white ring of \(\mathrm{NH}_{4} \mathrm{X}\) forms \(68.5 \mathrm{cm}\) from the ammonia flask. Identify element X.

Step-by-Step Solution

Answer

Answer: The unknown element X is Bromine (Br), making the compound NH4X ammonium bromide.

Step 1: Understand Graham's Law of Effusion

Graham's law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this can be expressed as: \(r_{1}/r_{2} = \sqrt{M_{2}/M_{1}}\) Where \(r_{1}\) and \(r_{2}\) are the rates of diffusion for gas 1 and gas 2, and \(M_{1}\) and \(M_{2}\) are their respective molar masses.

Step 2: Determine the rate of diffusion for ammonia and HX

The gases travel different distances to form NH4X. The rate of diffusion is the distance traveled divided by the time it takes for the gas to travel that distance. Let the rate of diffusion for ammonia be \(r_{\text{NH3}}\) and for the unknown acid HX be \(r_{\text{HX}}\). The distances traveled are given as \(68.5\,\text{cm}\) for ammonia and \((100 - 68.5)\,\text{cm} = 31.5\,\text{cm}\) for HX by using the length of the glass tube, which is \(1.00\,\text{m}\) or \(100\,\text{cm}\). // Note that we will use cm instead of m for distances in this problem, but it doesn't matter as long as we are consistent with units.

Step 3: Use Graham's Law to create an equation for the molar masses of ammonia and HX

Applying Graham's Law, we get: \(\frac{r_{\text{NH3}}}{r_{\text{HX}}} = \sqrt{\frac{M_{\text{HX}}}{M_{\mathrm{NH}_3}}}\) We can plug in the distances traveled to find the relation for the rates of the gases: \(\frac{68.5}{31.5} = \sqrt{\frac{M_{\text{HX}}}{M_{\mathrm{NH}_3}}}\)

Step 4: Calculate the molar mass of HX

Now, we can solve this equation for the molar mass of HX. First, square both sides of the equation: \(\left(\frac{68.5}{31.5}\right)^2 = \frac{M_{\text{HX}}}{M_{\mathrm{NH}_3}}\) The molar mass of ammonia is \(M_{\mathrm{NH}_3} = 14.01 + 3(1.01) = 17.03\,\text{g/mol}\). Replace \(M_{\mathrm{NH}_3}\) in the equation: \(\left(\frac{68.5}{31.5}\right)^2 = \frac{M_{\text{HX}}}{17.03}\) Calculate the molar mass of HX: \(M_{\text{HX}} = 17.03 \times \left(\frac{68.5}{31.5}\right)^2 = 83.96\,\text{g/mol}\)

Step 5: Identify element X

Now subtract the mass of hydrogen from the calculated molar mass of HX to get the mass of element X: \(M_{\text{X}} = M_{\text{HX}} - M_{\text{H}} = 83.96 - 1.01 = 82.95\,\text{g/mol}\) Based on the periodic table, element X is most likely Bromine, which has a molar mass closest to 82.95 g/mol. Therefore, the compound NH4X is ammonium bromide, and the unknown acid HX is hydrobromic acid (HBr).

Key Concepts

Understanding Molar Mass Calculation

Molar mass is a crucial concept in chemistry, particularly when dealing with gases. It is the mass of a given substance divided by the amount of substance, typically expressed in grams per mole (g/mol). When calculating molar mass, it is essential to consider each element within a compound and their corresponding atomic masses found on the periodic table.
For example, in ammonia (NH₃), we compute the molar mass by adding the mass of one nitrogen atom (14.01 g/mol) to three hydrogen atoms (3 x 1.01 g/mol), totaling 17.03 g/mol.
Understanding this process is vital in experiments involving gases and is used in our exercise to determine the molar mass of an unknown acid through Graham’s Law of Effusion.

Fundamentals of Chemical Diffusion

Chemical diffusion involves the movement of particles from a region of higher concentration to a lower concentration. It is a natural process resulting from the random thermal motion of particles. Diffusion rates vary among different substances and depend significantly on their molar masses.
Bigger and heavier molecules tend to diffuse more slowly than lighter ones. In our exercise, ammonia (a lighter molecule) travels a longer distance compared to the unknown HX (a heavier gas). This difference in distances traveled is critical for applying Graham’s Law, which relates the rates of diffusion to the molar masses of the gases.

• This demonstrates the inverse relationship between diffusion rate and molar mass.
• Knowing this helps explain the behavior of different gases when they are mixed or allowed to interact, such as during the formation of a white ring in the exercise.

Identifying Ammonium Bromide Formation

In this exercise, the challenge was to determine the identity of the unknown acid, HX, which reacted with ammonia to form the compound \(\text{NH}_4\text{X}\). Following the calculation of the molar mass using the distances traveled and Graham’s Law, the next logical step is to identify the specific identity of the element X in \(\text{NH}_4\text{X}\).
The calculated molar mass for HX is approximately 83.96 g/mol. Since \(\text{HX}\) is composed of hydrogen (with a known molar mass of about 1.01 g/mol), the mass of X can be found by subtracting the mass of hydrogen from the mass of HX, leaving approximately 82.95 g/mol for the element X.
By consulting the periodic table, we ascertain that bromine (Br) is the element with a molar mass closest to 82.95 g/mol, hence confirming that X is bromine. As a result, NH₄X is identified as ammonium bromide (NH₄Br), with the unknown acid being hydrobromic acid (HBr).

• This process demonstrates how calculations intersect with chemical knowledge to identify unknown substances.