Problem 98 - Chemistry The Central Science In Si Units

Question

An element \(X\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\) The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{XI}_{3}\) is treated with chlorine, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element \(\mathrm{X}\). (b) Identify the element X.

Step-by-Step Solution

Answer

The atomic weight of element X is approximately 52.08, identifying X as Chromium (Cr).

Step 1: Set up the equation for Element X

Let the atomic weight of element X be \( m \). The molar mass of \( \text{XI}_3 \) is \( m + 3\times 126.90 \) (since each iodine atom has a molar mass of \( 126.90 \)). The molar mass of \( \text{XCl}_3 \) is \( m + 3\times 35.45 \).

Step 2: Use given masses to set up an equation

According to the reaction, \( 2 \text{XI}_3 \) produces \( 2 \text{XCl}_3 \). Therefore, 0.5000 g of \( \text{XI}_3 \) will produce \( 0.2360 \) g of \( \text{XCl}_3 \). Using these relationships, we know:\[\frac{0.5000}{m + 3\times 126.90} = \frac{0.2360}{m + 3\times 35.45}\]

Step 3: Simplify and Solve for m

Cross-multiply to get the equation:\[0.5000 \times (m + 3 \times 35.45) = 0.2360 \times (m + 3 \times 126.90)\]This simplifies to:\[0.5m + 53.175 \times 3 = 0.236m + 89.1828 \times 3\]Rearranging gives:\[0.264m = 89.1828 \times 3 - 53.175 \times 3\]Solving for \( m \):\[m = \frac{89.1828 \times 3 - 53.175 \times 3}{0.264} \approx 52.08\]

Step 4: Identify the Element X

Based on the calculated atomic weight of \( 52.08 \), we check the periodic table to find that the element with an atomic weight close to this is Chromium (Cr) with an atomic weight of approximately 51.996.

Key Concepts

Chemistry Problem Solving

Approaching a chemistry problem can often seem daunting, but by methodically breaking it down, we can simplify the process significantly. Let's dive into solving this problem about element X and its compounds.One of the first steps in chemistry problem solving is to properly set up the reaction's chemical equation. Here, we're dealing with the conversion of iodide (\(\mathrm{XI}_{3}\)) to chloride (\(\mathrm{XCl}_{3}\)) using chlorine gas.We start by letting\(m\)represent the atomic weight of the element X. From the periodic table, we obtain the molar masses of iodine and chlorine, which are critical in calculating the molar masses of \(\mathrm{XI}_{3}\) and \(\mathrm{XCl}_{3}\):

• Each iodine atom has a molar mass of \(126.90\), so for \(\mathrm{XI}_{3}\), it's \(m + 3 \times 126.90\)
• Each chlorine atom has a molar mass of \(35.45\), thus making \(\mathrm{XCl}_{3}\) equal to \(m + 3 \times 35.45\)

This creates a sense of direction, allowing us to use the mass data given (0.5000 g of \(\mathrm{XI}_{3}\) producing 0.2360 g of \(\mathrm{XCl}_{3}\)) to set up an equation and solve for the atomic weight of X.By equating the ratios as expressed in the problem, we can cross-multiply and rearrange to solve for \(m\). The calculation shows that \(m\) is approximately 52.08, bringing us a step closer to identifying the element as chromium.

Iodide and Chloride Reactions

Understandably, entering the world of iodide and chloride reactions requires some groundwork. These reactions are essential, especially for the conversion between different halide states of an element. In our case, iodine is replaced by chlorine in the compound \(\mathrm{XI}_{3}\) to form \(\mathrm{XCl}_{3}\).The chemical equation for this conversion is:\[2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_2\]When you heat \(\mathrm{XI}_{3}\) in a chlorine stream, the chlorine atoms displace the iodine atoms. During this process:

• Iodine molecules (\(\mathrm{I}_2\)) are released as a byproduct.
• The stoichiometry of the equation shows that two molecules of \(\mathrm{XI}_{3}\) yield two molecules of \(\mathrm{XCl}_{3}\).

The efficiency of this reaction is captured through the masses of the reactants and products given. Understanding these stoichiometric relationships helps in calculating key variables like the atomic weight of the element involved.Overall, reactions such as these illustrate the beauty of chemistry in transforming substances, based on predictable scientific laws.

Periodic Table Identification

Once we have calculated the atomic weight of element X, the next step is to identify it using the periodic table. This is where the notion of the periodic table becomes invaluable. The periodic table is organized such that each element is listed with its atomic number and average atomic weight. When we calculated the atomic weight of X to be 52.08, we turned to the periodic table to find the element that most closely matches this value. In this case:

• We find that Chromium (Cr) has an atomic weight of approximately 51.996.
• Given that this is very close to our calculated atomic weight, it strongly suggests that the element X is indeed chromium.

This exercise not only helps in identifying the unknown element but also underscores the importance of the periodic table as a tool for chemists - making it simpler for them to draw connections between empirical data and theoretical knowledge. The periodic table not only serves as a reference guide but also provides insights into the chemical properties and similarities of elements, making chemistry problem-solving more intuitive.