An element \(\mathrm{X}\) forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right)\). Treatment of \(10.00 \mathrm{~g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{~g} \mathrm{XCl}_{4}\). Calculate the atomic mass of \(\mathrm{X}\), and identify \(\mathrm{X}\).

First, let's determine the number of moles for each compound based on the given masses and their molar mass: \(n_{XCl_{2}} = \frac{mass_{XCl_{2}}}{Molar \: mass_{XCl_{2}}}\) and \(n_{XCl_{4}} = \frac{mass_{XCl_{4}}}{Molar \: mass_{XCl_{4}}}\)

We can then set up a proportion between the amounts of X in each of the compounds, based on the fact that the amount of X remains the same in both XCl₂ and XCl₄: \(\frac{n_{XCl_{2}}}{n_{XCl_{4}}} = \frac{n_{X} \times n_{Cl_{2}}}{n_{X} \times n_{Cl_{4}}}\)

Substituting the number of moles calculated in step 1 and the known atomic masses for Chlorine (35.45 g/mol), we can now solve for the atomic mass of X: \(M_{X} = \frac{mass_{XCl_{2}} \times (4 \times 35.45) - mass_{XCl_{4}} \times (2 \times 35.45)}{mass_{XCl_{2}} - mass_{XCl_{4}}}\)

After calculating the atomic mass of X from the above equation, we can identify the element by matching it with the closest atomic mass in the periodic table. Following these steps, let's find the atomic mass of X and identify the element. Step 1: Determine the number of moles of XCl₂ and XCl₄ \(n_{XCl_{2}} = \frac{10.00 \, g}{M_{X} + 2 \times 35.45}\) and \(n_{XCl_{4}} = \frac{12.55 \, g}{M_{X} + 4 \times 35.45}\) Step 2: Set up a proportion \(\frac{\frac{10.00}{M_{X} + 2 \times 35.45}}{\frac{12.55}{M_{X} + 4 \times 35.45}} = \frac{2}{4}\) Step 3: Calculate the atomic mass of X Cross-multiplication and solving for the atomic mass of X: \(M_{X} = \frac{10.00 \times (4 \times 35.45) - 12.55 \times (2 \times 35.45)}{10.00 - 12.55}\) \(M_{X} \approx 37.21 \, g/mol\) Step 4: Identify element X The element X is the one with atomic mass closest to 37.21 g/mol in the periodic table, which is Chlorine (Cl) with an atomic mass of 35.45 g/mol.