Question

An element \(X\) forms an iodide \(\left(\mathrm{Xl}_{3}\right)\) 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 \(\mathrm{Xl}_{3}\) is treated, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.

Short answer

The atomic weight of the element X is approximately \(40.07 \text{ g/mol}\). The element X is Calcium (Ca) with a molar mass of \(40.08 \text{ g/mol}\).

Step-by-step solution

Calculate moles of Xl3 and XCl3 using given masses

First, we'll find the moles of Xl3 and XCl3 using the mass given in the problem. The mass of Xl3 is 0.5000 g, and the mass of XCl3 obtained from the reaction is 0.2360 g. To calculate moles, we need the molar mass (MM) of Xl3 and XCl3. We've given the molar mass of Iodine and Chlorine: 126.90 g/mol and 35.45 g/mol, respectively. Moles of Xl3 = mass of Xl3 / (MM of X + (3 * MM of I)) Moles of XCl3 = mass of XCl3 / (MM of X + (3 * MM of Cl))

Set up a proportion based on stoichiometry

From the balanced chemical equation: 2Xl3 + 3Cl2 -> 2XCl3 + 3I2 We can set up a proportion: moles of XCl3 / moles of Xl3 = 2 / 2 The number of moles of Xl3 should be equal to the number of moles of XCl3.

Solve for Molar Mass of X

Now we can solve for the molar mass (MM) of X: mm_XI3 = MM of X + (3 * MM of I) mm_XCl3 = MM of X + (3 * MM of Cl) moles_XI3 = 0.5000 g / mm_XI3 moles_XCl3 = 0.2360 g / mm_XCl3 Since moles_XI3 = moles_XCl3: 0.5000 g / mm_XI3 = 0.2360 g / mm_XCl3 Now rearrange and solve for MM of X: MM of X = ( 0.2360 g / 0.5000 g)*(mm_XI3 - 3*MM of I) + 3*MM of Cl Substitute MM of I and MM of Cl: MM of X = (0.2360 g / 0.5000 g) * (mm_XI3 - 3 * 126.90 g/mol) + 3 * 35.45 g/mol MM of X ≈ 40.07 g/mol

Identify the element X

Now that we have the approximate molar mass of element X, we can identify it. By looking at the periodic table, the element with a molar mass closest to 40.07 g/mol is: Element X = Calcium (Ca), with a molar mass of 40.08 g/mol. So the element X is Calcium.

Key concepts

Atomic Weight Calculation

Understanding atomic weight calculation is crucial for mastering the basics of chemistry. Atomic weight, also known as relative atomic mass, refers to the average mass of atoms of an element, measured in atomic mass units (amu). It reflects the weighted average of all the naturally occurring isotopes of an element.

To calculate atomic weight in exercises, you often deal with values that reflect the mass of a sample of a compound or element. By using the information provided in a balanced chemical equation, you can determine the mole ratio of the reactants and products. This mole ratio is key to relating masses of different compounds and finding out the atomic weight of an unknown element, as illustrated in the textbook solution.

For example, in the provided solution, we start with the known masses of two compounds, apply the mole concept, and use the balanced equation to establish a proportion between the compounds. From these proportions, we derive the atomic weight of the unknown element, following mathematical manipulations that involve the subtraction of the known weights of other elements in the compounds.

Reaction Stoichiometry

Reaction stoichiometry is the part of stoichiometry that deals with the proportions in which substances react and form products in a chemical reaction. It is grounded in the balanced chemical equation that shows the relationship between the quantities of reactants and products.

For instance, a balanced equation such as \(2 \mathrm{XI}_{3} + 3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3} + 3 \mathrm{I}_{2}\) indicates that two moles of \(\mathrm{XI}_{3}\) react with three moles of \(\mathrm{Cl}_{2}\) to produce two moles of \(\mathrm{XCl}_{3}\) and three moles of \(\mathrm{I}_{2}\). The numbers in front of the chemical formulas are known as stoichiometric coefficients, and they are essential for solving problems related to the amounts of reactants and products.

Using these coefficients, students can create a ratio and convert between masses of compounds using the molar mass concept. When dealing with a real-life problem, as in the exercise, the stoichiometric ratios lead to equations that help calculate unknown quantities such as atomic weights or the mass of products from a known mass of reactants.

Mole Concept

The mole concept is a fundamental principle in chemistry defining the quantity of substance. One mole of any substance contains the same number of entities (like atoms, molecules, or ions) as there are atoms in exactly 12 g of carbon-12, which is approximately \(6.022 \times 10^{23}\) entities, known as Avogadro's number.

The mole concept allows chemists to count particles by weighing them. A mole of any element has a mass in grams that is equal to its atomic weight. Therefore, when you know the mass and atomic or molecular weight, you can calculate the moles of a substance using the equation \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \).

This concept becomes especially powerful in stoichiometry because it enables the conversion of masses to moles, thus allowing to use balanced equations to relate the amounts in moles of each substance involved in a chemical reaction. Through the mole concept, the mass data can be translated into meaningful quantities leading to the discovery of unknowns such as the identity or mass of an element or compound, as done in solving the textbook exercise.