Question

An element 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 \(\mathrm{X}\) .

Short answer

The atomic mass of X is approximately 73.38 g/mol, and the element X is identified as Germanium (Ge) with an atomic mass of 72.63 g/mol.

Step-by-step solution

1. Identify the mass of chlorine in XCl2 and XCl4

First, let's find the mass of chlorine in 10.00 g of XCl2 and the mass of chlorine in 12.55 g of XCl4. We know the molar mass of Cl is approximately 35.45 g/mol. For XCl2, there are 2 moles of Cl for every 1 mole of XCl2. Therefore, the molar mass of XCl2 can be represented as: \[M_{XCl2} = M_X + 2 \times M_{Cl}\] For XCl4, there are 4 moles of Cl for every 1 mole of XCl4. Therefore, the molar mass of XCl4 can be represented as: \[M_{XCl4} = M_X + 4 \times M_{Cl}\]

2. Calculate the mass of chlorine in XCl2 and XCl4

For XCl2, let's calculate the mass of chlorine: \[mass_{Cl2} = \frac{mass_{XCl2}}{M_{XCl2}} \times 2 \times M_{Cl}\] \[mass_{Cl2} = \frac{10.00}{M_X + 2 \times 35.45} \times 2 \times 35.45\] For XCl4, let's calculate the mass of chlorine: \[mass_{Cl4} = \frac{mass_{XCl4}}{M_{XCl4}} \times 4 \times M_{Cl}\] \[mass_{Cl4} = \frac{12.55}{M_X + 4 \times 35.45} \times 4 \times 35.45\]

region#. Conservation of mass to find the atomic mass of X

As the reaction proceeds, the mass of X remains constant. Thus, the increase in the mass of chlorine in the reaction can be calculated to find the atomic mass of X: \[mass_{X} = mass_{XCl4} - mass_{XCl2} - (mass_{Cl4} - mass_{Cl2})\] Substitute the expressions for mass of chlorine in XCl2 and XCl4: \[M_X = \frac{2.55}{\frac{4 \times 35.45}{M_X + 4 \times 35.45} - \frac{2 \times 35.45}{M_X + 2 \times 35.45}}\]

4. Solve for atomic mass of X and identify the element

Rearrange and solve for M_X: \[M_X = \frac{2.55(M_X + 2 \times 35.45)(M_X + 4 \times 35.45)}{4 \times 35.45(M_X + 4 \times 35.45) - 2 \times 35.45(M_X + 2 \times 35.45)}\] After solving for M_X, we get M_X \(\approx\) 73.38 g/mol. Comparing this with known atomic masses of elements, we can conclude that the unknown element X is Germanium (Ge), which has an atomic mass of approximately 72.63 g/mol.

Key concepts

Chemical Compound

A chemical compound is a substance made of two or more different elements that are chemically bonded together. In this exercise, we're looking at two compounds: XCl₂ and XCl₄. These compounds form when element X combines with chlorine (Cl). The numbers in the formulas tell us how many chlorine atoms are present in each compound.

• XCl₂ has two chlorine atoms bonded to one atom of element X.
• XCl₄ has four chlorine atoms bonded to one atom of element X.

Understanding chemical compounds is crucial because it allows us to explore how different elements come together to create substances with new properties and uses.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's a fundamental concept that helps us calculate how much of each element is present in a compound.

To find the molar mass, you add together the atomic masses of all the atoms in a molecule. For example:

• The molar mass of XCl₂ is the mass of X plus twice the mass of chlorine (since there are two chlorine atoms).
• The molar mass of XCl₄ is the mass of X plus four times the mass of chlorine.

Understanding molar mass helps us move from theoretical chemistry to practical calculations, allowing us to find how much a substance weighs simply by knowing the amount of substance.

Element Identification

Element identification is the process of figuring out which chemical element is present based on given data like mass or atomic number. In our exercise, we calculate the atomic mass of element X using the mass of the compounds XCl₂ and XCl₄.

The steps involve finding the mass differences and using given concepts to solve for the unknown mass. Once we have the atomic mass of X, we compare it to known atomic masses of elements to find what element X is.

In this example, element X was calculated to have an atomic mass of approximately 73.38 g/mol. This information points to the element being Germanium (Ge), based on the periodic table.

Recognizing elements by their atomic mass is a crucial skill in chemistry and serves as a bridge between theoretical problems and real-world applications.

Stoichiometry

Stoichiometry is a branch of chemistry that involves using relationships between reactants and products in chemical reactions to perform calculations. It is driven by the law of conservation of mass, which states that mass is neither created nor destroyed.

In this problem, stoichiometry helps us maintain balance between the mass of chlorine and the mass calculated from XCl₂ to XCl₄.

• First, we identify the mass of chlorine in both compounds.
• Then, we use stoichiometry to track the mass of X throughout the reaction as it remains unchanged.
• Finally, we use this information to deduce the atomic mass of X.

Stoichiometry is vital in chemical calculations, ensuring that our chemical equations reflect the truth of the reactions occurring at the atomic level.