Question

An unidentified metal M reacts with an unidentified halogen \(\mathrm{X}\) to form a compound \(\mathrm{MX}_{2}\). When heated, the compound decomposes by the reaction: $$ 2 \mathrm{MX}_{2}(s) \longrightarrow 2 \mathrm{MX}(s)+\mathrm{X}_{2}(g) $$ When \(1.12 \mathrm{~g}\) of \(\mathrm{MX}_{2}\) is heated, \(0.720 \mathrm{~g}\) of \(\mathrm{MX}\) is obtained, along with \(56.0 \mathrm{~mL}\) of \(\mathrm{X}_{2}\) gas. Under the conditions used, \(1.00 \mathrm{~mol}\) of the gas has a volume of \(22.41 \mathrm{~L}\). (a) What is the atomic weight and identity of the halogen \(\mathrm{X} ?\) (b) What is the atomic weight and identity of the metal M?

Short answer

The halogen X is Bromine (Br), and the metal M is Copper (Cu).

Step-by-step solution

Determine the moles of X₂ gas

First, calculate the moles of \(\mathrm{X}_2\) gas produced. Use the volume of gas provided (56.0 mL) and convert it to liters (0.0560 L). Then use the molar volume to find moles:\[\text{Moles of } \mathrm{X}_2 = \frac{0.0560 \text{ L}}{22.41 \text{ L/mol}} = 0.00250 \text{ moles}\]

Relate moles of X₂ to moles of MX₂

From the balanced equation \(2 \mathrm{MX}_{2}(s) \rightarrow 2 \mathrm{MX}(s) + \mathrm{X}_2(g)\), 1 mole of \(\mathrm{X}_2\) is produced from 2 moles of \(\mathrm{MX}_2\). Therefore:\[\text{Moles of } \mathrm{MX}_2 = 2 \times 0.00250 = 0.00500 \text{ moles}\]

Calculate the molar mass of MX₂

Now, determine the molar mass of \(\mathrm{MX}_2\) using its mass (1.12 g) and the moles calculated:\[\text{Molar mass of } \mathrm{MX}_2 = \frac{1.12 \text{ g}}{0.00500 \text{ moles}} = 224 \text{ g/mol}\]

Calculate the mass and molar mass of X₂ in MX₂

From Step 1, 0.00250 moles of \(\mathrm{X}_2\) are formed, so the mass of the halogen in \(\mathrm{MX}_2\) is:\[\text{Mass of } \mathrm{X} = \text{mass of } \mathrm{X}_2 = \text{Moles} \times \text{Molar mass} = 0.00250 \text{ moles} \times 2X \text{ g/mol}\]Using the known total mass and the mass of \(\mathrm{MX}\) (0.720 g), the mass of \(\mathrm{X}\) in \(\mathrm{MX}_2\) must be 1.12 g - 0.720 g = 0.400 g.

Calculate the atomic weight of the halogen X

The molar mass of \(\mathrm{X}_2\) accounts for 0.400 g in 0.00250 moles:\[\text{Molar mass of } \mathrm{X}_2 = \frac{0.400 \text{ g}}{0.00250 \text{ moles}} = 160 \text{ g/mol}\]Thus, the atomic weight \(X\) is 80 g/mol. This corresponds to Bromine (Br).

Find the atomic weight of the metal M

Now use the molar mass of \(\mathrm{MX}_2\) to find the atomic weight of \(M\):\[\text{Molar mass of } \mathrm{MX}_2 = M + 2 \times 80 = 224 \text{ g/mol}\]Thus,\[M = 224 - 160 = 64 \text{ g/mol}\]This corresponds to Copper (Cu).

Key concepts

Molar Mass Calculation

Molar mass is a key concept in chemistry, representing the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Calculating the molar mass is essential when you need to determine how much of a substance you have in a given chemical reaction. When solving problems related to molar mass, you'll often start with known mass and moles values. From these values, you can use the formula:\[\text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}}\]For example, in the case of the compound \( \mathrm{MX}_2 \), the mass was given as 1.12 g, and we calculated the moles as 0.00500 mol. Using the formula above, we determined its molar mass to be 224 g/mol.To perform such calculations accurately:

• Always convert measurements to the appropriate units (e.g., volume to liters, if needed).
• Use the balanced chemical equation to relate the moles of different components involved.
• Double-check your calculations for consistency with given data.

Atomic Weight Determination

Atomic weight is another important concept, essential for identifying individual elements in chemical compounds. It's the weighted average of the masses of the isotopes of an element as they occur naturally.In chemical calculations such as this one, atomic weight is used to identify elements like the halogen \( \mathrm{X} \) and metal \( \mathrm{M} \) by comparing calculated weights to known atomic weights of elements. To find the atomic weight of an element in a chemical reaction:

• First, determine the mass associated with the element, as we did for \( \mathrm{X}_2 \) with a calculated mass of 0.400 g.
• Then calculate the molar mass using the relation \( \text{Molar mass} = \frac{\text{Mass}}{\text{Moles}} \).

In Step 5 of our example, we discovered the molar mass of the halogen \( \mathrm{X}_2 \) is 160 g/mol, leading us to an atomic weight of 80 g/mol for \( \mathrm{X} \). This value matched that of Bromine, identifying \( \mathrm{X} \) as \( \text{Br} \). Similarly, the atomic weight for the unidentified metal \( \mathrm{M} \) was calculated as 64 g/mol, corresponding to Copper, \( \text{Cu} \).

Balanced Chemical Equation

A balanced chemical equation is crucial for understanding how different substances interact and transform in a chemical reaction. It shows the relationships between the reactants and products in terms of their respective moles.For example, the equation\[2 \mathrm{MX}_{2}(s) \rightarrow 2 \mathrm{MX}(s) + \mathrm{X}_2(g)\]ensures that the number of atoms of each element is the same on both sides. This conservation of mass principle is fundamental in chemistry.Balanced equations help in several ways:

• They allow accurate calculations of reactants and products. Knowing that 2 moles of \( \mathrm{MX}_2 \) produce 1 mole of \( \mathrm{X}_2 \), you can determine the proportions needed for the reaction.
• Balancing ensures that no atoms are lost, indicating complete reactions.
• They provide a foundation for stoichiometry, which further helps in calculating other quantities like concentration and yield.

Understanding balanced chemical equations not only aids in solving textbook problems but also in experimental scenarios where precise reactant-product conversions are required.