Problem 181 - Chemistry An Atoms First Approach

Question

An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$\mathbf{M}+\mathbf{X}_{2} \longrightarrow \mathbf{M X}_{3}$$ A \(0.105-\mathrm{g}\) sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of \(54.47 \%\) X by mass. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\), and what is the correct name for \(\mathrm{MX}_{3} ?\) Starting with 1.00 g each of \(M\) and \(X_{2}\), what mass of \(M X_{3}\) can be prepared?

Step-by-Step Solution

Answer

We have identified M as Rubidium (Rb) and X as Chlorine (Cl), making the compound Rubidium Chloride (RbCl₃). Starting with 1.00 g each of Rb and Cl₂, we can prepare 2.20 g of RbCl₃.

Step 1: Calculate the molar mass of X₂

To calculate the molar mass of X₂, we'll use the mass of X₂ and the number of molecules available: Molar mass of X₂ = (Mass of X₂) / (Number of molecules of X₂ / Avogadro's number) Avogadro's number = 6.022 × 10²³ Number of moles of X₂ = (8.92 × 10²⁰ molecules) / (6.022 × 10²³) = 0.001482 moles Molar mass of X₂ = 0.105 g / 0.001482 moles = 70.81 g/mol

Step 2: Calculate the atomic mass of X

Since X₂ is a diatomic molecule, the atomic mass of X can be calculated by dividing the molar mass of X₂ by 2: Atomic mass of X = Molar mass of X₂ / 2 Atomic mass of X = 70.81 g/mol / 2 = 35.40 g/mol

Step 3: Calculate the molar mass of MX₃ and identify M and X

We know that the mass of X in MX₃ is 54.47%. Therefore, the mass of M in MX₃ will be 100% - 54.47% = 45.53%. Let molar mass of MX₃ = A g/mol Mass of X in 1 mol of MX₃ = 3 × atomic mass of X = 3 × 35.40 g/mol = 106.20 g/mol Mass of M in 1 mol MX₃ = mass of M in MX₃ = A × 45.53% Now, mass of X in MX₃ = A × 54.47% 106.20 g/mol = A × 54.47% We can solve for A: A = 106.20 g/mol / 54.47% = 194.95 g/mol Now that we have the molar mass of MX₃, we can determine the mass of M in 1 mol of MX₃ and identify M and X: Mass of M = A × 45.53% = 194.95 g/mol × 45.53% = 88.75 g/mol We can now identify M and X: M ≈ 89 g/mol (Rb, Rubidium) X ≈ 35.5 g/mol (Cl, Chlorine) Thus, we have found the elements in MX₃ to be Rubidium and Chlorine, so the compound is Rubidium Chloride (RbCl₃).

Step 4: Calculate the mass of MX₃ using 1.00g each of M and X₂

We start with 1.00 g of M (Rb) and 1.00 g of X₂ (Cl₂). Since M and X₂ react in a 1:1 ratio, we can find which reactant is the limiting reactant. Number of moles of M = 1.00 g / (atomic mass of Rb) = 1.00 g / 88.75 g/mol = 0.01127 moles Number of moles of X₂ = 1.00 g / (molar mass of Cl₂) = 1.00 g / 70.81 g/mol = 0.01412 moles Since the ratio of M:X₂ in the reaction is 1:1, Rb is the limiting reactant. Now, for each mole of Rb, one mole of RbCl₃ is formed. So, 0.01127 moles of Rb reacts completely forming 0.01127 moles of RbCl３. Let's calculate the mass of RbCl₃ formed: Mass of RbCl₃ = 0.01127 moles × (molar mass of RbCl₃) = 0.01127 moles × 194.95 g/mol = 2.20 g Hence, we can prepare 2.20 g of RbCl₃ using 1.00 g each of Rb and Cl₂.

Key Concepts

Chemical Equations

Chemical equations represent the substances involved in a chemical reaction. They show the reactants being transformed into products, and are crucial for understanding chemical processes. In the case of ionic compounds, such as in our exercise, the equation must balance, meaning the number of each type of atom must be equal on both sides.

For example, in the unbalanced equation given, \( \mathbf{M} + \mathbf{X}_2 \longrightarrow \mathbf{MX}_3 \), each molecule of \( X_2 \) reacts with atoms of \( M \) to form \( MX_3 \).

• The equation includes diatomic chlorine, \( \mathbf{Cl}_2 \), which naturally exists as pairs of atoms.
• You need to properly balance the number of molecules and atoms to make sure there is an equal number on each side.

The balanced equation becomes crucial when calculating the amounts of substances needed or produced, such as when determining the product yield based on available reactants.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol), and is essential for converting quantities in chemistry. Calculating molar mass involves adding up the atomic masses of all atoms in a molecule.

In our exercise, the molar mass of \( X_2 \) was calculated using:

• Mass of \( X_2 \) \( = 0.105\) g and number of molecules \( = 8.92 \times 10^{20} \).
• Use Avogadro’s number \( (6.022 \times 10^{23}) \) to find moles: \( 0.105 \text{ g} / (8.92 \times 10^{20} / 6.022 \times 10^{23}) \).
• This gave a molar mass of \( 70.81\) g/mol for \( X_2 \).

The atomic mass of \( X \) was half of \( X_2 \), since \( X_2 \) is diatomic. Therefore, \( 70.81\text{ g/mol} / 2 = 35.40 \text{ g/mol}\). Understanding these calculations helps in identifying unknown elements in compounds.

Limiting Reactants

In chemical reactions, the limiting reactant is the substance that is completely consumed first, limiting the amount of products formed. Identifying the limiting reactant is key to predicting reaction yields.

In the exercise:

• Two reactants were available: 1.00 g each of \( M \) and \( X_2 \).
• You compute moles for each: \( \text{moles of } M = \frac{1.00 \text{ g}}{88.75 \text{ g/mol}} = 0.01127 \text{ moles} \) and \( \text{moles of } X_2 = \frac{1.00 \text{ g}}{70.81 \text{ g/mol}} = 0.01412 \text{ moles} \).
• The reaction ratio is 1:1, making \( M \) the limiting reactant because it produces less product considering mole-to-mole comparison.

The moles of limiting reactant determine the amount of \( MX_3 \) that can be produced. Calculating this involves multiplying the moles of limiting reactant by the formula weight of the desired product, resulting in the mass of the product formed.