Question

A compound of formula \(\mathrm{XCl}_{3}\) reacts with aqueous \(\mathrm{AgNO}_{3}\) to yield solid AgCl according to the following equation: $$ \mathrm{XCl}_{3}(a q)+3 \mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{X}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3 \mathrm{AgCl}(s) $$ When a solution containing \(0.634 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) was allowed to react with an excess of aqueous \(\mathrm{AgNO}_{3}, 1.68 \mathrm{~g}\) of solid \(\mathrm{AgCl}\) was formed. What is the identity of the atom \(\mathrm{X} ?\)

Short answer

The atom \(\mathrm{X}\) is manganese (Mn).

Step-by-step solution

Determine Moles of AgCl Formed

To find the moles of AgCl formed, use its molar mass. The molar mass of AgCl is \(143.32 \, \text{g/mol}\). Use the mass of AgCl produced:\[m_{\text{AgCl}} = 1.68 \, \text{g}\]Calculate moles:\[\text{Moles of AgCl} = \frac{1.68 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.01172 \, \text{mol}\]

Relate Moles of AgCl to Moles of XCl3

The chemical equation shows a 1:3 reaction ratio between \(\mathrm{XCl}_3\) and \(\mathrm{AgCl}\). Therefore, 1 mole of \(\mathrm{XCl}_3\) forms 3 moles of \(\mathrm{AgCl}\):\[\text{Moles of } \mathrm{XCl}_3 = \frac{0.01172 \, \text{mol AgCl}}{3} = 0.003907 \, \text{mol}\]

Calculate Molar Mass of XCl3

Given the mass of \(\mathrm{XCl}_3\) is \(0.634 \, \text{g}\), use the moles of \(\mathrm{XCl}_3\) calculated to find its molar mass:\[M_{\mathrm{XCl}_3} = \frac{0.634 \, \text{g}}{0.003907 \, \text{mol}} \approx 162.23 \, \text{g/mol}\]

Find Molar Mass of X By Subtracting Chlorine Mass

The molar mass of \(\mathrm{XCl}_3\) is the sum of \(\mathrm{X}\) and 3 chlorines. The molar mass of chlorine \(\mathrm{Cl}\) is \(35.45 \, \text{g/mol}\). Calculate the molar mass of \(\mathrm{X}\):\[M_{\mathrm{X}} = 162.23 \, \text{g/mol} - 3 \times 35.45 \, \text{g/mol} = 55.88 \, \text{g/mol}\]

Determine the Element X

The element with a molar mass closest to \(55.88 \, \text{g/mol}\) is manganese (Mn), which has a molar mass of \(54.94 \, \text{g/mol}\). Therefore, \(\mathrm{X}\) is manganese, \(\mathrm{Mn}\).

Key concepts

Stoichiometry

Understanding stoichiometry is essential to analyzing chemical reactions. Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. In simple terms, it focuses on the measurement and calculation of these relationships, often using a balanced chemical equation as a foundation.
For example, the balanced chemical equation given in this exercise is: \[\mathrm{XCl}_3(aq) + 3 \mathrm{AgNO}_3(aq) \rightarrow \mathrm{X}(\mathrm{NO}_3)_3(aq) + 3 \mathrm{AgCl}(s)\] This equation shows a 1:3 ratio between \(\mathrm{XCl}_3\) and \(\mathrm{AgCl}\). Therefore, for every mole of \(\mathrm{XCl}_3\), three moles of \(\mathrm{AgCl}\) are produced.
This reactive relationship is crucial for calculating the number of moles in a reaction, which is a pivotal step to find the unknown element \(\mathrm{X}\) in the compound. Understanding the mole ratio allows us to link the mass of the product formed with the moles of the initial reactant compound, which in this case helps to determine the moles of \(\mathrm{XCl}_3\) used from the moles of \(\mathrm{AgCl}\) formed during the reaction. By mastering stoichiometry, we can predict the quantities of substances consumed and produced in a reaction, making it a powerful tool in chemistry.

Molar Mass Calculation

Molar mass calculation involves determining the mass of one mole of a substance, often expressed in grams per mole (g/mol). For this exercise, understanding molar mass is pivotal to find the identity of element \(\mathrm{X}\).
You start by calculating the molar mass of \(\mathrm{AgCl}\) from the mass of the sample produced in the reaction:

• Molar mass of \(\mathrm{AgCl}\) is known to be 143.32 g/mol.
• Using the measured mass of 1.68 g \(\mathrm{AgCl}\), calculate moles using the formula: \[\text{Moles of AgCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.68 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.01172 \, \text{mol}\]

Next, use stoichiometry to find moles of \(\mathrm{XCl}_3\): \[\text{Moles of } \mathrm{XCl}_3 = \frac{0.01172 \, \text{mol AgCl}}{3} = 0.003907 \, \text{mol}\] Finally, calculate the molar mass of \(\mathrm{XCl}_3\) given its mass is 0.634 g:

• Use the previously calculated moles: \[M_{\mathrm{XCl}_3} = \frac{0.634 \, \text{g}}{0.003907 \, \text{mol}} \approx 162.23 \, \text{g/mol}\]

Accurate molar mass calculations allow determination of unknown elements, like \(\mathrm{X}\), by subtracting the known masses (such as chlorine atoms). This helps in deducing the identity of elements based on their approximate molar mass.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two soluble salts in aqueous solution react to form an insoluble solid called a precipitate. The concept is crucial in this exercise as it involves AgCl, a common precipitate in chemical reactions.
The key features of a precipitation reaction include:

• The formation of a solid from aqueous solutions.
• The driving force is the formation of this insoluble product.

In the given chemical equation:\[\mathrm{XCl}_3(aq) + 3\mathrm{AgNO}_3(aq) \rightarrow \mathrm{X}(\mathrm{NO}_3)_3(aq) + 3\mathrm{AgCl}(s)\] \(\mathrm{AgCl}\) is the precipitate formed, appearing as a solid that settles out of solution. Understanding these reactions is important for predicting if a reaction will occur and how to separate or identify products. The precipitation of \(\mathrm{AgCl}\) from the reaction confirms the completion of the chemical reaction, and allows for calculations of the reaction yield and further identification of the unknown element in the compound.