Question

In a certain electrolysis experiment, \(1.44 \mathrm{~g}\) of Ag were deposited in one cell (containing an aqueous \(\mathrm{AgNO}_{3}\) solution), while \(0.120 \mathrm{~g}\) of an unknown metal \(\mathrm{X}\) was deposited in another cell (containing an aqueous \(\mathrm{XCl}_{3}\) solution) in series with the \(\mathrm{AgNO}_{3}\) cell. Calculate the molar mass of \(\mathrm{X}\).

Short answer

The molar mass of metal X is approximately \( 26.97 \, \text{g/mol} \).

Step-by-step solution

Determine Moles of Silver

First, we need to find the moles of Ag deposited. The molar mass of Ag is \( 107.87 \, \text{g/mol} \). Use the formula:\[ \text{Moles of Ag} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.44 \, \text{g}}{107.87 \, \text{g/mol}} \approx 0.01335 \, \text{mol} \]

Relate Moles of Electrons to Silver

The reaction at the Ag electrode is:\[ \text{Ag}^+ + \text{e}^- \rightarrow \text{Ag} \]So, 1 mole of Ag corresponds to 1 mole of electrons. Therefore, 0.01335 moles of electrons were also transferred.

Determine Moles of Metal X Using Electrons

For the reaction involving metal X:\[ \text{XCl}_3 + 3\text{e}^- \rightarrow \text{X} + 3\text{Cl}^- \]3 moles of electrons deposit 1 mole of X. The number of moles of X deposited can be calculated as:\[ \text{Moles of X} = \frac{\text{Moles of electrons}}{3} = \frac{0.01335}{3} \approx 0.00445 \, \text{mol} \]

Calculate Molar Mass of Metal X

Use the mass of the deposited X and the moles calculated:\[ \text{Molar Mass of X} = \frac{0.120 \, \text{g}}{0.00445 \, \text{mol}} \approx 26.97 \, \text{g/mol} \]

Key concepts

Molar Mass Calculation

Calculating molar mass is a fundamental step in understanding chemical equations and reactions. In the context of electrolysis, such calculations are crucial to determine the properties of unknown substances. The molar mass is defined as the mass of a given substance (in grams) divided by the amount of substance (in moles).

To find the molar mass, you'll use the formula:

• Molar mass, \( M = \frac{\text{mass of substance (g)}}{\text{moles of substance (mol)}} \)

Knowing the mass of the deposited substance and the moles of it lets us calculate the molar mass, thereby identifying the unknown metal in electrolysis experiments. Making these calculations accurately allows chemists to deduce important characteristics of the element, such as its position on the periodic table.

Chemical Reactions

Chemical reactions are processes where substances, the reactants, are transformed into different substances, known as products. In the context of electrolysis, different chemical reactions occur at the electrodes when an electric current is applied.

In a typical electrolysis setup, metallic ions gain electrons (a process known as reduction) to become metal atoms. For example, in the exercise given, we look at the reaction for silver:

• \( \text{Ag}^+ + \text{e}^- \rightarrow \text{Ag} \)

Here, silver ions gain electrons to form solid silver. Similarly, for the unknown metal \( \text{X} \) in the \( \text{XCl}_3 \) solution, the reaction occurs as:

• \( \text{XCl}_3 + 3\text{e}^- \rightarrow \text{X} + 3\text{Cl}^- \)

This illustrates how the electrons transferred during electrolysis aid the transformation of ions into their elemental form, helping us understand the underlying reaction mechanisms.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It involves using balanced equations to determine the ratios in which substances react or are produced.

In the electrolysis context, stoichiometry allows us to relate the amount of electrons transferred to the number of moles of substance deposited. This is crucial when calculating the moles of metal \( \text{X} \) deposited, as three moles of electrons are required for each mole of \( \text{X} \) in the reaction:

\( \text{XCl}_3 + 3\text{e}^- \rightarrow \text{X} + 3\text{Cl}^- \)

Understanding stoichiometry is key to solving many chemistry problems, as it helps you interpret the coefficients in balanced chemical equations and determine not just theoretical yields, but also the practical implications like the consumption of materials needed in reactions.