Problem 121 - General Chemistry

Question

A water-soluble compound of gold and chlorine is treated with silver nitrate to convert the chlorine completely to silver chloride, \(\mathrm{AgCl}\). In an experiment, \(328 \mathrm{mg}\) of the compound gave \(464 \mathrm{mg}\) of silver chloride. Calculate the percentage of \(\mathrm{Cl}\) in the compound. What is its empirical formula? 4.122 A solution of scandium chloride was treated with silver nitrate. The chlorine in the scandium compound was converted to silver chloride, AgCl. A \(58.9\) -mg sample of scandium chloride gave \(167.4 \mathrm{mg}\) of silver chloride. What are the mass percentages of \(\mathrm{Sc}\) and \(\mathrm{Cl}\) in scandium chloride? What is its empirical formula?

Step-by-Step Solution

Answer

Chlorine is 70.33% and scandium is 29.67% in ScCl3; empirical formula is ScCl3.

Step 1: Determine the Mass of Chlorine in Silver Chloride

Silver chloride, \( \text{AgCl} \), has a molar mass of \( 143.32 \text{ g/mol} \) (\( \text{Ag} = 107.87 \text{ g/mol} \), \( \text{Cl} = 35.45 \text{ g/mol} \)). From \( 167.4 \text{ mg} \) of \( \text{AgCl} \), we first find how much of that mass is due to chlorine: \( \frac{35.45}{143.32} \times 167.4 \text{ mg} = 41.42 \text{ mg of Cl} \).

Step 2: Calculate Mass Percentage of Chlorine

The mass percentage of chlorine in scandium chloride is found by \( \frac{41.42\text{ mg}}{58.9\text{ mg}} \times 100\% = 70.33\% \). Thus, chlorine constitutes 70.33% of the scandium chloride.

Step 3: Calculate Mass of Scandium

Subtract the mass of chlorine from the total mass of scandium chloride: \( 58.9 \text{ mg} - 41.42 \text{ mg} = 17.48 \text{ mg of Sc} \).

Step 4: Calculate Mass Percentage of Scandium

The mass percentage of scandium is \( \frac{17.48 ext{ mg}}{58.9 ext{ mg}} \times 100\% = 29.67\% \). So, scandium constitutes 29.67% of the compound.

Step 5: Determine Moles of Elements

Convert masses to moles: for \( \text{Sc} \), \( \frac{17.48 \text{ mg}}{44.96 \text{ g/mol}} = 0.388\text{ mmol} \); for \( \text{Cl} \), \( \frac{41.42 \text{ mg}}{35.45 \text{ g/mol}} = 1.168\text{ mmol} \).

Step 6: Find Empirical Formula

Divide moles by the smallest amount: \( \text{Sc:} \frac{0.388}{0.388} = 1 \), \( \text{Cl:} \frac{1.168}{0.388} \approx 3 \). The empirical formula is \( \text{ScCl}_3 \).

Key Concepts

Empirical Formula Determination

The empirical formula is a simple representation of a compound's composition by showing the smallest whole-number ratio of atoms. This process involves converting the masses of the elements present in the compound to moles. By dividing each element's mole quantity by the smallest amount, you find the ratio of atoms.

In the given exercise, scandium chloride was analyzed. The goal was to find the empirical formula by determining the mole ratio between scandium and chlorine. Initially, masses are calculated and then converted into moles using their respective molar masses:

• For scandium, the molar mass is 44.96 g/mol.
• For chlorine, it is 35.45 g/mol.

By following the steps, the moles of scandium were found to be 0.388 mmol and the moles of chlorine 1.168 mmol. Finally, by dividing both values by the smaller number of moles, 0.388, the ratio 1:3 is established, leading to the empirical formula ScCl₃.

Scandium Chloride

Scandium chloride is a compound consisting of scandium and chlorine atoms. Its chemical formula, based on empirical analysis, is ScCl₃. This indicates that one scandium atom is combined with three chlorine atoms to form this compound.

The formation and determination of scandium chloride's empirical formula is essential for understanding its stoichiometry. Scandium is a rare earth metal, which means compounds involving scandium, like scandium chloride, are often part of specialized applications in materials science.

• In this compound, scanning the role of scandium brings context to its atomic properties, such as promoting distinctive reactions when forming these salts.
• Chlorine adds to the compound's properties by providing charge balance and participating in chemical reactions when dissolved.

By understanding the elements and their mass ratios in scandium chloride, scientists and students can appreciate its composition and potential uses.

Silver Chloride Precipitation

Silver chloride (AgCl) precipitation is a common laboratory method used to determine the amount of chloride ions present in a compound. The process involves adding silver nitrate (AgNO₃) to a chlorine-containing solution, resulting in the formation of a solid precipitate of silver chloride.

This reaction is crucial for quantifying chlorine in unknown compounds.

• The chemical reaction can be described as: \[\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl (s, white precipitate)}\]
• Since silver chloride is insoluble in water, it precipitates out of the solution, allowing the measurement of its mass.

By measuring the mass and applying the known stoichiometry of AgCl, you can back-calculate the chloride's mass in the original solution. This method forms the foundational step in determining chlorine's percentage in compounds like scandium chloride.

Molar Mass Calculation

Calculating the molar mass is fundamental in converting between the mass of a substance and the amount of substance in moles. To find the molar mass, simply add up the atomic masses of all the atoms present in a molecule.

In the context of the exercise involving scandium chloride, we have the molar masses:

• Silver chloride (AgCl) has a molar mass of 143.32 g/mol.
• Scandium registers at 44.96 g/mol.
• Chlorine alone contributes 35.45 g/mol.

When determining the mass percentage composition of scandium chloride, or converting mass values to moles and vice versa, accurate molar masses are vital. They ensure that all subsequent calculations of percentages or determining the empirical formula are correct and reliable.