Problem 117 - General Chemistry

Question

You have a sample of a rat poison whose active ingredient is thallium(I) sulfate. You analyze this sample for the mass percentage of active ingredient by adding potassium iodide to precipitate yellow thallium(I) iodide. If the sample of rat poison weighed \(759.0 \mathrm{mg}\) and you obtained \(212.2 \mathrm{mg}\) of the dry precipitate, what is the mass percentage of the thallium(I) sulfate in the rat poison?

Step-by-Step Solution

Answer

The mass percentage of thallium(I) sulfate in the rat poison is 21.28%.

Step 1: Identify Relevant Compounds

First, identify the relevant compounds involved in the chemical reaction: Thallium(I) sulfate (\(\text{Tl}_2\text{SO}_4\)) and thallium(I) iodide (\(\text{TlI}\)). Potassium iodide (\(\text{KI}\)) is a reactant used to precipitate the thallium as thallium(I) iodide. The molar mass of thallium(I) iodide will be needed later for calculations.

Step 2: Calculate Molar Masses

Compute the molar masses:- Thallium (Tl): \(204.38 \, \text{g/mol}\)- Iodine (I): \(126.90 \, \text{g/mol}\)- Thallium(I) iodide (\(\text{TlI}\)): \[204.38 + 126.90 = 331.28 \, \text{g/mol}\].- Thallium(I) sulfate (\(\text{Tl}_2\text{SO}_4\)): \[(2 \times 204.38) + 32.07 + (4 \times 16.00) = 504.76 \, \text{g/mol}\].

Step 3: Determine Moles of Precipitate

Calculate the moles of thallium(I) iodide (\(\text{TlI}\)) precipitated using its mass: \[\frac{212.2 \, \text{mg}}{331.28 \, \text{g/mol}} = \frac{0.2122 \, \text{g}}{331.28 \, \text{g/mol}} \approx 0.000640 \, \text{mol}.\]

Step 4: Relate Precipitate to Thallium(I) Sulfate

Thallium in Thallium(I) iodide originates from Thallium(I) sulfate. Since each unit of thallium(I) sulfate produces two units of thallium, the number of moles of \(\text{Tl}_2\text{SO}_4\) is half the number of moles of **thallium in the precipitate**.Moles of thallium(I) sulfate: \[\frac{0.000640}{2} = 0.000320 \, \text{mol}.\]

Step 5: Calculate Mass of Thallium(I) Sulfate in Sample

Calculate the mass of thallium(I) sulfate using its molar mass:Mass of \(\text{Tl}_2\text{SO}_4\):\[(0.000320 \, \text{mol}) \times (504.76 \, \text{g/mol}) = 0.1615 \, \text{g} \text{ or } 161.5 \, \text{mg}.\]

Step 6: Calculate Mass Percentage of Thallium(I) Sulfate

Determine the mass percentage of thallium sulfate in the sample:\[\frac{161.5 \, \text{mg}}{759.0 \, \text{mg}} \times 100\% = 21.28\%.\]

Key Concepts

Thallium(I) sulfate

Thallium(I) sulfate, symbolized as \( \text{Tl}_2\text{SO}_4 \), is an important compound often used in various chemical processes. While it has applications in industry and research, it's infamously known for its effectiveness as a rat poison. This compound consists of two thallium atoms and one sulfate ion. Thallium is a heavy metal, hence compounds containing thallium should be handled with caution due to their toxicity.
Thallium(I) sulfate is a white crystalline powder. In solution, it dissociates to form thallium ions (\( \text{Tl}^+ \)) and sulfate ions (\( \text{SO}_4^{2-} \)). These thallium ions are the active ingredient in rat poison, affecting the metabolic processes of the rodents.

• Formula: \( \text{Tl}_2\text{SO}_4 \)
• A crystalline compound, consisting of thallium and sulfate
• Highly toxic and used in pesticides

Molar mass

Molar mass is a crucial concept in chemistry, relating to the mass of one mole of a compound. It's measured in grams per mole (g/mol). To find the molar mass of a compound, you need the atomic masses of each element present in the compound, as shown on the periodic table.
For thallium(I) sulfate (\( \text{Tl}_2\text{SO}_4 \)), we calculate its molar mass by adding the atomic masses of each atom:

• Thallium (\( \text{Tl} \)): 204.38 g/mol, multiplied by 2 because there are two thallium atoms.
• Sulfur (\( \text{S} \)): 32.07 g/mol
• Oxygen (\( \text{O} \)): 16.00 g/mol, multiplied by 4 for the four oxygen atoms.

Adding all these values gives the molar mass of \( \text{Tl}_2\text{SO}_4 \) as 504.76 g/mol. This value is used to convert between the mass of a substance and the number of moles.

Chemical reaction

A chemical reaction involves the transformation of substances into different substances, indicated by chemical changes. In the context of the rat poison exercise, the chemical reaction involves thallium(I) sulfate and potassium iodide.
Potassium iodide (\( \text{KI} \)) reacts with thallium(I) sulfate to form thallium(I) iodide (\( \text{TlI} \)) as a precipitate and potassium sulfate (\( \text{K}_2\text{SO}_4 \)). This reaction is key because it allows for the isolation of thallium in the form of a solid precipitate, which can be easily extracted and weighed:

• Initial compounds: \( \text{Tl}_2\text{SO}_4 \) and \( \text{KI} \)
• Products formed: \( \text{TlI} \) and \( \text{K}_2\text{SO}_4 \)
• Precipitation reaction aids in determining the composition of thallium in the sample

This reaction's product, thallium(I) iodide, separates from the solution because of its insolubility, allowing for mass measurement.

Precipitation

Precipitation in chemistry refers to the process where a soluble substance forms a solid. This process occurs when the concentration of ions in solution exceeds their solubility product. During a precipitation reaction, the solid formed is known as a precipitate.
The exercise with thallium(I) sulfate involves using a precipitation reaction to determine the mass percentage of thallium. By adding potassium iodide (\( \text{KI} \)) to the solution, thallium(I) iodide (\( \text{TlI} \)) precipitates as a yellow solid, leaving other ions in the solution.

• Formation of a solid precipitate from a solution
• Helps in isolating specific ions like thallium for analysis
• Thallium(I) iodide is the precipitate, distinct due to its yellow color

This process is valuable for analytical purposes in chemistry, allowing for the measurement and identification of specific components in a mixture.