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?

Short answer

The mass percentage of thallium(I) sulfate in the rat poison is approximately 21.29%.

Step-by-step solution

Understand the Reaction

Thallium(I) sulfate reacts with potassium iodide to form thallium(I) iodide precipitate. This is a stoichiometric reaction where all thallium(I) ions in thallium(I) sulfate will convert to thallium(I) iodide precipitate.

Calculate Molar Mass of Thallium(I) Iodide

Compute the molar mass of thallium(I) iodide (TlI). Using the periodic table, look up the atomic masses: Tl is approximately 204.38 g/mol, and I is 126.90 g/mol. So,\[ \text{Molar Mass of TlI} = 204.38 + 126.90 = 331.28 \text{ g/mol}. \]

Convert Precipitate Mass to Moles

Convert the given mass of the precipitate (212.2 mg) into grams and then to moles using the molar mass:\[ 212.2 \text{ mg} = 0.2122 \text{ g}, \]\[ \text{Moles of } TlI = \frac{0.2122 \text{ g}}{331.28 \text{ g/mol}} \approx 0.0006404 \text{ mol}. \]

Calculate Mass of Thallium(I) Sulfate That Produced the Precipitate

Determine the mass of thallium(I) sulfate (Tl₂SO₄) that corresponds to the moles of thallium in the precipitate. Molar mass of thallium(I) sulfate: Tl = 204.38 g/mol, S = 32.07 g/mol, and O = 16.00 g/mol:\[ \text{Molar Mass of Tl}_2\text{SO}_4 = 2(204.38) + 32.07 + 4(16.00) = 504.76 \text{ g/mol}. \]Since each molecule of Tl₂SO₄ yields 2 moles of Tl, moles of Tl₂SO₄ = 0.0006404 mol / 2:\[ \text{Moles of Tl}_2\text{SO}_4 = 0.0006404 / 2 \approx 0.0003202 \text{ mol}. \]Convert moles back to grams:\[ \text{Mass of Tl}_2\text{SO}_4 = 0.0003202 \text{ mol} \times 504.76 \text{ g/mol} \approx 0.1616 \text{ g}. \]

Calculate Mass Percentage of Thallium(I) Sulfate

Now compute the mass percentage of thallium(I) sulfate in the original sample of rat poison. Convert the original sample mass from mg to g (759.0 mg = 0.759 g):\[ \text{Mass Percentage} = \left( \frac{0.1616 \text{ g}}{0.759 \text{ g}} \right) \times 100\% \approx 21.29\%. \]

Key concepts

Precipitation Reaction

A precipitation reaction occurs when two soluble substances in solution react to form an insoluble product called a precipitate. In the analysis of thallium(I) sulfate, potassium iodide is added to the solution containing the sample. Upon reaction, thallium(I) iodide is formed, which is a yellow precipitate. This process is essential because it allows for the isolation and subsequent measurement of the thallium component from the original sample.

In our exercise, the reactants are thallium(I) ions from thallium(I) sulfate and iodide ions from potassium iodide. The reaction can be represented in the ionic form as:

\[ \text{Tl}^+ (aq) + \text{I}^- (aq) \rightarrow \text{TlI} (s) \]

Here, the thallium(I) iodide precipitate forms as a solid out of the aqueous solution, allowing for its collection and weighing. Understanding the stoichiometry helps in determining precisely how much of the reactant forms the precipitate.

Mass Percentage Calculation

Calculating the mass percentage of a component in a sample is a key part of many chemical analysis problems. This process helps quantify the amount of a specific substance within a mixture. In the given problem, our goal is to find the mass percentage of thallium(I) sulfate in a rat poison sample.

To calculate this, we first convert the mass of thallium(I) sulfate that we found from the precipitate mass into grams. Then, we express this mass as a percentage of the total original sample mass. The formula used is:

\[ \text{Mass Percentage} = \left( \frac{\text{Mass of Compound}}{\text{Total Mass of Sample}} \right) \times 100\% \]

This calculation provides us with a numerical value representing the concentration of thallium(I) sulfate in the rat poison sample, which is crucial for evaluating its potency.

Molar Mass Determination

Determining molar mass is an important step in converting between the mass of a compound and the number of moles, which is necessary for stoichiometric calculations. In our exercise, we needed to find the molar mass of both thallium(I) iodide and thallium(I) sulfate.

The molar mass of a compound can be calculated by adding up the atomic masses of all atoms in its molecular formula. For example, in finding the molar mass of thallium(I) iodide:

• Thallium (Tl) has an atomic mass of about 204.38 g/mol.
• Iodine (I) has an atomic mass of 126.90 g/mol.

This results in a molar mass for thallium(I) iodide of 331.28 g/mol.

For thallium(I) sulfate, the calculation involves two thallium atoms, one sulfur atom, and four oxygen atoms, leading to a larger molar mass. Understanding how to calculate molar mass allows us to accurately convert between mass and mole quantities, making it a critical tool in quantifying reactions.