Question

The thallium (present as \(\mathrm{Tl}_{2} \mathrm{SO}_{4}\) ) in a \(9.486-\mathrm{g}\) pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percent of \(\mathrm{Tl}_{2} \mathrm{SO}_{4}\) in the sample if \(0.1824 \mathrm{g}\) of TII was recovered.

Short answer

Moles of \(\mathrm{TlI}\) = \(0.1824 \, \mathrm{g} / 331.28 \, \mathrm{g/mol}\) = \(5.5 \times 10^{-4} \, \mathrm{mol}\) Moles of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = \(5.5 \times 10^{-4} \, \mathrm{mol} / 2\) = \(2.75 \times 10^{-4} \, \mathrm{mol}\) Mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = \(2.75 \times 10^{-4} \, \mathrm{mol} \times 600.82 \, \mathrm{g/mol}\) = \(0.165 \, \mathrm{g}\) Mass percent of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = \((0.165 \, \mathrm{g} / 9.486 \, \mathrm{g}) \times 100\) = \(1.74 \%\)

Step-by-step solution

Calculate the moles of thallium(I) iodide, \(\mathrm{TlI}\), obtained

To calculate the moles of \(\mathrm{TlI}\) we use the formula: moles = mass / molar mass. The molar mass of \(\mathrm{TlI}\) = 204.38 g/mol (Tl) + 126.9 g/mol (I) = 331.28 g/mol (TlI). Given the mass of thallium(I) iodide recovered: Moles of \(\mathrm{TlI}\) = mass of \(\mathrm{TlI}\) / molar mass of \(\mathrm{TlI}\) Moles of \(\mathrm{TlI}\) = \(0.1824 \, \mathrm{g} / 331.28 \, \mathrm{g/mol}\)

Calculate the moles of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) in the sample.

Since both thallium in thallium(I) iodide and thallium in thallium sulfate are in +1 oxidation state, the mole ratio between \(\mathrm{TlI}\) and \(\mathrm{Tl_2SO_4}\) is 2:1 (2 moles of \(\mathrm{TlI}\) for every 1 mole of \(\mathrm{Tl_2SO_4}\)). Moles of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = moles of \(\mathrm{TlI}\) / 2

Calculate the mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) in the sample.

Now we will multiply the moles of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) by its molar mass to get the mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) in the sample. The molar mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = 2 × 204.38 g/mol (Tl) + 96.06 g/mol (S) + 64 g/mol × 4 (O) = 600.82 g/mol. Mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = moles of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) × molar mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\)

Calculate the mass percent of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) in the sample.

Now just divide the mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) by the total sample mass and multiply by 100 to get the mass percent. Mass percent of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = (mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) / mass of sample) × 100 Let's do the calculations:

Key concepts

Mass Percent Calculation

Calculating mass percent is a key concept in stoichiometry, as it gives the proportion of a certain component within a mixture. Here, we seek to find the mass percent of thallium sulfate (\( \text{Tl}_2 \text{SO}_4 \)) in a pesticide sample. To do this, we first determine the mass of \( \text{Tl}_2 \text{SO}_4 \) in the sample. Then, we use the formula for mass percent:

• Mass percent = \( \frac{\text{mass of Tl}_2 \text{SO}_4 }{\text{mass of sample}} \times 100 \)

By dividing the mass of the thallium sulfate by the total mass of the sample, and multiplying the result by 100, we can express the proportion as a percentage. This calculation tells us how much of the sample’s weight is due to \( \text{Tl}_2 \text{SO}_4 \). Such insight is crucial to understand the effectiveness or concentration of a substance in a mixture.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is essential for converting between mass and number of moles. In the exercise, we calculate the molar masses of both thallium iodide (\( \text{TlI} \)) and thallium sulfate (\( \text{Tl}_2 \text{SO}_4 \)).
For \( \text{TlI} \):

• Thallium (Tl): 204.38 g/mol
• Iodine (I): 126.9 g/mol
• Molar mass of \( \text{TlI} \) = 204.38 + 126.9 = 331.28 g/mol

For \( \text{Tl}_2 \text{SO}_4 \):

• 2 Thallium (Tl): 2 x 204.38 g/mol
• Sulfur (S): 96.06 g/mol
• 4 Oxygen (O): 4 x 16 = 64 g/mol
• Molar mass of \( \text{Tl}_2 \text{SO}_4 \) = 2 x 204.38 + 96.06 + 64 = 600.82 g/mol

Knowing the molar mass allows us to convert grams of a substance to moles, an essential step in stoichiometric calculations.

Chemical Precipitation

Chemical precipitation is a process where soluble substances react to form an insoluble solid, known as a precipitate. In the exercise, thallium is precipitated as thallium(I) iodide (\( \text{TlI} \)). This process efficiently separates ions from a solution, enabling us to isolate and measure a specific compound.
In our example, when a thallium-containing substance is treated with an iodide source, \( \text{TlI} \) forms as a precipitate. We can collect and weigh this solid to determine how much thallium was present in the original mixture. This method is instrumental in analyzing the composition of complex samples, ensuring accurate quantitative chemical analysis.

Chemical Formula Interpretation

Understanding a chemical formula is critical to identifying the elements and their ratios in a compound. A chemical formula denotes the number of each type of atom in a molecule. For example, the formula \( \text{Tl}_2 \text{SO}_4 \) tells us:

• There are 2 thallium (Tl) atoms
• 1 sulfur (S) atom
• 4 oxygen (O) atoms

This interpretation is pivotal when analyzing and calculating the substance's molar mass, involved in many of the calculations in stoichiometry. By reading the formula \( \text{Tl}_2 \text{SO}_4 \), we categorically know it consists of these elements in defined proportions, helping us accurately predict and measure the compound's mass and moles. Such comprehension is essential for balanced chemical equations and reactions.