Problem 27 - Chemistry and Chemical Reactivity

Question

A pesticide contains thallium(I) sulfate, \(\mathrm{TI}_{2} \mathrm{SO}_{4}\). Dissolving a \(10.20-\mathrm{g}\) sample of impure pesticide in water and adding sodium iodide precipitates \(0.1964 \mathrm{g}\) of thallium(I) iodide, TII. $$ \mathrm{TI}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \rightarrow 2 \mathrm{TII}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) $$ What is the mass percent of \(\mathrm{TI}_{2} \mathrm{SO}_{4}\) in the original \(10.20-\mathrm{g}\) sample?

Step-by-Step Solution

Answer

The mass percent of \(\mathrm{TI}_{2} \mathrm{SO}_{4}\) is 1.47\%.

Step 1: Find Moles of Thallium(I) Iodide

First, let's determine the moles of thallium(I) iodide, TII, produced. The molar mass of TII is approximately:\[ \text{TI: } 204.38\,\text{g/mol}, \quad \text{I: } 126.90\,\text{g/mol} \]Thus, the molar mass of TII = 204.38 + 126.90 = 331.28 g/mol.Use the mass of TII precipitated to find moles:\[ \text{Moles of TII} = \frac{0.1964\,\text{g}}{331.28\,\text{g/mol}} = 5.93\times10^{-4}\,\text{mol} \]

Step 2: Determine Moles of Thallium(I) Sulfate

From the reaction, \(1\text{ mole of TI}_2\text{SO}_4\) reacts to produce \(2\text{ moles of TII}\). Thus, \(2\times5.93\times10^{-4}\,\text{mol of TII}\) corresponds to \[ \text{Moles of TI}_2\text{SO}_4 = \frac{5.93\times10^{-4}\,\text{mol}}{2} = 2.965\times10^{-4}\,\text{mol} \]

Step 3: Calculate Mass of Thallium(I) Sulfate

Now, calculate the mass of \(\text{TI}_2\text{SO}_4\) in the sample. Its molar mass is calculated as follows:\[ \text{TI}_2\text{SO}_4: (2\times204.38) + 32.06 + (4\times16.00) = 504.76\,\text{g/mol} \]Now find the mass:\[ \text{Mass of TI}_2\text{SO}_4 = 2.965\times10^{-4}\,\text{mol} \times 504.76\,\text{g/mol} = 0.1495\,\text{g} \]

Step 4: Determine Mass Percent of Thallium(I) Sulfate

Finally, calculate the mass percent of \(\text{TI}_2\text{SO}_4\) in the original sample:\[ \text{Mass percent} = \left( \frac{0.1495\,\text{g}}{10.20\,\text{g}} \right) \times 100\% = 1.47\% \]

Key Concepts

Mass Percent

Understanding mass percent is essential in determining the concentration of a particular component in a mixture. In our exercise with thallium(I) sulfate, we are interested in how much of this compound is present in the pesticide sample.
Mass percent is calculated using the formula:

• Mass Percent = \( \left( \frac{\text{Mass of component}}{\text{Total mass of sample}} \right) \times 100\% \)

For our specific example, the mass of thallium(I) sulfate in the sample is 0.1495 g, and the total mass of the pesticide is 10.20 g.
This means our mass percent calculation becomes:

• Mass Percent = \( \left( \frac{0.1495}{10.20} \right) \times 100\% \approx 1.47\% \)

Such a small mass percent indicates that the pure thallium(I) sulfate makes up only a fraction of the total sample's mass.

Stoichiometry

In chemical reactions, stoichiometry helps us understand the quantitative relationships between reactants and products. It allows chemists to predict how much product will form from given reactants and vice versa.
For the problem involving thallium(I) sulfate reacting with sodium iodide, we rely on stoichiometry to connect the moles of thallium(I) iodide produced back to the amount of thallium(I) sulfate present.
The balanced chemical equation from our exercise:

• \( \mathrm{TI}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaI} \rightarrow 2 \mathrm{TII} + \mathrm{Na}_{2} \mathrm{SO}_{4} \)

indicates that one mole of thallium(I) sulfate produces two moles of thallium(I) iodide.
Thus, to find the thallium(I) sulfate from the moles of TII, the relationship is:

• Moles of \( \mathrm{TI}_2\mathrm{SO}_4 = \frac{1}{2} \times \text{Moles of TII} \)

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, guided by a balanced chemical equation. An equation that captures these transformations ensures atoms are conserved across the reaction.
In our example, we have:

• \( \mathrm{TI}_{2} \mathrm{SO}_{4} + 2 \mathrm{NaI} \rightarrow 2 \mathrm{TII} + \mathrm{Na}_{2} \mathrm{SO}_{4} \)

Here, thallium(I) sulfate and sodium iodide react to form precipitated thallium(I) iodide (TII) and sodium sulfate.
This reaction illustrates how specific products, like TII, can be separated out from a mixture of substances using selective precipitation. This technique is particularly useful for determining the composition of a complex mixture.

Molar Mass Calculation

Calculating the molar mass of compounds is a foundational skill in chemistry, allowing us to transition between the mass of compounds and the number of moles present. It's crucial for stoichiometric calculations.
For thallium(I) iodide (TII), we calculated its molar mass by adding the atomic masses of thallium (TI) and iodine (I):

• TI: 204.38 g/mol
• I: 126.90 g/mol

The molar mass of TII = 204.38 + 126.90 = 331.28 g/mol.
Similarly, for thallium(I) sulfate (\(\mathrm{TI}_{2} \mathrm{SO}_{4} \)), we add up the masses of its constituent atoms:

• 2 TI: \(2 \times 204.38 = 408.76 \text{ g/mol} \)
• S: 32.06 g/mol
• 4 O: \(4 \times 16.00 = 64.00 \text{ g/mol} \)

Giving us a total of 504.76 g/mol for \(\mathrm{TI}_{2} \mathrm{SO}_{4} \).
Understanding these calculations allows us to relate mass to moles, which is essential in quantitative chemical analysis.