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Chapter 6: Problem 99

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

Expert verified
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

01

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 \(\mathrm{TlI}\) = \(0.1824 \, \mathrm{g} / 331.28 \, \mathrm{g/mol}\)
02

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
03

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 脳 204.38 g/mol (Tl) + 96.06 g/mol (S) + 64 g/mol 脳 4 (O) = 600.82 g/mol. Mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) = moles of \(\mathrm{Tl}_2\mathrm{SO}_{4}\) 脳 molar mass of \(\mathrm{Tl}_2\mathrm{SO}_{4}\)
04

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) 脳 100 Let's do the calculations:

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. It is composed solely of \(\mathrm{C}, \mathrm{F},\) and \(\mathrm{B} ;\) it is \(42.23 \%\) C and \(55.66 \%\) F by mass. a. What is the empirical formula of BARF? b. A \(2.251-\mathrm{g}\) sample of BARF dissolved in \(347.0 \mathrm{mL}\) of solution produces a 0.01267-M solution. What is the molecular formula of BARF?

How would you prepare 1.00 L of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 \mathrm{M})\) sulfuric acid b. HCl from "concentrated" (12 M) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO_ from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 \(M\) hydrochloric acid, \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and 200.0 mL. of 0.100 \(M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A \(0.256-\mathrm{g}\) sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g. a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.
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