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Chapter 4: Problem 111

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

Short Answer

Expert verified
The mass percent of \(\mathrm{Tl}_{2} \mathrm{SO}_{4}\) in the pesticide sample is approximately 2.926 %.

Step by step solution

01

Find the molar mass of TlI and Tl2SO4

First, calculate the molar mass of thallium(I) iodide (TlI) and thallium sulfate (Tl2SO4) using the molar masses of their constituent elements: Molar mass of TlI = Molar mass of Tl + Molar mass of I = 204.38 g/mol + 126.90 g/mol = 331.28 g/mol Molar mass of Tl2SO4 = (2 × Molar mass of Tl) + Molar mass of S + (4 × Molar mass of O) = (2 × 204.38 g/mol) + 32.07 g/mol + (4 × 16.00 g/mol) = 408.76 g/mol + 32.07 g/mol + 64.00 g/mol = 504.83 g/mol
02

Calculate moles of TlI

Now, calculate the moles of thallium(I) iodide (TlI) using its mass and molar mass: Moles of TlI = mass / molar mass = 0.1824 g / 331.28 g/mol = 5.5 × 10^{-4} mol
03

Determine mole ratio in the precipitate

For every two moles of Tl in Tl2SO4, there are two moles of Tl in TlI, as well. Therefore, the mole ratio between Tl2SO4 and TlI in the precipitate is 1:1.
04

Calculate moles of Tl2SO4

Using the mole ratio from Step 3, we can find the moles of Tl2SO4 present in the sample: Moles of Tl2SO4 = Moles of TlI = 5.5 × 10^{-4} mol
05

Calculate mass of Tl2SO4

To obtain the mass of Tl2SO4 in the sample, multiply its moles by its molar mass: Mass of Tl2SO4 = moles × molar mass = 5.5 × 10^{-4} mol × 504.83 g/mol = 0.2777 g
06

Calculate mass percent of Tl2SO4

Lastly, use the mass of Tl2SO4 and the mass of the sample to determine the mass percent of Tl2SO4: Mass percent of Tl2SO4 = (Mass of Tl2SO4 / Mass of sample) × 100 = (0.2777 g / 9.486 g) × 100 = 2.926 % The mass percent of \(\mathrm{Tl}_{2} \mathrm{SO}_{4}\) in the pesticide sample is approximately 2.926 %.

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Most popular questions from this chapter

Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)$$ Balance this equation.

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\( with \)\mathrm{I}_{3}-$ in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. $$\operatorname{copper}(\mathrm{II}) \text { sulfate }(a q)+\mathrm{iron}(s) \longrightarrow (s)+\operatorname{iron}(\mathrm{II}) \text { sulfate }(a q)$$ $$\operatorname{copper}(\mathrm{II}) \text { sulfate }(a q)+\mathrm{iron}(s) \longrightarrow (s)+\text { iron (III) sulfate }(a q) $$ You place 87.7 mL of a 0.500-M solution of copper(II) sulfate in a beaker. You then add 2.00 g of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate 2.27 g of copper. Which equation above describes the reaction that occurred? Support your answer

A 500.0-mL sample of 0.200 M sodium phosphate is mixed with 400.0 mL of 0.289 M barium chloride. What is the mass of the solid produced?

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$\mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ The ion formed as a product, \(X^{2-},\) was shown to have 36 total electrons. What is element X? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\) . To completely neutralize a sample of $\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{mL}\( of 0.175 \)\mathrm{M}\( \)\mathrm{OH}^{-}$ solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?
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