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Chapter 8: Problem 18

A \(516.7-\mathrm{mg}\) sample that contains a mixture of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) and \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) is dissolved in water and treated with \(\mathrm{BaCl}_{2},\) precipitating the \(\mathrm{SO}_{4}^{2-}\) as \(\mathrm{BaSO}_{4}\). The resulting precipitate is isolated by filtration, rinsed free of impurities, and dried to a constant weight, yielding \(863.5 \mathrm{mg}\) of \(\mathrm{BaSO}_{4} .\) What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

Short Answer

Expert verified
The \(\%\) w/w of \(\mathrm{K}_2\mathrm{SO}_4\) in the sample is approximately 22.34\%.

Step by step solution

01

Write the chemical equations

First, write the chemical reactions involved: Each mole of \(\mathrm{K}_2\mathrm{SO}_4\) and \((\mathrm{NH}_4)_2\mathrm{SO}_4\) produces a mole of \(\mathrm{SO}_4^{2-}\), which reacts with \(\mathrm{BaCl}_2\) to form \(\mathrm{BaSO}_4\): 1. \(\mathrm{K}_2\mathrm{SO}_4 \rightarrow \mathrm{K}^+ + \mathrm{SO}_4^{2-}\)2. \((\mathrm{NH}_4)_2\mathrm{SO}_4 \rightarrow 2 \mathrm{NH}_4^+ + \mathrm{SO}_4^{2-}\)3. \(\mathrm{SO}_4^{2-} + \mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4 + 2\mathrm{Cl}^-\)
02

Calculate moles of \(\mathrm{BaSO}_4\)

Use the mass of \(\mathrm{BaSO}_4\) precipitate to calculate the moles. The molar mass of \(\mathrm{BaSO}_4\) is about 233.39 g/mol. Convert the given mass into grams and calculate moles:\[\text{Moles of } \mathrm{BaSO}_4 = \frac{863.5 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}}}{233.39 \text{ g/mol}}\approx 0.0037 \text{ mol}.\]
03

Calculate moles of sulfate in original mixture

Each mole of \(\mathrm{BaSO}_4\) corresponds to one mole of \(\mathrm{SO}_4^{2-}\), so the moles of \(\mathrm{SO}_4^{2-}\) in the mixture is also approximately 0.0037 mol.
04

Express moles of sulfates as a function of components

Let \(x\) be the moles of \(\mathrm{K}_2\mathrm{SO}_4\) and \(y\) the moles of \((\mathrm{NH}_4)_2\mathrm{SO}_4\). From Step 3: \[x + y = 0.0037\]
05

Setup equation with sample mass

The total mass of the sample mixture can be expressed as:\[174.26x + 132.14y = 0.5167\]where 174.26 g/mol is the molar mass of \(\mathrm{K}_2\mathrm{SO}_4\) and 132.14 g/mol is the molar mass of \((\mathrm{NH}_4)_2\mathrm{SO}_4\).
06

Solve the system of equations

Now we have:1. \(x + y = 0.0037\)2. \(174.26x + 132.14y = 0.5167\)From the first equation, \(y = 0.0037 - x\).Substitute \(y\) in the second equation:\[174.26x + 132.14(0.0037 - x) = 0.5167\]Expand and simplify:\[174.26x + 0.488818 - 132.14x = 0.5167\]\[42.12x + 0.488818 = 0.5167\]\[42.12x = 0.5167 - 0.488818\]\[42.12x = 0.027882\]\[x \approx 0.000662\]
07

Calculate mass and percentage of \(\mathrm{K}_2\mathrm{SO}_4\)

Calculate the mass of \(\mathrm{K}_2\mathrm{SO}_4\) in the mixture:\[\text{Mass of } \mathrm{K}_2\mathrm{SO}_4 = 0.000662 \times 174.26 \approx 0.1154 \text{ g}\]Now calculate the percentage:\[\%\ \mathrm{w} / \mathrm{w} \ \mathrm{K}_2\mathrm{SO}_4 = \frac{0.1154}{0.5167} \times 100\%\approx 22.34\%\]

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Key Concepts

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

Gravimetric Analysis
Gravimetric analysis is a technique used in analytical chemistry to determine the quantity of an analyte based on the mass of a solid. In the given exercise, we dissolved a sample containing potassium sulfate oindent and ammonium sulfate in water, then added barium chloride to it. This process caused the sulfate ions, oindent sulfate ions (\(\mathrm{SO}_4^{2-}\) ), to precipitate as barium sulfate (\(\mathrm{BaSO}_4\) ). The procedure follows straightforward steps:
  • A chemical reaction causes the analyte to precipitate. The precipitate is then collected, dried, and weighed.In this exercise, barium chloride (\(\mathrm{BaCl}_2\) ) was added, which reacted with the sulfate ions to form \(\mathrm{BaSO}_4\) precipitate.
  • Next, we ensured that all impurities were rinsed away before weighing the final solid precipitate.
  • The mass of the precipitate allowed us to calculate back to find the moles of the original analytes present in the sample.
The key advantage of gravimetric analysis is its high precision and accuracy when conducted meticulously.
Chemical Reactions
Chemical reactions convert reactants into products through processes like precipitation in this exercise. When the sample dissolved in water is treated with \(\mathrm{BaCl}_2\), it undergoes multiple reactions:
  • \(\mathrm{K}_2\mathrm{SO}_4 \rightarrow \mathrm{K}^+ + \mathrm{SO}_4^{2-}\)
  • \((\mathrm{NH}_4)_2\mathrm{SO}_4 \rightarrow 2 \mathrm{NH}_4^+ + \mathrm{SO}_4^{2-}\)
  • \(\mathrm{SO}_4^{2-} + \mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4 + 2\mathrm{Cl}^-\)
These reactions show how \(\mathrm{SO}_4^{2-}\) ions are freed from \(\mathrm{K}_2\mathrm{SO}_4\) and \((\mathrm{NH}_4)_2\mathrm{SO}_4\), present in the sample. Each sulfate ion then reacts with \(\mathrm{BaCl}_2\) to produce \(\mathrm{BaSO}_4\). The precipitate's mass is directly related to the reactants' quantity, influencing our further stoichiometric calculations. Understanding these reactions is crucial as they form the backbone of identifying and quantifying the substances involved.
Stoichiometry
Stoichiometry is the branch of chemistry involving calculating the relative quantities of reactants and products in chemical reactions. In the exercise, stoichiometry helped us find out how much potassium sulfate was in the sample.First, we calculated the moles of precipitated \(\mathrm{BaSO}_4\) using its mass: \[\text{Moles of } \mathrm{BaSO}_4 = \frac{863.5 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}}}{233.39 \text{ g/mol}} \approx 0.0037 \text{ mol}.\]We then aligned the need to find \(\mathrm{SO}_4^{2-}\) moles in our original mixture because each mole of \(\mathrm{BaSO}_4\) corresponds to a mole of \(\mathrm{SO}_4^{2-}\). Using stoichiometry, we set up equations involving the moles from \(\mathrm{K}_2\mathrm{SO}_4\) and \((\mathrm{NH}_4)_2\mathrm{SO}_4\): \[x + y = 0.0037\]\[174.26x + 132.14y = 0.5167\]Solving these provided us the proportion of both compounds, enabling us to finally determine the percentage mass of \(\mathrm{K}_2\mathrm{SO}_4\) in the sample. Understanding stoichiometry is essential to effectively link chemical equations with practical, real-world measurements accurately.

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Most popular questions from this chapter

If a precipitate of known stoichiometry does not form, a gravimetric analysis is still feasible if we can establish experimentally the mole ratio between the analyte and the precipitate. Consider, for example, the precipitation gravimetric analysis of \(\mathrm{Pb}\) as \(\mathrm{PbCrO}_{4}{ }^{14}\) (a) For each gram of \(\mathrm{Pb}\), how many grams of \(\mathrm{PbCrO}_{4}\) will form, assuming the reaction is stoichiometric? (b) In a study of this procedure, Grote found that \(1.568 \mathrm{~g}\) of \(\mathrm{PbCrO}_{4}\) formed for each gram of \(\mathrm{Pb}\). What is the apparent stoichiometry between \(\mathrm{Pb}\) and \(\mathrm{PbCrO}_{4} ?\) (c) Does failing to account for the actual stoichiometry lead to a positive determinate error or a negative determinate error?

The water content of an \(875.4-\mathrm{mg}\) sample of cheese is determined with a moisture analyzer. What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{H}_{2} \mathrm{O}\) in the cheese if the final mass was found to be \(545.8 \mathrm{mg} ?\)

After preparing a sample of alum, \(\mathrm{K}_{2} \mathrm{SO}_{4} \cdot \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 24 \mathrm{H}_{2} \mathrm{O},\) an ana- lyst determines its purity by dissolving a \(1.2931-\mathrm{g}\) sample and precipitating the aluminum as \(\mathrm{Al}(\mathrm{OH})_{3}\). After filtering, rinsing, and igniting, \(0.1357 \mathrm{~g}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is obtained. What is the purity of the alum preparation?

Aluminum is determined gravimetrically by precipitating \(\mathrm{Al}(\mathrm{OH})_{3}\) and isolating \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) A sample that contains approximately \(0.1 \mathrm{~g}\) of \(\mathrm{Al}\) is dissolved in \(200 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O},\) and \(5 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) and a few drops of methyl red indicator are added (methyl red is red at pH levels below 4 and yellow at \(\mathrm{pH}\) levels above 6 ). The solution is heated to boiling and \(1: 1 \mathrm{NH}_{3}\) is added dropwise until the indicator turns yellow, precipitating \(\mathrm{Al}(\mathrm{OH})_{3} .\) The precipitate is held at the solution's boiling point for several minutes before filtering and rinsing with a hot solution of \(2 \%\) \(\mathrm{w} / \mathrm{v} \mathrm{NH}_{4} \mathrm{NO}_{3} .\) The precipitate is then ignited at \(1000-1100^{\circ} \mathrm{C},\) form- ing \(\mathrm{Al}_{2} \mathrm{O}_{3}\) (a) Cite at least two ways in which this procedure encourages the formation of larger particles of precipitate. (b) The ignition step is carried out carefully to ensure the quantitative conversion of \(\mathrm{Al}(\mathrm{OH})_{3}\) to \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) What is the effect of an incomplete conversion on the \(\% \mathrm{w} / \mathrm{w}\) Al? (c) What is the purpose of adding \(\mathrm{NH}_{4} \mathrm{Cl}\) and methyl red indicator? (d) An alternative procedure foraluminum involves isolating and weighing the precipitate as the 8 -hydroxyquinolate, \(\mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} .\) Why might this be a more advantageous form of Al for a gravimetric analysis? Are there any disadvantages?

Calcium is determined gravimetrically by precipitating \(\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) and isolating \(\mathrm{CaCO}_{3}\). After dissolving a sample in \(10 \mathrm{~mL}\) of water and \(15 \mathrm{~mL}\) of \(6 \mathrm{M} \mathrm{HCl}\), the resulting solution is heated to boiling and a warm solution of excess ammonium oxalate is added. The solution is maintained at \(80^{\circ} \mathrm{C}\) and \(6 \mathrm{M} \mathrm{NH}_{3}\) is added dropwise, with stirring, until the solution is faintly alkaline. The resulting precipitate and solution are removed from the heat and allowed to stand for at least one hour. After testing the solution for completeness of precipitation, the sample is filtered, rinsed with \(0.1 \% \mathrm{w} / \mathrm{v}\) ammonium oxalate, and dried for one hour at \(100-120^{\circ} \mathrm{C}\). The precipitate is transferred to a muffle furnace where it is converted to \(\mathrm{CaCO}_{3}\) by drying at \(500 \pm 25^{\circ} \mathrm{C}\) until constant weight. (a) Why is the precipitate of \(\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) converted to \(\mathrm{CaCO}_{3} ?\) (b) In the final step, if the sample is heated at too high of a temperature some \(\mathrm{CaCO}_{3}\) is converted to \(\mathrm{CaO}\). What effect would this have on the reported \(\% \mathrm{w} / \mathrm{w}\) Ca? (c) Why is the precipitant, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) added to a hot, acidic solution instead of a cold, alkaline solution?
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