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Chapter 4: Problem 141

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The BaSO\(_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

Short Answer

Expert verified
The mass of \(\mathrm{SO}_{4}^{2-}\) ion in the 0.205-g sample is 0.1227 g. The mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample is 59.85%. The combined percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample is 40.15%.

Step by step solution

01

Find moles of BaSO\(_{4}\) formed

To find the moles of BaSO\(_{4}\) formed, we use the mass of BaSO\(_{4}\) and its molar mass. The molar mass of BaSO\(_{4}\) is calculated as follows: \[M_{\text{BaSO}_4} = M_{\text{Ba}} + M_{\text{S}} + 4 \cdot M_{\text{O}}\] Using the atomic masses from the periodic table: \[M_{\text{BaSO}_4} = 137.33 \,(\text{Ba}) + 32.07 \,(\text{S}) + 4 \cdot 16.00 \,(\text{O}) = 233.43\, \text{g/mol}\] Now we can find the moles of BaSO\(_{4}\) formed: \[n_{\text{BaSO\(_4\)}}= \frac{m_{\text{BaSO\(_4\)}}}{M_{\text{BaSO}_4}} = \frac{0.298 \, \text{g}}{233.43\, \text{g/mol}} = 1.277 \times 10^{-3}\, \text{mol}\]
02

Find moles of \(\mathrm{SO}_{4}^{2-}\) ion in the sample

Since each mole of BaSO\(_{4}\) contains one mole of \(\mathrm{SO}_{4}^{2-}\) ion, the moles of \(\mathrm{SO}_{4}^{2-}\) ion in the sample are equal to the moles of BaSO\(_{4}\) formed. Therefore, the moles of \(\mathrm{SO}_{4}^{2-}\) ion are: \[n_{\text{SO\(_4^{2-}\)}} = n_{\text{BaSO\(_4\)}} = 1.277 \times 10^{-3}\, \text{mol}\]
03

Find the mass of \(\mathrm{SO}_{4}^{2-}\) ion in the sample

To find the mass of \(\mathrm{SO}_{4}^{2-}\) ion in the sample, we use its molar mass and the moles we found in step 2. The molar mass of \(\mathrm{SO}_{4}^{2-}\) ion is: \[M_{\text{SO\(_4^{2-}\)}} = M_{\text{S}} + 4 \cdot M_{\text{O}} = 32.07 \,(\text{S}) + 4 \cdot 16.00 \,(\text{O}) = 96.07\, \text{g/mol}\] Now, we can find the mass of \(\mathrm{SO}_{4}^{2-}\) ion in the sample as follows: \[m_{\text{SO\(_4^{2-}\)}} = n_{\text{SO\(_4^{2-}\)}} \cdot M_{\text{SO\(_4^{2-}\)}} = (1.277 \times 10^{-3}\, \text{mol}) (96.07\, \text{g/mol}) = 0.1227\, \text{g}\]
04

Calculate mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample

To find the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample, we use the mass of \(\mathrm{SO}_{4}^{2-}\) ion found in step 3 and the given mass of the sample (0.205 g), then apply the mass percent formula: \[\text{Mass percent of SO\(_4^{2-}\)} = \frac{m_{\text{SO\(_4^{2-}\)}}}{m_{\text{sample}}} \cdot 100\%\] \[= \frac{0.1227\, \text{g}}{0.205\, \text{g}} \cdot 100\% = 59.85\%\]
05

Calculate percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\)

Since the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample is 59.85%, the remaining mass percent (i.e., 100% - 59.85% = 40.15%) must be contributed by the \(\mathrm{Na}^+\) and \(\mathrm{K}^+\) ions combined. The percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) are not given separately. We can find their combined mass percent by adding their mass percents, which is equal to the mass percent of \(\mathrm{SO}_{4}^{2-}\) we found in step 4. Therefore, the combined percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample is 40.15%.

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Key Concepts

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

Stoichiometry
Stoichiometry is one of the fundamental concepts in chemistry that focuses on the relative quantities of reactants and products in chemical reactions. It is like a recipe that tells us how much of each ingredient we need to make a desired product. In our exercise, the stoichiometry revolves around the formation of barium sulfate (BaSO\(_4\)) from the mixture containing sodium sulfate (\(\mathrm{Na}_{2} \mathrm{SO}_{4}\)) and potassium sulfate (\(\mathrm{K}_{2} \mathrm{SO}_{4}\)).

In particular, stoichiometry helps us determine the relationship between the moles of the solid mixture and the moles of barium sulfate formed. Each mole of BaCl\(_2\) reacts with one mole of sulfate ions (\(\mathrm{SO}_{4}^{2-}\)) to produce a mole of BaSO\(_4\).

Using the molar ratios established by the balanced chemical equation, we found that the moles of \(\mathrm{SO}_{4}^{2-}\) ions are equivalent to the moles of BaSO\(_4\) formed during the reaction. This principle of equivalent moles serves as the basis for further calculations.
Molar Mass Calculation
Calculating molar mass is a key part of solving problems in chemistry, especially when dealing with reactions and mixtures. Molar mass can be thought of as the weight of one mole of a substance. It helps us convert between grams and moles.

To solve the original exercise, we needed to find the molar mass of barium sulfate (BaSO\(_4\)) and sulfate ions \(\mathrm{SO}_{4}^{2-}\). This was done using the atomic masses of barium (Ba), sulfur (S), and oxygen (O) from the periodic table, which provide:
  • Molar mass of BaSO\(_4\) was found to be 233.43 g/mol, calculated by adding up the atomic masses of Ba (137.33), S (32.07), and O (16.00 each for four oxygen atoms).
  • For \(\mathrm{SO}_{4}^{2-}\), the molar mass was calculated as 96.07 g/mol, comprising of one sulfur and four oxygen atoms.
These calculations allowed us to convert the mass of BaSO\(_4\) formed in the reaction to moles, which was crucial for understanding the quantities involved in the reaction.
Mass Percent Composition
Understanding mass percent composition is essential in determining what percentage of a component is present in a mixture. Mass percent composition tells us how much of each element or compound is present in a sample relative to the whole.

In our exercise, we first calculated the mass of \(\mathrm{SO}_{4}^{2-}\) ions in the sample, which was found to be 0.1227 g. Using this value, along with the total mass of the sample (0.205 g), allowed us to calculate the mass percent of sulfate ions in the initial mixture. The formula used was:\[\text{Mass percent of \(\mathrm{SO}_{4}^{2-}\)} = \frac{m_{\text{SO}$$_{4^{2-}}\}}{m_{\text{Sample}}} \times 100\%\]This resulted in a 59.85% mass percent of \(\mathrm{SO}_{4}^{2-}\) ions.

The remaining mass percent, 40.15%, accounted for the rest of the mixture, made up of \(\mathrm{Na}_2 \mathrm{SO}_4\) and \(\mathrm{K}_2 \mathrm{SO}_4\). This knowledge helps in analyzing and understanding the sample's composition on a fundamental level.

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Most popular questions from this chapter

Chromium has been investigated as a coating for steel cans. The thickness of the chromium film is determined by dissolving a sample of a can in acid and oxidizing the resulting \(\mathrm{Cr}^{3+}\) to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) with the peroxydisulfate ion: $$\begin{aligned} \mathrm{S}_{2} \mathrm{O}_{\mathrm{x}}^{2-}(a q)+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \\ &+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q)(\text { Unbalanced }) \end{aligned} $$ After removal of unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) an excess of ferrous ammonium sulfate \(\left[\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right]\) is added, reacting with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produced from the first reaction. The unreacted \(\mathrm{Fe}^{2+}\) from the excess ferrous ammonium sulfate is titrated with a separate \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution. The reaction is: $$\begin{array}{c}{\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)} \\\ {+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \text { (Unbalanced) }}\end{array}$$ a. Write balanced chemical equations for the two reactions. b. In one analysis, a \(40.0-\mathrm{cm}^{2}\) sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2}-, 3.000 \mathrm{g}\) of \(\operatorname{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) was added. It took 8.58 \(\mathrm{mL}\) of 0.0520 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\) . Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 \(\mathrm{g} / \mathrm{cm}^{3} .\) )

When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 5.0 M HCl is required to react completely with 3.00 g of magnesium?

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{SrCr}_{2} \mathrm{O}_{7} \quad\) g. \(\mathrm{PbSO}_{3}\) b. \(\mathrm{CuCl}_{2} \quad \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\quad\quad\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\quad \mathrm{j} . \mathrm{CO}_{2}\) e. \(\mathrm{MgCO}_{3} \quad\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) f. \(\mathrm{Ag} \quad\quad\quad \)l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Citric acid, which can be obtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) . A 0.250 -g sample of citric acid dissolved in 25.0 mL of water requires 37.2 mL of 0.105 M NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?

A solution of permanganate is standardized by titration with oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) .\) It required 28.97 \(\mathrm{mL}\) of the permanganate solution to react completely with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{CO}_{2}(g)$$ What is the molarity of the permanganate solution?
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