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Chapter 4: Problem 22

Black smokers are found in the depths of the oceans (page 110 ). Thinking that the conditions in these smokers might be conducive to the formation of organic compounds, two chemists in Germany found the following reaction could occur in similar conditions. $$ 2 \mathrm{CH}_{3} \mathrm{SH}+\mathrm{CO} \rightarrow \mathrm{CH}_{3} \mathrm{COSCH}_{3}+\mathrm{H}_{2} \mathrm{S} $$ If you begin with \(10.0 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{SH}\) and excess \(\mathrm{CO}\) (a) What is the theoretical yield of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3} ?\) (b) If \(8.65 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{COSCH}_{3}\) is isolated, what is its percent yield?

Short Answer

Expert verified
(a) Theoretical yield: 9.39 g; (b) Percent yield: 92.09%.

Step by step solution

01

Calculate Molar Mass

First, calculate the molar mass of the compounds involved. For \(\mathrm{CH}_{3}\mathrm{SH}\): \(\mathrm{C} = 12.01\, \mathrm{g/mol}, \ \mathrm{H} = 1.01\, \mathrm{g/mol}\) will contribute \(3 \times 1.01\, \mathrm{g/mol}, \ \mathrm{S} = 32.07\, \mathrm{g/mol}\). The molar mass of \(\mathrm{CH}_{3}\mathrm{SH}\) is thus \(12.01 + 3.03 + 32.07 = 48.11\, \mathrm{g/mol}\). The molar mass for \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\) is calculated similarly: \(2 \times (12.01 + 3.03) + 32.07 + 16.00 = 90.20\, \mathrm{g/mol}\).
02

Calculate Moles of Reactant

Determine the moles of \(\mathrm{CH}_{3}\mathrm{SH}\): \[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0\, \mathrm{g}}{48.11\, \mathrm{g/mol}} = 0.208\, \mathrm{mol}\]
03

Determine Theoretical Yield of Product

Using the balanced reaction, \(2 \mathrm{CH}_{3}\mathrm{SH}\) produces \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\), therefore half the amount of \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\) moles is produced: \[0.208\, \mathrm{mol} \times \frac{1}{2} = 0.104\, \mathrm{mol}\] Now, calculate the mass of \(\mathrm{CH}_{3}\mathrm{COSCH}_{3}\) produced theoretically: \[0.104\, \mathrm{mol} \times 90.20\, \mathrm{g/mol} = 9.39\, \mathrm{g}\]
04

Calculate Percent Yield

Percent yield is calculated using \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\] Substitute the known values: \[\frac{8.65\, \mathrm{g}}{9.39\, \mathrm{g}} \times 100 = 92.09\%\]

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Key Concepts

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

Molar Mass
Understanding molar mass is crucial when performing chemical reactions involving specific amounts of substances. Molar mass, often expressed in grams per mole (g/mol), represents the mass of one mole of a substance. Each element has a specific atomic mass contributing to the total molar mass of a compound.
For example, in the given reaction, you need to know the molar mass of both reactants and products to determine yields.
  • The molar mass of a compound is calculated by summing the atomic masses of all atoms present in its formula.
  • For instance, to find the molar mass of \(\mathrm{CH}_3\mathrm{SH}\), you add the carbon (C), hydrogen (H), and sulfur (S) contributions. The calculation is \(12.01 + 3 \times 1.01 + 32.07 = 48.11\, \mathrm{g/mol}\).
Molar mass is essential for determining the number of moles, a critical step in calculating theoretical and percent yields.
Percent Yield
Percent yield is a measure of efficiency in a chemical reaction. It compares the actual yield鈥攖he amount of product actually obtained from a reaction鈥攖o the theoretical yield, which is the maximum amount possible under perfect conditions.
  • To calculate percent yield, use the formula: \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]
  • For example, if your theoretical yield of \(\mathrm{CH}_3\mathrm{COSCH}_3\) is \(9.39\, \mathrm{g}\) and the actual yield is \(8.65\, \mathrm{g}\), your percent yield would be \(\frac{8.65}{9.39} \times 100 = 92.09\%\).
Understanding percent yield helps identify how effective a reaction was, taking into account losses due to incomplete reactions or side reactions.
Chemical Reaction
A chemical reaction involves the transformation of reactants into products, typically represented by a balanced chemical equation. In these equations, substances (molecules, atoms, or ions) undergo bond formation or breakage, resulting in new substances.
  • The equation provided, \(2 \mathrm{CH}_3\mathrm{SH} + \mathrm{CO} \rightarrow \mathrm{CH}_3\mathrm{COSCH}_3 + \mathrm{H}_2\mathrm{S}\), shows how two molecules of \(\mathrm{CH}_3\mathrm{SH}\) react with one molecule of carbon monoxide (CO) to produce one molecule of \(\mathrm{CH}_3\mathrm{COSCH}_3\) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)).
  • A balanced equation ensures that the number of each type of atom is equal on both sides, reflecting the law of conservation of mass.
Understanding the stoichiometry of a chemical reaction allows you to calculate the theoretical yield and assess how much of each reactant is needed.

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Most popular questions from this chapter

Which of the following methods would you use to prepare \(1.00 \mathrm{L}\) of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) Dilute \(20.8 \mathrm{mL}\) of \(6.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) to a volume of \(1.00 \mathrm{L}\) (b) Add \(950 .\) mL. of water to 50.0 mL of \(3.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

At higher temperatures, NaHCO is converted quantitatively to \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) $$ 2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ Heating a \(1.7184-\mathrm{g}\) sample of impure NaHCO \(_{3}\) gives \(0.196 \mathrm{g}\) of \(\mathrm{CO}_{2} .\) What was the mass percent of \(\mathrm{NaHCO}_{3}\) in the original 1.7184 -g sample?

Iodine is made by the following reaction \(2 \mathrm{NaIO}_{3}(\mathrm{aq})+5 \mathrm{NaHSO}_{3}(\mathrm{aq}) \rightarrow\) $$ 3 \mathrm{NaHSO}_{4}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{I}_{2}(\mathrm{aq}) $$ (a) Name the two reactants. (b) If you wish to prepare \(1.00 \mathrm{kg}\) of \(\mathrm{I}_{2},\) what masses of \(\mathrm{NalO}_{3}\) and \(\mathrm{NaHSO}_{3}\) are required? (c) What is the theoretical yield of \(I_{2}\) if you mixed \(15.0 \mathrm{g}\) of \(\mathrm{NaIO}_{3}\) with \(125 \mathrm{mL}\) of \(0.853 \mathrm{M} \mathrm{NaHSO}_{3} ?\)

A Copper metal can be prepared by roasting copper ore, which can contain cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right)\) and copper (11) sulfide. $$ \begin{aligned} \mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{CuS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) & \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \end{aligned} $$ Suppose an ore sample contains \(11.0 \%\) impurity in addition to a mixture of CuS and Cu \(_{2} \mathrm{S}\). Heating \(100.0 \mathrm{g}\) of the mixture produces \(75.4 \mathrm{g}\) of copper metal with a purity of \(89.5 \% .\) What is the weight percent of CuS in the ore? The weight percent of \(\mathrm{Cu}_{2} \mathrm{S} ?\)

A A compound has been isolated that can have either of two possible formulas: (a) \(\mathrm{K}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) or (b) \(\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right] .\) To find which is correct, you dissolve a weighed sample of the compound in acid, forming oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). You then titrate this acid with potassium permanganate, \(\mathrm{KMnO}_{4}\) (the source of the \(\left.\mathrm{MnO}_{4}-\text { ion }\right) .\) The balanced, net ionic equation for the titration is $$ \begin{aligned} 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+& 2 \mathrm{MnO}_{4}-(\mathrm{aq})+6 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow \\ & 2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_{2}(\mathrm{g})+14 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned} $$ Titration of \(1.356 \mathrm{g}\) of the compound requires \(34.50 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{KMnO}_{4} .\) Which is the correct formula of the iron-containing compound: (a) or (b)?
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