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The number of ethoxy groups \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}-\right)\) in an organic compound is determined by the following two reactions. $$ \begin{array}{c} \mathrm{R}\left(O \mathrm{CH}_{2} \mathrm{CH}_{3}\right)_{x}+x \mathrm{HI} \longrightarrow \mathrm{R}(\mathrm{OH})_{x}+x \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \\ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{Ag}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{AgI}(s)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \end{array} $$ A 36.92 -mg sample of an organic compound with an approximate molecular weight of 176 is treated in this fashion, yielding \(0.1478 \mathrm{~g}\) of \(\mathrm{AgI}\). How many ethoxy groups are there in each molecule of the compound?

Short Answer

Expert verified
There are approximately 3 ethoxy groups per molecule of the compound.

Step by step solution

01

Understanding the Problem

We are trying to determine the number of ethoxy groups in a compound using two chemical reactions. These reactions indicate that for each ethoxy group, one mole of silver iodide (AgI) is formed.
02

Calculate Moles of Silver Iodide

First, calculate the number of moles of AgI formed. Using the given mass of AgI (0.1478 g) and the molar mass of AgI (234.77 g/mol), we calculate the moles of AgI produced. \[ \text{Moles of AgI} = \frac{0.1478 \text{ g}}{234.77 \text{ g/mol}} \approx 6.293 \times 10^{-4} \text{ moles} \]
03

Determine Moles of Ethoxy Groups

Since each ethoxy group leads to the formation of one mole of AgI, the moles of AgI calculated are equal to the moles of ethoxy groups in the 36.92 mg of the compound. Thus, the moles of ethoxy groups are approximately \(6.293 \times 10^{-4}\).
04

Find Number of Ethoxy Groups in One Molecule

First, calculate the number of moles of the organic compound in the 36.92 mg sample: \[ \text{Moles of compound} = \frac{36.92 \text{ mg}}{176 \text{ g/mol}} = \frac{36.92 \times 10^{-3} \text{ g}}{176 \text{ g/mol}} \approx 2.098 \times 10^{-4} \text{ moles} \]Since these moles contain \(6.293 \times 10^{-4}\) moles of ethoxy groups, calculate the number of ethoxy groups per molecule by dividing moles of ethoxy groups by moles of compound:\[ \text{Number of ethoxy groups per molecule} = \frac{6.293 \times 10^{-4}}{2.098 \times 10^{-4}} \approx 3 \]
05

Conclusion

We have determined that there are approximately 3 ethoxy groups per molecule of the compound.

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Key Concepts

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

Ethoxy Groups Determination
Determining ethoxy groups in an organic compound involves finding out how many ethoxy groups (\(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{O}-\)) are present in each molecule. This is crucial in understanding the compound's composition and functionality. The procedure typically involves using chemical reactions to encourage the release or transformation of ethoxy groups into a measurable product.
In the provided problem, a sample compound reacts to form silver iodide (AgI) whenever an ethoxy group is displaced. Hence, by measuring the amount of AgI produced, it becomes possible to reverse-calculate the number of ethoxy groups. Such quantitative analysis is invaluable in both research and industry for accurate compound characterization.
Chemical Reactions in Organic Compounds
Analyzing organic compounds involves using specific chemical reactions to dissect complex molecules into measurable parts. Two key reactions help in determining ethoxy groups:
  • First Reaction: \(\mathrm{R}(\mathrm{OCH}_2\mathrm{CH}_3)_x + x \mathrm{HI} \to \mathrm{R}(\mathrm{OH})_x + x \mathrm{CH}_3\mathrm{CH}_2\mathrm{I} \)
    Here, iodine from HI replaces the ethoxy group, converting it to ethyl iodide (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{I}\)).
  • Second Reaction: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I} + \mathrm{Ag}^+ + \mathrm{H}_2 \mathrm{O} \to \mathrm{AgI}(s) + \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \)
    This reaction further pushes the transformation, with silver ions (\(\mathrm{Ag}^+\)) aiding in precipitating silver iodide as a solid for quantitative analysis.
This systematic use of reactions highlights the precision with which reactions manipulate and reveal structured details of organic compounds.
Mole Calculations
Using mole calculations allows chemists to link measurable masses to atomic levels, bridging macroscopic and molecular scales. Moles give a count of particles based on their atomic or molecular weights, expressing them in terms compatible with macroscopic observations.
For the silver iodide formed: \[ \text{Moles of AgI} = \frac{0.1478 \text{ g}}{234.77 \text{ g/mol}} \approx 6.293 \times 10^{-4} \text{ moles} \]
These moles directly correspond to the number of moles of ethoxy groups in the compound since every ethoxy group leads to an equivalent mole of AgI formed.
Molecular Weight in Analysis
Molecular weight plays a crucial role in quantitative analysis. It serves as the bridge linking molecular structure to measurable mass. With molecular weight, one can determine the moles of a substance when its mass is known. This is crucial for calculations in analytical chemistry.
For example, in our compound sample with a molecular weight of 176 g/mol, it is possible to calculate how many moles exist in a known mass:\[ \text{Moles of compound} = \frac{36.92 \text{ mg}}{176 \text{ g/mol}} = \frac{36.92 \times 10^{-3} \text{ g}}{176 \text{ g/mol}} \approx 2.098 \times 10^{-4} \text{ moles} \]
With moles calculated, we determine molecular composition, such as how many functional groups per molecule are present. This is essential for understanding chemical reactivity and properties.

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Most popular questions from this chapter

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?

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?

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 concentration of arsenic in an insecticide is determined gravimetrically by precipitating it as \(\mathrm{MgNH}_{4} \mathrm{AsO}_{4}\) and isolating it as \(\mathrm{Mg}_{2} \mathrm{As}_{2} \mathrm{O}_{7}\) Determine the \(\% \mathrm{w} / \mathrm{w} \mathrm{As}_{2} \mathrm{O}_{3}\) in a \(1.627-\mathrm{g}\) sample of insecticide if \(\mathrm{it}\) yields \(106.5 \mathrm{mg}\) of \(\mathrm{Mg}_{2} \mathrm{As}_{2} \mathrm{O}_{7}\)

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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