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Chapter 8: Problem 17

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

The earliest determinations of elemental atomic weights were accomplished gravimetrically. To determine the atomic weight of manganese, a carefully purified sample of \(\mathrm{MnBr}_{2}\) weighing \(7.16539 \mathrm{~g}\) is dissolved and the \(\mathrm{Br}^{-}\) precipitated as \(\mathrm{AgBr}\), yielding \(12.53112 \mathrm{~g}\). What is the atomic weight for \(\mathrm{Mn}\) if the atomic weights for \(\mathrm{Ag}\) and \(\mathrm{Br}\) are taken to be 107.868 and 79.904 , respectively?

The amount of iron and manganese in an alloy is determined by precipitating the metals with 8 -hydroxyquinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NO}\). After weighing the mixed precipitate, the precipitate is dissolved and the amount of 8-hydroxyquinoline determined by another method. In a typical analysis a 127.3 -mg sample of an alloy containing iron, manganese, and other metals is dissolved in acid and treated with appropriate masking agents to prevent an interference from other metals. The iron and manganese are precipitated and isolated as \(\mathrm{Fe}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}\) and \(\mathrm{Mn}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2},\) yielding a total mass of \(867.8 \mathrm{mg}\). The amount of 8 -hydroxyquinolate in the mixed precipitate is determined to be \(5.276 \mathrm{mmol}\). Calculate the \(\% \mathrm{w} / \mathrm{w} \mathrm{Fe}\) and \(\% \mathrm{w} / \mathrm{w} \mathrm{Mn}\) in the alloy.

A sample of an impure iron ore is approximately \(55 \% \mathrm{w} / \mathrm{w} \mathrm{Fe}\). If the amount of Fe in the sample is determined gravimetrically by isolating it as \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) what mass of sample is needed to ensure that we isolate at least \(1.0 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3} ?\)

In the presence of water vapor the surface of zirconia, \(\mathrm{ZrO}_{2}\), chemically adsorbs \(\mathrm{H}_{2} \mathrm{O},\) forming surface hydroxyls, \(\mathrm{ZrOH}\) (additional water is physically adsorbed as \(\mathrm{H}_{2} \mathrm{O}\) ). When heated above \(200^{\circ} \mathrm{C}\), the surface hydroxyls convert to \(\mathrm{H}_{2} \mathrm{O}(g),\) releasing one molecule of water for every two surface hydroxyls. Below \(200^{\circ} \mathrm{C}\) only physically absorbed water is lost. Nawrocki, et al. used thermogravimetry to determine the density of surface hydroxyls on a sample of zirconia that was heated to \(700^{\circ} \mathrm{C}\) and cooled in a desiccator containing humid \(\mathrm{N}_{2}{ }^{15}\) Heating the sample from \(200^{\circ} \mathrm{C}\) to \(900^{\circ} \mathrm{C}\) released \(0.006 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) for every gram of dehy- droxylated \(\mathrm{ZrO}_{2}\). Given that the zirconia had a surface area of \(33 \mathrm{~m}^{2} / \mathrm{g}\) and that one molecule of \(\mathrm{H}_{2} \mathrm{O}\) forms two surface hydroxyls, calculate the density of surface hydroxyls in \(\mu \mathrm{mol} / \mathrm{m}^{2}\).

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?
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