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Chapter 1: Problem 66

A polystyrene of formula \(\mathrm{Br}_{3} \mathrm{C}_{6} \mathrm{H}_{2}\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)_{n}\) was prepared by heating styrene with tribromobenzyl peroxide in the absence of air. It was found to contain \(10.46 \%\) bromine, by mass. The value of \(n\) is \((\mathrm{Br}=80)\) (a) 20 (b) 21 (c) 19 (d) 22

Short Answer

Expert verified
n = 21

Step by step solution

01

Understand the Problem

We need to determine the value of 'n' in the polymer formula by using the given percentage of bromine by mass. The molecular weight of bromine (Br) is given as 80. The polymer is composed of a tribromobenzyl group and 'n' repeating units of styrene.
02

Calculate the Mass of Bromine in One Unit of the Polymer

Since there are three bromine atoms in the tribromobenzyl group, the total mass of bromine in one unit of the polymer is the molecular weight of bromine times three: \(3 \times 80 = 240\) g/mol.
03

Calculate the Mass of the Repeating Styrene Unit

Styrene has a formula \(C_8H_8\), and its molecular weight is \(8 \times 12 + 8 \times 1 = 96\) g/mol where the atomic weight of carbon (C) is 12 and of hydrogen (H) is 1.
04

Set Up the Equation

The mass percent of bromine in the polymer can be used to set up the equation: \(\frac{240}{240 + n \times 96} = \frac{10.46}{100}\). Solving for 'n' will give us the number of repeating styrene units in the polymer.
05

Solve for 'n'

Cross-multiply and solve the equation for 'n': \(240 \times 100 = 10.46 \times (240 + 96n) \Rightarrow 24000 = 2510.4 + 1006.56n \Rightarrow 24000 - 2510.4 = 1006.56n \Rightarrow 21489.6 = 1006.56n \Rightarrow n = \frac{21489.6}{1006.56} \approx 21.35\). Since 'n' needs to be a whole number, and it is closer to 21 than 22, the value of 'n' is 21.

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Key Concepts

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

Physical Chemistry and Polymer Synthesis
Physical chemistry offers a bridge between the theoretical world of chemistry and its practical applications, such as the synthesis of polymers. In the problem at hand, polystyrene synthesis involves a chemical reaction that chains together monomers of styrene in the presence of a tribromobenzyl peroxide initiator. This creation of large macromolecules or polymers is a cornerstone of materials chemistry, a sub-discipline of physical chemistry.

A clear understanding of the principles of physical chemistry, such as reaction mechanisms, stoichiometry, and the behavior of molecules, informs the process of synthesizing and characterizing new materials like the given polystyrene. The objective of calculating the degree of polymerization, indicated by the variable 'n', is a practical application of these principles that underpin the structure-property relationships in polymer chemistry.
Percent Composition Calculation
Understanding the percent composition by mass is crucial when working with complex compounds like polymers. It is the percentage by mass of each individual element within a compound. For the given polymer, the percent composition calculation was employed to find the amount of bromine in the compound, which speaks volumes about its properties and potential applications.

The step-by-step solution shows us how to use the percent composition of bromine, given as 10.46%, to calculate the number of styrene units in the polymer chain. By establishing the relationship between the mass of bromine atoms in a tribromobenzyl group and the total mass of the repeating styrene units, we can set up a proportionality equation to solve for 'n', the number of styrene units. Calculating the percent composition of chemical compounds is an integral part of chemistry that allows scientists to deduce empirical formulas and understand the quantitative makeup of substances.
Molecular Weight Determination
Molecular weight determination is a fundamental concept in chemistry and particularly in the synthesis and analysis of polymers. It encompasses the calculation of the mass of all atoms in a molecule. We apply this principle to determine the molecular weight of both the tribromobenzyl group and the repeating styrene unit in the polymer.

In the context of our polystyrene synthesis problem, it's critical to calculate the molecular weight of a single repeating unit, which is used in conjunction with its percent composition to find the degree of polymerization. The completed step-by-step solution efficiently employs molecular weights to set up and solve for the value of 'n', reflecting the number of styrene monomer units present in the polymer chain. This illustrates the significance of accurately determining molecular weights to deduce structural aspects of polymers.

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Most popular questions from this chapter

Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{P}_{t}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)

To determine soluble (free) \(\mathrm{SiO}_{2}\) in a rock. an alkaline extraction was carried out, as a result of which there was found \(1.52 \%\) of \(\mathrm{SiO}_{2}\) in the extract and also \(1.02 \%\) of \(\mathrm{Al}_{2} \mathrm{O}_{3} .\) Considering that, apart from the free \(\mathrm{SiO}_{2}\), the extract also contained the \(\mathrm{SiO}_{2}\) that had passed into it from Kaolin \(\left(2 \mathrm{SiO}_{2} \cdot \mathrm{Al}_{2} \mathrm{O}_{3}\right)\), the percentage of free \(\mathrm{SiO}_{2}\) in the rock being analysed is \((\mathrm{Si}=28\), \(\mathrm{Al}=27\) ) (a) \(1.20\) (b) \(0.32\) (c) \(0.50\) (d) \(1.52\)

The legal limit for human exposure to \(\mathrm{CO}\) in the work place is 35 ppm. Assuming that the density of air is \(1.3 \mathrm{~g} / 1\), how many grams of \(\mathrm{CO}\) are in \(1.0 \mathrm{l}\) of air at the maximum allowable concentration? (a) \(4.55 \times 10^{-5} \mathrm{~g}\) (b) \(3.5 \times 10^{-5} \mathrm{~g}\) (c) \(2.69 \times 10^{-5} \mathrm{~g}\) (d) \(7.2 \times 10^{-5} \mathrm{~g}\)

Iodobenzene is prepared from aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) in a two-step process as shown here: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{HNO}_{2}+\mathrm{HCl} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+2 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+\mathrm{KI} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}+\mathrm{N}_{2}+\mathrm{KCl}\) In an actual preparation, \(9.30 \mathrm{~g}\) of aniline was converted to \(16.32 \mathrm{~g}\) of iodobenzene. The percentage yield of iodobenzene is \((\mathrm{I}=127)\) (a) 8\% (b) \(50 \%\) (c) \(75 \%\) (d) \(80 \%\)

A mixture is made equal volume of \(\mathrm{CO}\) and air. A spark passed through so that all the oxygen is converted to carbon dioxide. What will be fractional decrease in the total volume of system assuming pressure and temperature remain constant? Air contains \(20 \%\) oxygen by volume. (a) \(0.1\) (b) \(0.2\) (c) \(0.15\) (d) \(0.3\)
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