   Chapter 3, Problem 152AE

Chapter
Section
Textbook Problem

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile (C3H3N), butadiene (C4H6), and styrene (C8H8).a. A sample of ABS plastic contains 8.80% N by mass. It took 0.605 g of Br2 to react completely with a 1.20-g sample of ABS plastic. Bromine reacts 1: 1 (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer?b. What are the relative numbers of each of the monomer units in this polymer?

(a)

Interpretation Introduction

Interpretation: The mass percent of acrylonitrile and butadiene is to be calculated. The relative numbers of each of the monomer units in the given polymer is to be calculated.

Concept introduction: Mass percent of an element is defined as the way by which the concentration of an element in a compound is expressed and it is calculated when the molar mass of the element is divided by the total molar mass of the compound and multiplied by hundred.

To determine: The mass percent of acrylonitrile and butadiene in the given polymer.

Explanation

Explanation

Given

Mass percent of nitrogen is 8.80% .

Mass of bromine is 0.605 g .

Mass of ABS plastic sample is 1.20 g .

Bromine reacts 1:1 (by moles) with butadiene that is 1 mol of bromine reacts with 1 mol of butadiene.

Therefore, the mass of butadiene is calculated by using the expression,

Mass of Br2Molar Mass of Br2=Mass of butadieneMolar Mass of butadiene

Molar mass of bromine is 160 g/mol .

Molar mass of butadiene is 54 g/mol .

Substitute the values of mass and molar mass of bromine and molar mass of butadiene in the above expression to calculate the mass of butadiene.

Mass of Br2Molar Mass of Br2=Mass of butadieneMolar Mass of butadiene0.605 g160 g/mol=Mass of butadiene54 g/molMass of butadiene=0.605×54160=0.204 g

Mass of nitrogen present is 8.80% of 1.20 g .

Mass of nitrogen=8.8100×1.2 g=0.1056 g .

14 Nitrogen means it contains 53 g of C3H3N .

1 Nitrogen means it contains 5314 g of C3H3N .

0.1056 Nitrogen means it contains C3H3N =5314×0.1056 g

Mass of C3H3N present is calculated as,

Mass of C3H3N=5314×0.1056=0.400 g

Mass of C8H8 present is calculated as,

Mass of C8H8 present=Total Mass of sampleMass of C3H3NMass of C4H6=1

(b)

Interpretation Introduction

Interpretation: The mass percent of acrylonitrile and butadiene is to be calculated. The relative numbers of each of the monomer units in the given polymer is to be calculated.

Concept introduction: Mass percent of an element is defined as the way by which the concentration of an element in a compound is expressed and it is calculated when the molar mass of the element is divided by the total molar mass of the compound and multiplied by hundred.

To determine: The relative numbers of each of the monomer units in the given polymer.

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