Chapter 3, Problem 152AE

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?

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.

Answer to Problem 152AE

The mass percent of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is $\underset{\_}{33.33\%}$.

The mass percent of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ is $\underset{\_}{17.0\%}$.

Explanation of Solution

Explanation

Given

Mass percent of nitrogen is $8.80\%$.

Mass of bromine is $0.60\text{5 g}$.

Mass of ABS plastic sample is $1.2\text{0 g}$.

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,

$\frac{{\text{Mass of Br}}_{\text{2}}}{{\text{Molar Mass of Br}}_{\text{2}}}=\frac{\text{Mass of butadiene}}{\text{Molar Mass of butadiene}}$

Molar mass of bromine is $16\text{0 g/mol}$.

Molar mass of butadiene is $\text{54 g/mol}$.

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.

$\begin{array}{c}\frac{{\text{Mass of Br}}_{\text{2}}}{{\text{Molar Mass of Br}}_{\text{2}}}=\frac{\text{Mass of butadiene}}{\text{Molar Mass of butadiene}}\\ \frac{0.60\text{5 g}}{16\text{0 g/mol}}=\frac{\text{Mass of butadiene}}{5\text{4 g/mol}}\\ \text{Mass of butadiene}=\frac{0.605\times 54}{160}\\ =0.20\text{4 g}\end{array}$

Mass of nitrogen present is $8.80\%$ of $1.2\text{0 g}$.

$\begin{array}{c}\text{Mass of nitrogen}=\frac{8.8}{100}\times 1.\text{2 g}\\ =0.105\text{6 g}\end{array}$.

$14$ Nitrogen means it contains $5\text{3 g}$ of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$.

$1$ Nitrogen means it contains $\frac{53}{14}\text{ g}$ of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$.

$0.1056$ Nitrogen means it contains ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ $=\frac{53}{14}\times 0.105\text{6 g}$

Mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ present is calculated as,

$\begin{array}{c}{\text{Mass of C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}=\frac{53}{14}\times 0.1056\\ =0.40\text{0 g}\end{array}$

Mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$ present is calculated as,

$\begin{array}{c}{\text{Mass of C}}_{\text{8}}{\text{H}}_{\text{8}}\text{ present}=\text{Total Mass of sample}-{\text{Mass of C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}\\ -{\text{Mass of C}}_{\text{4}}{\text{H}}_{\text{6}}\\ =1.\text{2 g}-0.40\text{0 g}-0.2\text{04 g}\\ =0.596\text{ g}\end{array}$

Now, mass percent of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is calculated by using the expression,

${\text{Mass percent of C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}=\frac{{\text{ Mass of C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}{\text{Total Mass of ABS sample}}\times 100$.

Substitute the value of mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ and mass of sample in the above expression to calculate the mass percent.

$\begin{array}{c}{\text{Mass percent of C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}=\frac{{\text{ Mass of C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}}{\text{Total Mass of ABS sample}}\times 100\\ =\frac{0.40\text{0 g}}{1.\text{2 g}}\times 100\\ =\underset{\_}{33.33\%}\end{array}$

Hence, the mass percent of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is $\underset{\_}{33.33\%}$.

The mass percent of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ is calculated by using the expression,

${\text{Mass percent of C}}_4{\text{H}}_6=\frac{{\text{ Mass of C}}_4{\text{H}}_6\text{N}}{\text{Total Mass of ABS sample}}\times 100$.

Substitute the value of mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ and mass of sample in the above expression to calculate the mass percent.

$\begin{array}{c}{\text{Mass percent of C}}_4{\text{H}}_6=\frac{{\text{ Mass of C}}_4{\text{H}}_6}{\text{Total Mass of ABS sample}}\times 100\\ =\frac{0.204\text{ g}}{1.\text{2 g}}\times 100\\ =\underset{\_}{17.0\%}\end{array}$

Hence, the mass percent of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ is $\underset{\_}{17.0\%}$.

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.

Answer to Problem 152AE

The relative numbers of each of the monomer units in the given polymer is $\underset{\_}{2:1:1.5}$.

Explanation of Solution

Explanation

Given

Mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is $0.400\text{ g}$

Mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ is $0.204\text{ g}$

Mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$ is $0.59\text{6 g}$,

Molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$ is $5\text{3 g/mol}$.

Molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ is $\text{54 g/mol}$

Molar mass of ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$ is $\text{104 g/mol}$

The number of moles is calculated by using the expression,

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$.

Substitute the value of mass and molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$, ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$ and ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$ to calculate their number of moles in the above expression.

For ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$,

$\begin{array}{c}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ =\frac{0.40\text{0 g}}{5\text{3 g/mol}}\\ =75\text{5 mol}\end{array}$

For, ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$

$\begin{array}{c}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ =\frac{0.204\text{ g}}{54\text{ g/mol}}\\ =378\text{ mol}\end{array}$

For, ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$

$\begin{array}{c}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ =\frac{0.596\text{ g}}{\text{104 g/mol}}\\ =573\text{ mol}\end{array}$

The calculated values are divided by the smallest number of moles to determine the simplest whole number ratio of moles of each constituent.

For ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}$,

${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}\text{N}=\frac{755\text{mol}}{378\text{mol}}\approx 2$ w

For ${\text{C}}_{\text{4}}{\text{H}}_{\text{6}}$,

${\text{C}}_4{\text{H}}_6=\frac{378\text{mol}}{378\text{mol}}=1$

For ${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}$,

${\text{C}}_{\text{8}}{\text{H}}_{\text{8}}=\frac{573\text{mol}}{378\text{mol}}=1.5$

The relative numbers of each of the monomer units in the given polymer is $\underset{\_}{2:1:1.5}$.