Chapter 23, Problem 8QAP

Interpretation Introduction

Interpretation:

From the given table, the polymer with the largest molar mass needs to be determined. Assuming the number of monomer units are same in all the polymers.

Concept introduction:

A polymer is a long chain consists of large number of monomer units. In a polymer, the monomers are linked to each other in a continuous or repetitive manner. These monomer units are linked to each other either through the formation of peptide linkage, glycosidic linkage or by removal of any moiety such as a water molecule. Polyvinyl chloride, Bakelite and polystyrene are some of the examples of polymers.

Mass percent of an atom present in the sample can be determined by dividing mass of atoms present in the monomer to the overall mass of the monomer unit and multiplying the overall result with 100%

For example, the mass percent of x g of an atom present in the y g of monomer unit can be determined as:

$M\text{ \% = }\left(\frac{x\text{ g}}{y\text{ g}}\right)\times \text{100\%}$

Answer to Problem 8QAP

Styrene is present with highest percentage by mass of carbon.

Explanation of Solution

If all the polymers contain the same number of monomer units, the molar mass of polymers can be compared by comparing the molar mass of the monomers in it.

(1)

In the table mentioned in the problem, the 1^st polymer contains ethylene as a monomer unit. Ethylene consists of 2 carbon and 4 hydrogen atoms. The molar mass can be calculated as follows:

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of H}\right)\left(\text{molar mass of H}\right)\end{array}\right)$

Putting the values,

$\begin{array}{c}\text{Molar mass of monomer unit = 2 }\times \text{ 12 g/mol + 4 }\times \text{ 1 g/mol}\\ \text{= 28 g/mol}\end{array}$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{ =}\left(\text{2 }\times \frac{\text{12 g/mol}}{\text{28 g/mol}}\right)\times \text{ 100 = 85}\text{.72 \%}$

(2)

The 2^nd polymer shown in table consists of propylene as a monomer unit. Propylene consists of 3 carbon and 6 hydrogen atoms.

The molar mass can be calculated as follows:

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of H}\right)\left(\text{molar mass of H}\right)\end{array}\right)$

Putting the values,

$\text{Molar mass of monomer unit = 3 }\times \text{ 12 + 6 }\times \text{ 1 = 42 g /mol}$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{ = }\left(\text{3 }\times \frac{\text{12 g/mol}}{\text{42 g/mol}}\right)\times \text{ 100 = 85}\text{.71 \%}$

(3)

The 3^rd polymer shown in table consists of vinyl chloride as monomer unit. Vinyl chloride consists of 2 carbon and 3 hydrogen and 1 chlorine atoms.

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of H}\right)\left(\text{molar mass of H}\right)\\ \text{+ }\left(\text{Number of Cl}\right)\left(\text{molar mass of Cl}\right)\end{array}\right)$

Putting the values,

$\begin{array}{c}\text{Molar mass of monomer unit = 2 }\times \text{ 12 + 3 }\times \text{ 1 + 1 }\times \text{ 35 }\\ \text{= 62 g /mol}\end{array}$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{= }\left(\text{2 }\times \frac{\text{12 g/mol}}{\text{62 g/mol}}\right)\times \text{100\% = 38}\text{.76 \%}$

(4)

The 4^th polymer shown in table consists of acrylonitrile as monomer unit. Acrylonitrile consists of 3 carbon, 3 hydrogen and 1 nitrogen atoms. The molar mass can be calculated as follows:

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of H}\right)\left(\text{molar mass of H}\right)\\ \text{+ }\left(\text{Number of N}\right)\left(\text{molar mass of N}\right)\end{array}\right)$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{= }\left(\text{3 }\times \frac{\text{12 g/mol}}{\text{53 g/mol}}\right)\times \text{100\% = 67}\text{.92 \%}$

(5)

The 5^th polymer shown in table consists of styrene as monomer unit. Styrene consists of 8 carbon and 8 hydrogen atoms. The molar mass can be calculated as follows:

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of H}\right)\left(\text{molar mass of H}\right)\end{array}\right)$

$\begin{array}{c}\text{Molar mass of monomer unit = 8 }\times \text{ 12 + 8 }\times \text{ 1 }\\ \text{= 104 g /mol}\end{array}$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{ =}\left(\text{8 }\times \frac{\text{12}}{\text{104}}\right)\times \text{ 100 = 95}\text{.68 \%}$

(6)

The 6^th polymer shown in table consists of methyl methacrylate as monomer unit. Methacrylate consists of 2 carbon and 2 oxygen atoms. The molar mass can be calculated as follow:

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of H}\right)\left(\text{molar mass of H}\right)\\ \text{+ }\left(\text{Number of O}\right)\left(\text{molar mass of O}\right)\end{array}\right)$

Putting the values,

$\begin{array}{c}\text{Molar mass of monomer unit = 5 × 12+ 8 × 1 + 2 × 16 }\\ \text{= 101 g /mol}\end{array}$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{ =}\left(\text{5 }\times \frac{\text{12}}{\text{101}}\right)\times \text{ 100 = 59}\text{.86 \%}$

(7)

The 7^th polymer shown in table consists of tetrafluoroethylene as monomer unit. Tetrafluoroethylene consists of 2 carbon and 4 fluorine atoms. The molar mass can be calculated as follows:

$\text{Molar mass of monomer unit = }\left(\begin{array}{l}\left(\text{Number of C}\right)\left(\text{molar mass of C}\right)\\ \text{+ }\left(\text{Number of F}\right)\left(\text{molar mass of F}\right)\end{array}\right)$

Putting the values,

$\begin{array}{c}\text{Molar mass of monomer unit= 2 }\times \text{ 12 + 4 }\times \text{ 18 }\\ \text{= 100 g /mol}\end{array}$

The mass percentage of carbon can be calculated as follows:

$\%m_C\text{ = }\left(\frac{M_{\text{C}}}{M_{\text{monomer unit}}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{ = }\left(\frac{\text{24}}{\text{100}}\right)\times \text{ 100 = 24 \%}$

Here, styrene consists of highest percentage by mass of carbon that is around 96 %. Thus, styrene is present with highest percentage by mass of carbon.

Conclusion

Thus, styrene is present with highest percentage by mass of carbon.