Chapter 23, Problem 17QAP

Interpretation Introduction

(a)

Interpretation:

The mass percent of C, H and O in cellulose needs to be determined.

Concept introduction:

Cellulose is present in plant cell wall and is a polysaccharide consists of several glucose units bind together.

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 17QAP

Mass percent of C in polymer is 42.46 %.

Mass percent of H in polymer is 7.66 %.

Mass percent of O in polymer is 54.34 %.

Explanation of Solution

Cellulose is a polyssaccharide which consists of several units of glucose joined through glycosidic linkage.

Molecular formula of glucose is C₆ H₁₂ O₅.

Molar mass of monomer unit glucose in cellulose can be calculated as follows:

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

Putting the values,

$\text{M= 6 }\times \text{ 12 + 12 }\times \text{ 1 + 5 }\times \text{ 16 = 164 g /mol}$

There are 6 C in a monomer unit of cellulose.

Now, molar mass of C in a monomer $\text{= 6 }\times \text{ 12 = 72 g /mol}$

Thus, the mass percent of C can be calculated as:

$\%m_C\text{ = }\left(\frac{M_C\text{ g}}{M_{\text{monomerunit}}\text{ g}}\right)\times \text{100\%}$

Putting the values,

$\%m_C\text{= }\frac{\text{72 g/mol}}{\text{164 g/mol}}\times \text{100\% = 42}\text{.46 \%}$

Thus, mass percent of C in polymer is 42.46 %.

There are 12 H in a monomer unit of cellulose.

Now, molar mass of H in a monomer $\text{= 12 }\times \text{ 1 = 12 g /mol}$

Mass percent of H can be calculated as follows:

$\%m_H\text{ = }\left(\frac{M_H\text{ g}}{M_{\text{monomerunit}}\text{ g}}\right)\times \text{100\%}$

Putting the values,

$\%m_H\text{ = }\frac{\text{12 }}{\text{164}}\times \text{ 100 = 7}\text{.66 \%}$

Thus, mass percent of H in polymer is 7.66 %.

There are 6 O in a monomer unit of cellulose.

Now, molar mass of O in a monomer $\text{= 6 }\times \text{ 16 = 96 g /mol}$

The mass percent of O can be calculated as:

$\%m_O\text{ = }\left(\frac{M_O\text{ g}}{M_{\text{monomerunit}}\text{ g}}\right)\times \text{100\%}$

Putting the values,

$\%m_O\text{= }\frac{\text{96 g/mol }}{\text{164 g/mol}}\times \text{ 100\% = 54}\text{.34 \%}$

Thus, mass percent of O in polymer is 54.34 %.

Interpretation Introduction

(b)

Interpretation:

The molar mass of the cellulose needs to be determined.

Concept introduction:

Cellulose is present in plant cell wall and is a polysaccharide consists of several glucose units in it.

The molar mass of any compound can be calculated by taking sum of molar masses of all the atoms present in that compound.

For a molecular formula of compound C_x H_y O_z, the molar mass can be calculated as follows:

Molar mass of compound = (Number of C) $\times$ (Molar mass of C) + (Number of H) $\times$ (Molar mass of H) + (Number of O) $\times$ (Molar mass of O)

Answer to Problem 17QAP

Molar mass of cellulose is $\text{1}\text{.8 }\times {\text{ 10}}^{\text{6}}\text{g /mol}$

Explanation of Solution

The molecular formula of glucose is C₆ H₁₂ O₆ which is linked with other glucose molecule through glycosidic linkage to form polysaccharides. The cellulose molecule is formed from the linkage of more than 100 glucose units.

Cellulose is a polyssaccharide which consists of several units of glucose joined through glycosidic linkage.

Molecular formula of glucose is C₆ H₁₂ O₆

Thus, its molar mass can be calculated as follows:

Molar mass of monomer unit glucose in cellulose = (Number of C) $\times$ (Molar mass of C) + (Number of H) $\times$ (Molar mass of H) + (Number of O) $\times$ (Molar mass of O)

Putting the values,

$\text{= 6 }\times \text{ 12 + 12 }\times \text{ 1 + 6 }\times \text{ 16 = 180 g /mol}$

Now,

Molar mass of cellulose = 10000 $\times$ Molar mass of monomer unit glucose in cellulose

Thus, the molar mass of cellulose will be:

$\text{Molar mass of cellulose= 10000 }\times \text{ 180 = 1}\text{.8 }\times {\text{ 10}}^{\text{6}}\text{g /mol}$

Thus, the molar mass of cellulose is $\text{1}\text{.8 }\times {\text{ 10}}^{\text{6}}\text{g /mol}$