Chapter 4, Problem 27P

4-43 Calculate the molecular weight of:

(a) Sucrose, C₁₂H₂₂O₁₁

(b) Glycine, C₂H₅NO₂

(C) DDT, C₁₄H₉Cl₅

Interpretation Introduction

(a)

Interpretation:

To determine the molecular weight of the compound sucrose ${\mathrm{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - ${\text{H}}_2\text{O}$

Molecular weight is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=2+16\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$.

Answer to Problem 27P

The Molecular weight of sucrose ${\mathrm{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$

= 342amu.

Explanation of Solution

Molecular weight is the sum of masses of the atoms present in a molecule.

Sucrose ${\mathrm{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$

Molecular weight is calculated as:

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=Mass\text{ 12 }of\text{C}\text{a}rb\text{O}n\text{ atom }+\text{ 22 }\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}m+\text{ 1}1oxyg\text{e}n\text{a}t\text{O}m\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=\left(12\times 12\right)+\left(22\times 1\right)+\left(11\times 16\right)\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=342\text{ amu}\hfill \end{array}$.

Interpretation Introduction

(b)

Interpretation:

To determine the masses of the compound glycine ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NO}}_{\text{2}}$

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - ${\text{H}}_2\text{O}$

Molecular weight is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=2+16\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$.

Answer to Problem 27P

The Molecular weight of glycine, ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NO}}_{\text{2}}$

=75amu.

Explanation of Solution

Molecular weight is the sum of masses of the atoms present in a molecule.

Glycine ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NO}}_{\text{2}}$

Molecular weight is calculated as:

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_2{\text{H}}_5\text{N}{\text{O}}_2=Mass\text{ 2 }of\text{C}\text{a}rb\text{O}n\text{ atom }+\text{ 5 }\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}m+\text{ 1 }\text{N}\text{i}tr\text{O}g\text{e}n\text{a}t\text{O}m+\text{ 2 }oxyg\text{e}n\text{a}t\text{O}m\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_2{\text{H}}_5\text{N}{\text{O}}_2=\left(2\times 12\right)+\left(5\times 1\right)+\left(1\times 14\right)+\left(2\times 16\right)\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_2{\text{H}}_5\text{N}{\text{O}}_2=75\text{ amu}\hfill \end{array}$.

Interpretation Introduction

(c)

Interpretation:

To determine the masses of the compound DDT, ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}$

Concept Introduction:

Molecular weight is the sum of masses of the atoms present in a molecule.

When we calculate the sum of masses of the elements present in a compound in 1 mole, we calculate it for the Avogadro number, molar mass comes out in grams.

For e.g. - ${\text{H}}_2\text{O}$

Molecular weight is calculated as sum of 2 Hydrogen atoms and 1 Oxygen atom.

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=Mass2\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}ms+1 oxyg\text{e}n\text{a}t\text{O}m.\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=2+16\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{H}}_2\text{O}=18\text{ amu}\hfill \end{array}$.

Answer to Problem 27P

The Molecular weight of DDT, ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}$

= 354.5 amu.

Explanation of Solution

Molecular weight is the sum of masses of the atoms present in a molecule.

DDT, ${\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}$

Molecular weight is calculated as:

$\begin{array}{l}M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_{14}{\text{H}}_9\text{C}{\text{l}}_5=Mass\text{ 14 }of\text{C}\text{a}rb\text{O}n\text{ atom }+\text{ 9 }\text{H}ydr\text{O}g\text{e}n\text{a}t\text{O}m+\text{ 5 }\text{C}\text{H}\text{l}\text{O}rin\text{e}\text{a}t\text{O}m\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_{14}{\text{H}}_9\text{C}{\text{l}}_5=\left(14\times 12\right)+\left(9\times 1\right)+\left(5\times 35.5\right)\hfill \\ M\text{O}\text{l}\text{e}\text{C}u\text{l}\text{a}rw\text{e}\text{i}g\text{H}tof{\text{C}}_{14}{\text{H}}_9\text{C}{\text{l}}_5=354.5\text{ amu}\hfill \end{array}$.