Chapter 6, Problem 6.160AP

(a)

Interpretation Introduction

Interpretation: A cotton ball soaked in either hydrochloric acid or acetic acid is given to be placed at one end and another cotton ball soaked in one of three amines placed in the other end. The given questions based upon the above statement are to be answered.

Concept introduction: Graham’s law of diffusion states that. “the rate of diffusion of gases is inversely proportional to the square root of their molar masses”.

To determine: The acid and amine used to form a white ring exactly halfway between the two ends.

Answer to Problem 6.160AP

Solution

The combination of ${\text{CH}}_{\text{3}}\text{COOH}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}$ that is used to form the ring exactly halfway between the two ends.

Explanation of Solution

Explanation

The molecular weight of given gases is calculated as,

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}\text{HCl}=\text{H}+\text{Cl}\\ =1.0079\text{g/mol}+35.453\text{g/mol}\\ =\text{36}\text{.460}\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=2\text{C}+\text{4H}+\text{2O}\\ =\left(\begin{array}{l}2\times 12.0107\text{g/mol}+4\times 1.0079\text{g/mol}\\ +\text{2}\times \text{15}\text{.999}\text{g/mol}\end{array}\right)\\ =60.049\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}=1\text{C}+\text{5H}+1\text{N}\\ =\left(\begin{array}{l}12.0107\text{g/mol}+\text{5}\times \text{1}\text{.0079}\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =31.0569\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}=2\text{C}+\text{7H}+1\text{N}\\ =\left(\begin{array}{l}2\times 12.0107\text{g/mol}+7\times 1.0079\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =45.0834\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}=3\text{C}+9\text{H}+1\text{N}\\ =\left(\begin{array}{l}3\times 12.0107\text{g/mol}+7\times 1.0079\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =59.1097\text{g/mol}\end{array}$

If the molecular masses of acid and amine are equal, then the ring will be observed almost exactly halfway between the two ends.

Hence, the combination of acid and amine is ${\text{CH}}_{\text{3}}\text{COOH}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$.

(b)

Interpretation Introduction

To determine: The combination of acid and amine that would produce a ring closest to the amine end of the tube.

Answer to Problem 6.160AP

Solution

The combination of $\text{HCl}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}$ is used to form the ring closest to the amine end.

Explanation of Solution

Explanation

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}\text{HCl}=1.0079\text{g/mol}+35.453\text{g/mol}\\ =\text{36}\text{.460}\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_2\text{N}=\left(\begin{array}{l}2\times 12.0107\text{g/mol}+7\times 1.0079\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =45.0834\text{g/mol}\end{array}$

The lighter molecules diffuse more rapidly than heavier ones. The combination of $\text{HCl}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}$ is used to form the ring closet to the amine end.

(c)

Interpretation Introduction

To determine: The two combinations that result in the formation of product at the same position.

Answer to Problem 6.160AP

Solution

The two combinations that result in the formation of product at the same position are ${\text{CH}}_{\text{3}}\text{COOH}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ and ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}\text{HCl}$.

Explanation of Solution

Explanation

The molecular weight of given gases is calculated by the formula,

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}\text{HCl}=\text{H}+\text{Cl}\\ =1.0079\text{g/mol}+35.453\text{g/mol}\\ =\text{36}\text{.460}\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=2\text{C}+\text{4H}+\text{2O}\\ =\left(\begin{array}{l}2\times 12.0107\text{g/mol}+4\times 1.0079\text{g/mol}\\ +\text{2}\times \text{15}\text{.999}\text{g/mol}\end{array}\right)\\ =60.049\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}=1\text{C}+\text{5H}+1\text{N}\\ =\left(\begin{array}{l}12.0107\text{g/mol}+\text{5}\times \text{1}\text{.0079}\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =31.0569\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}=2\text{C}+\text{7H}+1\text{N}\\ =\left(\begin{array}{l}2\times 12.0107\text{g/mol}+7\times 1.0079\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =45.0834\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molecular}\text{weight}\text{of}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}=3\text{C}+9\text{H}+1\text{N}\\ =\left(\begin{array}{l}3\times 12.0107\text{g/mol}+7\times 1.0079\text{g/mol}\\ +14.0067\text{g/mol}\end{array}\right)\\ =59.1097\text{g/mol}\end{array}$

According to grahams’ law of diffusion, the rate of diffusion is inversely proportional to the molar mass of the compound.

As the molar mass of two compounds is close to each other; therefore, the rate of diffusion for those compounds is equal and a ring formation is observed at the middle of the tube.

Hence, when ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ react with each other the ring is formed at the middle of the tube because the molecular masses of both compounds are comparatively same.

Similarly, when ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}\text{HCl}$ react with each other, the ring is formed at the middle of the tube due to close molecular masses.

Therefore, the two combinations that result in the formation of product at the same position are ${\text{CH}}_{\text{3}}\text{COOH}$ and ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ and ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ and $\text{HCl}$.

Conclusion

1. a. The combination of ${\text{CH}}_{\text{3}}\text{COOH}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}$ that is used to form the ring exactly halfway between the two ends.
2. b. The combination of $\text{HCl}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_3\text{N}$ is used to form the ring closet to the amine end.
3. c. The two combinations that result in the formation of product at the same position are ${\text{CH}}_{\text{3}}\text{COOH}\text{and}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{N}$ and ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{and}\text{HCl}$