Chapter 10, Problem 10.141QA

Interpretation Introduction

To find:

a) The combination of an acid and an amine, which gives a white ring halfway between the two ends.

b) The combination of an acid and an amine pair, which produces a ring closest to the amine end of the tube.

c) Which two of the six possible combinations produce rings in the same position, assuming the measurements can be made to the nearest centimeter?

Answer to Problem 10.141QA

Solution:

a) Acetic acid ($C{H}_{3}COOH$) and ${\left({CH}_3\right)}_{3}N$ give a white ring halfway between the two ends.

b) Hydrochloric acid ($HCl$) and ${\left({CH}_3\right)}_{3}N$ produce a ring closest to the amine end of the tube.

c) No two of the six possible combinations produce the rings in the same position.

Explanation of Solution

1) Concept:

We are given two acids and three amines. We calculate molar mass of each acid and amine. Then using Graham’s law of effusion, we can calculate the ratio of the rate of effusion between any two of the six possible combinations. If the rates of effusion calculated for a combination are the same, then they will travel equal distance from two ends and will meet at the center of the tube.

2) Formula:

$\frac{{effusion rate}_{acid}}{{effusion rate}_{amine}}= \sqrt{\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}}$

where $\mathcal{M}$ represents molar mass.

3) Given:

The molar mass of each molecule is calculated using the periodic table given in book.

Molecule	Molar mass $(g/mol)$
$HCl$	$36.46$
$C{H}_{3}COOH$	$60.05$
$\left({CH}_3\right){NH}_2$	$31.06$
${\left({CH}_3\right)}_{2}NH$	$45.08$
${\left({CH}_3\right)}_{3}N$	$59.11$

4) Calculations:

a) The combination of an acid and an amine which gives a white ring halfway between the two ends:

The combination of an acid and an amine gives a white ring halfway between the two ends. This indicates that both the acid and the amine travel equal distance in the same time. Therefore, the rate of effusion of the acid and rate of effusion of the amine is the same.

$\frac{{effusion rate}_{acid}}{{effusion rate}_{amine}}=1$

$1= \sqrt{\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}}$

Squaring both sides of the rightmost equality yields

$\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}=1$

${\mathcal{M}}_{acid}={\mathcal{M}}_{amine}$

This indicates that the molar mass of the acid and the amine should be the same or very close. From the given list of acids and amines, the molar mass of acetic acid ($60.05 g/mol$) and ${\left({CH}_3\right)}_{3}N$($59.11 g/mol$) are very close to each other. Therefore, the combination of acetic acid ($C{H}_{3}COOH$) and ${\left({CH}_3\right)}_{3}N$ gives a white ring halfway between the two ends.

b) The combination of an acid and an amine that will produce a ring closest to the amine end of the tube:

The result shows a ring closest to the amine end of the tube. This indicates that the amine travels a smaller distance while the acid travels a larger distance in the tube. Therefore, the rate of effusion of the acid is more than the rate of effusion of the amine. This also means that the molar mass of the amine should be greater than that of the acid. From this, we can write the equation

$\frac{{effusion rate}_{acid}}{{effusion rate}_{amine}}>1$

$\sqrt{\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}}>1$

Squaring both sides of the rightmost equality yields

$\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}>1$

${\mathcal{M}}_{amine}>{\mathcal{M}}_{acid}$

This shows that molar mass of the amine should be larger than the molar mass of the acid. From the given list of acid and amines, we select an acid and an amine such that the molar mass of the amine is greater than that of the acid. Therefore, hydrochloric acid ($HCl=36.46 g/mol$) and ${\left({CH}_3\right)}_{3}N$($59.11 g/mol$) produce a ring closest to the amine end of the tube. There is no amino acid that has the higher molar mass than acetic acid. So, there would be no combination with acetic acid.

c) Calculations for the two of the six possible combinations that produce rings in the same position is as follows:

The six possible combinations and their effusion rates are listed as

Combination	Molar mass (g/mol)	Ratio $\sqrt{\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}}$
$HCl$ and $\left({CH}_3\right){NH}_2$	$36.46$ and $31.06$	$\sqrt{31.06 / 36.46}=0.923$
$HCl$ and ${\left({CH}_3\right)}_{2}NH$	$36.46$ and $45.08$	$\sqrt{45.08 / 36.46}=1.11$
$HCl$ and ${\left({CH}_3\right)}_{3}N$	$36.46$ and $59.11$	$\sqrt{59.11 / 36.46}=1.27$
$C{H}_{3}COOH$ and $\left({CH}_3\right){NH}_2$	$60.05$ and $31.06$	$\sqrt{31.06 / 60.05}=0.517$
$C{H}_{3}COOH$ and ${\left({CH}_3\right)}_{2}NH$	$60.05$ and $45.08$	$\sqrt{45.08 / 60.05}=0.866$
$C{H}_{3}COOH$ and ${\left({CH}_3\right)}_{3}N$	$60.05$ and $59.11$	$\sqrt{59.11 / 60.05}=0.992$

No two values in the ratio for $\sqrt{\frac{{\mathcal{M}}_{amine}}{{\mathcal{M}}_{acid}}}$  are the same. Therefore, no two of the six possible combinations produce rings in the same position.

Conclusion:

Using Graham’s law of effusion, the rate of effusion can be calculated. The higher the molar mass, the slower will be the effusion rates and vice versa.