Chapter 3

Question 1

At any given temperature, liquid A has a higher vapor pressure than liquid B. Which liquid has the higher boiling boint?

Answer 1

Liquid B has a higher boiling point than liquid A. Boiling points and equilibrium vapor pressure are inversely proportional.

Question 2

Explain why the boiling point at 760 Torr of a solution of water, bp 100°C (760 Torr), and ethylene glycol (HOCH₂CH₂OH), bp 196-198°C (760 Torr), exceeds 100°C. For purposes of your answer, consider ethylene glycol as a nonvolatile liquid.

Answer 2

If you have pure water, its mole fraction would be one. In a solution of water and ethylene glycol, the mole fraction of water becomes less than one, which lowers the partial pressure of water in the mixture (Raoult’s Law). The partial pressure of ethylene glycol is zero (b/c it is nonvolatile), which makes the total equilibrium vapor pressure of the misture lower than that of pure water.

Question 3

Raoult’s Law

Answer 3

Raoult’s Law: P_mix = N_xP⁰_x

Pure Water: P_H2O = N_H2OP⁰_H2O

N = 1 for pure water, then P_{<span>H2O = </span>}P⁰_H2O

Say we had a 1:1 mixture of water with ethylene glycol (EG):

P_mix = (P⁰_H20 x 0.5) + (P⁰_EG x 0.5) = (P⁰_H20 x 0.5) because P⁰_EG = 0 (remember it’s nonvolatile). Since P⁰_H20 > (P⁰_H20 x 0.5) → P⁰H20 > P⁰_mix

The above mathematical analysis would be true for whatever ratio of water and EG we consider, b/c it would always make the mole fraction of water less than one.

Question 4

Why should you never heat a closed heat system, and how does this rule apply to distillation?

Answer 4

Heating increases the equilibrium vapor pressure of a liquid. In a closed heat system, this increased pressure can exceed the limits of the glasssware, causing it to explode. If you are performing a vacuum distillation, you should only use glassware designed to withstand pressure, and you should ensure there are no cracks in the glassware before you put the system under pressure

Question 5

In a miniscale distillation, the top of the mercury bulb of the thermometer placed at the head of a distillation apparatus should be adjacent to the exit opening of the condenser. Explain the effect on the observed temperature reading if the bulb is placed (a) below the opening of the condenser or (b) above the opening.

Answer 5

a) If the thermometer is placed below the opening to the condenser, it will record a higher temperature than if it is positioned near the opening since the bulb is contact with vapor and liquid that has a higher temp. than that entering the condenser.
b) If the thermometer is placed above the opening to the condenser, it will record a lower temperature than if it is positioned near the opening since the bulb is not in full contact with vapor and liquid entering the condenser.

Question 6

Using Raoult’s and Dalton’s laws, explain why an aqueous NaCl solution will have a higher boiling point than pure water

Answer 6

If you have pure water, its mole fraction would be one. In a solution of water and NaCl, the mole fraction of water becomes less than one, which lowers the partial pressure of water in the mixture (see Raoult’s Law). The partial pressure of NaCl is zero (b/c it is nonvolatile), which makes the total equilibrium vapor pressure of the mixture lower than that of pure water. Since equilibrium vapor pressure is inversely proportional to boiling point, the mixture has a higher boiling point than pure water.

Question 7

At 100°C, the vapor pressures for water, methanol, and ethanol are 760, 2625, and 1694 Torr, respectively. Which compound has the highest normal boiling point and which the lowest?

Answer 7

Boiling points and equilibrium vapor pressure are inversely proportional, because less heat is required to raise P^o to reach atmospheric pressure. Therefore, since water has the lowest P^o water it has the highest boiling point and since methanol has the highest P^o it has the lowest boiling point.

Question 8

Stationary Phase

Answer 8

Solid or liquid adsorbent that is either bonded to support particles and/or packed in column tubing used during column, thin-layer or gas chromatography. In our GC the stationary phase is Carbowax®, a polyethylene glycol liquid polymer

Question 9

Mobile Phase

Answer 9

Gas or liquid that causes the analyte to elute (move) through the column, TLC, or GC. In our GC the mobuile phase is Helium gas.

Question 10

Carrier Gas

Answer 10

Inert gas, such as hellium or nitrogen, that elutes an analyte through a gas chromatography column

Question 11

Retention Time

Answer 11

The characteristic time it takes for a particular analyte to pass through the system (from the column inlet to the detector) under set conditions

Question 12

Solid Support

Answer 12

In GC, it is the solid within the column that the liquid-phase is adsorbed to. In our GC, the solid-support is CHromosorb®, a form of diatomaceous earth.

Question 13

Thermal Conductivity

Answer 13

A measure of the ability of a substance to conduct heat

Question 14

Benzene (1 g, 12.5 mmol) is allowed to react with 1-chloropropane (1 g, 12.5 mmol) and AlCl3. The product (1.2 g) is subjected to analysis on a GLC equipped with a thermal conductivity detector. The chromatogram shows two product peaks identified as n-propylbenzene (area = 65 mm² ; W_f = 1.06) and isopropylbenzene (area = 113 mm² ; W_f = 1.09). Calculate the percent yield of each of the two isomeric products obtained in the reaction. Note that since each of the products has the same molar mass of 120, the use of weight factors gives both weight and mole percent compositions.

Answer 14

Composition of the product, based on GC analysis

% n-propylbenzene = (65 x 1.06)/ ((65 x 1.06) + (113 x 1.09))

%n-propylbenzene = 0.36 (36%)

% isopropylbenzene = (113 x 1.09) / ((65 x 1.06) +(113 x 1.09))

%isopropylbenzene = 0.64 (64%)

Theoretical Mass

grams n-propylbenzene = 12.5mmol benzene x (1 mol n-propylbenzene / 1 mol benzene) x (120g n-propylbenze / mol n-propylbenzene)

grams n-propylbenzene = 1.5g (theoretical)

grams isopropylbenzene = 12.5mmol benzene x (1 mol isoropylbenzene / 1 mol benzene) x (120g isopropylbenze / mol isopropylbenzene)

grams isopropylbenzene = 1.5g (theoretical)

Actual Mass

g n-propylbenzene = 1.2g product mixture x (0.36g n-propylbenzene / 1g product mixture)

g n-propylbenzene = 0.432g (actual)

g isopropylbenzene = 1.2g product mixture x (0.64g isopropylbenzene / 1g product mixture)

g isopropylbenzene = 0.768g (actual)

Percent Yield

% yield n-propylbenzene: (0.432g / 1.5g) x 100% = 28.8%

% yield isopropylbenzene: (0.768g / 1.5g) x 100% = 51.2%

Question 15

A gas chromatogram of the organic components of a sample of beer using a column that separates compounds on the basis of their relative boiling points provides a GLC trace with several peaks. Two of the smaller peaks, with retention times of 9.56 and 16.23 minutes, are believed to be ethyl acetate and ethyl butyrate, respectively.

a) From the above information, which components of the sample, ethyl acetate or ethyl butyrate, elutes faster? What are the reported boiling points of these two substances?
b) Suggest two ways in which you could adjust the condition of the experiment so as to reduce the retention time for all components in the sample.

Answer 15

a) Ethyl acetate and ethyl butyrate have very similar structures, and as a consequence very similar polarities. Therefore in a normal-phase GC, the major determining factor for which compound will elute first would be the vapor pressure (or boiling points). Because ethyl acetate has fewer atoms than ethyl butyrate, it would make fewer intermolecular attractions, and therefore have a higher equilibrium vapor pressure, or lower boiling point. The opposite is true for ethyl butyrate. Therefore ethyl acetate would elude faster from the GC.

• b) Reduce the amount of stationary phase (shorten or narrow the column)
• Increase the carrier gas flow rate
• Reduce polarity of the stationary phase
• Increase column temperature