Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3},\) to be larger or smaller than that of 1 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\) ? Explain. [Section 11.3\(]\)

Short Answer

Expert verified
The viscosity of glycerol \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\) is expected to be larger than that of 1-propanol \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\) because glycerol has three -OH groups compared to the single -OH group in 1-propanol, which results in stronger intermolecular forces, including hydrogen bonding, and therefore greater resistance to flow.

Step by step solution

01

Identifying molecular structures of glycerol and 1-propanol

First, we need to identify the molecular structures of glycerol and 1-propanol to determine their hydrogen bonding capabilities. - Glycerol: \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), also known as glycerin, is a triol compound with three alcohol functional groups (-OH). Hydrogen bonding is present in glycerol molecules because of the -OH groups. - 1-propanol: \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\), also known as propyl alcohol, is an alcohol with one alcohol functional group (-OH). We can expect hydrogen bonding between 1-propanol molecules due to the -OH group.
02

Comparing the hydrogen bonding potential of glycerol and 1-propanol

We can compare the hydrogen bonding potential in both compounds. Glycerol has three alcohol functional groups (-OH) compared to the single -OH group in 1-propanol. As a result, the glycerol molecules will have stronger intermolecular forces, including hydrogen bonding, than 1-propanol molecules.
03

Relating hydrogen bonding potential to viscosity

Viscosity is related to intermolecular forces. Compounds with stronger intermolecular forces tend to have higher viscosity as the forces between molecules create resistance to flow. Since glycerol has stronger intermolecular forces due to the presence of three -OH groups, it is expected to have a higher viscosity compared to 1-propanol, which has only one -OH group.
04

Concluding the comparison

In conclusion, based on the presence of three -OH groups in glycerol and one -OH group in 1-propanol, we can infer that glycerol will have stronger intermolecular forces and thus, a higher viscosity compared to 1-propanol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intermolecular Forces
When exploring the properties of liquids, one key factor to consider is intermolecular forces. These forces are the attractions between molecules, and they play a significant role in determining a substance's physical properties, such as boiling point, melting point, and in this case, viscosity. Viscosity is a measure of a fluid's resistance to flow and can be influenced by the type of intermolecular forces at play.

There are different types of intermolecular forces, including dispersion forces, dipole-dipole interactions, and hydrogen bonding. Dispersion forces are weak and occur between all molecules, while dipole-dipole interactions occur between polar molecules. Hydrogen bonding, on the other hand, is a special type of dipole-dipole interaction and is particularly strong; it occurs when a hydrogen atom is bonded to a highly electronegative atom, such as oxygen, nitrogen, or fluorine.

When comparing substances, those with stronger intermolecular forces generally exhibit a higher viscosity. This correlation is due to the increased attraction between molecules, which requires more energy to overcome when the molecules move past one another. Hence, understanding the nature and strength of intermolecular forces is crucial when predicting viscosity differences.
Hydrogen Bonding
Hydrogen bonding plays a pivotal role in the context of viscosity. This specific intermolecular force occurs when a hydrogen atom covalently bonded to an electronegative atom, such as oxygen, is electrostatically attracted to another electronegative atom nearby. This bond is stronger than other types of dipole-dipole interactions because the small size of the hydrogen atom allows it to get very close to the electronegative atom, creating a strong attraction.

Molecules with hydrogen bonding capability will experience significant attractions with neighboring molecules, leading to a higher viscosity. This is clearly illustrated when comparing compounds like glycerol and 1-propanol. Glycerol, with its three hydroxyl groups, can form more hydrogen bonds than 1-propanol, which only has one. Consequently, glycerol's ability to create a network of hydrogen bonds will lead to a substance that is more viscous and flows less readily than 1-propanol.
Molecular Structure
The molecular structure directly influences a compound's physical properties, such as viscosity, by dictating the nature and strength of intermolecular interactions. The arrangement of atoms within a molecule, and the functional groups present, determine how molecules will interconnect and the types of intermolecular forces they can form.

For instance, consider the comparison of glycerol and 1-propanol. Glycerol's molecular structure includes three hydroxyl (-OH) groups attached to a three-carbon backbone, permitting a dense network of intermolecular hydrogen bonding. On the other hand, 1-propanol has a similar backbone but only one hydroxyl group, limiting its capacity for hydrogen bonding. This structural difference results in distinct intermolecular interactions for each substance, leading to a higher viscosity in glycerol versus 1-propanol.

Therefore, the presence and number of certain functional groups, like hydroxyl groups, are of great significance. They are responsible for the formation of strong intermolecular forces like hydrogen bonds which, in turn, greatly affect a molecule's viscosity.

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Most popular questions from this chapter

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the heat required to convert \(35.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00^{\circ} \mathrm{C}\) to a gas at \(105.00^{\circ} \mathrm{C}\).

Explain the following observations: (a) The surface tension of \(\mathrm{CHBr}_{3}\) is greater than that of \(\mathrm{CHCl}_{3} .\) (b) As temperature increases, oil flows faster through a narrow tube. (c) Raindrops that collect on a waxed automobile hood take on a nearly spherical shape. (d) Oil droplets that collect on a waxed automobile hood take on a flat shape.

As the intermolecular attractive forces between molecules increase in magnitude, do you expect each of the following to increase or decrease in magnitude? (a) vapor pressure, (b) heat of vaporization, (c) boiling point, (d) freezing point, (e) viscosity, (f) surface tension, (g) critical temperature.

Liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is stored in cylinders to be used as a fuel. The normal boiling point of butane is listed as \(-0.5^{\circ} \mathrm{C}\). (a) Suppose the tank is standing in the sun and reaches a temperature of \(35^{\circ} \mathrm{C}\). Would you expect the pressure in the tank to be greater or less than atmospheric pressure? How does the pressure within the tank depend on how much liquid butane is in it? (b) Suppose the valve to the tank is opened and a few liters of butane are allowed to escape rapidly. What do you expect would happen to the temperature of the remaining liquid butane in the tank? Explain. (c) How much heat must be added to vaporize \(250 \mathrm{~g}\) of butane if its heat of vaporization is \(21.3 \mathrm{~kJ} / \mathrm{mol} ?\) What volume does this much butane occupy at 755 torr and \(35^{\circ} \mathrm{C} ?\)

The boiling points, surface tensions, and viscosities of water and several alchohols are as follows: $$ \begin{array}{lrcc} & \begin{array}{l} \text { Boiling } \\ \text { Point }\left({ }^{\circ} \mathbf{C}\right) \end{array} & \begin{array}{l} \text { Surface } \\ \text { Tension }\left(\mathbf{J} / \mathbf{m}^{2}\right) \end{array} & \begin{array}{l} \text { Viscosity } \\ (\mathbf{k g} / \mathbf{m}-\mathbf{s}) \end{array} \\ \hline \text { Water, } \mathrm{H}_{2} \mathrm{O} & 100 & 7.3 \times 10^{-2} & 0.9 \times 10^{-3} \\ \text {Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & 78 & 2.3 \times 10^{-2} & 1.1 \times 10^{-3} \\ \text {Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 97 & 2.4 \times 10^{-2} & 2.2 \times 10^{-3} \\ n \text { -Butanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 117 & 2.6 \times 10^{-2} & 2.6 \times 10^{-3} \\\ \text {Ethylene glycol, } \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 197 & 4.8 \times 10^{-2} & 26 \times 10^{-3} \end{array} $$ (a) For ethanol, propanol, and \(n\) -butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights \((60\) versus \(62 \mathrm{amu}),\) yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?

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