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Ch.11 - Liquids and Intermolecular Forces
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 11, Problem 91

Acetone [(CH3)2CO] is widely used as an industrial solvent. (d) 1-Propanol (CH3CH2CH2OH) has a molecular weight that is very similar to that of acetone, yet acetone boils at 56.5 °C and 1-propanol boils at 97.2 °C. Explain the difference.

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1
Identify the molecular structures of acetone and 1-propanol. Acetone is a ketone with the formula (CH3)2CO, while 1-propanol is an alcohol with the formula CH3CH2CH2OH.
Recognize the types of intermolecular forces present in each compound. Acetone primarily exhibits dipole-dipole interactions due to its polar carbonyl group, while 1-propanol exhibits hydrogen bonding due to the presence of the hydroxyl (OH) group.
Understand that hydrogen bonds are generally stronger than dipole-dipole interactions. This means that more energy (in the form of heat) is required to break the hydrogen bonds in 1-propanol compared to the dipole-dipole interactions in acetone.
Relate the strength of intermolecular forces to boiling points. Since 1-propanol has stronger intermolecular forces (hydrogen bonds), it requires a higher temperature to transition from liquid to gas, resulting in a higher boiling point compared to acetone.
Conclude that the difference in boiling points is primarily due to the presence of hydrogen bonding in 1-propanol, which is absent in acetone, leading to stronger intermolecular forces and a higher boiling point for 1-propanol.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Intermolecular Forces

Intermolecular forces are the attractions between molecules that influence physical properties like boiling points. Acetone, being a polar molecule, exhibits dipole-dipole interactions and some hydrogen bonding, while 1-propanol has stronger hydrogen bonding due to its hydroxyl group. The strength and type of these forces significantly affect the energy required to change a substance from liquid to gas.
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Molecular Structure and Polarity

The molecular structure and polarity of a compound determine how molecules interact with each other. Acetone has a carbonyl group (C=O) that contributes to its polarity, while 1-propanol has a hydroxyl group (–OH) that enhances its ability to form hydrogen bonds. This difference in structure leads to variations in boiling points, as stronger interactions require more energy to overcome.
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Boiling Point and Energy Requirements

The boiling point of a substance is the temperature at which its vapor pressure equals atmospheric pressure, allowing it to transition from liquid to gas. The energy required to reach this point is influenced by the strength of intermolecular forces. In this case, the stronger hydrogen bonding in 1-propanol compared to the weaker interactions in acetone results in a higher boiling point for 1-propanol.
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Related Practice
Textbook Question
Liquid butane (C4H10) is stored in cylinders to be used as a fuel. The normal boiling point of butane is listed as -0.5 °C. (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.
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Textbook Question

Naphthalene (C10H8) is the main ingredient in traditional mothballs. Its normal melting point is 81 °C, its normal boiling point is 218 °C, and its triple point is 80 °C at 1000 Pa. Using the data, construct a phase diagram for naphthalene, labeling all the regions of your diagram.

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Textbook Question

The vapor pressure of ethanol (C2H5OH) at 19 °C is 40.0 torr. A 1.00-g sample of ethanol is placed in a 2.00 L container at 19 °C. If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain?

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Textbook Question

Using information in Appendices B and C, calculate the minimum grams of propane, C3H8(g), that must be combusted to provide the energy necessary to convert 5.50 kg of ice at -20 °C to liquid water at 75 °C

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Textbook Question

In Table 11.3, we saw that the viscosity of a series of hydrocarbons increased with molecular weight, doubling from the six-carbon molecule to the ten-carbon molecule.

(a) The eight-carbon hydrocarbon, octane, has an isomer, isooctane. Would you predict that isooctane would have a larger or smaller viscosity than octane? Why?

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