Question

Using the following list of normal boiling points for a series of hydrocarbons, estimate the normal boiling point for octane, \(\mathrm{C}_{8} \mathrm{H}_{18}:\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8},-42.1{ }^{\circ} \mathrm{C}\right)\), bu- tane \(\left(\mathrm{C}_{4} \mathrm{H}_{10},-0.5^{\circ} \mathrm{C}\right)\), pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}, 36.1^{\circ} \mathrm{C}\right)\), hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}, 68.7^{\circ} \mathrm{C}\right)\), heptane \(\left(\mathrm{C}_{7} \mathrm{H}_{16}, 98.4{ }^{\circ} \mathrm{C}\right)\). Explain the trend in the boiling points.

Short answer

The estimated normal boiling point of octane (\(C_8H_{18}\)) is approximately \(133.52^\circ C\). This estimation is based on analyzing the trend in boiling points of the given hydrocarbons, which shows that boiling points increase as the number of carbon atoms in the hydrocarbon chain increases. The trend can be explained by the increasing molecular size and intermolecular forces, specifically van der Waals forces, as the length of the carbon chain increases.

Step-by-step solution

List boiling points and number of carbon atoms in each hydrocarbon

We begin by listing the boiling points and the number of carbon atoms in the given hydrocarbons. 1. Propane: \(C_3H_8\), boiling point = \(-42.1^\circ C\) 2. Butane: \(C_4H_{10}\), boiling point = \(-0.5^\circ C\) 3. Pentane: \(C_5H_{12}\), boiling point = \(36.1^\circ C\) 4. Hexane: \(C_6H_{14}\), boiling point = \(68.7^\circ C\) 5. Heptane: \(C_7H_{16}\), boiling point = \(98.4^\circ C\)

Identify the trend

Next, we will observe the trend in the boiling points with respect to the increase in the hydrocarbon chain's carbon atoms. From the given list, it can be clearly observed that the boiling point increases as the number of carbon atoms in the hydrocarbon chain increase. This indicates that the boiling point of octane, which has 8 carbon atoms in the chain, should be higher than the boiling point of heptane.

Estimate the boiling point of octane

To estimate the boiling point of octane, we will consider the difference in boiling points between consecutive hydrocarbons and calculate an average. Then, we will add the average difference to the boiling point of heptane to find the boiling point of octane. 1. Difference between butane and propane boiling points: \((-0.5) - (-42.1) = 41.6^\circ C\) 2. Difference between pentane and butane boiling points: \(36.1 - (-0.5) = 36.6^\circ C\) 3. Difference between hexane and pentane boiling points: \(68.7 - 36.1 = 32.6^\circ C\) 4. Difference between heptane and hexane boiling points: \(98.4 - 68.7 = 29.7^\circ C\) Average difference: \(\frac{41.6 + 36.6 + 32.6 + 29.7}{4} = 35.12^\circ C\) Now, we can estimate the boiling point of octane by adding the average difference to the boiling point of heptane: Estimated boiling point of octane: \(98.4 + 35.12 = 133.52^\circ C\) Therefore, the estimated normal boiling point of octane, \(C_8H_{18}\), is \(133.52^\circ C\).

Explain the trend in boiling points

The observed trend in the boiling points can be explained by the incremental increase in molecular size and intermolecular forces as the size of the hydrocarbon chain increases. As the number of carbon atoms in a hydrocarbon molecule increases, so does the size and surface area of the molecule. This results in a stronger dispersion of forces (also known as van der Waals forces) between the molecular surfaces, which requires more energy to separate the molecules and causes a higher boiling point. In summary, the normal boiling point of hydrocarbons increases as the length of their carbon chain increases, due to the corresponding increase in molecular size and intermolecular forces.

Key concepts

Intermolecular Forces

Intermolecular forces are the attractive forces that occur between molecules. They are critical for determining the boiling points of substances, including hydrocarbons. As molecules get larger, the surface area available for these forces to act upon increases, leading to stronger intermolecular attractions.

Different types of intermolecular forces include dipole-dipole attractions, hydrogen bonding, and London dispersion forces, which are particularly relevant for hydrocarbons. The increase in boiling points of hydrocarbons, as seen in the series from propane to heptane, can be attributed to the rise in strength of dispersion forces with the increase in molecular size.

Van der Waals Forces

Van der Waals forces encompass various weak forces, including London dispersion forces, which are the key intermolecular forces in nonpolar molecules like hydrocarbons. The strength of these forces depends on how easily the electrons within a molecule can move, creating temporary dipoles that induce attractions with neighboring molecules.

Why London Dispersion Forces Matter in Hydrocarbons

Hydrocarbons are primarily nonpolar molecules, making the London dispersion forces the most significant type of van der Waals force among them. These forces become stronger as the length of the carbon chain increases due to more electrons and a larger area over which the temporary dipoles can interact. This is one of the reasons the boiling point of octane is higher than that of heptane.

Molecular Size

The size of a molecule has direct implications on its physical properties, including its boiling point. Larger molecules like those of higher alkanes have long carbon chains that contribute to an increased molecular weight and more extensive electron cloud.

Impact of Increased Molecular Size

When molecular size increases, the London dispersion forces also increase because there are more electrons to create stronger temporary dipoles. From propane to heptane, and estimately for octane, the trend of increasing molecular size matches the trend of increasing boiling points. This correlation is vital for predicting the boiling points of unknown hydrocarbons based on their molecular size, given the data of known substances.