Chemistry » Organic Molecules » Physical Properties And Structure

# Melting and Boiling Points

## Melting and Boiling Points

Intermolecular forces affect the boiling and melting points of substances. Substances with weak intermolecular forces will have low melting and boiling points as less energy (heat) is needed to overcome these forces. Those with strong intermolecular forces will have high melting and boiling points as more energy (heat) is required to overcome these forces. When the temperature of a substance is raised beyond it’s melting or boiling point the intermolecular forces are not weakened. Rather, the molecules have enough energy to overcome those forces.

#### Fact:

An ether is a compound that contains two alkyl chains (e.g. methyl, ethyl) joined at an angle by an oxygen atom ($$-$$O$$-$$).

 Name Mainintermolecularforces MolecularMass ($$\text{g.mol^{-1}}$$) Melting point (℃) Boiling point (℃) Phase(at 25℃) ethane induced-dipole $$\text{30.06}$$ $$-\text{183}$$ $$-\text{89}$$ gas dimethyl ether dipole-dipole $$\text{46.06}$$ $$-\text{141}$$ $$-\text{24}$$ gas chloroethane dipole-dipole $$\text{64.5}$$ $$-\text{139}$$ $$\text{12.3}$$ gas pentane induced-dipole $$\text{72.12}$$ $$-\text{130}$$ $$\text{36}$$ liquid propan-1-ol hydrogen bonds $$\text{60.08}$$ $$-\text{126}$$ $$\text{97}$$ liquid ethanol hydrogen bonds $$\text{46.06}$$ $$-\text{114}$$ $$\text{78.4}$$ liquid butan-1-ol hydrogen bonds $$\text{74.1}$$ $$-\text{90}$$ $$\text{118}$$ liquid ethanoic acid hydrogen bonds $$\text{60.04}$$ $$\text{16.5}$$ $$\text{118.5}$$ liquid

Table: Relationship between intermolecular forces and melting point, boiling point and physical state.

The structural formula of dimethyl ether.

As the intermolecular forces increase (from top to bottom in the table above) the melting and boiling points increase. The stronger the intermolecular forces the more likely a substance is to be a liquid or a solid at room temperature.

## Example: Comparing Physical Properties

### Question

Given:

 Melting point (℃) Boiling point (℃) propane $$-\text{188}$$ $$-\text{42}$$ butanoic acid $$-\text{7.9}$$ $$\text{163.5}$$ bromoethane $$-\text{118}$$ $$\text{38.5}$$ diethyl ether $$-\text{116.3}$$ $$\text{34.6}$$

Fill in the table below:

 Name Main intermolecularforces Molecularformula Molecularmass ($$\text{g.mol^{-1}}$$) Phase(at 25 ℃) propane butanoic acid bromoethane diethyl ether

### What are the intermolecular forces that each molecule will experience?

• propane has only single carbon-carbon bonds and no other functional group. It will therefore have induced-dipole forces only.

• butanoic acid has the carboxylic acid functional group. It can therefore form hydrogen bonds.

• bromoethane has the highly electronegative bromine atom. This means that it will form dipole-dipole interactions with neighbouring molecules.

• diethyl ether will have induced-dipole forces, however due to the flexible nature of the molecule is can also have dipole-induced-dipole (stronger van der Waals interactions) and dipole-dipole forces.

### Do these forces make sense with the melting and boiling points provided?

Propane has the lowest melting and boiling points and the weakest interactions. The next lowest melting and boiling points are for bromoethane and diethyl ether, which both have dipole-dipole interactions, the next strongest intermolecular forces. The highest melting and boiling points are for butanoic acid which has strong hydrogen bonds. Therefore these forces do make sense.

### Calculate the molecular mass of these molecules

• propane – $$\text{C}_{3}\text{H}_{8}$$, molecular mass = $$\text{44.08}$$ $$\text{g.mol^{-1}}$$

• butanoic acid – $$\text{C}_{4}\text{H}_{8}\text{O}_{2}$$, molecular mass = $$\text{88.08}$$ $$\text{g.mol^{-1}}$$

• bromoethane – $$\text{C}_{2}\text{H}_{5}\text{Br}$$, molecular mass = $$\text{108.95}$$ $$\text{g.mol^{-1}}$$

• diethyl ether – $$\text{C}_{4}\text{H}_{10}\text{O}$$, molecular mass = $$\text{74.10}$$ $$\text{g.mol^{-1}}$$

### What will the phase of each compound be at $$\text{25}$$ $$\text{℃}$$?

To determine the phase of a molecule at $$\text{25}$$ $$\text{℃}$$ look at the melting and boiling points:

• The melting and boiling points of propane are both below $$\text{25}$$ $$\text{℃}$$, therefore the molecule will be a gas at $$\text{25}$$ $$\text{℃}$$.

• The melting points of butanoic acid, bromoethane and dimethyl ether are below $$\text{25}$$ $$\text{℃}$$, however the boiling points of all three molecules are above $$\text{25}$$ $$\text{℃}$$. Therefore these molecules will be liquids at $$\text{25}$$ $$\text{℃}$$.

### Fill in the table

 Name Main intermolecularforces Molecularformula Molecularmass ($$\text{g.mol^{-1}}$$) Phase(at 25 ℃) propane induced-dipole $$\text{C}_{3}\text{H}_{8}$$ $$\text{44.08}$$ gas butanoic acid hydrogen bonds $$\text{C}_{4}\text{H}_{8}\text{O}_{2}$$ $$\text{88.08}$$ liquid bromoethane dipole-dipole $$\text{C}_{2}\text{H}_{5}\text{Br}$$ $$\text{108.95}$$ liquid diethyl ether dipole-dipole $$\text{C}_{4}\text{H}_{10}\text{O}$$ $$\text{74.10}$$ liquid

## Optional Experiment: Investigation of boiling and melting points

### Aim

To investigate the relationship between boiling points and intermolecular forces

### Apparatus

• butan-1-ol ($$\text{CH}_{2}(\text{OH})\text{CH}_{2}\text{CH}_{2}\text{CH}_{3}$$), propanoic acid ($$\text{CH}_{3}\text{CH}_{2}\text{COOH}$$) and ethyl methanoate ($$\text{HCOOCH}_{2}\text{CH}_{3}$$), cooking oil

• three test tubes, a beaker, a thermometer, a hot plate

### Method

#### Warning:

Ethyl methanoate can irritate your eyes, skin, nose and lungs. Keep open flames away from your experiment and make sure you work in a well ventilated area.

1. Label the test tubes 1, 2 and 3. Place $$\text{20}$$ $$\text{ml}$$ of butan-1-ol into test tube 1, $$\text{20}$$ $$\text{ml}$$ of propanoic acid into test tube 2, and $$\text{20}$$ $$\text{ml}$$ of ethyl methanoate into test tube 3.

2. Half-fill the beaker with cooking oil and place it on the hot plate.

3. Place the thermometer and the three test tubes in the beaker.

4. Make a note of the temperature when each substance starts to boil.

### Results

Fill in the gaps in the table below. Do the values you obtained match those reported in literature?

 Compound ethyl methanoate butan-1-ol propanoic acid Molecularformula Molecularmass Main intermolecularforces Literatureboiling point (℃) $$\text{54}$$ $$\text{118}$$ $$\text{141}$$ Experimentalboiling point (℃)

Draw the structural representation of ethyl methanoate, butan-1-ol and propanoic acid.

### Discussion and conclusion

You should have found that the ethyl methanoate boiled first, then the butan-1-ol and then the propanoic acid. Ethyl methanoate has some dipole-dipole interactions, but cannot form a hydrogen bond. The alcohol (butan-1-ol) can form hydrogen bonds and so has a higher boiling point. This strong intermolecular force needs more energy to break and so the boiling point is higher. For propanoic acid hydrogen bonds form between the carbonyl group on one acid and the hydroxyl group on another. This means that each molecule of propanoic acid can be part of two hydrogen bonds (this is called dimerisation, see figure below) and so the boiling point is even higher for propanoic acid than for butan-1-ol.

A carboxylic acid hydrogen bonding dimer.

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