Why naphthalene has a low melting point




















At the first drop of liquid in the capillary, record the temperature of the oil bath. Keep watching the capillary tube as you slowly increase the temperature. Record the temperature when the last piece of solid melts.

Carefully remove the capillary and thermometer from the oil bath and set them aside to cool. Weigh roughly 1 mg of urea and transfer it to a clean watch glass. Crush the urea to a powder. Load the powder into a capillary tube as before. Attach the capillary to the thermometer with the rubber band. Insert the thermometer and capillary tube into the oil bath. Record the temperature when the first drop of liquid appears and when the last bit of solid melts. Turn off the hotplate and remove the capillary and thermometer from the oil bath to allow them to cool.

Test the sample of urea mixed with an unknown impurity, which was prepared by your instructor. Perform two runs of this sample, so weigh 2 mg of the sample on the balance and transfer it to a watch glass.

Crush the sample to powder and load two capillaries by tapping them on the powder and dropping them through the glass tube like before. Attach one of the loaded capillaries to the thermometer and clamp them in the oil bath. Record the temperature when you see the first drop of liquid form and the last drop of liquid form to determine the rough melting point range. Turn off the hotplate and allow it to cool. When you have finished measuring the melting point of the urea mixed with the unknown substance, turn off the hotplate and detach the capillary tube from the thermometer.

Dispose of your capillary tubes in the glass waste container. Return the mineral oil to your instructor, and clean all of your glassware using detergent and water. First, look at the melting point range of naphthalene. These differences may be caused by experimental errors, such as heating the sample too quickly, or by impurities in the sample.

Next, check the results for the pure urea sample. Why is urea's melting point so much higher than naphthalene's? Looking at the structure of naphthalene, the intermolecular forces must be primarily London dispersion forces.

Urea is capable of hydrogen bonding, so the intermolecular forces in solid urea are much stronger than the forces in solid naphthalene. Finally, look at the results from the urea sample with the impurity. The melting point range should be broader and lower than the range for pure urea because of freezing point depression. Please enter your institutional email to check if you have access to this content.

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As a broad generalisation the melting of solids is more dependant on the nature of the repulsive forces between molecules than on their attractive ones. Experimentally the enthalpy of melting Latent Heat or just Heat of Fusion is approximately ten times less than the corresponding heat of vaporisation which indicates that repulsive forces play a greater role in melting than in vaporisation: in a solid molecules tend to be slightly closer together than in a liquid, thus may experience larger repulsive forces.

Van-der-Waals forces tend to have weak orientational dependence, but repulsive forces do not as they are short range and depend on the asymmetric shape of the molecules and can have a large effect. Thus melting points are rather dependent on the geometry of the molecules, if they can pack well into a lattice then a high melting point is expected.

If the molecule has a dipole this can cause both attractive and repulsive interactions depending on the relative positioning of molecules; head to tail can be attractive if molecules are face to face, but repulsive if in line one behind the other and vice versa if their relative orientation is changed. In the case of naphthalene and the amine, clearly the naphthalene can stack nicely and so increase its attractive interaction relative to repulsive ones, whereas the amine is twisted and cannot so easily pack meaning that repulsion can be stronger and attractive interactions weaker even with dipole and melting points lower.

Attractive forces largely determine the enthalpy of vaporisation, and by Trouton's rule their boiling points. The enthalpy of vaporisation is closely related to the attractive cohesive energy in a liquid.

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Like, why does high symmetry and pi-stacking increase melting point from a molecular perspective?



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