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Glycerin's Melting Point Lab Report

702 Words3 Pages

1) Sodium Chloride’s melting point is 800.7° which matches the observation acquired for this compound. Sucrose’s melting point is 185.5° which also resembles the results from the experiment. Glycerin’s melting point is 18.1° which is very logical since Glycerin is a liquid at room temperature. The only discrepancy is sodium hydrogen carbonate’s melting point, which is 50°, this does not match the recorded results, because sodium hydrogen carbonate was still in solid form when sucrose was changing state (suggesting that it possesses a higher melting point than 186°). The reason behind this discrepancy is because sodium hydrogen carbonate does not melt or change into a liquid, it decomposes, separating into carbon dioxide (gaseous state), dihydrogen oxide (water vapour, gaseous state), and sodium carbonate (solid state). Furthermore, the reason why …show more content…

Firstly, unknown B has a low melting point, a prominent characteristic among covalent compounds. This is due to the attraction between the atoms not being as reinforced as an ionic compound, thus it takes less energy to separate. In addition, unknown B has a very low solubility and conductivity, this is due to the atoms sharing electrons, therefore they cannot have the ability to separate and form an ion. However, it should be noted that covalent compounds should most definitely not be conductive or soluble, but the results have shown otherwise, thus it should be concluded that there may have been contamination between the scoopula’s used. Conclusion All in all, the experiment has provided much insight into the topic of ionic and covalent bonds regarding compounds. In conclusion, Unknown A is an ionic compound due to a high melting point, solubility and conductivity while Unknown C is a covalent compound due to a low melting point, low conductivity and

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