Discussion and Scientific Explanations
The main goals of this experiment was to find the unknown compound, to discover the compounds chemical and physical properties and create two synthesis of the compound and compare the two compounds. We found out our compound was ionic because the compound dissolved in water, but not in toluene or acetone by using a solubility test, results on Table 1 of the group lab report. On page 59 of the Cooperative Chemistry Laboratory Manual is a chart on suggested liquids and the inference if the compound is soluble [1]. When then checked the pH of the solution by using pH strip. The solution had a pH>7. Then we did an anion test and discovered the compound reacted with chloride by producing a white precipitate. The precipitate that formed was an insoluble solid because Cl of NaCl reacted with Ag of AgNO3, creating a Cl- ion above the aqueous solution [3], results on Table 2. Finally we used a cation flame test, results on Table 4. When we burned the unknown substance the flame was a bright orange, yellow color. This was because the flame reacted with the Cl of NaCl causing …show more content…
Both compounds dissolved in the water solution. Both flame tests created a high intensity yellow flame and both compounds reacted with AgNO3. By looking at the Material Safety Data Sheet for NaCl I saw that both compounds have similar physical properties [2]. Both unknown and known NaCl appeared in a solid, crystallized state. Both compounds were white and had a slight odor. Also on the Data Sheet it states that NaCl is dissolvable in water. The Data sheet supports our results. Both unknown and known compounds had similar amounts of AgCl on the filtration paper. From the results I can conclude that the unknown NaCl and the known NaCl produces the same product, AgCl, when reacting with AgNO3. The AgCl forms because Cl reacts with Ag. The formula below shows this interaction between the two
The main goals of this experiment was to find the unknown compound, to discover the compounds chemical and physical properties and create two synthesis of the compound and compare the two compounds. We found out our compound was ionic because the compound dissolved in water, but not in toluene or acetone by using a solubility test, results on Table 1 of the group lab report. On page 59 of the Cooperative Chemistry Laboratory Manual is a chart on suggested liquids and the inference if the compound is soluble [1]. When then checked the pH of the solution by using pH strip. The solution had a pH>7. Then we did an anion test and discovered the compound reacted with chloride by producing a white precipitate. The precipitate that formed was an insoluble solid because Cl of NaCl reacted with Ag of AgNO3, creating a Cl- ion above the aqueous solution [3], results on Table 2. Finally we used a cation flame test, results on Table 4. When we burned the unknown substance the flame was a bright orange, yellow color. This was because the flame reacted with the Cl of NaCl causing …show more content…
Both compounds dissolved in the water solution. Both flame tests created a high intensity yellow flame and both compounds reacted with AgNO3. By looking at the Material Safety Data Sheet for NaCl I saw that both compounds have similar physical properties [2]. Both unknown and known NaCl appeared in a solid, crystallized state. Both compounds were white and had a slight odor. Also on the Data Sheet it states that NaCl is dissolvable in water. The Data sheet supports our results. Both unknown and known compounds had similar amounts of AgCl on the filtration paper. From the results I can conclude that the unknown NaCl and the known NaCl produces the same product, AgCl, when reacting with AgNO3. The AgCl forms because Cl reacts with Ag. The formula below shows this interaction between the two