The goal of this laboratory project was to identify the unknown compound and thoroughly investigate it to find its physical and chemical properties which reveal the way it might behave. This was done by performing many anion, cation, solubility, and pH tests to help us to determine the identity of our compound. The list of salts in the lab manual made it possible to narrow down the possibilities of our unknown compound. After identifying the compound, the goal was to synthesize the unknown compound. To determine the initial properties of the unknown compound, a qualitative solubility test was performed by mixing five milliliters of water with one gram of the unknown compound. After placing the solution in a test tube, the test tube was studied
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NaCl2 was soluble in water. Although CaCO3 was not soluble in water, there are 3 unpaired non-bonding electrons which are moving freely throughout the solution which is why it did not form a precipitate. Lastly, we added MgCl2 to the compound and the reactants (CaCl2 and MgCl) were both soluble in solution, which means that they did not form a …show more content…
According to Herman and Schafer, it is important to remember that precipitates are ionic solid products of a reaction and are formed when cations and anions intermix in an aqueous solution.3 Some reactions vary based on temperature or solute concentration. The determination of precipitates also come from solubility rules. According to Cooper, almost all sodium, potassium, and ammonium salts are soluble which mean that they will not form a precipitate.1 All chlorides, bromides, iodides, and fluorides, nitrates, chlorates, perchlorates, and acetates, and sulfates are also soluble. The only exceptions are as follows: Ag+, Hg2+, Pb2+, Mg2+, Ca2+, Sr2+, Ba2+, Pb2+. For week three, the goal was to determine the mass of the precipitate of the unknown compound and of the reference compound by using vacuum filtration. This test was run twice in triplicate which means that it was ran 6 times total. First, it was run with the unknown compound and then with the reference compound. As seen in Table 4, the results didn’t very much between the unknown compound and the reference compound which helps to prove that the unknown compound is
NaCl2 was soluble in water. Although CaCO3 was not soluble in water, there are 3 unpaired non-bonding electrons which are moving freely throughout the solution which is why it did not form a precipitate. Lastly, we added MgCl2 to the compound and the reactants (CaCl2 and MgCl) were both soluble in solution, which means that they did not form a …show more content…
According to Herman and Schafer, it is important to remember that precipitates are ionic solid products of a reaction and are formed when cations and anions intermix in an aqueous solution.3 Some reactions vary based on temperature or solute concentration. The determination of precipitates also come from solubility rules. According to Cooper, almost all sodium, potassium, and ammonium salts are soluble which mean that they will not form a precipitate.1 All chlorides, bromides, iodides, and fluorides, nitrates, chlorates, perchlorates, and acetates, and sulfates are also soluble. The only exceptions are as follows: Ag+, Hg2+, Pb2+, Mg2+, Ca2+, Sr2+, Ba2+, Pb2+. For week three, the goal was to determine the mass of the precipitate of the unknown compound and of the reference compound by using vacuum filtration. This test was run twice in triplicate which means that it was ran 6 times total. First, it was run with the unknown compound and then with the reference compound. As seen in Table 4, the results didn’t very much between the unknown compound and the reference compound which helps to prove that the unknown compound is