Digesting The Precipitate Lab Report

The purpose of the lab was to determine the percent of water in the hydrate MgSO4 * N2H0 through the process of heating the hydrate, releasing the water molecules, and leaving only the anhydrous magnesium sulfate and also determine the number of water molecules in the hydrate. Throughout the lab, various masses were tooki, including the mass of the evaporating glass with the watch glass, the mass of the evaporating glass, and watch glass with the additional hydrate sample, and the mass of the the evaporating glass, and the watch glass with the dehydrated sample, at two different times, each separated by a five minute interval. After, calculations based on the mass of the left over anhydrous solution were made where the mass of the either the original recorded number for the hydrate was used, or the aftermath dehydrated value of the hydrate. Soon the mass converted

We then proceeded in testing for excess Ca2+ by adding two drops of .5 M K2C2O4 to test tube two and attentively observed to see if a precipitate formed, which it did. This meant that Ca2+ was in excess and C2O42- was the limiting reactant in the original salt mixture. We then cleaned up. Upon returning to our next class, we took the filter paper, with the precipitate on it, and took its mass.

As a group, we obtained our salt mixture of calcium chloride and potassium oxalate, and weighed the mixture. We were able to make an aqueous solution from the mixture and distilled water. We boiled and filtered off the solution, leaving the precipitate. Once the precipitate was dried overnight, it was weighed and the mass was measured. Then we calculated the moles of the precipitate.

The two most obvious formation of the precipitate were the combinations with the MgSO4. The MgSO4 and NH3 solution became very opaque and the MgSO4 and Na2CO3 turned from liquid to a full solid white substance. The Na2CO3 and CH3COOH did not have as strong of a reaction, however, the precipitates were able to be visualized with in the clear

.73 mol x (1 L/1000 mL) x (1 mL/1.08g) x (74.4 g NaClO/1 mol) x 100% =5.03%

to the unknown solution in order to completely precipitate the cations in part A. 2 A compound

In reference to the analysis of anions, Table 1 shows that a precipitate was formed when our unknown was combined with HNO3 and AgNO3, thus indicating the presence of a chloride ion. Because our unknown did not form a precipitate due to HCl and BaCl2, separate, effervesce, or smell, we concluded that neither sulfate, nitrate, carbonate nor

3. When the anhydrous sample was rehydrated, only 93.4% of the sample could be recovered. This was because some of the mass of the sample remained stuck to the filter paper and could not be measured in the final mass calculation. This automatically resulted in less mass and did not allow for 100% of the mass to be recovered.

Finally ensure that your results are accurate. You should have a light blue precipitate remaining when the sodium carbonate reacts with the copper chloride and when sodium sulphate ionises with copper chloride no precipitate should have formed.

The precipitate was observed to adhere to the glass on the inside of the beaker, and even by scraping with the rubber policeman and rinsing, it cannot be ensured that all of the precipitate was gathered. This is a minor random error, and it causes a decrease in the final mass of the filter paper and precipitate, which would explain the percent yield of 92.2%.

For example, silver nitrate formed a white precipitate when it was tested with ammonium chloride. In contrast, unknown 3 did not formed any precipitate with ammonium chloride. Ammonium chloride change the color of unknown 3 to a light green while the solution of silver nitrate and ammonium chloride was cloudy white solution. Likewise, the metal in unknown 3 could have been Calcium neither. Data and observation shows that calcium nitrate whether formed a white precipitate or did not react at all while unknown 3 formed an orange precipitate. Therefore, silver and calcium are not the two metal present in unknown

3. Leong, O. H. (2009). Qualitative Analysis. In Know Your 'O' Level Chemistry (pp. 101-103). Singapore: Panpac Education Pte Ltd.

In this experiment, a saturated calcium sulfate was already made and ready to use. 25.00 mL of this solution was then mixed with 10 mL of an ammonia buffer and 1 drop of

This is because the Law of Conservation of Mass states, in any given chemical reaction, the total mass of the reactants equal the total mass of the products. In Part B the total mass of the reactants did not equal the total mass of the products. There was in fact a difference of 0.62grams between the reactants and the produced mass.

6. The precipitate may have not dried up properly making it so there was water adding weight on it.