Green Crystals Lab

with a solid oxidant that is composed of KMnO4 and CuSO4. 1.0mL of diphenylmethanol was placed into

The next step was the titration. 30 drops of diluted peroxide was added to a 30 mL plastic beaker. To acidify the peroxide solution, 35 drops of 0.1 M HCl was added into the same cup. We then added 0.010 M potassium permanganate solution into the acidified peroxide and recorded the amount of drops needed to turn the peroxide solution light-pink to brown color. When the color changed, it indicated that all the peroxide has reacted and now there is an excess amount of KMnO4.

Introductory Paragraph- In this lab, we tested how much water was in a 3.00 gram amount of hydrated MgSO4 (Magnesium sulphate). MgSO4 is a salt; salts are often associated with water being in their structure. This lab was conducted so that we would be able to find the formula for the amount of water in the sample of magnesium sulphate. The goal was find the formula of a hydrous substance if you are given two masses.

The solution is then refluxed for five minutes. Reflux results in purification of the mixture as well as initiates the reaction mechanism. After reflux, the solution is placed on ice to begin crystallization. Following crystallization, the resultant solid is isolated via vacuum filtration and cold DI water on the filter paper. The end product results in a molecule that is highly excited and goes to a ground state quickly once a catalyst, such as potassium ferricyanide solution is added to it.

By following the above procedure, it was determined that my hypothesis was accepted. Initially, I predicted that sugar will have the best solution for making crystals because the sugar dissolves better in water and will be able to make the best crystals in size and shape. During the experiment, I investigated which solution made the best crystals. However, there was some inconsistencies and consistencies. The inconsistencies and consistencies were finding the perfect spot to put the solutions,the crystals forming, and the right materials. The first inconsistency was finding the perfect spot to put my solutions. It was very difficult for me

In the ion lab elements and compounds were combined in mixing wells in order to observe their reactions. The purpose of the ion lab was to gain an understanding of the reactions and solubility of ionic compounds, to practice writing balanced chemical equations, and to practice observational skills. During the lab it was found that when ionic solids dissolve the compound separates into its individual ions. This state of ions being dispersed within a solution when dissolved is what we describe as aqueous, and the act of being able to dissolve is called being soluble. When a combination of ions is insoluble, or not able to dissolve it results in a precipitate, because the ions in the compound could not separate or dissolve. This insolubility is

The percent composition by mass of oxalate was calculated to be 53.69% while potassium, iron, and water were 20.28%, 11.36% and 14.67% respectively. With these values it was estimated that there was 0.609975 mol of oxalate, 0.51867 mol of potassium, 0.2034 mol of iron, and 0.81409 mol of water. By using the lowest value, 0.234, each value was divided and then rounded to be 3 potassium ions, 1 iron (III), 3 oxalate, and 4 hydrates. These values were mostly consistent with what was originally hypothesized, however the number of hydrates was too high. Before the experiment was conducted, it was hypothesized that the empirical formula for the crystalline solid would be K3Fe(ox)3*3H2O, but the experiment alluded to the empirical formula being K3Fe(ox)3*4H2O. This evident discrepancy in the two formulas infers that the experiment had error within it and the values obtained were skewed in either direction which lead to the incorrect percent composition by mass of water. It is also possible that during the first step of this lab, the crystalline solid that was created was not potassium ferrioxalate trihydrate as expected, but rather potassium ferrioxalate

Then, 5 mL of the catalase was put in a test tube and placed in a beaker of boiling water. Then 1 mL of the boiled catalase was added to 10 mL of hydrogen peroxide. The boiled catalase was denatured, or no longer functional, so it did not aid the reaction. In section 2B, the procedure was repeated, except without the addition of the catalase. Instead of the catalase, 1 mL of H2O was added. Then 10 mL of H2SO4 was added. KMnO4 was added to the solution until it changed to a light pink or brown color. The amount of KMnO4 added will equal the amount of H2O2 in the solution, because the KMnO4 will react with the H2O2 until there is no more H2O2 left. In section 2C, 15 mL of H2O2 was put in a beaker and stored uncovered at room temperature for 24 hours. In section 2D, 10 mL of H2O2 was put in a beaker. 1 mL of catalase extract was added. After 10 seconds, H2SO4 was added, which stops the reaction. This procedure was repeated for 30, 60, 90, 120, 180, and 360

My hypothesis was correct. By calculating the average amount of KMnO4 used to titrate (0.01803 L and 0.01262 L), I was able to determine it's amount of moles (0.003606 and 0.003606). From that point, I used the balanced equation (2 KMnO4 + 5 H2O2 + 3 H2SO4 → K2SO4 + 2 MnSO4 + 8H2O + 5 O2) to find the ratio between the KMnO4 and H2O2 (2:5) to get the amount of moles for H202 (0.009015 and 0.00631). Then I simply plugged in that information into the equation for morality (M=m/L) to find the morality of the H2O2 which was 0.9015 and 0.631 respectively. This outcome was to be expected as my hypothesis was based upon the relationship of the KMnO4 and H2O2 in the reaction, so I had to do was find the characteristics of the KMnO4 in the solution

In the experiment, the majority of the times tested, the concentration of KMnO4 in the water had increased from the previous time it had been tested. But, in a few circumstances

Procedure: 1. A solution of NaOH (100 mL of 0.100 M) was prepared. 2. Burette was washed with 2-3 mLs of NaOH solution 3. Solution of NaOH was placed in the burette.

We left the solution to sit ther for two weeks, and then after two weeks, we added 1 μl of Izit dye to the drops, then waited two weeks for the crystals to absorb the dye in-order to observe the crystal under the microscope. Results for Row A with NaCl being the changing variable. The concentration of PEG for all of the wells was 20%. Well A1: Crystals produced in 1 M NaCl Well A2: Crystals produced 1.2M NaCl Well A3: Crystals produced 1.4 M NaCl

6. Immediately titrate this solution with the standardized MnO4 solution from Part 1. Record both the initial and final buret readings in the Part 2 Data Table.

solution was stirred with a glass stirring rod. It was important to remember that diluting the

To start the experiment, a 150 mL beaker was weighed. By using the tare function, 6.619 g of nickel sulfate was added to 150 mL beaker. The substance was in the shape of crystals and had a teal green color. Then, 15 mL of deionized water was added to the beaker containing the nickel sulfate hexahydrate. The beaker was then set atop a hot plate and heated on setting four while constantly being stirred until the solid had dissolved. After the heating had occurred, the color of the liquid in the beaker was a light green color. The 150 mL beaker was taken off of the hot plate and set on a ring stand base to cool slightly. Next, on the same scale as before, 4.425 g of potassium sulfate was added to a 200 mL beaker. Then 50 mL of deionized water was added to the 200 beaker. It was heated at setting four on the hot plate and stirred until dissolved. Then warm nickel sulfate (150 mL beaker) was poured into the solution of hot potassium sulfate (200 mL beaker) and mixed