• Iodine Solution Weigh 7.7g of potassium iodide into a 50cm3 beaker. Use distilled water to help the iodide dissolve. Swirl for a few minutes until the iodide has completely dissolved. Using a funnel to help, pour the potassium iodide into a 500cm3 volumetric flask, make sure all traces of the solution is in the volumetric flask. Using distilled water would be a good method in order to rinse the beaker. Make the solution up to the 500cm3 mark with iodine (1% concentration) • Starch Indicator Solution Weigh 0.25g of soluble starch and add it to 50cm3 of near boiling water in a 50cm3 beaker. Stir it in order to dissolve and wait for it to cool before using. Procedure Safety • Before conducting the experiment, be sure to put on safety goggles and …show more content…
Using the equation of the titration in the experiment (Equation 1), calculate the number of moles of the ascorbic acid reacting. 4. Then proceed to calculating the concentration in mol. dm3 of the ascorbic acid in the solutions that was obtained. Data Collection and Processing Raw Data Table 1: Amount of Iodine Reacting (± 0.05cm3) in the titration FT 1 FT 2 FT 3 FT 4 FT 5 Trial 1 1.9 1.8 1.8 1.5 1.6 Trial 2 2.1 1.8 1.7 1.5 1.6 Trial 3 2.1 2.0 1.8 1.5 1.5 Trial 4 2.5 2.3 1.7 1.6 1.5 Trial 5 1.9 1.8 1.6 1.6 1.4 Average 2.1 1.94 1.72 1.54 1.52 Analysis The results of Table 1 are shown graphically in Figure 2 where the calculated and measured Iodine Solution are plotted with respect to the amount of days. The graph can be observed, showing that the measured value of the iodine is very close to the theoretical one. It can be seen that each time the Vitamin C is frozen, the iodine solution reacting with the Acid decreases. This means that the Ascorbic Acid is being destroyed every time it is frozen. Volumetric Analysis Calculating the moles of iodine reacting Since the potassium iodide was mixed with the iodine to create triiodide so that the starch indicator could form the complex, it does not have to be
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
Next, about 10 mL of both solutions, Red 40 and Blue 1, were added to a small beaker. The concentration of the stock solution were recorded, 52.1 ppm for Red 40 and 16.6 ppm for Blue 1. Then, using the volumetric pipette, 5 mL of each solution was transferred into a 10 mL volumetric flask, labelled either R1 or B1. Deionized water was added into the flask using a pipette until the solution level reached a line which indicated 10 mL. A cap for the flask was inserted and the flask was invented a few times to completely mix the solution. Then, the volumetric pipette was rinsed with fresh deionized water and
In order to find the amount of a product made during a double displacement reaction, the product has to be separated from the solution. From this number of moles of precipitate can be calculated. From there the number of moles of reactants can be calculated using the mole ratios of the particular reaction that occurred. As seen in Table 5 it is shown that by finding out the number of moles of the unknown, the molar mass of the unknown can be calculated. From the found mass of the unknown compound, the mound of the original ion can be found.
6. Rinse a 500 ml volumetric flask with deionized water. 7. Label the volumetric flask so you know which solution is in it. 8.
Using the equation m = ΔTf/Kf , the molality of the unknown solution was found. Then, moles of unknown were calculated, which was used to calculate the average molar mass of unknown. Theory: After the experiment was completed, the data
Procedure: As follows in Mrs. Lubin’s “Finding Moles within a Sample” Lab, found on https://docs.google.com/document/d/1m36zfJCLjGNxoswIMUya_U1HJ4agrHv2yhF6cv1JJU/edit. IV. Data & Calculations: See attached data table. V. Analysis: 1. Define the three identities of the mole.
Starch solution is then placed into the test tube at a quantity of 5 mL. 5 drops of Lugol’s Iodine solution is added to the test tube. If the color changes, then it is known that starches are present in the solution. Proteins are next tested. In order to do this, 5 mL of gelatin solution is added to the test tube. 10 drops of Biuret’s reagent are added to test for protein.
The investigation was carried out to identify the presence or absence of biological molecules in serum 2216. If the concentration in each test tube of the dilutions carried out will be more concentrated then the concentration of the test tube before it, then the color will be at an equal concentration with the other dilutions performed. The hypothesis was wrong because of the difference in concentrations due to the different measurements within the dilutions done. The test for starch was to add a drop of iodine solution to the pipette in the spotting tile. A reducing sugar solutions is add inside a test tube with 3 drops to then add 3 drops of benedicts and plane in a water bath.
= 10^-3 M = 1,000 mL Here C1,C2; are the first and second concentrations of solution V1 and V2 ; are the required and current volumes. The impeller turned on and DDA, and tap water left to be mixed properly with water for 2 minutes. Approximately 150 grams of quartz added into the solution.
Weighed 1 gram of NaC2H3O2 and mixed it with ionized water. Boiled 12 mL of 1.0M Acetic Acid added into a beaker containing the sodium carbonate on a hot plate until all the liquid is evaporated
The iodine test determines the presence of starch in biological materials. It is predicted that, if starch is not present, the solution with iodine remains yellow. However, if starch is present the solution with iodine becomes a blue-black colour. Plants have starch as the storage polysaccharide (glucose units held together by glycosidic bonds) while animals have the equivalent of glycogen. In this experiment, the dark blue colour is visible because of the helical amylose and amylopectin reacting with iodine (Travers et al., 2002).
Take the NaClO and the sodium thiosulfate solutions and measure the temperature of each solution. Record in the data table. 2. Mix a combination of the NaClO and the thiosulfate solution equal to 50 ml in a Styrofoam cup, stir with thermometer, and record temperature in the data table. Dispose of solutions and rinse cup well.
Uncontrolled Environmental conditions Atmospheric conditions The controlled variable Concentration of amylase was kept under control by measuring the amount of amylase used and also it was made sure the percentage of amylase used was 1%. The Amount of amylase/starch used were kept to 5cm3 at all times. Materials needed Beakers Bunsen burner Test tube Thermometer Stopwatch Test plate Glass rod Starch Amylase solution Water bath Iodine solution. Test tube holder Labels Marker Procedure First 5 test tubes were taken and labeled with numbers from 1 to
Decomposition of Aspirin Studied with UV/Visible Absorption Spectroscopy Aims: To determine the concentration of salicylic acid, formed from the hydrolysis of Aspirin, at regular intervals using the UV/Visible Absorption Spectroscopy From the concentration of salicylic acid, concentration of Aspirin to be determined using an equation Calculate the rate constant of this reaction and its order from a plot of graph of ln(aspirin) vs time Discuss the overall flaws and improvements to the experiment Results: As per schedule1, 0.212g of aspirin was added to 50 ml boiling water to form salicylic acid in a 100 ml flask, of which 1 ml was then pipetted to a 50 ml volumetric flask at the 5th min. Following an ice bath, the solution was mixed
The mass of vinegar used during the experiment was 4.108 grams. It was determined that there were .003129 moles of CH3COOH in the vinegar sample. Using this information and the molar mass of CH3COOH, which was 60.05 g/mol, the mass of acetic acid in the vinegar was calculated: 4.Vinegar is a 5% aqueous solution of acetic acid. Since the mass of acetic acid within the vinegar was calculated as .18789 g in step 3, the percent of CH3COOH was calculated using the following equation: To calculate the percent error, the experimental value of 4.5% acetic acid in vinegar was subtracted by the theoretical value of 5% and divided by 5% to yield a percent error of 8.54%. The following is a copy of the calculations done using decimals: 5.The equivalence point of the titration curve measured in step 1 was 25.25 mL of NaOH.