Sunday, September 15, 2019
Lab Report I
Determining the Stoichiometry of Chemical Reactions Mrs. Farrales Nikita Pandya October 23, 2012 December 3, 2012 INRODUCTION In the method of continuous variations the total number of moles of reactants is kept constant for the series of measurements. Each measurement is made with a different mole ratio of reactants. A mole ratio is ratio between the amounts in moles of any two compounds involved in a chemical reaction.Mole ratios are used as conversion factors between products and reactants in many chemistry problems. The optimum ratio, which is the stoichiometric ratio in the equation, form the greatest amount of product, and, if the reaction is exothermic, generate the most heat and maximum temperature change, Double replacement reactions are generally considered to be irreversible. The formation of an insoluble precipitate provides a driving force that makes the reaction proceed in one direction only.In a double reaction, the two reactants which are aqueous solutions (which can be broken down), can form two products one also an aqueous solution, and another which can be a precipitate, water, or a gas, which cannot be broken down, therefore making the reaction irreversible. The objective/goal of this laboratory is to find the optimum mole ratio for the formation of a precipitate in a double replacement reaction and use this information to predict the chemical formula of the precipitate. How can the products of a double reaction be predicted?How will it be determined if a product is aqueous or a precipitate? How will the method of continuous variations help determine the mole ratio of the two reactants? METHODS Materials: 1. Copper (II) chloride solution, CuCl2, 0. 05 M, 210 mL 2. Iron (III) nitrate solution, Fe(NO3) 3, 0. 1 M. 110 mL 3. Sodium hydroxide solution, NaOH, 0. 1 M, 320 mL 4. Sodium phosphate, tribasic, solution, Na3PO4, 0. 05 M, 210 ml 5. (14+) Test tubes (some graduated, some without graduations) 6. Black Marker 7. Marking tape 8. (2) Stirring rods, large 9. Pipets 10. (2) Test tube racks 1. (2) Pairs of gloves 12. Lab goggles 13. Lab apron 14. Timer 15. Para film Procedure The lab was set up, as it is seen in figure 1, with seven test tubes in a test tube rack. Using a clean 10mL graduated cylinder, the appropriate volume of iron(III) nitrate solution was taken from its container and transferred/added to each test tube using a pipet. Using a another clean 10mL graduated cylinder, the appropriate volume of sodium hydroxide solution was taken from its container and transferred/added to each test tube, which already contained iron(III) nitrate, using a pipet.Before the timer was started, each of the solutions in the test tube was stirred/mixed with a large stirring rod. This ensured that both the reactants mixed properly. After stirring the solutions, observations were noted for any signs of chemical changes. The mixtures were to be left for 10 minutes (a timer was used) to sit undisturbed, because any movement of the test tube could cause a hindrance in the settlement of the precipitate. Though each test tube was left undisturbed for 10 minutes, final observations were made after the solutions were left to sit undisturbed for 24 hours.After the 24 hours of settling, the volume of the precipitate in each test tube was measured and recorded. For test tube with graduations, seeing the numbers at eye level made the calculations, but for test tubes with no graduations a different method was used to measure the volume of the precipitate. First another test tube of the same size was found, then using a pipet, 1ml of water was measured in a 10mL graduated cylinder, and then poured into the similar size test tube. Using a black marker graduations were written on the test tube. Graduations up to 5mL only were made.After the graduations were complete the Ã¢â¬Å"graduatedÃ¢â¬ test tube was held side by side with the similar size test tube with no graduations, and the precipitate was measured using this metho d. The same procedures were repeated with the reactants of the second table, CuCl2 and Na3PO4 Figure 1: The set up of the lab, the test tubes were labeled 1-7 RESULTS Data Table 1: Ratio between Fe(NO3)3 and NaOH are presented in this table along with the amount of precipitate that was produced in each of the test tube. Test Tube | 1| 2| 3| 4| 5| 6| 7| Fe(NO3)3, 0. 1M, mL| 1| 2| 4| 3| 2| 5| 4|NaOH, 0. 1M, mL| 11| 10| 16| 9| 5| 10| 6| Fe:OH Mole Ratio| 1:11| 1:5| 1:4| 1:3| 2:5| 1:2| 2:3| Volume of Precipitate (mL)| 1 mL| 2. 8 mL| 3 mL| 0 mL| 0 mL| 0 mL| 0 mL| Data Table 2: Ratio between CuCl2 and Na3PO4 are presented in this table along with the amount of precipitate that was produced in each of the test tube. Test Tube | 1| 2| 3| 4| 5| 6| 7| CuCl2, 0. 05 M, mL| 1| 4| 4| 6| 6| 8| 5| Na3PO4, 0. 05 M, mL| 5| 8| 6| 6| 4| 4| 1| Cu:PO4 Mole Ratio| 1:5| 1:2| 2:3| 1:1| 3:2| 2:1| 5:1| Volume of Precipitate (mL)| 1 mL| 4 mL| 3. 75 mL| 4 mL| 2. 5 mL| 3. 2 mL| 1 mL|RESULTS PARAGRAPH Ã¢â¬â PO ST LAB QUESTIONS Observations: Fe(NO3)3 and NaOH 1. Start time: 12:07 Separated instantly. After 3 minutes separated halfway End time: 12:17 Precipitate is 1/5 of test tube color is light orange 24 hours later: Same results 2. Start time: 12:11 Separated a little End time: 12:21 The precipitate takes up 2/5 of the test tube & is orange 24 hours later: Same results 3. Start time: 12:15 Instant separation End time: 12:25 Liquid still a little cloudy. Precipitate is ? of test tube color is light/dark orange 24 hours later: Same results 4.Start time: 12:19 Separating VERY slowly End time: 12:29 There is no precipitate just yet. Very cloudy. 24 hours later: Same results 5. Start time: 12:22 Separation did not occur instantly End time: 12:32 There is no precipitate just yet. Very cloudy. 24 hours later: Same results 6. Start time: 12:24 Separation did not occur instantly End time: 12:34 There is no precipitate. Very Cloudy. 24 hours later: Same results 7. Start time: 12:27 Separation did not occur instantly End time: 12:37 There is no precipitate just yet. 24 hours later: Same results CuCl2 and Na3PO4 1.Start time: 12:36 Separated quickly End time: 12:46 Precipitate ? of test tube. Color is light blue 24 hours later: Same results 2. Start time: 12:37 Separated quickly End time: 12:47 Precipitate is ? of test tube. Color is regular blue 24 hours later: Same results 3. Start time: 12:39 Separated quickly End time: 12:49 Precipitate is ? of test tube. Color is regular blue. 24 hours later: Same results 4. Start time: 12:41 Separated only a little bit in the first two minutes End time: 12:51 Precipitate is ? of test tube and color is light blue 24 hours later: Same results 5.Start time: 12:42 Separated only a little bit in the first two minutes End time: 12:52 Precipitate is 2/5 of test tube and color is light blue 24 hours later: Same results 6. Start time: 12:43 Separated only a little bit in the first two minutes End time: 12:53 Precipitate is 3/5 of test tube and co lor is light blue 24 hours later: Same results 7. Start time: 12:45 Separated only a little bit in the first two minutes End time: 12:55 Precipitate is 1/5 of test tube and color is light blue In the observations mentioned above, estimates using numbers (fractions) were made.These fractions basically estimate the amount of precipitate that was formed in each test tube, or the lack of a precipitate. Observations were made after the ten-minute mark, and then left under the fume hood for 24hrs due to the fact that time fell short; observations were made then also. The observations also show that in the test tube where it was recorded that the separation between the compounds was instant, there was a precipitate formed. Respectively the observations also show that in test tubes where it was recorded that separation between the compounds was not instant, there was no precipitate formed.These observations describe the color of the solution/precipitate, and tell the transparency of the sol ution. Lastly these observations elaborate on the slow or fast process of how each solution separated into a precipitate, or didnÃ¢â¬â¢t, based on their specific mole ratio. It justifies how the different mole ratio produced the different precipitate amount. Figure 2: Fe(NO3)3 and NaOHFigure 3: CuCl2 and Na3PO4 These pictures show a visual of the seven test tubes in each experiment. In some of them the precipitates are present, in other test tubes there are no precipitates present, which means that they are still solutions.The test tubes with graduations, that had precipitates present were measured by reading the number at eye level. But test tube with no graduations, that had precipitates present, a special method that was mentioned in the procedures were used. Since in experiment two, all of the test tubes had a precipitate present there was a clear distinction in colors, the blue and clear, they were heterogonous mixtures. But in experiment one, only three of the seven test tub es had precipitates present, in those three test tubes there is a distinction in color, the red-ish orange and clear, they were heterogonous mixtures.But in the other four test tubes, since they are solutions it is a homogenous mixture where the entire solution has one consistency and color. DISCUSSION By conducting the experiment, and analyzing the results, the optimum mole ratio for the formation of the precipitate in a double replacement reaction was found, and the chemical formula of the precipitate was found, the initial purpose of the experiment. At the beginning of the experiment two questions were proposed. ANSWER QUESTIONS ERRORS CONCLUSION LITTLE BIT FROM DATA AND DISCUSSION PARAGRAPH.