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 seawalls of reactants is kept constant for the series of measurements. Each measurement is made with a divergent mole dimension of reactants. A mole ratio is ratio between the amounts in moles of whatever 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 grea discharge amount of product, and, if the reaction is exothermic, generate the most heat and maximum temperature change, Double deputy reactions are generally considered to be irreversible. The shaping of an insoluble go down provides a driving force that makes the reaction proceed in one direction only.In a double reaction, the two reactants which are sedimentary solutions (which can be broken down), can form two products one also an aqueous solution, and a nonher which can be a hasty, 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 renders (some graduated, some without graduations) 6. Black Marker 7. Marking ta pe 8. (2) Stirring rods, large 9. Pipets 10. (2) Test tubing racks 1. (2) Pairs of gloves 12. Lab goggles 13. Lab apron 14. whilekeeper 15. Para film Procedure The lab was set up, as it is seen in figure 1, with seven mental ladder tubes in a test tube rack. use 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 clippingr 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 classifications were to be left for 10 minutes (a ager 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 aft(prenominal) 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 aim 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 homogeneous size was found, then using a pipet, 1ml of water was measured in a 10mL graduated cylinder, and then poured into the like 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 graduated test tube was held side by side with the similar size test tube with no graduati ons, and the precipitate was measured using this method. 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 board 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 4NaOH, 0. 1M, mL 11 10 16 9 5 10 6 FeOH Mole Ratio 111 15 14 13 25 12 23 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 CuPO4 Mole Ratio 15 12 23 11 32 21 51 Volume of Precipitate (mL) 1 mL 4 mL 3. 75 mL 4 mL 2. 5 mL 3. 2 mL 1 mLRESULTS PARAGRAPH POST LAB QUESTIONS Observations Fe(NO3)3 and NaOH 1. lift off time 1207 Separated instantly. After 3 minutes separated halfway land up time 1217 Precipitate is 1/5 of test tube color is softly orange 24 hours later Same results 2. Start time 1211 Separated a little death time 1221 The precipitate takes up 2/5 of the test tube & is orange 24 hours later Same results 3. Start time 1215 Instant legal separation End time 1225 Liquid still a little cloudy. Precipitate is ? of test tube color is light/dark orange 24 hours later Same results 4.Start time 1219 Separating VERY slowly End time 1229 on that point is no precipitate just yet. Very cloudy. 24 hours later Same results 5. Start time 1222 Separation did not occur instantly End time 1232 There is no precipitate just yet. Very cloudy. 24 hours later Same results 6. Start time 1224 Separation did not occur instantly End time 1234 There is no precipitate. Very Cloudy. 24 hours later Same results 7. Start time 1227 Separation did not occur instantly End time 1237 There is no precipita te just yet. 24 hours later Same results CuCl2 and Na3PO4 1.Start time 1236 Separated quickly End time 1246 Precipitate ? of test tube. Color is light blue 24 hours later Same results 2. Start time 1237 Separated quickly End time 1247 Precipitate is ? of test tube. Color is regular blue 24 hours later Same results 3. Start time 1239 Separated quickly End time 1249 Precipitate is ? of test tube. Color is regular blue. 24 hours later Same results 4. Start time 1241 Separated only a little numeral in the first of all two minutes End time 1251 Precipitate is ? of test tube and color is light blue 24 hours later Same results 5.Start time 1242 Separated only a little bit in the first two minutes End time 1252 Precipitate is 2/5 of test tube and color is light blue 24 hours later Same results 6. Start time 1243 Separated only a little bit in the first two minutes End time 1253 Precipitate is 3/5 of test tube and color is light blue 24 hours later Same results 7. Start time 1245 Separated only a little bit in the first two minutes End time 1255 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 solution. Lastly these observations elaborate on the slow or fast process of how ea ch solution separated into a precipitate, or didnt, 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 tubes had precipitates present, in those three test tubes there is a distinction in color, th e 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 selective information AND DISCUSSION PARAGRAPH.