LABORATORY REPORT I. PURPOSE (10 POINTS) The purpose of this laboratory is to learn the proper use of an analytical technique based on the measurement of mass. Gravimetric analyses depend on comparing the masses of two compounds containing the analyte. The principle behind gravimetric analysis is that the mass of an ion in a pure compound can be determined and then used to find the mass percent of the same ion in a known quantity of an impure compound. II. TEST DATA (15 POINTS) Trial 1 Mass of weigh boat: 2.9 g Mass of same weigh boat and CaCl2 (Note: This sample will be set aside and exposed to air for 24 hours): 6.9 Mass of same weigh boat and CaCl2 after 24 hours: 7.1 g Observations of CaCl2: It became to a liquid form Mass of 50 mL beaker: 6.9 g Mass of same beaker and K2CO3: 11.9 g Amount of time beaker solution stirred: 4 min Amount of time beaker solution set: 15 min Mass of filter paper and watch glass: 34.9 g Mass of same filter paper watch glass and dried product: 44.1 g Trial 2 Mass of 250 mL beaker: 114.9 g Mass of same beaker and CaCl2: 118.9 g Observations of CaCl2: little white spheres Mass of 50 mL beaker: 6.9 g Mass of same beaker and K2CO3: 11.9 g Amount of time beaker solution stirred: 4 min Amount of time beaker solution set: 15 min Mass of filter paper and watch glass: 34.9 Mass of same filter paper watch glass and dried product: 43.9 III. CALCULATIONS (15 POINTS) Trial 1 Mass of CaCl2: 4 g Mass of K2CO3: 5 g Mass of product: 9.2 Mass of water absorbed with the CaCl2 after 24 hours: .2 Trial 2 Mass of CaCl2: 4 g Mass of K2CO3: 5 g Mass of product: 9 g IV. RESULTS (20 POINTS) Trial 1 Theoretical yield (CaCO3): Actual yield (CaCO3): Percent yield: Moles of Ca 2 present in original solution: 3.57 g Mass of CACl2 present in original solution: 4 g Percent water absorbed by CaCl2: .05 g Trial 2 Theoretical yield (CaCO3): Actual yield (CaCO3): Percent yield: Moles of Ca 2 present in original solution: 3.57 g Mass of CACl2 present in original solution: 4 g Show calculations 9 g CaCO3 x1 mol CaCO3/100 089g x 1 mol Ca /1 mol CaCO3= 3.47 g V. CONCLUSION (10 POINTS) Gravimetric analysis is important because we can determinate a substance based upon the measurement of mass. This involves isolation of an ion in solution by a precipitation reaction filtering washing the precipitate free of contaminants conversion of the precipitate to a product of known composition and finally weighing the precipitate and determining its mass by difference. From the mass and known composition of the precipitate the amount of the original ion can be determined. VI. QUESTIONS (30 POINTS) 1. Write the balanced reaction equation for the precipitation of calcium carbonate from potassium carbonate and calcium chloride. (6 points) K2CO3 CaCl2 —–> CaCO3 2KCl 2. Using this balanced equation determine the limiting reactant if 15 grams of calcium chloride reacted with 15 grams of potassium carbonate.(7 points) 15 g of CaCl2 Moles= 15/111= .135 15 g of K2CO3 Moles= 15/138.2=.1085 Limiting reagent= K2CO3 3. Using your answer for Question 2 determine the mass of potassium carbonate needed to fully precipitate all the calcium from a 25 mL sample of 15% calcium chloride solution. (Note: this is the same concentration as the solution you used in Experiment 1.)(7 points) 25*15/100)= 3.75 g of Calcium Chloride. Moles of CaCl2 = 3.75/110.98=0.0338 moles The balanced reaction is K2CO3 CaCl2 = CaCO3 2KCl So 1 mole of CaCl2 requires 1 mole of K2CO3 for complete precipitation therefore 0.0338 moles of CaCl2 requires 0.0338 moles of K2CO3 Mass of K2CO3 required= 0.0338*138.21=4.67 g 4. Create a pie chart representing the percent by mass of calcium chloride and water in the original solution for each trial. (10 points) Trial 1 .gif”> Trial 1 .gif”>
