1. Analytical Gravimetric Determination Suh Kwon

2. Purpose To measure the number of a given substance in a solution by precipitation, filtration, drying, and weighing To measure the number of a given substance in a solution by precipitation, filtration, drying, and weighing – Approximating the amount of SO 4 2- ions in a sample of alum, KAl(SO 4 ) 2 · 12H 2 0

3. Background Information Gravimetric analysis is used to determine the amount of a substance by finding its mass, and then using the mass to find the quantity of the substance Gravimetric analysis is used to determine the amount of a substance by finding its mass, and then using the mass to find the quantity of the substance – Example: to find the measurement of solids suspended in a water sample, a known volume of water is filtered and the collected solids are weighed

4. For this experiment… The precipitate that will be formed in this experiment is barium sulfate, BaSO 4 The precipitate that will be formed in this experiment is barium sulfate, BaSO 4 – Forms very fine crystals http://www.jinyakc.com/jy/My%20Pictures/9.jpg Beforehand, calculate how much 0.200 M Ba(NO 3 ) 2 you would need to totally precipitate all of the sulfate ion present in solution

5. !!! Make sure you add twice the calculated amount of Ba(NO 3 ) 2 because for every 1 mole of alum, there are 2 moles of Ba(NO 3 ) 2 Make sure you add twice the calculated amount of Ba(NO 3 ) 2 because for every 1 mole of alum, there are 2 moles of Ba(NO 3 ) 2 KAl(SO 4 ) 2 · 12H 2 O + 2Ba(NO 3 ) 2  KNO 3 + Al(NO 3 ) 3 + 2BaSO 4 + 12H 2 O

6. Materials Buchner Funnel Buchner Funnel A piece of Whatman No. 42 filter paper A piece of Whatman No. 42 filter paper Filter flask Filter flask 1.059 g of alum compounds 1.059 g of alum compounds 50.0 mL of distilled water 50.0 mL of distilled water 22.3 mL (calculated) of 0.200 M Ba(NO 3 ) 2 22.3 mL (calculated) of 0.200 M Ba(NO 3 ) 2 Heating Plate Heating Plate Analytical Balance Analytical Balance Drying oven set at 50°C Drying oven set at 50°C

7. Filter Flask Filter Paper Analytical Balance Buchner Funnel http://www.laboratoryequipmentworld.com/gifs/filter-flasks.jpg http://www.usdoj.gov/dea/photos/lab/analytic al_balance_mettler_ae-260.jpg http:// www.isss.biz/prodImages/Whatman-Filter-Paper-Grade-2.jpg http://upload.wikimedia.org/wikipedia/commons/5/52/Buchner_funnel.jpg http://www.rickly.com/sai/images/BUCHNER.JPG An analytical balance is an instrument that's used to measure mass at a very high degree of precision.

8. Procedures 1.Dissolve the 1.059 g of alum in the 50.0 mL of distilled water 2.Add twice the amount of 0.200 M Ba(NO 3 ) 2 to alum solution, stirring constantly 3.Heat this solution for 15 minutes and allow the solution to stand overnight Next Day: You will discover that fine crystals had appeared in the solution

9. Procedures (2) Weigh the filter paper separately and determine its mass on the analytical balance 4. Weigh the filter paper separately and determine its mass on the analytical balance 5. Filter the solution through the Buchner funnel, containing the filter paper to collect the crystals 6. Remove the filter paper and allow it to dry in the 50°C drying oven 7. Weigh the paper and crystals together and determine their total mass

10. http://depts.washington.edu/chem/courses/labs/162labs/images/PA230280.JPG http://www.creative- chemistry.org.uk/activities/images/buchner.gif

11. Data INITIAL Mass of alum: 1.059 g Mass of alum: 1.059 g Volume of distilled water: 50.0 mL Volume of distilled water: 50.0 mL Amount of 0.200 M Ba(NO 3 ) 2 : 22.3 mL Amount of 0.200 M Ba(NO 3 ) 2 : 22.3 mLCOLLECTED Mass of filter paper (alone): 1.675 g Mass of filter paper (alone): 1.675 g Mass of filter paper + filtered crystals: 2.715 g Mass of filter paper + filtered crystals: 2.715 g

12. Equations Balanced Chemical Formula: Balanced Chemical Formula: *** KAl(SO 4 ) 2 · 12H 2 O + 2Ba(NO 3 ) 2  KNO 3 + Al(NO 3 ) 3 + 2BaSO 4 + 12H 2 O *** Ba 2+ + SO 4 2-  BaSO 4 (s) % Sulfate = (Mass of Sulfate/Mass of Sample) * 100% % Sulfate = (Mass of Sulfate/Mass of Sample) * 100% Percent Error = (Theoretical Value – Actual Value) * 100% Percent Error = (Theoretical Value – Actual Value) * 100% Theoretical Value

13. Calculations How much 0.200 M Ba(NO 3 ) 2 would be needed to totally precipitate all of the sulfate ion present in the alum solution: How much 0.200 M Ba(NO 3 ) 2 would be needed to totally precipitate all of the sulfate ion present in the alum solution: 1.059 g alum · 1 mol of alum · 2 mol Ba(NO 3 ) 2 · 1 L of Ba(NO 3 ) 2 · 1000 mL = 22.3 mL 474.46 g alum 1 mol of alum 0.200 mol Ba(NO 3 ) 2 1 L 474.46 g alum 1 mol of alum 0.200 mol Ba(NO 3 ) 2 1 L

14. The percent of sulfate ion in the alum based upon the experiment: The percent of sulfate ion in the alum based upon the experiment: 2.715 g filter paper + barium sulfate 2.715 g filter paper + barium sulfate - 1.675 g paper 1.040 g BaSO 4 1.040 g BaSO 4 1.040 g BaSO 4 · 1 mol BaSO 4 · 1 mol SO 4 2- · 96.04 g SO 4 2- = 0.4281 g SO 4 2- 233.37 g BaSO 4 1 mol BaSO 4 1 mol SO 4 2- 233.37 g BaSO 4 1 mol BaSO 4 1 mol SO 4 2- % sulfate = 0.4281 g SO 4 2- · 100% = 40.43% 1.059 g alum 1.059 g alum 1 2 3

15. The theoretical % of sulfate ion in the alum: The theoretical % of sulfate ion in the alum: 192.08 g SO 4 2- · 100% = 40.48 % 192.08 g SO 4 2- · 100% = 40.48 % 474.46 g alum 474.46 g alum The percent error in this investigation: The percent error in this investigation: % error = 40.43 – 40.48 · 100% = -0.15% 40.48 40.48

16. Conclusion With the help of this experiment, you were able to find the amount of sulfate ions in a sample of alum compound, comparing that value to the theoretical amount necessary With the help of this experiment, you were able to find the amount of sulfate ions in a sample of alum compound, comparing that value to the theoretical amount necessary A negative percent error signifies that there was a production of a solid throughout the reaction A negative percent error signifies that there was a production of a solid throughout the reaction