Chemistry 305 Laboratory Exercise #6
Purpose: In this experiment you will determine the molar quantities of two reagents; the moles of one product (a precipitate); and the limiting reagent in this chemical process.
Discussion: Known quantities of two water soluble substances (copper (II) chloride dihydrate, CuCl2·2H2O, and silver nitrate, AgNO3) are combined and a precipitate forms (silver chloride, AgCl). The balanced equation is:
2 AgNO3(aq) + CuCl2·2H2O(aq) 2 AgCl(s) + Cu(NO3)2(aq) + 2 H2O(l)
The precipitate is filtered from the solution, dried and weighed. From the balanced chemical reaction and the molecular weights, you will be able to determine how many grams of silver combined with how many grams of chloride, and which of the original compounds was present in excess.
|Equipment:||Beakers||Stirring Rod||Filter Paper||Buchner Funnel|
|Clamp||Ring Stand||Hose||Vacuum flask|
|Test Tubes||Watch Glass||Spatula||Rubber policeman|
|0.1M CuCl2·2H2O||0.1M AgNO3|
Safety: Wear goggles at all times. Wear gloves to prevent silver nitrate from staining your hands. These stains are harmless, but can take several days to wear off. Dispose of all wastes as described by your instructor.
In this experiment, note any observations that you make while carrying out the procedure (i.e., note cloudiness, colors, losses, films on glass, speed of reactions, etc.). Be sure to include units.
- Using two small, dry beakers, weigh out about 0.500 gram each of CuCl2·2H2O and AgNO3. Record exact mass. Add approximately 10 ml of water to each beaker; stir with a glass stirring rod until they dissolve completely. Once they are completely dissolved, combine the two solutions by pouring one completely into the other. Rinse any remaining droplets from the empty beaker into the collection beaker with a distilled water squirt bottle. A white cloudiness should develop. AgCl is insoluble and will form a precipitate. The other components (ions) of the solution (Cu2+ and NO3-) are soluble and will remain dissolved in the solution (these are called spectator ions).
- Obtain a piece of No. 6 filter paper, weigh it and record the weight.
- Folsom Lake College Students:Using a ring stand and clamp, fix a side-arm vacuum flask firmly in place and connect the vacuum hose to the vacuum line on the lab bench. El Dorado Center Students: Using a ring stand and clamp, fix a side-arm vacuum flask firmly in place and connect the vacuum hose to the side-arm of the flask and the side-arm of the aspirator on the water faucet or use one of the small vacuum pumps set up in various locations in the lab.
- Fix a Buchner funnel firmly in the top of the vacuum flask. Wet the funnel with a squirt or two from your wash bottle filled with distilled water. Drop the weighed, circular piece of filter paper into the funnel, and make sure that it is centered in the funnel. Wet the paper with a squirt or two of water. Turn on the vacuum and make sure the water in the funnel is sucked through the filter paper. Be careful to not aspirate water into the vacuum pump.
- Slowly and carefully pour the “milky” precipitate mixture from the small beaker onto the center of the filter paper. It is best if you do not form a puddle of water in the funnel. When all the solution has been poured, wash the small beaker with your squirt bottle and slowly pour that water through your filter. Continue until all the precipitate material has been transferred onto the filter paper. Use a spatula or rubber policeman if necessary to harvest every bit of precipitate. You may use as much water as you like. When all the particulate matter has been transferred to the filter paper and the water passed through it, continue to run the filter apparatus for a minute or two more to remove as much water as possible from the filter paper. Turn off the filter apparatus.
- If the liquid in the vacuum filter flask (the supernatant) is cloudy, then remove the funnel (with the filter paper intact) and place carefully on the benchtop. Pour all of the supernatant into a clean beaker. Rinse the filter flask with a little water and pour this water into the beaker as well. Then put the funnel back onto the filter flask and turn the filter apparatus back on. Slowly pour the supernatant in the beaker into the funnel, rinsing the beaker with water. This is a 2nd filtration, and now the supernatant in the filter flask should be clear. Turn off the filter apparatus after a few minutes.
- Carefully remove the filter paper and any solid from the funnel, and lay it on a watch glass that you have previously weighed. Collect any little pieces of precipitate clinging to the sides of the filter on the filter paper. El Dorado Center students:Place it on a weighed watch glass and put it in the drying oven for 45 minutes to dry. Be sure that all of the precipitate stays on the filter paper. Remove from oven when dry, allow to cool and weigh and record mass. Then dispose of filter cake in labeled waste container. Folsom Lake College students:See #9 below.
- Using stockroom solutions of 0.1M CuCl2·2H2O and 0.1M AgNO3 test your supernatant to see which reagent was limiting and which was in excess. Pour a few mL of the supernatant (the liquid solution that went through the filter paper) into each of two clean test tubes. To the supernatant in one test tube, add a few drops of the stock 0.1M silver nitrate solution and note results. To the supernatant in the other test tube, add a few drops of the stock 0.1M copper chloride solution and note results.
- Folsom Lake College students:Come to the other lab period after your product has dried a few days, reweigh the dried filter with the precipitate on it. Record the mass, and dispose of the filter paper in the appropriate waste container.
1) Weight of AgNO3 = 0.5000g
2) Weight of CuCl2·2H2O = 0.5000g
3) Weight of unused, marked filter paper = 0. 2g
3b) Weight of watch glass (optional) = 50.49 g
4) Weight of dried filter paper and filtrate (and watch glass if used) = 51.11g
5) 0.1M AgNO3 in supernatant observation: A white precipitate was formed
6) 0.1M CuCl2·2H2O in supernatant observation: No precipitate was formed
Data Analysis (Show your work and put answer on the line.)
Calculate the net weight and the number of moles of AgCl(s) that you obtained (that is, the net weight of the filtrate divided by the molecular weight of AgCl). Determine the number of moles of each of the starting components [that is, the actual weight of each AgNO3 and CuCl2·2H2O divided by their molecular weights]. Using your results, determine what percentage of the moles of original silver and what percentage of the moles of original chloride were used to form your precipitate AgCl(s).
1) Molar Mass of AgNO3 = 169.87g/mol
2) Molar Mass of CuCl2·2H2O= 170.48g/mol
3) Molar Mass of AgCl=143.32 g/mol
4) Moles of original AgNO3 = 0.5000g/(169.87g/mol)=0.002943427 moles
Moles of original Ag+: = 0.002943427 moles
5) Moles of original CuCl2·2H2O = 0.5000g/170.48g/mol= 0.002932895 moles
Moles of original Cl-: 0.002932895*2=0.005865791 moles
6) Net weight of AgCl precipitate = 0.36g
7) Moles of new AgCl=0.36g/(143.32g/mol) = 0.002511862 moles
8) Percent of original Ag+ used= (0.002511862/0.002943427)*100 =85.34%
(mol AgCl ÷ mol original Ag+) x 100
9) Percent of original Cl- used =(0.002511862/0.005865791)*100 = 42.82%
(mol AgCl ÷ mol original Cl-) x 100
- Which original compound was nearly all used, and which was present in excess?
Silver chloride, AgCl, was nearly used all used. The percent of the original Ag+ used was 85.34%.on the other hand, the percent of original Cl- used was 42.82 %, meaning approximately 57.18% of the CuCl2·2H2O was unused after the chemical reaction .Therefore, the copper (II) chloride dihydrate, CuCl2·2H2O was in excess.
- How does your supernatant test support your answer to question #1?
The supernatant test supports my answer to question #1. After adding drops of 0.1M AgNO3 solution into supernatant, a precipitate was formed. On adding few drops and more drops of 0.1M CuCl2·2H2O solution into the supernatant, no precipitate was formed.
- Based on your limiting reagent (reactant), calculate the theoretical yield of AgCl in g.
2 AgNO3(aq) + CuCl2·2H2O(aq) 2 AgCl(s) + Cu(NO3)2(aq) + 2 H2O(l)
The limiting reagent (reactant is) AgNO3 . The mole ratioAgNO3 to AgCl of is 1:1. Therefore, theoretical yield is: 0.002943427moles*143.32 g/mol =0.4219 g
- Calculate your %-Yield [(Actual Yield ÷ Theoretical Yield) x 100].
%-Yield =(0.36g/0.4219 g)*100 =85.35%
What did you learn from this experiment?
I learned how to check which reagent is in excess and which is the limiting reagent, in a precipitation reaction, by using supernatant test. Also, I learned how to find the percent yield in a laboratory experiment.
Also see: The Mole and Avogadro’s Number
Last Updated on December 15, 2020 by EssayPro