2 PurposeThe purpose of this lab is to find out how much calcium carbonate is in an eggshell.
3 ProcedureObtain an egg. Break the egg into a beaker. Add water to the egg and stir before pouring the egg down the drain.Wash the shell with deionized water and peel off all the membranes from the inside of the shell. These membranes (proteins), if left, will react with the NaOH. Dry the eggshell with a paper towel and put into a labeled beaker. Wash your hands with soap and water.Dry the shell for about 10 minutes in the ovenGrind the shell to a fine powder in a mortar.Accurately weigh between 0.45 and 0.55g of dried eggshell into each of 3 labeled 250 mL Erlenmeyer flasks. Be sure you know how much eggshell is in each flask.Add several drops of ethanol to each flask. This acts as a wetting agent and helps the HCl dissolve the CaCO3Carefully pour 10.0 mL of 1.0M of HCl into each of the Erlenmeyer flasks with the eggshell in it. Swirl the flask to wet the entire solid.Heat the solutions in the 250 mL flasks until they begin to boil and then allow them to cool. Do not let them boil to dry. Rinse the walls of the flask with water from a wash bottle.Add 3-4 drops of phenolphthalein indicator to each flask.Prepare a buret with 0.1M NaOH.Titrate one sample of eggshell to the first persistent pink colour. When you are close to the endpoint the colour will fade slowly. Add the remaining NaOH dropwise or by half drops until the color remains for at least 30 seconds.Repeat the titration for the other two samples.Calculate the percentage calcium carbonate in each sample.Find the literature value for the percentage of CaCO3 in the typical chicken egg.
5 Observations When we poured the HCl into the eggshell it bubbled Clumps of white eggshell leftWhen we dropped the correct amount of NaOH it turned fluorescent pink because of the phenolphthalein indicator
6 Calculations First Trial Second Trial Third Trial 2.86mL NaOH∗ 0.1 M NaOH 1000 mL ∗ 1 mol HCl 1 mol NaOH =2.86∗ 10 −410 mL HCl∗ 1 M HCl 1000 mL =0.01 mol HCl−2.86∗ 10 −4 mol HCl= mol HCl usedmol HCl∗ 1mol Ca mol HCl ∗ 1 mol CaC O 3 1 mol Ca 2+ = mol CaCO 3 ∗ g CaC O 3 1 mol CaC O 3 = g CaC O g CaC O 3 ∗100=107%Second Trial2.35mL NaOH∗ 0.1 M NaOH 1000 mL ∗ 1 mol HCl 1 mol NaOH =2.35∗ 10 −410 mL HCl∗ 1 M HCl 1000 mL =0.01 mol HCl−2.35∗ 10 −4 mol HCl= mol HCl usedmol HCl∗ 1mol Ca mol HCl ∗ 1 mol CaC O 3 1 mol Ca 2+ = mol CaCO 3 ∗ g CaC O 3 1 mol CaC O 3 = g CaC O g CaC O 3 ∗100=108%Third Trial3.00mL NaOH∗ 0.1 M NaOH 1000 mL ∗ 1 mol HCl 1 mol NaOH =3.00∗ 10 −410 mL HCl∗ 1 M HCl 1000 mL =0.01 mol HCl−3.00∗ 10 −4 mol HCl= mol HCl usedmol HCl∗ 1mol Ca mol HCl ∗ 1 mol CaC O 3 1 mol Ca 2+ = mol CaCO 3 ∗ g CaC O 3 1 mol CaC O 3 = g CaC O g CaC O 3 ∗100=107%
8 Conclusion Continued.. Just kidding~! We were able to find out how much calcium carbonate is in an eggshell through the process of volumetric analysis. Applying significant figures to our calculations would skew our results from 107% and 108% to 100%. That still doesn’t make sense because calcium carbonate isn’t the only component of eggshells. Also, we had excess eggshell after adding the HCl which would have skewed our results in the opposite direction than they were skewed. So our results aren’t too reliable. One theorized issue might have been the molarity of the HCl. We repeated the calculations using 0.8 M instead of the labeled 1 M and got a percent calcium carbonate of 85% which makes sense. The theoretical results after research were found to be between 77% and 95% calcium carbonate in the eggshell. Cool!