Composition of calcium carbonate in egg shell free essay sample

Wear eye goggle during an experiment because of the strong hydrochloric acid In case, if there is an acid contact your skin, quickly wash your skin with water Pour acid on body level so it does not interact with you face Independence variable: Mass of egg shell: we can use different mass of egg shell to find our result Dependence variable: Purity of calcium carbonate in egg shell Controlled variable: Concentration of Hydrochloric acid Uncertainty and error: Measuring cylinder scale is plus minus 0.08 cm cube Egg shell may have little bit of membrane which effect weight Little bit of egg shell may not all dissolve Our vision while pouring or recording data Egg shell may contain some water Collecting data Take off egg shell and clean it with water Take off egg membrane that attach in the inner shell Dry it using tissue paper or dryer Put it in mortar and slowly crush it with pastel Weigh them with a balance a record it as gram Pour egg shell into a conical flask Add the amount of HCl (in the theoretical volume) to conical flask (Warning: slowly pour it because the reaction is rapid due to less surface area of egg shell) Wait for the reaction to stop (a days) Set up clamp which hold a burette Now we have to titrate calcium carbonate with sodium hydroxide to find how many mole we have to use to neutralize Before titrate pour 2 two drops of phenolphthalein on to calcium carbonate Titrate until it look light pink Titrate at least three times Now use your record to calculate (see the calculation below) When use back titration Result of how much NaOH is used: 1st time 5.3 cm cube 2nd time 5.5 cm cube 3rd time 5.4 cm cube Calculate the percent composition of calcium carbonate Equation: CaCO3 +2HCl → CaCl2 + CO2 + H2O Mass of egg shell: 8.34 g Hydrochloric acid: 170 cm cube Find mole of calcium carbonate: n=m/mr: So: n= 8.34/100 = 0.0834 and *2 (because of molar ratio 1:2 to HCl) Now n=cv: So, 0.1668=concentration (1 mole/dm cube)*volume V=166.8 cm cube 1 mol dm cube acid 200 cm cube to fully dissolve the shell 25 cm cube and titrate NaOH +HCl → NaCl + H2O Average volume: (5.3+5.5+5.4)/5.5 = 5.4cm cube = 0.0054 dm cube n=cv → n = 0.0054*1= 0.0054 mol for 25 cm cube moles needed to neutralize = 0.2- (0.0054*8) = 0.1568 mole CaCO3 + 2HCl → CaCl2 + CO2 + H2O Molar ratio of CaCO3 : HCl 1:2 → m=n*mr = 0.0784*100 = 7.84 Ratio 0.0784:0.1568 Percent of CaCO3 in egg shell is = (7.84/8.34)*100 = 94 % Conclusion: There is 94% composition of calcium carbonate which is a good result we got. The percentage range should be around 90% 95%. However egg shell also contains other element that is not calcium carbonate so the final result maybe different but should be in the range. Improvement: Try measure or collect data by looking at the equipment on a parallel level Make sure that all egg shell have no membrane inside Make sure that all egg shell is dry Should leave egg shell to react with HCl for two days to let egg shell dissolve