Candidate’s name: Eunika Orlowska Candidate’s number: School’s name: School’s number: Determining the amount of CaCO3 in eggshell of hen’s egg | Design | DCP | CE | Aspect 1 | | | | Aspect 2 | | | | Aspect 3 | | | | Introduction: The back titration is a method used in determining the amount of excess of the reagent. The calcium carbonate is a substance which gives the eggshell stiffness. Research question: What is the amount of calcium carbonate in the eggshell measured by back titration? Table 1. Variables.
Type of variable | Variable | Unit | Dependent | Amount of calcium carbonate in eggshell | % by mass | Independent | Volume of titrated excess of hydrochloric aced | cm3 | Controlled | Volume of hydrochloric acid Weight of eggshell Temperature Amount of phenolophateine | cm3 g oC drop | Uncontrolled | Purity of solutions Biological diversity of eggs Pressure | – – hPa | Equipment: buret 5 beakers 50 cm3 baguette 1 plastic pipette balance clamp 2,5 g of eggshell mortar 100 cm3 of 1moldm-3 hydrochloric acid ap. 70 cm3 of 1moldm-3 sodium hydroxide 20 cm3 pipette Risk assessment: you have to remember to wear gloves, goggles and apron. Solutions may be irritating. Method: Crush to dust eggshell in the mortar. Fill each of the 5 beakers with 20 cm3 of hydrochloric acid measured by glass pipette. Add 0. 5 g of eggshell dust to each beaker, measured by balance. While the reaction of eggshell with acid occurs, prepare the buret and clamp for titration. Make sure they are clean. Pour NaOH solution into the buret to the ‘0’ level. Make sure all of the eggshell reacted with the HCl. If not, you can help the reaction by using the baguette.
Put two drops of phenolophateine into each beaker using the plastic pipette. Take the first beaker and titrate the excess of hydrochloric acid. When the solution starts to be pinkish, record the volume of titrated NaOH. Refill the buret to the ‘0’ level and repeat the procedure for each beaker. Remember to record the results. Remember to be careful and to leave your workplace clean! Data Collection Table 2. Raw data. The weight of eggshell reacting with HCl and titrated NaOH. Number of trial | Weight of eggshell [g±0,01g] | Volume of HCl [cm3±0,05cm3] | Volume of titrated NaOH [cm3±0,05cm3] | 1 | 0. 9 | 20. 00 | 9. 60 | 2 | 0. 50 | 20. 00 | 11. 50 | 3 | 0. 51 | 20. 00 | 11. 60 | 4 | 0. 50 | 20. 00 | 9. 90 | 5 | 0. 50 | 20. 00 | 10. 30 | Mean | 0. 50±0,01 | 20. 00±0,05 | 9. 93±0,05 | Standard deviation | 0. 00047 | 0. 00 | 0. 29 | Uncertanties were taken as in measurments, not calculated by formula, to avoid too large and unreliable uncertainties in further calculations in which they’re calculated according to formulas: in case of division and multiplication: =dA/A+dB/B, where d is overall uncertainty, dA is uncertainty of A and dB is uncertainty of B in case of addition and subtraction: d=dA+dB, where where d is overall uncertainty, dA is uncertainty of A and dB is uncertainty of B Trials 2 and 3 were rejected because of too large differentiation of results. Data Processing Two reactions occured in the experiment. Firstly, the HCl reacted with CaCO3 and secondly, the excess of HCl was neutralized by NaOH. 2HCl + CaCO3 > CaCl2 + CO2 + H2O HCl + NaOH > NaCl + H2O 1. Calculating the amount of HCl at the beginning of reaction CmHCl = 1. 0 moldm-3 VHCl = 20. 00 cm3 = 0. 20 dm3 ± 0. 00005 n = Cm * V n = 1. 0* 0. 020 = 0. 020 mol ± 0. 00025 2. Calculating mean amount of NaOH which neutralized the excess of HCl CmNaOH = 1. 0 moldm-3 VNaOH = 9. 93 cm3 = 0. 0099 dm3 ± 0. 00005 n = Cm * V n = 1. 0 * 0. 0099 = 0. 0099 mol ± 0. 0005 3. Calculating the amount of HCl which reacted with CaCO3 0. 020 mol – 0. 0099dm3 = 0. 010 mol ± 0. 00075 4. Calculating the amount of CaCO3 which was in the eggshell We know that the molar ratio in the reaction between HCl and CaCO3 is 2/1, which means that 2 moles of HCl react with 1 mole of CaCO3. If 0. 010 moles of HCl reacted with CaCO3 then there was 0. 05 mole of CaCO3 in the eggshell. nCaCO3 = 0. 010/2 = 0. 005±0. 0075 5. Calculating the percentage of CaCO3 in the eggshell. MCaCO3 = 40. 09 + 12. 01 + 3 * 16,00 = 100. 1 gmol-1 nCaCO3 = 0. 005 mol ± 0. 0075 m = M * n m = 0. 005*100. 1 = 0. 50 g ± 0. 0075 meggshell=0. 50 ± 0. 01 CCaCO3 in eggshell = 0. 50 / 0. 50 = 100 % ± 3. 5% Table 3. The results and uncertaities Calculated value | Value | Uncertainty | Number of moles of HCl at the beginning of reaction | 0. 020 mol | 0. 00025 mol | Mean amount of NaOH which neutralized HCl | 0. 0099 mol | 0. 005 mol | Mean amount of HCl which reacted with CaCO3 in the eggshell | 0. 010 mol | 0. 00075 mol | Number of moles of CaCO3 in the eggshell | 0. 005 mol | 0. 0075 mol | Molar mass of CaCO3 | 100,1 gmol-1 | – | Percentage of CaCO3 | 100% | 3. 5% | The eggshell consists of 94-97% of calcium carbonate, meanly 95. 5 %. From collected data it is 100%, which suggests that percentage error was not big and remains in accepted value of 20%. percentage error = (|theoretical value – experimental value| ? theoretical value) x 100% = (95. 5 – 100 ? 95. 5) = 4 %
Conclusion and Evaluation The eggshell consists of 94 – 97% of calcium carbonate. The experiment suggests that the eggshell has 100% of calcium carbonate and the percentage error is 4 % which means calculations and results where accurate. The fact that the result has shown 100% of calcium carbonate in the eggshell may lay in the construction of eggshell which is biological “machine” to give hen’s embryo the best possible environment for development. Apart from calcium carbonate, there are other components of eggshell, for example magnesium carbonate and calcium phosphate which also react with hydrochloric acid.
Reaction of calcium phosphate with hydrochloric acid: Ca3(PO4)2 + 6HCl > 3CaCl2 + 2H3PO4 this reaction should’t have influence on titration that much, because as a result there’s the same number of hydrogen ions which can be neutralized, but the second reaction, of magnesium carbonate and hydrochloric acid: MgCO3 + 2HCl > MgCl2 + CO2 + H2O also takes HCl to it’s reaction, decreases number of H+ ions and so suggests that more HCl reacted with calcium carbonate. This is the uncontrolled variable, the impurity of the eggshell, which affected the result.
Other factor, which may have had influence on the results is the human’s imprecision. The used equipment was as accurate as possible in school laboratory, but still, the titration is made by hand and by eye, which may make mistakes although back titration is the best possible way in school laboratory to check the amount of particular components in mixtures. Other method, which would distinguish between magnesium carbonate and calcium carbonate is gas spectrometry – mass spectrometry. This method uses combined gas chromatography and mass spectrometry. First, the substance is put into gas chromatograph.
The mobile phase, in which the particles of substances move towards the stationary phase is made of unreactive gas, such as nitrogen or helium. In this, the substance is separated into particular components and then, the mass spectrometer analyses the components to identify them. This method is commonly used to determine the ingredients of substances, of food, beverages, perfumes. Also, it is useful in medicine and and in exploring Universe, for example, one GC-MS was taken by Curiosity to examine the surface of Mars. Summing up: in school laboratory, accuracy of experiments is limited by equipment.
Better accuracy can be obtained by using more accurate balance, as no other equipment can be changed in used method. Back titration can’t be replaced by other methods of determining the percentage of CaCO3 in the eggshell, as it is the best way to do it in school laboratory, although generally more accurate methods are created, using machines which exclude the human factor from experiments, for example gas chromatography – mass spectrometry. Bibliography http://antoine. frostburg. edu/chem/senese/101/consumer/faq/eggshell-composition. shtml 18th November 2012 J. Green, S. Damji “Chemistry” IBID 2008