Composition of calcium carbonate in egg shell

5 May 2016

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:

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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
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 %

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

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