To Investigate the Percentage of CaCO3 in a Sample of Marble
- Pages: 2
- Word count: 396
- Category: Chemistry College Example
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Order NowMaterials used:
* Beaker
* Conical Flask
* Measuring Cylinder of 15 cc
* Burette
* Indicator
* Clamp Stand
* Pipette
* Beam Balance
* HNO3 of 2M concentration
* NaOH of 2M concentration
Procedure :
* Crush the sample of marble in to small pieces, and take 1 gram of it in a beaker
* Add excess of acid into the beaker, allow to react completely
* Pipette 25 cc of the mixture into the conical flask
* Add 2 drops of indicator (phenolphthalein) in the conical flask
* Rinse the burette with NaOH, then fill it with the latter up to the 0 cc mark
* Titrate the mixture with the base until color changes to purple. In the process, shake the conical flask thoroughly
* Calculate the amount of base reacted
* Repeat the titration again to improve accuracy.
Data Collection :
Burette reading (NaOH) / cc
HNO3 + marble used
(cc)
Initial
Final
Used
Mass of marble used = 3 grams
Amount of acid used to dissolve the marble, 100 cc = 0.1 dm3
Volume of mixture taken = 0.025 dm3
Mean volume of NaOH used = 18.76 cm3 = 0.01876 dm3
CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O(l) + CO2(g)
Mol of HNO3 used = 0.1 x 2M = 0.2 mol
Molar ratio of NaOH ~ 0.01876 x 2 = 0.03752 mol
Mol in 0.1 dm3 of NaOH = 0.03752 x 4 = 0.15008 mol
Mol reacted with HNO3 = 0.2 – 0.15008 = 0.04992 mol
Mol of CaCO3 = 0.04992 = 0.02496 mol
Mass of CaCO3 = 0.02496 x 100 = 2.496 grams
% of CaCO3 = 2.496/3 x 100% = 83.2 %
Precautions taken :
Make sure that the marble reacted completely with the acid, it may take quite a long time, but we are not allowed to proceed with the titration until the marble is completely reacted, because it will give us incorrect readings.
Care should be taken when titrating the mixture of acid and marble with NaOH. Because color changes may occur more than once, so we need to be attentive in the process.