Aim : To investigate the percentage of CaCO3 in a sample of marble

Data Collection :

        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.