Typical eggshells are composed mostly of calcium carbonate. A CAMP060 chemistry

Question

Typical eggshells are composed mostly of calcium carbonate. A CAMP060 chemistry student attempted to determine the percentage of calcium carbonate in different brands of eggs. The student collected three different eggs of each brand. The eggs were then crushed and the shell was separated, washed and dried. Then the shells were weighed and were placed in excess 1.00M hydrochloric acid (400mL) solution for two hours. The solutions were then filtered and titrated with 1.00M sodium hydroxide solution. Determine the percentage of calcium carbonate in each egg shell sample and suggest why chickens eat gravel. The results are given below.

Explanation / Answer

2HCl(aq) excess + CaCO3(s) limiting --------> CaCl2(aq) + H2O(l) + CO2(g) + HCl(aq) unreacted

adding an excess amount of hydrochloric acid to react with all of the CaCO3 and then titrating the remaining unreacted HCl with NaOH solution to determine the amount of acid which did not react with the calcium carbonate.

The difference between the moles of the acid (HCl) initially added and the moles of HCl left unreacted after the reaction, is equal to the moles of HCl that did react with CaCO3. The reaction used to determine the amount of unreacted acid.

HCl(unreacted) + NaOH ----> NaCl + H2O

moles of HCl used = (1.0 M x 0.4 L) = 0.4 moles HCl

brand A

moles of NaOH = (0.03868 L x 1.0 M) = 0.03868 moles

HCl react with NaOH

(0.4 moles HCl - 0.03868 moles NaOH) = 0.36132 moles HCl

0.36132 moles HCl react with CaCO3

2 moles of HCl react with 1 mole of CaCO3 so

0.36132 moles react with = (1/2) x 0.36132 = 0.18066 moles CaCO3

mass of CaCO3 = (100.0869 g/mol x 0.18066 moles) = 18.08167 g

mass percentage of CaCO3 =  (18.08167 g / 18.31 g) x 100 = 98.75 %

Brand B

similarly all process.

moles of NaOH = (1.0 M x 0.03745 L) = 0.03745 moles NaOH

0.4 mole HCl - 0.03745 moles NaOH = 0.36255 moles HCl

0.36255 moles HCl react with = (1/2) x 0.36255 moles = 0.181275 moles CaCO3

mass of CaCO3 = (100.0869 g/mol x 0.181275 moles) = 18.143 g CaCO3

mass percent of CaCO3 = (18.143 g / 17.98 g) x 100 = 100.9 %

Brand C

moles of NaOH = 0.038 moles

0.4 - 0.038 = 0.362 moles HCl react with CaCO3

moles of CaCO3 = 0.181 moles

mass of CaCO3 = (100.0869 g/mol x 0.181 moles) = 18.1157 g

mass percent = (18.1157 /17.93) x 100 = 101 %

brand D

moles of NaOH = 0.03764 moles

0.4 - 0.03764 = 0.36236 moles

CaCO3 = (0.36236 /2) = 0.18118 moles

mass of CaCO3 = 18.133 g

mass percent = (18.133 / 17.67 ) x 100 = 102.6 %

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