Question

A 0.0528g sea shell sample is treated with 20.00 mL of a 1.00M HCl to dissolve all of the sample. The resulting solution is titrated with 8.75 mL of 1.00 M NaOH. What is the percent, by mass, of the calcium carbonate in the seashell?

Answer 1

Something is wrong with this problem because % CaCO3 exceeds 100%. So check the numbers and repost.

Answer 2

A seashell is crushed, and a 0.0528g sample is treated with 20.00 mL of a 1.00M HCl to dissolve all of the sample. The resulting solution is titrated with 8.75 mL of 1.00 M NaOH. What is the percent, by mass, of the calcium carbonate in the seashell.

This is the full question.

Answer 3

Something still wrong.
millimols HCl added initially = mL x M = 20.00 mL x 1.00 M = 20.00
millimols NaOH to determine excess HCl = 8.75 mL x 1.00 M = 8.75
millimols HCl used for the CaCO3 in the sea shell = 20.00 - 8.75 = 11.25
Convert to mols = 0.01125
Equation for CaCO3. CaCO3 + 2HCl ==> CaCl2 + CO2 + H2O
mols CaCO3 = 1/2 * mol HCl = 1/2*0.01125 = 0.005625
grams CaCO3 = mols x molar mass = 0.005625 x 100 = 0.5625
You can see something is wrong. You have more CaCO3 that is in the sample initially. Can't be.
% CaCO3 = (mass CaCO3/mass sample)*100 = (0.5625/0.05628)*100 = 1,065.3 %. NOTE; YOU HAD BETTER GET A PATENT ON THIS IF YOU CAN GET MORE THAN 1000%. YOU'RE A BETTER CHEMIST THAT I AM. I've worked all my life and I've never been able to do that. I've looked at several scenarios and the most likely error is the 0.0528 g sample might be a 5.28 g sample or 52.8 g sample in which case it would be approximately 10% or 1% CaCO3.

Answer 4

I looked on the web and found that the percent CaCO3 in a sea shell is 95-99%.

Answer 5

A seashell is crushed, and a 5.28g sample is treated with 20.00 mL of a 1.00M HCl to dissolve all of the sample. The resulting solution is titrated with 8.75 mL of 1.00 M NaOH. What is the percent, by mass, of the calcium carbonate in the seashell.

This is the full question.