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?
Something is wrong with this problem because % CaCO3 exceeds 100%. So check the numbers and repost.
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.
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.
I looked on the web and found that the percent CaCO3 in a sea shell is 95-99%.
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.
To find the percent, by mass, of calcium carbonate in the seashell, we need to calculate the number of moles of calcium carbonate in the sample.
First, let's calculate the number of moles of HCl used in the reaction.
The molarity of the HCl solution is 1.00 M, and we used 20.00 mL of it:
moles of HCl = (molarity of HCl) * (volume of HCl in L)
moles of HCl = 1.00 M * 0.02000 L
moles of HCl = 0.0200 moles
Next, we need to determine the stoichiometry of the reaction between HCl and calcium carbonate. From the balanced chemical equation:
HCl + CaCO3 → CaCl2 + H2O + CO2
We see that one mole of HCl reacts with one mole of calcium carbonate, which means the number of moles of calcium carbonate is also 0.0200 moles.
Now, let's calculate the number of moles of NaOH used in the titration.
The molarity of the NaOH solution is 1.00 M, and we used 8.75 mL of it:
moles of NaOH = (molarity of NaOH) * (volume of NaOH in L)
moles of NaOH = 1.00 M * 0.00875 L
moles of NaOH = 0.00875 moles
Using the stoichiometry of the balanced chemical equation between NaOH and calcium carbonate:
NaOH + CaCO3 → Ca(OH)2 + NaCl
We can see that each mole of NaOH reacts with one mole of calcium carbonate. Therefore, the number of moles of calcium carbonate is again 0.00875 moles.
Since the moles of calcium carbonate are the same in both the reaction with HCl and the titration with NaOH, we can conclude that the molar mass of calcium carbonate is the same as the molar mass of the sample.
To calculate the molar mass of calcium carbonate, we need to determine the number of moles present in the sample.
mass of calcium carbonate = 0.0528 g
number of moles of calcium carbonate = mass of calcium carbonate / molar mass of calcium carbonate
Now, we can rearrange the equation to solve for the molar mass of calcium carbonate:
molar mass of calcium carbonate = mass of calcium carbonate / number of moles of calcium carbonate
Substituting the given values:
molar mass of calcium carbonate = 0.0528 g / 0.0200 moles
Now we know the molar mass of calcium carbonate.
Finally, to find the percent, by mass, of calcium carbonate in the seashell, we use the following formula:
percent, by mass, of calcium carbonate = (mass of calcium carbonate / mass of seashell sample) * 100
Substituting the values:
percent, by mass, of calcium carbonate = (molar mass of calcium carbonate / 0.0528 g) * 100
By following these calculations, you can find the percent, by mass, of calcium carbonate in the seashell sample.