We have a investigation in my chemistry class in which we must calculate the amount of calcium carbonate in seashells. We can have as many sea shells as we wish, lab equipment, and hydrochloric acid of varying concentration from $\pu{0.5 mol dm-3}$ to $\pu{2 mol dm-3}$. How is it to be done?
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2 Answers
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$\ce{CaCO3}$ is basic so it will neutralize the acid.
$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O +CO2}$$
In your experiment you need to devise a way of finding out what the end point of the reaction is (using a $\mathrm{pH}$ indicator to find when the mixture is neutral). You need to be able to measure the amount of hydrochloric acid used in moles (you should measure the volume and you know the concentration. The number of moles of water is twice the amount of moles of the calcium carbonate (from the chemical equation above).
You should measure the mass of the amount of shell you use at the start. You will be able to figure out the mass of the calcium carbonate in that sample of shells that you use (since you now know the number of moles of calcium carbonate). Then:
$$\left(\frac{m({\ce{CaCO3})}}{m(\mathrm{shells})}\right) \times 100\% = \mathrm{Percent}~\ce{CaCO3}$$
You should be able to do it with just one concentration but your teacher probably wants to talk to you about reliability and experimental errors.
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$\ce{CaCO3}$ reacts with $\ce{HCl}$ in the following reaction:
$$\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) +CO2(g)}$$
In my opinion, the easiest way to measure the results of this reaction would be to collect the $\ce{CO2}$. Grind up the seashell into a powder, record the mass of this. Assemble an apparatus like this, and react the seashell with an excess of $\ce{HCl}$ in the reaction chamber to collect and measure the volume of gas evolved (docbrown.info):
As this is a gas law calculation, you will need the atmospheric pressure and ambient air temperature at the time you perform the reaction. After obtaining the final volume of the gas in the reaction, you will then need to account for the partial pressure due to water vapor, and subtract this from the atmospheric pressure accordingly. Now this is simply a matter of stoichiometry, calculating how much $\ce{CaCO3}$ would be required to obtain this volume of gas at this temperature and pressure.
You can then use the equation RobChem stated:
$$\left(\frac{m({\ce{CaCO3})}}{m(\mathrm{shells})}\right) \times 100\% = \mathrm{Percent}~\ce{CaCO3}$$
to find the percent $\ce{CaCO3}$ by mass in the seashell.
