SBTL, swim, chem, baseball, I talk with my hands

I can’t fight this feeling…

Good Day

So, you are confused and frustrated by the lab report? Maybe you are lost on what the heck the lab was all about? And to top it off, the guy who is SUPPOSED to help, well, just isn’t getting it done. GRRRR… Let’s take a deep breath and take this one step at a time. First, let’s talk about the idea behind the lab.


The goal of the lab is to determine the formula of a solid salt. That’s it. But we know more! The salt is a carbonate and the metal is an alkali metal (Group 1). This means the formula is M2CO3. Since all alkali metal salts are soluble then we dissolve this stuff in water, which results in separating the carbonate ions from the two metal ions. Why is this important? Well, if we can precipitate all of the carbonate with a different metal then we might be able to calculate backwards to find the identity of the original metal. So let’s make a calcium chloride solution that can be added to the original one, resulting in a precipitate.

CaCl2(aq) + M2CO3(aq) –> CaCO3(s) + 2MCl(aq)

Question – does it matter how much water is used to make the solutions? NO! The water is just a vehicle for the ions to swim and find each other so that they can react (hook up) and form a solid. So while it is advantageous to use less water in an effort to shorten the lab (filtration time) it really doesn’t matter as long as both salts are dissolved to start.

Question – does it matter how much of each salt is used? YES! By examining the reaction above we will measure the mass of the precipitate formed and then convert to the moles of this substance. Then, we will use that to find the moles of original carbonate used. That only works if the original carbonate is the limiting reactant. (Uh oh…) How the heck do we know this? We need to do some calculations once we get started. To be clear – if the calcium chloride is limiting then the lab cannot be performed because the amount of precipitate is linked to that and NOT the original carbonate. We need to look at some calculations to see this better.


  • Mass of Precipitate = 20.0563 g – 19.6827 g = 0.3736 g CaCO3
  • Moles of Precipitate = (0.3736 g CaCO3)(1 mole CaCO3/100.05 g CaCO3) = 0.003734 moles CaCO3
  • Moles CaCl2 used = (0.4113 g CaCl2)(1 mole CaCl2/110.98 g CaCl2) = 0.003706 moles CaCl2


Yikes – we have a small problem because, as stated above, the lab only works if the M2CO3 is the limiting reactant. Yet the moles of calcium chloride and calcium carbonate are the same (within 1% of each other). There is only one thing we can conclude from this – the limiting species was THE WRONG REACTANT!

What do I do now?

Well, this means that I cannot draw a conclusion for the lab. The good news is that your grade is NOT dependent on getting a perfect answer. Just process the lab and report your answer – the formula cannot be determined from this data and the lab needs to be repeated. In your write up be sure and discuss (paragraph two) what caused this. In other words, what step in the procedure could have caused this?

In Conclusion

I have your back! Let me repeat – I HAVE YOUR BACK! Relax, do your best, keep working (maybe harder than you planned) and it will all be OK.


This entry was posted on 2013-10-02 by in APChem and tagged , .


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