Calculate the mass of CaCO 3 present in the original sample.

Question

A mixture of calcium oxide, CaO, and calcium carbonate, CaCO 3, that had a mass of 3.454 g was heated until all the calcium carbonate was decomposed according to the following equation. After heating, the sample had a mass of 3.102 g. Calculate the mass of CaCO 3 present in the original sample.

CaCO 3 (solid) ® CaO (solid) + CO 2 (gas)

Answer 1

Hello, Kent,

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I've updated this after J.R. S. commented on my answer. I did not fully appreciate the question and produced a ridiculous answer. I'm leaving my original response, below the "====" to remind me that I waste a lot of time if I don't fully understand the question before getting out my calculator..

As J.R. S. points out, "How could my answer of 55.4 grams of CaCO3 make any sense if we start with only 3.454 grams of a mixture of CaCO3 and CaO?" I apologize for not taking the time to check the logic and then understand the question. If we start with a mixture of CaCO3 and CaO (3.454 grams) and heat to decompose the CaCO3 to CaO, the final mass of 3.102 grams is a result of the CaCO3 losing the mass of CO2 in the decomposition.

The balanced equation:

CaCO3 → CaO + CO2

The mass loss of (3.454g - 3.102 g) = 0.352 grams is due to just the CO2 being released. The molar mass of CO2 is 44 g/mole, which means that (0.352g/(44 g/mole)) = 0.00800 moles of CO2 were released. The balanced equation tells us that we need the same number of moles of CaCO3, since the molar ratio of CO2 to CaCO3 is 1:1. Therefore, the original sample had 0.00800 moles of CaCO3. With a molar mass of 100.1 g/mole, this means we have (0.00800 moles)*(100.1 g/mole) = 0.081 grams CaCO3.

The original sample of 3.454 grams is composed of:

CaCO3 0.081 grams

CaO 3.373 g

Total 3.454 g

This makes more sense. Thank J.R. S. for catching my mistake.

Ignore everything below. It is a reminder of what can go wrong if a question isn't fully understood before bringing out the calculator/spreadsheet.

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Given the balanced equation:

CaCO₃ (solid) → CaO (solid) + CO₂ (gas)

we can see that the CO₂ is removed from the CaCO3 upon heating. We can also see that that 1 mole of CaO is produced for every 1 mole CaCO₃ that is consumed. We are given the mass of CaO, so let's determine the number of moles contained in 3.102 grams of the compound.

The molar mass of CaO is 51.6 g/mole. 3.102 grams is thus (3.102 grams/(51.6 g/mole)) = 0.5532 moles CaO. That means we consumed 0.5532 moles of CaCO₃. The molar mass of CaCO₃ is 100.1 grams/mole. Thus, we have (0.5532 moles CaCO₃)*(100.1 g/mole) = 55.4 grams CaCO₃.