Using the mass of calcium carbonate $m(\ce{CaCO3}) = \pu{0.500 g}$, what volume of carbon dioxide $V(\ce{CO2})$ (in $\pu{mL})$ could theoretically be obtained if the reaction progressed to completion? Assume that the pressure was $\pu{101.3 kPa}$ and the temperature was $\pu{22.5 ^\circ C}$ and that the gas was behaving ideally $(R = \pu{8.314 kPa L mol-1 K-1})$.

I know that I will work with $$pV = nRT$$ and to bring the $n$ (amount of substance), I have to divide the mass by the molar mass and I assumed that $\ce{CO2}$ mass will be $\pu{0.5 g}$ as well because $\ce{CaCO3}$ will give $\ce{CaO + CO_2}$ so the ratio is $1:1$ so $\ce{CO2}$ mass is $\pu{0.5 g}$, but my problem is that when I bring the amount of substance it should be for $\ce{CO_2}$ or $\ce{CaCO3}$? Because it will differ, $\frac{0.5}{44.01}$ or $\frac{0.5}{100.09}$, that's what I am stuck on.

  • $\begingroup$ It seems that you perform a little calculation without thinking on its meaning. You are interested on the number of mole of CO2. There is a 1:1 correspondence between CO2 and the carbonate, ie one mole of the latter gives out one mole of the gas when it totally decomposes. Thus you are ultimately interested on how many moles of carbonate you are starting with. So you need to plug in the molar mass of the carbonate,because it is what you are weighing at start. $\endgroup$
    – Alchimista
    Commented Oct 9, 2020 at 9:59
  • $\begingroup$ so you mean in order to bring the volume, when i calculate the n (moles number) so i should divide the mass (0.5) by the molar mass of the co2 not caco3? $\endgroup$
    – lizzy
    Commented Oct 9, 2020 at 10:44
  • $\begingroup$ it is getting puzzling $\endgroup$
    – Alchimista
    Commented Oct 9, 2020 at 10:55
  • $\begingroup$ well it was an experiment and the experiment ends with 73ml of co2 but i donot think that it is the answer because the question says calulate it theroitically $\endgroup$
    – lizzy
    Commented Oct 9, 2020 at 11:11
  • $\begingroup$ Consider question titles to be like book titles that provide rather sufficient content hints than story summaries. E.g. "Computation of CO2 volume evolved from carbonates" $\endgroup$
    – Poutnik
    Commented Oct 9, 2020 at 13:04

2 Answers 2


You are confused about chemical equations. The stoichiometric coefficient (the numbers before a compound) do not indicate the mass of the compound, they indicate the amount of substance. For example $$\ce{CaCO3 -> CaO + CO2}.$$ This equation means that $\pu{1 mol}$ of calcium carbonate will decompose to give $\pu{1 mol}$ of $\ce{CaO}$ and $\pu{1 mol}$ of $\ce{CO2}$.

It does not mean that if 1 gram calcium carbonate were to decompose, 1 gram of calcium oxide and 1 gram of carbon dioxide will be obtained.

In your question the mass of $\ce{CaCO3}$ is given. From it you can calculate the moles of $\ce{CaCO3}$ involved. Same number of moles of $\ce{CO2}$ will be obtained and you can put that in the ideal gas equation. Just so you know, $$n_{\ce{CO_2}} = n_{\ce{CaCO_3}} = \frac{\pu{0.5 g}}{\pu{100.09 g/mol}}.$$

I recommend you study the chapter about the mole concept again.

As a side note: You mentioned in your question that the ratio of products is $1 : 1$. That's actually the ratio of the amounts of substances, not the mass ratio.

  • 1
    $\begingroup$ Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". $\endgroup$ Commented Oct 9, 2020 at 20:14
  • $\begingroup$ I'd upvote that, but I'll do that once it is complete, because I'd still think that OP will struggle with the completeness of this answer, too. $\endgroup$ Commented Oct 9, 2020 at 20:31
  • $\begingroup$ @Martin I used the word number of moles because some people, including me, confuse amount of susbtance with mass of the compound $\endgroup$
    – Eyy boss
    Commented Oct 10, 2020 at 5:34
  • 1
    $\begingroup$ @Eyyboss which is indeed reasonable $\endgroup$
    – Alchimista
    Commented Oct 10, 2020 at 10:27
  • 1
    $\begingroup$ I am not saying that it is correct. I am saying that it is a reasonable mistake at least in four languages. @Martin..... $\endgroup$
    – Alchimista
    Commented Oct 12, 2020 at 17:00

You are mentioning $0.500$ g $\ce{CaCO3}$. Whatever the use of this substance later on, it contains $\ce{0.500 g/(100 g/mol CaCO3) = 5.00·10^{-3} mol CaCO3}$. Of course, if this sample reacts with an acid later on, it will produce $\ce{5.00·10^{-3}}$ mol $\ce{CO2}$. And these $\ce{5.00·10^{-3}}$ mol $\ce{CO2}$ occupies a volume $V$ = $\ce{5.00·10^{-3}} RT/p$.

  • $\begingroup$ yeah i know this but thats not my question my question is when i want to bring the n i shoud bring the n by dividing the mass by the molar mass my question is which molar mass i should put the molar mass of co2 or caco3 ? $\endgroup$
    – lizzy
    Commented Oct 9, 2020 at 9:39
  • $\begingroup$ @lizzy you need to understand that the molar mass is irrelevant to the idea used in your answer.. the number of moles in this reaction is what is formed in the ratio given above. $\endgroup$ Commented Oct 9, 2020 at 10:22
  • $\begingroup$ If you know how to calculate it, this answer is easy to understand. If you are struggling with the concept of what react to what and which molar masses to use when, then this answer is indeed impossible to understand. @Safdar Obviously the molar mass is needed to calculate the initial amount of substance, so it is relevant to know which one to use. $\endgroup$ Commented Oct 9, 2020 at 20:25

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