How to determine the equivalent weight of carbon dioxide correctly? I just started learning about equivalent weight and encountered this solved example.


  Problem Statement:
  
  $4.215~\mathrm{g}$ of a metallic carbonate was heated in a hard glass tube and the $\ce{CO2}$ evolved was found to measure $1336~\mathrm{mL}$ at $27~^\circ\mathrm{C}$ and $700~\mathrm{mmHg}$ of pressure. What is the equivalent weight of the metal?


My attempt:

First of all lets find the volume occupied by $\ce{CO2}$ at NTP.

We know,
$$\frac{P_1 V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Now, as it is given that 
$$\begin{align}
&amp;P_1 = 700~\mathrm{mmHg}\\
&amp;V_1 = 1336~\mathrm{mL} = 1.336~\mathrm{L}\\
&amp;T_1 = 27~^\circ\mathrm{C} = 300~\mathrm{K}\\
&amp;P_2 = 760~\mathrm{mmHg}\\
&amp;T_2 = 273~\mathrm{K}
\end{align}$$

Hence, $$V_2=\frac{P_1\times V_1\times T_2}{T_1\times P_2}=\frac{700\times 1.336\times 273}{300\times 760}=1.12~\mathrm{L}$$

Also, according to the law of equivalence, we get 

$$\text{equivalents of metal carbonate} = \text{equivalents of }\ce{CO2}$$

Now comes the part where I am really stuck, i.e. how to determine the volume occupied by $1$ equivalent of $\ce{CO2}$.

I started with finding the equivalent weight of $\ce{CO2}$ by considering the reaction of carbonation of water which produces $\ce{H2CO3}$
$$\ce{CO2 + H2O -&gt; H2CO3}$$
As we know the equivalent weight of $\ce{H2CO3}$ is $\frac{M}{2}$ because it has a basicity of $2$.

Hence, according to the law of equivalence,

$$\text{Equivalent of }\ce{CO2} = \text{Equivalent of }\ce{H2CO3}$$
which gives us the equivalent weight of $\ce{CO2}$ to be $22~\mathrm{g/equivalent}$. Hence, the equivalent volume of $\ce{CO2}$ at NTP is $11.2~\mathrm{L}$

Now consider this another method.

The equivalent weight of $\ce{C}$ in $\ce{CO2}$ is $3~\mathrm{g/equivalent}$ as $1 \ce{C}$ combines with $4$ equivalents of $\ce{O}$ to form $\ce{CO2}$, hence we get the equivalent weight of $\ce{CO2}$ as $3+2\times8=19\text{ g/equivalent}$ which is super weird.



My deal with the question:

What am I doing wrong in the second method for finding the equivalent weight of $\ce{CO2}$ and if you wanna know why I added the equivalent weights of the elements was because I had read that for finding the equivalent weight of an electrolyte we add the equivalent weights of the cations and anions. Hence, I tested this method of finding the equivalent weight on $\ce{SO4^2-}$.

Here is my attempt of finding the equivalent weight of $\ce{SO4^2-}$ using the method for electrolytes. We know that equivalent weight of $\ce{S}$ is $\frac{32}{2}=16$ and that of $\ce{O}$ is $8$, so we have 

$$\text{Eq. } \ce{SO4^2-} = \text{Eq. }\ce{S} + 4 \times \text{Eq. }\ce{O} = 16 + 32 = 48~\mathrm{g/equivalent}$$



For those who are attempting to solve the problem:-

The solution provided by the book is as follows:-


   Volume of $\ce{CO2}$ at NTP=$\dfrac{1336\times273}{300}\times\dfrac{700}{760}=1120~\mathrm{mL}$
 Suppose the equivalent weight of the metal is $\ce{E}$.
So, Equivalent Weight of metal Carbonate $=\ce{E}+30$
Because, Eq. wt. of ${\ce{CO3}}^{2-}=\frac{60}{2}=30$
Now, 
 $$\begin{align}\text{equivalents of metal carbonate} &amp;= \text{equivalents of }\ce{CO2}\\\frac{4.215}{E+30}&amp;=\frac{1120\text{(vol. at NTP)}}{11200\text{(vol. of 1 eq. at NTP)}}\\\text{So, } \text{E}&amp;=12.15~\mathrm{g/equivalent}\end{align}$$
Metallic Carbonate means $\ce{M^_x(CO3)_y}$ .

For example there are $\ce{Li^_2(CO3)}$, $\ce{Ca(CO3)}$, and $\ce{Al^_2(CO3)_3}$  

and $\ce{M^_x(CO3)_y -&gt; M^_xO_y + yCO2}$

$PV=nRT$

n= 0.0500 moles carbon dioxide evolved

mass of the evolved carbon dioxide is 0.0500 moles $\times$ 44.0 g/mol = 2.20 grams

mass of the remaining metal oxide is 4.215 g - 2.20 g = 2.015  grams. 

Of the 2.015 grams of the remaining metal oxide, 0.0500 moles $\times$ 16.00 g/mol = 0.800 grams are oxide, while the mass of metal is 1.215 grams.

There are 24.3 grams of metal per mole of carbon dioxide evolved.

Because the ratio x:y is unknown, it is unknown whether 1.215 grams corresponds to 0.05 moles, 0.1 moles or some other number of moles.  

However, since there is one mole of carbon dioxide evolved per carbonate ion, and carbonate has a charge of -2, the molar mass of metal per charge of the metal ion will be 24.3/2 = 12.15 grams.  This is what is meant by equivalent weight of the metal in the question.