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 $$\begin{align} &P_1 = 700~\mathrm{mmHg}\\ &V_1 = 1336~\mathrm{mL} = 1.336~\mathrm{L}\\ &T_1 = 27~^\circ\mathrm{C} = 300~\mathrm{K}\\ &P_2 = 760~\mathrm{mmHg}\\ &T_2 = 273~\mathrm{K} \end{align}$$
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 $\ce{H2CO3}$ $$\ce{CO2 + H2O -> H2CO3}$$ 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, $$\begin{align}\text{equivalents of metal carbonate} &= \text{equivalents of }\ce{CO2}\\\frac{4.215}{E+30}&=\frac{1120\text{(vol. at NTP)}}{11200\text{(vol. of 1 eq. at NTP)}}\\\text{So, } \text{E}&=12.15~\mathrm{g/equivalent}\end{align}$$
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