# 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} &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}

• The whole concept of equivalent weight is somewhat obsolete. It is much easier to use molar mass instead as that is independent of the actual chemistry that is happening. A note on you MathJax: It is better to not use \dfrac{}{} as it does not render on mobile devices, it is also redundant, since $$...$$ sets fractions already in display mode. – Martin - マーチン Nov 30 '16 at 13:04
• @Martin-マーチン-Yeah, after reading all the other posts regarding equivalent weight on Chemistry.SE it seems that everyone is saying that, but isn't the concept of equivalent mass makes it easier to understand electrochemistry and redox reaction and solve them with a lot of speed and accuracy. – user350331 Nov 30 '16 at 13:46
• @Martin-マーチン-I would still appreciate it if you could tell me what is wrong in my thinking in the above post because I am a student who is doing self study and it has left me in tears after searching for it through all the books that I have and material regarding it on the internet but nothing is there that answers my doubt that I have posted here. – user350331 Nov 30 '16 at 13:51

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 $\ce{M^_x(CO3)_y -> M^_xO_y + yCO2}$

$PV=nRT$

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.

• I have edited my post to include the answer provided by the book you might wanna have a look. – user350331 Nov 30 '16 at 22:53
• @user350331 so you have 4.125g in the question, but they have 4.215g in answer. Is the typo yours or the book's? – DavePhD Dec 1 '16 at 0:18
• @DavePhD-I am extremely sorry in the question too it is $4.215~\mathrm{g}$. I apologise again for the inconvenience. – user350331 Dec 1 '16 at 3:04
• @user350331 so then the mass of metal is 1.215 grams. There are 24.3 grams of metal per mole of CO2 evolved. (This is consistent with the metal being Mg.) – DavePhD Dec 1 '16 at 3:13
• Although your solution does provide the correct equivalent weight of the metal but it doesn't answer the concern that I have written in the last section of my post. Can you shed some light on what is wrong in the second method by which I determined the equivalent weight of $\ce{CO2}$ – user350331 Dec 1 '16 at 4:37

You began with

4.125 g of $\ce{M^_x(CO3)_y}$

And you finished with

1.925 g of $\ce{M^_xO_y}$

The subscripts on the general formulas are the same because the charge on oxygen and carbonate are the same.

Since 0.5 mol * 44 g/mol = 2.2 g of $\ce{CO2}$ evolved then 16/44 of that mass (0.8 g) of oxygen remains. (16/44 being the mass ratio of the one oxygen atom remaining for each one molecule of carbon dioxide evolved).

1.925 g metal oxide minus 0.8 g oxygen leaves 1.125 grams of metal

• There is no such metal oxide with a ratio of 1.125 masses of metal per 1.925 masses of metal oxide. The metal in the problem would have the an atomic mass corresponding to charge of: – Joseph Hirsch Nov 30 '16 at 20:37
• +1=11.25 amu +2=22.5 amu +3=33.75 amu +4=45 amu +5=56.25 amu +6=67.5 amu +7=78.75 amu It is possible, however that the sample contained a mixture of a given metal element with some proportion of atoms in different oxidation states – Joseph Hirsch Nov 30 '16 at 20:40
• I see that the values in the original question changed, indicating that the metal is Mg – Joseph Hirsch Dec 1 '16 at 15:35