Question

Given,$4.215\;g$ of a metallic carbonate was heated in a hard glass tube and the $C{O_2}$ evolved was found to measure $1336\;ml$ at ${27^ \circ }C$ and $700\;mm$ pressure. What is the equivalent weight of the metal?

Answer 1

Hint: In chemistry, equivalent weight is the amount of any substance that precisely tends to have reactions with an arbitrarily set amount of another substance in a given reaction. This weight has a relation with other factors like number of moles and molecular mass, so using the ideal gas equation we can compute the exact equivalent weight.

Complete answer:
Let us note down what’s given in the question:
There is an unknown metal’s carbonate, so let the metal be denoted as ‘$M$’.
Every metallic carbonate will take the form: ${M_2}C{O_3}$
The weight of the metallic carbonate is said to be $4.215\;g$.
It is said that $C{O_2}$ evolved so its volume, temperature and pressure are given as $1336\;ml$, ${27^ \circ }C$ and $700\;mm$ respectively.
Also the mass or weight of $C{O_2}$ $ \Rightarrow {M_{C{O_2}}} = 44$
Our aim is to calculate the equivalent weight of unknown metal ‘$M$’.
Let us start by writing the balanced chemical equation of the given reaction:
${M_2}C{O_3}\xrightarrow{\Delta }{M_2}O + C{O_2}$
Since the given equation is balanced, it shows that the number of moles of $C{O_2}$ is the same as ${M_2}C{O_3}$
Now since we know the pressure, volume, temperature and molecular weight of $C{O_2}$ we will use the ideal gas equation to compute the weight of $C{O_2}$.
(But remember to convert all the given values into their standard forms.)
The equation is $ \Rightarrow PV = nRT$, where $R = 0.0821\;atmL{K^{ - 1}}mo{l^{ - 1}}$
$ \Rightarrow PV = \dfrac{W}{{{M_{C{O_2}}}}}RT$
Converting each of the quantities to their standard form:
$1336\;ml = \dfrac{{1336}}{{1000}}\;L$, $700\;mm = \dfrac{{700}}{{760}}\;atm$, ${27^ \circ }C = (27 + 273)K$
Substituting the values:
$ \Rightarrow \dfrac{{700}}{{760}}atm \times \dfrac{{1336}}{{1000}}L = \dfrac{W}{{44}} \times 0.0821\;atmL{K^{ - 1}}mo{l^{ - 1}} \times 300K$
Rearranging we get:
$ \Rightarrow \dfrac{{\dfrac{{700}}{{760}}atm \times \dfrac{{1336}}{{1000}}L \times 44}}{{0.0821\;atmL{K^{ - 1}}mo{l^{ - 1}} \times 300K}} = W$
Simplifying we get:
$ \Rightarrow W = 2.198\;g$ of $C{O_2}$
Now we can equate both number of moles of metal carbonate and carbon dioxide (since they are equal from the balanced chemical equation):
$ \Rightarrow {n_{{M_2}C{O_3}}} = {n_{C{O_2}}}$
$\Rightarrow \dfrac{{Weight\;of\;{M_2}C{O_3}}}{{Molecular\;mass\;of\;{M_2}C{O_3}}} = \dfrac{{Weight\;of\;C{O_2}}}{{Molecular\;mass\;of\;C{O_2}}}$
Substituting the values we have;
$ \Rightarrow \dfrac{{4.215}}{{Molecular\;mass\;of\;{M_2}C{O_3}}} = \dfrac{{2.198}}{{44}}$
$ \Rightarrow 4.215 \times \dfrac{{44}}{{2.198}} = Molecular\;mass\;of\;{M_2}C{O_3} \to (i)$
But we also have:
$Molecular\;mass\;of\;{M_2}C{O_3} = 2 \times Equivalent\;weight\;of\;M + (12 + 3 \times 16)$
$Molecular\;mass\;of\;{M_2}C{O_3} = 2 \times Equivalent\;weight\;of\;M + 60 \to (ii)$
So equating the equation (i) and (ii) we get:
$ \Rightarrow 4.215 \times \dfrac{{44}}{{2.198}} = 2 \times Equivalent\;weight\;of\;M + 60$
Solving we get:
$ \Rightarrow Equivalent\;weight\;of\;M = 12.188\;g$
Therefore the final answer for the equivalent weight of the unknown metal $M$ is $12.188\;g$.

Note:
Similar to the equivalent weight, we also commonly use the molecular weight. In simple words, a molecule's molecular weight is its mass. For a molecule, when each constituent element’s atomic weights are added then multiplied with the total number of those atoms present in that element (within the molecular formula) it becomes the formula used to compute the molecular weight. Basically the molecular weight is made up of equivalent weights.