Question

What is the equivalent weight of Calcium carbonate ${\mathrm{CaCO}}_3$?

Answer 1

Step $1$: Formula for Equivalent weight and finding n factor:

$\mathrm{Equivalent}\mathrm{weight}=\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{n}\mathrm{factor}}$

• Here n factor is the valency of the cation and anion or the change in the oxidation state of the element in the reactant and product side.
• As calcium belongs to $\mathrm{group}-2$ elements, these elements have the oxidation state of $+2$ and carbonate has an oxidation state of $(-2)$.
• The salt calcium carbonate will be as: ${\mathrm{Ca}}^{2+}{{\mathrm{CO}}_3}^{2-}$
• so the n factor value for ${\mathrm{CaCO}}_3$ is $2$.

Step $2$: Finding the molecular mass of ${\mathrm{CaCO}}_3$:

• To calculate the equivalent weight first we have to calculate the molecular mass of the compound

$$\begin{gathered} \mathrm{Molecular}\mathrm{mass}: \\ \mathrm{Ca}-40,\mathrm{C}-12\&\mathrm{O}-16 \\ \mathrm{Molecular}\mathrm{mass}=40+12+\left(3\mathrm{x}16\right) \\ =40+12+48 \\ =100\mathrm{u} \end{gathered}$$

Step $3$: Finding equivalent weight of ${\mathrm{CaCO}}_3$:

• On substituting the values of equivalent weight we get,

$$\begin{gathered} \mathrm{Equivalent}\mathrm{weight}=\frac{100}{2} \\ =50\mathrm{u} \end{gathered}$$