Question

A mineral consists of an equimolar mixture of carbonates of two bivalent metals$\left(MC{O}_3\right)$. One metal is present to the extent of 13.2 % by weight. 2.58 g of mineral on heating lost 1.233 g of $C{O}_2$. Calculate the percentage by weight of the other metal.

• A. 65.12 %
• B. 21.68 %
• C. 13.20 %
• D. 42.80 %

Answer 1

The correct option is B 21.68 %

Let us consider $MC{O}_3$ and $M^{{}^{\prime }}C{O}_3$ be two carbonates with respect to $M$ and $M^{{}^{\prime }}$ bivalent metals respectively.

On heating
(a) $MC{O}_3\to MO+C{O}_2\left(g\right)$.........(i)
(b) $M^{{}^{\prime }}C{O}_3\to M^{{}^{\prime }}O+C{O}_2\left(g\right)$ .......(ii)

1 mole of $C{O}_2\left(g\right)$ i.e. 44 g is produced from 1 mole of carbonate $\left(C{O}_3^{2-}\right)$ ion i.e. 60 g (using equation (i) and (ii))

1.233 g of $C{O}_2$ is produced from $\frac{60\times 1.233}{44}=1.68$ g of carbonate $\left(C{O}_3^{2-}\right)$ ion.

Percentage of $\left(C{O}_3^{2-}\right)$ ion = $\frac{MassofC{O}_3^{2-}ion}{Totalmass}\times 100$

= $\frac{1.68g}{2.58g}\times 100$

= 65.12%

Percentage of one of the metals = 13.2
Percentage of the other metal
= 100 - percentage of $C{O}_3^{2-}$ ion - percentage of one metal
= 100 - 65.12 - 13.2
= 21.68%