Question

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of $15.0\%$ by weight. The percentage by weight of the other metal if $3.0$ g of the mineral on heating lost $1.10$ g of $C{O}_2$ is __________.

[If the answer is X, then report the answer as $\frac{X}{5}$].

Answer 1

Let the two metal carbonates be $MC{O}_3$ and $M^{\prime }C{O}_3$. On heating the metal carbonates, the reactions that occur are:
$MC{O}_3\to MO+C{O}_2\uparrow$
$M^{\prime }C{O}_3\to M^{\prime }O+C{O}_2\uparrow$
Equivalents of $C{O}_2=$ Equivalents of carbonates of metals
$1$ mol of $C{O}_2=1$ mol of $C{O}_3^{2-}$
$44$ g of $C{O}_2=60$ g of $C{O}_3^{2-}$
$1.10$ g of $C{O}_2=\frac{60}{44}\times 1.10=1.5$ g of $C{O}_3^{2-}$

Total mass of carbonate is 3 grams. Hence,
$\%$ of $C{O}_3^{2-}=\frac{1.5\times 100}{3}=50\%$
$\%$ of one metal $=15\%$
therefore, $\%$ of another metal $=100-50-15=35\%$.