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. $3.0$ g of the mineral on heating lost $1.10$ g of ${\text{C}}{{\text{O}}_{\text{2}}}$ . The percent by weight of the other metal is:
A. 65
B. 25
C. 75
D. 35

Answer 1

Hint: When metal carbonates undergo decomposition, the carbonate ion gets split into carbon dioxide and oxide ions.
The combined weight of the two metals can be found out from the difference between the weight of the mineral and the weight of the carbonate ions.

Complete step by step answer:
Given that, a mineral consists of an equimolar mixture of the carbonates of two bivalent metals.
Also given, one metal is present to the extent of $15.0\% $ by weight. Moreover, $3.0$ g of the mineral on heating loses $1.10$ g of ${\text{C}}{{\text{O}}_{\text{2}}}$ .
We need to find out the percent by weight of the other metal.
Let the two bivalent metal carbonates be ${\text{MC}}{{\text{O}}_{\text{3}}}$ and ${\text{M'C}}{{\text{O}}_{\text{3}}}$ .
On heating the first metal carbonate, we will get the corresponding metal oxide and carbon dioxide gas.
${\text{MC}}{{\text{O}}_{\text{3}}} \to {\text{MO + 2C}}{{\text{O}}_{\text{2}}}$
Similarly, on heating the second metal carbonate also, we will get its corresponding oxide and carbon dioxide gas.
${\text{M'C}}{{\text{O}}_{\text{3}}} \to {\text{M'O + 2C}}{{\text{O}}_{\text{2}}}$
They are present as an equimolar mixture in the mineral. This can be expressed by the combination of the two metal carbonates as:
${\text{MC}}{{\text{O}}_{\text{3}}} + {\text{M'C}}{{\text{O}}_{\text{3}}} \to {\text{MO + M'O + 2C}}{{\text{O}}_{\text{2}}}$
From here, we can see that 2 moles of carbonate ion gives us 2 moles of carbon dioxide. Or, in other words, 1 mole of carbon dioxide is equivalent to 1 mole of carbonate ion.
Now, the molar mass of carbon dioxide is 44 g per mole and that of carbonate ion is 60 g per mole. So, 44 g of carbon dioxide is equal to 60 g of carbonate ion.
Thus, $1.10$ g of ${\text{C}}{{\text{O}}_{\text{2}}}$ is equal to
\[
   = \dfrac{{60}}{{44}} \times 1.10 \\
   = 1.5{\text{g}} \\
 \]
of carbonate ions.
Thus, mass of carbonate required to produce $1.10$ g of ${\text{C}}{{\text{O}}_{\text{2}}}$ is equal to $1.5$ g.
Total combined weight of the two metals is equal to the difference between the weight of the mineral and the weight of the carbonate ions. So, we have, weight of metals
$
   = 3.0 - 1.5 \\
   = 1.5{\text{g}} \\
 $
Now, weight of metal M
\[
   = \dfrac{{15.0}}{{100}} \times 3.0 \\
   = 0.45{\text{g}} \\
 \]
Therefore, the weight of the other metal is $ = 1.5 - 0.45 = 1.05{\text{g}}$ .
So, the percentage of the other metal is
$
   = \dfrac{{1.05}}{{3.0}} \times 100 \\
   = 35 \\
 $

So, the percent by weight of the other metal is 35 %. So, the correct option is D.

Note:
Minerals may contain low concentration of metal. They may also contain impurities. All these may be difficult to remove and thus every mineral is not suitable for the extraction of that metal.
Minerals from which the extraction of metal is conveniently and economically possible are called ores of that metal. Thus, all ores are considered to be minerals but all minerals cannot be ores.