Question

1.64g of a mixture of calcium carbonate and magnesium carbonate were dissolved in 50ml of 0.8N hydrochloric acid. The excess of the acid required 16ml N/4 sodium hydroxide solution for neutralisation. Find out the percentage composition of the mixture of two carbonates.

Answer 1

Hint: We can calculate the percentage composition by dividing the mass of a component by the total mass of the mixture. Then the ratio is to be multiplied by 100. It is also called the mass percent.

Complete step by step solution:
- We will write the reaction first:
\[CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O\]
Here, we will consider the moles of calcium carbonate as X mole and hydrochloric acid as 2x.
\[MgC{{O}_{3}}+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O\]
Here, we will consider moles of magnesium carbonate as Y mole and hydrochloric acid as 2y.
Now, we can write
\[\dfrac{{{N}_{1}}{{V}_{1}}-{{N}_{2}}{{V}_{2}}}{1000}\]
where, \[{{N}_{1}}\]= Number of moles of HCl
               \[{{V}_{1}}\]= Volume of HCl
              \[{{N}_{2}}\]= Number of moles of NaOH
              \[{{V}_{2}}\]= Volume of NaOH
By putting all values in above equation, we get:
\[\begin{align}
  & 2x+2y=\dfrac{\left( 0.8\times 50-0.25\times 16 \right)}{1000} \\
 & 2x+2y=\dfrac{\left( 40-4 \right)}{1000} \\
 & 2x+2y=\dfrac{36}{1000} \\
 & 2x+2y=0.036 \\
 & x+y=\dfrac{0.036}{2} \\
 & x+y=0.018 \\
\end{align}\]
Consider this equation as equation 1,
Now, we will find the molecular weight of calcium carbonate:
\[\begin{align}
  & CaC{{O}_{3}} \\
 & =40+12+3\times 16 \\
 & =40+12+48 \\
 & =40+60 \\
 & =100 \\
\end{align}\]
Now, we will find the molecular weight of magnesium carbonate:
\[\begin{align}
  & MgC{{O}_{3}} \\
 & =24+12+3\times 16 \\
 & =24+12+48 \\
 & =24+60 \\
 & =84 \\
\end{align}\]
We can write:
\[x\times 100+y\times 84=1.64\]
Consider this equation as equation 2
- From equation 1 and 2 we can find the value of x as:
from equation 1 we can write: x + y =0.018
so, y = x + 0.018
Now substitute the value of y in equation 2:
\[\begin{align}
  & 100\left( x \right)+84\left( 0.018-x \right)=1.64 \\
 & 100x+1.512-84x=1.64 \\
 & 100x-84x+1.512=1.64 \\
 & 16x=1.64-1.512 \\
 & 16x=0.128 \\
 & x=\dfrac{0.128}{16} \\
 & x=0.008 \\
\end{align}\]
Now, we will write the percentage of calcium carbonate as:
Weight of calcium carbonate x =0.008 gm
$\% \,of\, CaCO=\dfrac{x\times molecular\text{ }weight}{1.64}\times 100$
$=\dfrac{0.008\times 100}{1.64}\times 100$
$=\dfrac{0.8}{1.64}\times 100$
$=0.4878\times 100$
$=48.78%$

Now, we will write the percentage of magnesium carbonate as:
100 - percentage of calcium carbonate
= 100- 48.78
=51.22%

Hence, the percentage composition of the mixture of two carbonates that is of calcium carbonate is 48.78% and of magnesium carbonate is51.22%.

Note:If we know the information of the percent composition of a compound, then it will allow us to determine the mole to mole ratio of the elements present in a compound. And once, we know this, we can find the ratio of ions in a given compound.