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

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:

& 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{aligned}$$ Consider this equation as equation 1, Now, we will find the molecular weight of calcium carbonate: $$\begin{aligned} & CaC{{O}_{3}} \\\ & =40+12+3\times 16 \\\ & =40+12+48 \\\ & =40+60 \\\ & =100 \\\ \end{aligned}$$ Now, we will find the molecular weight of magnesium carbonate: $$\begin{aligned} & MgC{{O}_{3}} \\\ & =24+12+3\times 16 \\\ & =24+12+48 \\\ & =24+60 \\\ & =84 \\\ \end{aligned}$$ We can write: $$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{aligned} & 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{aligned}$$ 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.