Question

A 1.50 g sample of an unknown alkali metal carbonate was dissolved in water, and 31.10 mL of 0.350 M CaCl₂ was required to precipiate all the carbonate ions as CaCO₃. What is the name and formula of the unknown compound?

Answer 1

Answer : The formula of unknown compound is $K_2CO_3$ and the name of the compound is potassium carbonate.

Explanation :

First we have to calculate the moles of $CaCl_2$

$\text{Moles of }CaCl_2=\text{Concentration of }CaCl_2\times \text{Volume of solution}=0.350M\times 0.0311L=0.0109mole$

Now we have to calculate the moles of unknown alkali metal carbonate.

Let the alkali metal be, 'M'.

The balanced chemical reaction will be:

$M_2CO_3+CaCl_2\rightarrow CaCO_3+2MCl$

From the balanced reaction we conclude that,

As, 1 moles of $CaCl_2$ react with 1 mole of $M_2CO_3$

So, 0.0109 moles of $CaCl_2$ react with 0.0109 mole of $M_2CO_3$

Now we have to calculate the molar mass of unknown alkali metal carbonate.

$\text{Molar mass of }M_2CO_3=\frac{\text{Given mass of }M_2CO_3}{\text{Moles of }M_2CO_3}$

$\text{Molar mass of }M_2CO_3=\frac{1.50g}{0.0109mole}=137.6g/mole$

Thus, the molar mass of $M_2CO_3$ is 137.6 g/mole

Now we have to calculate the mass of alkali metal.

$137.6=(2\times \text{Mass of alkali metal})+(12)+(3\times 16)$

$\text{Mass of alkali metal}=38.8g$

Thus, we conclude that the alkali metal will be potassium (K) that has 38.8 grams of mass.

So, the formula of unknown compound is $K_2CO_3$ and the name of the compound is potassium carbonate.