Question

4) A mixture of Cao and Caco, is analysed by TGA. The result indicates that mass of the sample decreases from 250.6 mg to 190.8 mg only between 600 °C and 900 °C. Calculate the percentage of calcium carbonate in the mixture (1 x 4 - 4 marks)

Answer 1

The possible decomposition of
Cac
and $CaCO_{3}$ mixture is given by,

$CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)$.................................................1

So, the mass loss in the mixture is due to formation of  $CO_{2}(g)$

Now,

Molecular weight of $CaCO_{3}$ = 100.1 g/mol

Molecular weight of = 56.1 g/mol

Cac

Molecular weight of = 44 g/mol

со

From the given data, mass loss in the mixture is = $(250.6-190.8)mg=59.8mg$

$\therefore$ Loss of  $CO_{2}(g)$ is 59.8 mg

$\therefore$ Loss of $CO_{2}(g)$ = $\frac{59.8}{44}$ mmol

=1.359 mmol

From the above chemical equation 1,

1 mol of  $CaCO_{3}$ give 1 mol of

со

$\therefore$ the amount of $CaCO_{3}$ in the mioxture is 1.359 mmol

or, the amount of $CaCO_{3}$ in the mioxture is,  $1.359\times 100.1mg=136.04mg$

$\therefore$ the percentage of $CaCO_{3}$ in the mixture is given by =$\frac{136.04}{250.6}\times 100$

= 54.29%

Answer : The percentage of calcium carbonate in the mixture is 54.29%.