Question

4.430g mixture containing CaCO₃ (s) is heated until all CaCO₃ (s) is decomposed completly.

CaCO_{3 (s)} >>>> CaO _(s) + CO_{2 (g)} (Molar mass: CaCO₃= 100.0g/mol CaO= 56.1g/mol CO₂= 44.0g/mol)

If the mass of the mixture remaining after heating is 3.580g, calculate the percent CaCO₃ in the mixture from the mass loss of the sample.

Answer 1

CaCO₃---------> CaO + CO₂

Initial mass of the mixture = 4.430 g (Containing CaCO₃)

After heating the CaCO₃ decomposes to CaO.

After decompostion,the mass is 3.580g.

Assuming all the CaCO₃ is decomposed the mass difference is due to CaO

Amount of CaO = 4.430 - 3.580 = 0.85g

100g of CaCO₃ gives 56g of CaO

0.85g of CaO corresponds to (100 x 0.85/56) 1.517 g of CaCO₃

So, 4.430g of sample contained 1.517 g of CaCO₃

so the percentage of CaCO₃ in the original sample is (1.517/4.430 ) x 100 = 34.24 percent