Question

If 88.0 g of CO2 is produced from the complete decomposition of calcium carbonate in 250 g of impure sample, what is the purity in terms of the carbonate content?

Answer 1

`\text{Percentage (\%) purity = }\frac{mass\text{ of pure compound }in\text{ sample}}{\text{total mass of impure sample}}`

Now purity is the effective percentage of pure reactant in the total mass.

The rest is inert impurities.

They ask you to calculate the purity of the carbonate, so you have to focus on CaCO3

Let's calculate with the reaction:

CaCO3 (s) -----------------> CaO (s) + CO2 (g)

x 88.0 g

100 g CaCO3 ------- 44 g CO2

x ------- 88 g CO2

x = 200.16 g

Sample has 250 g of impurities. So:

Purity % = (200.16 g / 250 g impure) x 100 = 80.0 % of purity