Question

What is the percentage purity of a sample of caco3 if 150 g of the sample gives out 44 g of co2 on heating?

Answer 1

No of moles = given mass of substance/ molar mass of substance

Moles of CO₂ emitted $= \frac{44}{44} = 1$ mole

So, moles of CaCO₃ = 1

So, mass of CaCO₃ = 40 + 12 + 48 = 100

Percentage purity = $\frac{100 x 100}{50} = 66.67%$

Therefore 66.67% is your answer.

Answer 2

Answer: The percentage purity of calcium carbonate is 66.73 %

Explanation:

We know that:

Molar mass of calcium carbonate = 100.1 grams

Molar mass of carbon dioxide = 44 grams

We are given:

Mass of carbon dioxide formed = 44 g

The chemical equation for the decomposition of calcium carbonate follows:

$CaCO_3\rightarrow CaO+CO_2$

By Stoichiometry of the reaction:

44 grams of carbon dioxide is formed by 100.1 grams of calcium carbonate

To calculate the percentage purity of sample, we use the equation:

$\%\text{ purity of sample}=\frac{\text{Mass of pure sample}}{\text{Mass of impure sample}}\times 100$

Mass of pure calcium carbonate = 100.1 grams

Mass of impure calcium carbonate = 150 grams

Putting values in above equation, we get:

$\%\text{ purity of calcium carbonate}=\frac{100.1g}{150g}\times 100\\\\\%\text{ purity of calcium carbonate}=66.73\%$

Hence, the percentage purity of calcium carbonate is 66.73 %