A 5.00\text{ g} sample of impure limestone (\text{CaCO}_3) is strongly heated until decomp… — JAMB UTME

A $5.00\text{ g}$ sample of impure limestone ( $\text{CaCO}_3$ ) is strongly heated until decomposition is complete. If the loss in mass due to the escape of carbon dioxide ( $\text{CO}_2$ ) gas is $1.76\text{ g}$ , what is the percentage purity of the limestone sample? $[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16]$

Answer: 80 %

Answer

The percentage purity of the limestone sample is 80%.

Thermal decomposition of calcium carbonate yields calcium oxide and carbon dioxide. The mass loss of 1.76 g corresponds to the evolved CO2. From the molar masses (CaCO3 = 100 g/mol, CO2 = 44 g/mol), 44 g of CO2 is released by 100 g of pure CaCO3. Thus, 1.76 g of CO2 is released by 4.00 g of pure CaCO3. Dividing the pure mass (4.00 g) by the original sample mass (5.00 g) and multiplying by 100 yields a percentage purity of 80%.

Step-by-Step Solution

Write the balanced equation for the decomposition reaction.

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

The decrease in mass is entirely due to the evolved carbon dioxide gas.

Calculate the relative formula mass of calcium carbonate and carbon dioxide.

\text{Molar mass of } \text{CaCO}_3 = 100\text{ g/mol}, \quad \text{Molar mass of } \text{CO}_2 = 44\text{ g/mol}

Required to relate the mass of evolved gas to the mass of reacting calcium carbonate.

Calculate the mass of pure calcium carbonate in the sample.

\text{Mass of pure } \text{CaCO}_3 = \left(\frac{100}{44}\right) \times 1.76\text{ g} = 4.00\text{ g}

Direct stoichiometric ratio derived from 1 mol CaCO3 producing 1 mol CO2.

Calculate percentage purity.

\text{Percentage purity} = \left(\frac{4.00\text{ g}}{5.00\text{ g}}\right) \times 100 = 80\%

Ratio of pure reactant mass to total sample mass expressed as a percentage.

Key Concept

Calculating percentage purity using stoichiometry and gravimetric decomposition data.

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