A 10.0\text{ g} sample of impure calcium carbonate (\text{CaCO}_3) is completely decompose… — JAMB UTME

10.0\text{ g}

sample of impure calcium carbonate (

\text{CaCO}_3

) is completely decomposed by strong heating according to the chemical equation:

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

The carbon(IV) oxide gas evolved is passed into an excess solution of sodium hydroxide, causing the mass of the solution to increase by

3.52\text{ g}

. Assuming the impurities present in the sample do not react or produce any gas, what is the percentage purity of the calcium carbonate sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16]

Answer: 80 %

The percentage purity of the calcium carbonate sample is 80%.

Sodium hydroxide absorbs carbon(IV) oxide (

\text{CO}_2

) gas released during the thermal decomposition of calcium carbonate (

\text{CaCO}_3

). The

3.52\text{ g}

mass gain of the solution equals the mass of

\text{CO}_2

evolved. Dividing this mass by the molar mass of

\text{CO}_2

(

44\text{ g/mol}

) yields

0.08\text{ mol}

\text{CO}_2

. According to the 1:1 stoichiometric relationship,

0.08\text{ mol}

of pure

\text{CaCO}_3

decomposed. Multiplying by the molar mass of

\text{CaCO}_3

(

100\text{ g/mol}

) gives

8.00\text{ g}

of pure

\text{CaCO}_3

. The percentage purity is

(8.00\text{ g} / 10.0\text{ g}) \times 100\% = 80\%

Calculate the molar masses of carbon(IV) oxide and calcium carbonate

Molar mass of CO2 = 44 g/mol, Molar mass of CaCO3 = 100 g/mol

Molar masses are required to convert between mass and moles.

Determine the moles of carbon(IV) oxide gas evolved

Moles of CO2 = 3.52 g / 44 g/mol = 0.08 mol

Sodium hydroxide reacts with and absorbs acidic carbon(IV) oxide, so mass increase equals the mass of CO2.

Determine the mass of pure calcium carbonate in the sample

Mass of pure CaCO3 = 0.08 mol * 100 g/mol = 8.00 g

The mole ratio of CaCO3 to CO2 in the thermal decomposition reaction is 1:1.

Calculate the percentage purity of the sample

(8.00 g / 10.0 g) * 100 = 80%

Percentage purity is the ratio of pure reactive substance mass to total sample mass expressed as a percentage.

Thermal decomposition of trioxocarbonates and percentage purity stoichiometry

Question