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Difficulty: HardOxides of Carbon and Trioxocarbonate(IV) Salts

A 10.0 g10.0\text{ g} sample of impure calcium trioxocarbonate(IV), CaCO3\text{CaCO}_3, was strongly heated until decomposition was complete. If the volume of carbon(IV) oxide gas evolved at STP was 1.792 dm31.792\text{ dm}^3, what is the percentage purity of the CaCO3\text{CaCO}_3 sample? [Molar volume of gas at STP = 22.4 dm3mol−122.4\text{ dm}^3\text{mol}^{-1}; relative atomic masses: Ca=40,C=12,O=16\text{Ca}=40, \text{C}=12, \text{O}=16]

Answer: 80 %

Answer

The percentage purity of the calcium trioxocarbonate(IV) sample is 80%80\%.
Thermal decomposition of pure calcium trioxocarbonate(IV) releases carbon(IV) oxide gas according to CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}). Dividing the gas volume (1.792 dm31.792\text{ dm}^3) by the molar gas volume at STP (22.4 dm3mol−122.4\text{ dm}^3\text{mol}^{-1}) yields 0.08 mol0.08\text{ mol} of CO2\text{CO}_2. Due to the 1:1 stoichiometry, 0.08 mol0.08\text{ mol} of pure CaCO3\text{CaCO}_3 reacted. Multiplying by the molar mass of CaCO3\text{CaCO}_3 (100 g/mol100\text{ g/mol}) gives 8.0 g8.0\text{ g} of pure CaCO3\text{CaCO}_3. The percentage purity is calculated as (8.0 g10.0 g)×100%=80%\left(\frac{8.0\text{ g}}{10.0\text{ g}}\right) \times 100\% = 80\%.

Step-by-Step Solution

1
Write the balanced chemical equation for the thermal decomposition of calcium trioxocarbonate(IV).
CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})
Establishes the 1:1 stoichiometric mole ratio between CaCO3\text{CaCO}_3 and CO2\text{CO}_2.
2
Calculate the number of moles of CO2\text{CO}_2 gas produced at STP.
Moles of CO2=1.792 dm322.4 dm3mol−1=0.08 mol\text{Moles of CO}_2 = \frac{1.792\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.08\text{ mol}
Uses the molar gas volume relationship at standard temperature and pressure (V/VmV / V_m).
3
Calculate the mass of pure CaCO3\text{CaCO}_3 in the original sample.
Mass of CaCO3=0.08 mol×100 g/mol=8.0 g\text{Mass of CaCO}_3 = 0.08\text{ mol} \times 100\text{ g/mol} = 8.0\text{ g}
Because 1 mol1\text{ mol} of CaCO3\text{CaCO}_3 produces 1 mol1\text{ mol} of CO2\text{CO}_2, 0.08 mol0.08\text{ mol} of pure CaCO3\text{CaCO}_3 reacted.
4
Calculate the percentage purity of the sample.
Percentage purity=(8.0 g10.0 g)×100%=80%\text{Percentage purity} = \left(\frac{8.0\text{ g}}{10.0\text{ g}}\right) \times 100\% = 80\%
Compares the mass of active pure reactant to the total mass of the impure sample.

Key Concept

Stoichiometry of thermal decomposition of trioxocarbonate(IV) salts and gas molar volume calculations at STP.
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Topics

Oxides of Carbon and Trioxocarbonate(IV) Salts

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