A 10.0\text{ g} sample of impure limestone containing calcium trioxocarbonate(IV), \text{C… — JAMB UTME

10.0\text{ g}

sample of impure limestone containing calcium trioxocarbonate(IV),

\text{CaCO}_3

, was strongly heated until decomposition was complete according to the equation:

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

1.792\text{ dm}^3

of carbon(IV) oxide gas was evolved at s.t.p., what is the percentage purity of the limestone sample?

[\text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

The percentage purity of the limestone sample is 80.0%.

From the balanced chemical equation, 1 mole of calcium trioxocarbonate(IV) produces 1 mole of carbon(IV) oxide gas. At s.t.p.,

1.792\text{ dm}^3

\text{CO}_2

equals

\frac{1.792}{22.4} = 0.08\text{ mol}

. Consequently,

0.08\text{ mol}

of pure

\text{CaCO}_3

was present. With a molar mass of

100\text{ g mol}^{-1}

, the mass of pure

\text{CaCO}_3

8.00\text{ g}

. Dividing this pure mass by the

10.0\text{ g}

total sample mass yields a purity of

80.0\%

Calculate the number of moles of carbon(IV) oxide gas evolved at s.t.p.

\text{Moles of CO}_2 = \frac{\text{Volume at s.t.p.}}{\text{Molar volume at s.t.p.}} = \frac{1.792\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.08\text{ mol}

Gas volume at standard temperature and pressure directly determines mole quantity using standard molar gas volume.

Determine the molar mass of pure calcium trioxocarbonate(IV),

\text{CaCO}_3

\text{Molar mass of CaCO}_3 = 40 + 12 + (3 \times 16) = 100\text{ g mol}^{-1}

Required to convert moles of pure reactant into mass.

Use the mole ratio from the balanced chemical equation to find the mass of pure

\text{CaCO}_3

Mole ratio of

\text{CaCO}_3 : \text{CO}_2 = 1 : 1

. Moles of pure

\text{CaCO}_3 = 0.08\text{ mol}

. Mass of pure

\text{CaCO}_3 = 0.08\text{ mol} \times 100\text{ g mol}^{-1} = 8.00\text{ g}

Stoichiometry dictates that 1 mole of

\text{CaCO}_3

produces 1 mole of

\text{CO}_2

upon complete decomposition.

Calculate the percentage purity of the limestone sample.

\text{Percentage purity} = \left( \frac{\text{Mass of pure CaCO}_3}{\text{Mass of impure sample}} \right) \times 100 = \left( \frac{8.00\text{ g}}{10.0\text{ g}} \right) \times 100 = 80.0\%

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

Percentage Purity and Stoichiometry from Gas Volumes

Estimated Time:1m 30s

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