A sample containing 8.0\text{ g} of impure calcium trioxocarbonate(IV) reacts completely w… — JAMB UTME

A sample containing

8.0\text{ g}

of impure calcium trioxocarbonate(IV) reacts completely with excess dilute hydrochloric acid according to the equation:

\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

1.344\text{ dm}^3

of carbon(IV) oxide gas measured at s.t.p. is liberated, what is the percentage purity of the calcium trioxocarbonate(IV) 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 calcium trioxocarbonate(IV) sample is

75.0\%

First, find the moles of carbon(IV) oxide produced at s.t.p.:

1.344 / 22.4 = 0.060\text{ mol}

. According to the balanced chemical equation,

1\text{ mol}

of calcium trioxocarbonate(IV) reacts to yield

1\text{ mol}

of carbon(IV) oxide. Therefore,

0.060\text{ mol}

of pure calcium trioxocarbonate(IV) reacted. The mass of pure calcium trioxocarbonate(IV) is

0.060 \times 100\text{ g mol}^{-1} = 6.0\text{ g}

. Calculating percentage purity gives

(6.0\text{ g} / 8.0\text{ g}) \times 100\% = 75.0\%

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

\text{Moles of CO}_2 = \frac{1.344\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.060\text{ mol}

Molar gas volume at s.t.p. equals

22.4\text{ dm}^3\text{ mol}^{-1}

Determine the mole relation and calculate the mass of pure calcium trioxocarbonate(IV).

Molar mass of

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

. From the

1:1

stoichiometric ratio,

\text{moles of pure CaCO}_3 = 0.060\text{ mol}

\text{Mass} = 0.060\text{ mol} \times 100\text{ g mol}^{-1} = 6.0\text{ g}

The balanced chemical equation shows a 1:1 mole ratio between calcium trioxocarbonate(IV) and carbon(IV) oxide.

Calculate the percentage purity of the sample.

\text{Percentage purity} = \frac{\text{Mass of pure CaCO}_3}{\text{Total mass of impure sample}} \times 100\% = \frac{6.0\text{ g}}{8.0\text{ g}} \times 100\% = 75.0\%

Percentage purity expresses the mass fraction of pure active component relative to total sample mass.

Mass-volume stoichiometric calculations and percentage purity determination

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