Carbonate Purity by Gas Collection over Water
A technician assays a 0.400 g sample of impure calcium carbonate (the impurities are insoluble and do not react with acid) by adding 25.0 mL of 3.00 M hydrochloric acid, which is in excess. The CO₂ evolved is collected in a graduated eudiometer over water. Once the reaction has finished and the apparatus has stood until the gas reaches room temperature, the following are recorded: gas volume 84.5 mL, room temperature 22.0 °C, barometric pressure 99.2 kPa. The vapour pressure of water at 22.0 °C is 2.64 kPa. A second student repeats the work but omits the vapour-pressure correction. A third student reads the volume while the gas is still warm from the reaction (about 35 °C, vapour pressure of water 5.6 kPa) but uses 22.0 °C in the calculation. Ar(Ca) = 40.1, Ar(C) = 12.0, Ar(O) = 16.0.
1. Calculate the mass percentage of CaCO₃ in the sample, stating and justifying each correction you apply.
Read the worked answer
The measured pressure is the total pressure of a mixture: P(CO₂) = P(barometer) − P(H₂O) = 99.2 − 2.64 = 96.56 kPa = 96 560 Pa. The gas is saturated with water vapour, so neglecting this over-counts CO₂.
T = 22.0 + 273.15 = 295.15 K; V = 84.5 mL = 8.45 × 10⁻⁵ m³.
n(CO₂) = PV/RT = (96 560 × 8.45 × 10⁻⁵)/(8.314 × 295.15) = 8.159/2454 = 3.325 × 10⁻³ mol.
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so n(CaCO₃) = 3.325 × 10⁻³ mol and m = 3.325 × 10⁻³ × 100.1 = 0.333 g.
Mass % = 0.333/0.400 × 100 = 83.2 %.
Acid check: 0.0250 × 3.00 = 7.50 × 10⁻² mol HCl available, but 2 × 3.325 × 10⁻³ = 6.65 × 10⁻³ mol is consumed, so the acid is indeed in excess and the carbonate is the limiting reagent - the assumption behind the calculation holds.
2. Quantify the error made by the second student who ignored the water-vapour correction, and say whether the reported purity would be too high or too low.
Read the worked answer
Using the total pressure as if it were all CO₂:
n = (99 200 × 8.45 × 10⁻⁵)/(8.314 × 295.15) = 8.382/2454 = 3.416 × 10⁻³ mol
m(CaCO₃) = 3.416 × 10⁻³ × 100.1 = 0.342 g → 0.342/0.400 = 85.5 %.
The reported purity is too high by 2.3 percentage points (85.5 vs 83.2, a relative error of about 2.8 %). The reason is that the 2.64 kPa of water vapour was counted as CO₂, i.e. about 2.7 % of the total pressure was converted into extra moles. Note that the error is systematic and always in the same direction, so it cannot be averaged away by repeating the measurement - the correction must be applied every time, and its relative size grows as the collected volume shrinks or the temperature rises.
3. Predict the direction of the error made by the third student, who read the volume while the gas was still at 35 °C but used 22.0 °C in the calculation. Justify your answer quantitatively if possible.
Read the worked answer
The reported purity would again be too high, for two independent reasons that reinforce each other.
- Thermal expansion. At 35 °C the same amount of gas occupies more volume: V(308 K)/V(295 K) ≈ 1.044, so the recorded volume is about 4.4 % too large. Using T = 295.15 K with a volume that actually belongs to 308 K inflates n by roughly that factor. In n = PV/RT, using a temperature that is too small makes n too large.
- Larger water-vapour pressure. At 35 °C, P(H₂O) = 5.6 kPa, not 2.64 kPa. Subtracting the 2.64 kPa figure leaves ~3 kPa of water vapour counted as CO₂, adding a further positive bias.
The two effects both push the calculated mole count (and hence apparent purity) upward. To avoid the problem, the eudiometer must be allowed to stand, with the levels equalized, until the gas is at room temperature before the volume is read - otherwise a good student can report a purity above 100 % for a sufficiently wet, warm gas sample.