Identifying an Unknown Group 2 Carbonate
A student is given an unlabelled bottle of a white solid, suspected to be a Group 2 carbonate, MCO3. To identify the metal, 1.250 g of the solid is reacted completely with excess dilute hydrochloric acid, and the carbon dioxide evolved is collected over water. At 25.0 °C and a total pressure of 101.3 kPa, the gas volume collected is 296 mL. The vapour pressure of water at 25.0 °C is 3.17 kPa. The student notices that a few bubbles escaped before the collection tube was properly sealed, and that the solid had absorbed some moisture from the air. The molar masses of candidate metals are: Mg = 24.3, Ca = 40.1, Sr = 87.6, Ba = 137.3 g/mol.
1. Calculate the molar mass of the unknown carbonate from the data and identify the most likely metal. Show each step of your reasoning.
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Step 1 - Correct the gas pressure. The CO2 is collected over water, so it is saturated with water vapour. By Dalton's law: P(CO2) = P(total) − P(H2O) = 101.3 − 3.17 = 98.13 kPa.
Step 2 - Moles of CO2 using the ideal gas law. n = PV/RT = (98.13 × 10³ Pa)(296 × 10⁻⁶ m³) / [(8.314)(298.15 K)] = 0.02905 / 2478.8 ≈ 1.172 × 10⁻² mol.
Step 3 - Relate moles of CO2 to moles of carbonate. From MCO3 + 2HCl → MCl2 + H2O + CO2, the ratio is 1:1, so n(MCO3) = 1.172 × 10⁻² mol.
Step 4 - Molar mass. M = m/n = 1.250 / 1.172 × 10⁻² ≈ 106.7 g/mol.
Step 5 - Subtract the carbonate part. M(CO3) = 12.0 + 48.0 = 60.0 g/mol, so M(metal) ≈ 46.7 g/mol.
This is closest to calcium (40.1), but the value is noticeably high. The escaped bubbles mean the true CO2 amount was slightly larger, so the computed molar mass is an overestimate - the real metal mass is below 46.7, moving it toward Ca. Moisture in the solid would make the weighed mass too high, which also inflates the calculated molar mass, again consistent with Ca. Conclusion: the metal is calcium, CaCO3, with the discrepancies explained by procedural losses.
2. Explain how each of the two stated errors would shift the calculated metal molar mass, and state whether they could together change the identification.
Read the worked answer
Escaped CO2 (gas loss). If some CO2 escapes before collection, the measured volume is smaller than the true amount produced. Since n is calculated from V, n is underestimated, and M = m/n becomes too large. The apparent metal mass is therefore too high.
Moisture in the solid. The 1.250 g weighed includes water that does not produce CO2. The effective mass of carbonate is lower than assumed, so the true n is smaller than the mass suggests. Using m = 1.250 g overestimates n, which makes M too small. So this error pushes the result in the opposite direction to the gas loss.
Combined effect. The two errors oppose each other, so they may partially cancel. Neither is large enough to shift the answer from Ca (40.1) to the next candidate, Mg (24.3) or Sr (87.6) - the gap to either neighbour exceeds the plausible error. A careful answer notes that the identification is robust because the candidate molar masses are widely spaced, but the precise value of M cannot be trusted. Best practice: reseal the apparatus, dry the solid, and repeat, or use a mass-loss method (weigh before and after reaction) which is insensitive to gas collection losses.