Why the Product Doesn't Match the Formula
Students precipitate a metal carbonate by mixing 25.0 mL of 0.20 M CaCl₂ with 25.0 mL of 0.20 M Na₂CO₃. They filter, dry, and weigh the solid, expecting CaCO₃ (molar mass 100.09 g/mol). The predicted mass is 0.50 g. They obtain 0.42 g and are asked whether the reaction failed. The drying oven was set to 105 °C, and the filter paper was weighed before use. One student recalls that CaCO₃ can decompose on strong heating, but 105 °C is far below its decomposition temperature.
1. Calculate the theoretical yield and determine whether 0.42 g indicates a failed reaction or a recoverable loss.
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Limiting reagent / theoretical yield:
- Moles CaCl₂ = 0.0250 L × 0.20 mol/L = 5.0 × 10⁻³ mol
- Moles Na₂CO₃ = 0.0250 L × 0.20 mol/L = 5.0 × 10⁻³ mol
They are in 1:1 stoichiometry and exactly equimolar, so neither is limiting - the yield is set by both equally:
- Theoretical CaCO₃ = 5.0 × 10⁻³ mol × 100.09 g/mol ≈ 0.50 g
Interpretation of 0.42 g: Percent yield = 0.42/0.50 × 100 ≈ 84%. This is a recovery shortfall, not a failed reaction. The reaction clearly occurred (solid formed). At 105 °C, CaCO₃ does not decompose (its decomposition is near 800 °C), so thermal loss is not the explanation.
Likely causes: incomplete precipitation/transfer, product lost during filtration or rinsing, or residual moisture making the mass low is less likely (moisture would raise it). The most probable losses are mechanical (product left on glassware or filter) and incomplete drying/transfer. An 84% yield is normal for a first-attempt gravimetric procedure.
2. What single procedural improvement would most likely raise the yield, and how would you verify the improvement rather than assume it?
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Improvement: Use quantitative transfer and a digestion step - after adding the precipitant, warm and let the precipitate stand (digest) so crystals grow larger and filter more cleanly, then rinse the beaker and filter with small portions of distilled water, transferring all solid to the pre-weighed filter.
Verification (don't assume):
- Run the procedure in triplicate and report mean yield ± spread. A single 0.42 g result cannot distinguish random loss from a systematic problem.
- Check the filtrate with a drop of Na₂CO₃ or CaCl₂ to confirm no unprecipitated ions remain - if more precipitate forms, the loss was incomplete precipitation, not transfer.
- Weigh to constant mass (dry, cool in a desiccator, reweigh until stable) to rule out moisture.
If triplicate yields cluster near 95–98% with the digestion step, the improvement is confirmed; if not, the shortfall lies elsewhere (e.g., solubility of CaCO₃ in rinse water, which is small but nonzero).
3. Suppose the student had instead used 25.0 mL of 0.20 M CaCl₂ with only 15.0 mL of 0.20 M Na₂CO₃. How would the analysis change?
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Now the reagents are not equimolar, so a limiting reagent appears:
- CaCl₂: 0.0250 L × 0.20 = 5.0 × 10⁻³ mol
- Na₂CO₃: 0.0150 L × 0.20 = 3.0 × 10⁻³ mol
Na₂CO₃ is limiting (1:1 reaction), so maximum CaCO₃ = 3.0 × 10⁻³ mol × 100.09 ≈ 0.30 g. CaCl₂ is in excess (2.0 × 10⁻³ mol left over).
Two consequences:
- The target yield drops to 0.30 g, so the earlier 0.42 g would be impossible and would signal contamination (e.g., unreacted/excess salt or moisture) rather than good recovery.
- Identification of the limiting reagent is essential before judging yield. A student who still expected 0.50 g would wrongly conclude the reaction was superb (140% yield) when the real issue is excess reagent or poor drying.
This is exactly the judgment the exam tests: yield percentages are meaningless until the limiting reagent and stoichiometry are established.