The Over-100% Yield Precipitate
During a stoichiometry lab, students mix 25.0 mL of 0.200 M calcium chloride with 25.0 mL of 0.200 M sodium carbonate and collect the calcium carbonate precipitate by vacuum filtration. The teacher tells them to dry the precipitate to constant mass. Group A obtains 0.62 g. Group B obtains 0.41 g. A third group records 0.50 g but used 30.0 mL of calcium chloride and 20.0 mL of sodium carbonate, then reports the same theoretical yield as the other groups. The accepted theoretical yield for the original procedure is 0.500 g. Group A's filter paper felt cool and damp when weighed; Group B admits some precipitate passed through the filter during decanting. The class must decide how to report results and what to revise before the lab practical.
1. Calculate the theoretical yield for the original 25.0 mL + 25.0 mL procedure, showing the limiting-reactant reasoning. Then explain whether Group A's or Group B's result is more trustworthy.
Read the worked answer
Moles of CaCl2 = 0.0250 L x 0.200 mol/L = 0.00500 mol. Moles of Na2CO3 = 0.0250 L x 0.200 mol/L = 0.00500 mol. The balanced reaction is CaCl2 + Na2CO3 -> CaCO3 + 2NaCl, a 1:1:1 mole ratio. Since the reactants are present in equal moles, neither is in excess; both are limiting, and 0.00500 mol CaCO3 can form.
Theoretical mass = 0.00500 mol x 100.09 g/mol = 0.500 g.
Neither result is trustworthy as an accurate yield. Group A's 0.62 g exceeds the theoretical maximum, which is physically impossible for pure dry product. The damp, cool filter paper indicates trapped water or incomplete drying, so the mass includes solvent or paper. Group B's 0.41 g is below theoretical and is explained by precipitate passing through the filter, a negative error from product loss. Group B's result is at least physically possible, but it is not accurate. The class should not average the two; both trials should be repeated with corrected technique, and the report should distinguish the measured mass from the accepted theoretical yield.
2. The third group used 30.0 mL of CaCl2 and 20.0 mL of Na2CO3. Identify the limiting reactant and calculate the expected mass. Explain why reusing 0.500 g is wrong.
Read the worked answer
Moles of CaCl2 = 0.0300 L x 0.200 mol/L = 0.00600 mol. Moles of Na2CO3 = 0.0200 L x 0.200 mol/L = 0.00400 mol. The reaction consumes them in a 1:1 ratio, so the carbonate is limiting: 0.00400 mol Na2CO3 can produce at most 0.00400 mol CaCO3.
Expected mass = 0.00400 mol x 100.09 g/mol = 0.400 g.
The group cannot reuse 0.500 g because that value was calculated for equimolar 0.00500 mol amounts. Changing the volumes changes the mole ratio and the limiting reactant. Reporting 0.500 g would make the percent yield 0.50/0.40 = 125 percent, which is impossible for pure product and would hide the stoichiometry error. The teacher should require students to show moles of each reactant, identify the limiting reactant, and calculate the theoretical yield from the limiting moles before dividing by the theoretical value. This case also shows why percent yield comparisons across groups are only valid when the recipe is the same.
3. Propose three specific procedural changes to reduce the discrepancy between groups, and explain how each change addresses a source of error.
Read the worked answer
Dry to constant mass and cool in a desiccator before weighing. Group A's damp paper shows that a single drying period can leave water. Repeating heat-dry-cool-weigh cycles until the mass changes by a small amount ensures that the reported mass is dry CaCO3, not CaCO3 plus water.
Use a pre-weighed filter or crucible and rinse with small portions of distilled water. Group B lost precipitate through the filter or during decanting. A finer filter, careful vacuum filtration, and a rubber policeman to transfer the solid reduce mechanical loss. Rinsing with small amounts of cold distilled water removes soluble NaCl without dissolving significant CaCO3.
Check precipitation completeness and standardize the recipe. Adding a small excess of one reactant can drive precipitation, but the class must then recalculate the limiting reactant. For fair comparisons, all groups should use the same volumes and concentrations, verify that the supernatant gives no further precipitate, and record the actual molarities. This reduces discrepancies caused by different recipes and incomplete reaction.