Question: You are given three mixtures that include compounds containing various forms of the carbon — Laboratory Practical / Experimental Design Chemistry Question

Sample plan, up to 5 pts
1. Label slider plastic bag to identify sample
2. Measure “mass” 1.0 g of sample, carefully transfer to the bag. Preferably in a corner opposite to the closing end of the slider. Push out as much air as possible and insert the funnel into the corner near where the slide fully closes.
3. Measure enough 0.8M acetic acid solution to ensure complete reaction of the carbonate sample.
4. Support volume choice with reaction equations and theoretic calculations in Q₂
5. Pour the acetic acid solution into the plastic bag using the funnel, immediately slide it close.
6. Make sure the carbonate sample has completely reacted, and none of the gases produced leak out of the bag.
7. Set aside to compare with next sample.
8. Compare the relative sizes of the inflated bags.
Titration methods will not differentiate carbonate content.
While it was not intended that students use electronic balances to weigh zipper bags of solutions, there were no instructions indicating that they could not do this. This was method that yielded accurate results.

4 total pts or 2 pts each for showing or indicating stoichiometry difference with carbonate and bicarbonate
(carbonate) X₂CO₃ (s) + 2 CH₃CO₂H (aq) → 2 CH₃CO₂X (aq) + H₂O (l) + CO₂ (g)
(bicarbonate) XHCO₃ (s) + CH₃CO₂H (aq) → CH₃CO₂X (aq⁺ H₂O (l) + CO₂ (g)
If using 1.00 g of solid that contains the carbonate ion, the amount of vinegar needed and CO₂ produced can be shown using common carbonates. Showing variations, up to 6 pts.
1.00 g NaHCO₃ * (1 mol / 84.007 g) * (1 mol H⁺ / 1 mol HCO^3-) * (1 L / 0.83 mol H⁺) = 0.014 L, 14 mL acid
1.0 g NaHCO₃ * (1 mol / 84.007 g) * (1 mol CO₂ / 1 mol HCO^3-) * RT/P = 0.291 L CO₂(g)
1.00 g Na₂CO₃ * (1 mol / 105.99 g) * (2 mol H⁺ / 1 mol CO₃ 2-) * (1 L / 0.83 mol H⁺) = 0.023 L, 23 mL acid
1.00 g Na₂CO₃ * (1 mol / 105.99 g) * (1 mol CO₂ / 1 mol CO₃ 2-) * RT/P = 0.231 L CO₂(g)
1.00 g Li₂CO₃ * (1 mol / 73.89 g) * (2 mol H⁺ / 1 mol CO₃ 2-) * (1 L / 0.83 mol H⁺) = 0.033 L, 33 mL acid
Probably unlikely that salt is pure lithium carbonate, should use excess acid to completely react all carbonates, therefor should use at least 35 mL vinegar per gram of solid unknown.

Sample data table. Masses of unknowns should be consistent within a set. Volume of vinegar should be consistent and in excess. More than one data set preferred. 6 total pts

Sample Data for qualitative observations of gas filling bags
Unknown Solid | Mass of Unknown Solid | Volume of Vinegar | Bag Inflation Volume
A | 1.009 g | 40 mL | Mostly full
A | 1.011 g | 40 mL | Mostly full
B | 0.995 g | 40 mL | Minimally full
B | 1.004 g | 40 mL | Minimally full
C | 1.011 g | 40 mL | Moderately full
C | 0.998 g | 40 mL | Moderately full

Sample Data for quantitative observations of mass lost
Unknown Solid | Mass of Unknown Solid | Mass Vinegar | Final Mass | Mass CO₂
A | 1.009 g | 40.012 g | 40.493 g | 0.528 g
A | 1.011 g | 41.234 g | 41.715 g | 0.530 g
B | 0.995 g | 40.349 g | 41.084 | 0.260 g
B | 1.004 g | 40.253 g | 40.994 | 0.263 g
C | 1.011 g | 41.045 g | 41.703 | 0.353 g
C | 0.998 g | 40.874 g | 41.524 | 0.348 g

Might show a drawing or give a description of the inflated bag being rolled and clipped. This allows for better visualization of the differences in the inflation volumes
If attempting to titrate, data for similar mass samples of A^-C should give similar volume of vinegar. Not able to differentiate.

Student answer may not be correct but should be based on their collected data/observations. Students should comment on sources of error 4 total points.
___ Mixture B (36%) < Mixture C (48%) < Mixture A (71%)