D. Quantities in Chemical Reactions · Quantities in Chemical ReactionsInvestigation: Measuring Actual, Theoretical and Percentage Yield
D2.7 D2.6 A1.1 A1.4 A1.5 A1.6 A1.8 A1.10 A1.11 A1.13About 75 minutes
Learning goals- I can carry out (or follow) a precipitation investigation to find an actual yield.
- I can calculate the theoretical yield and percentage yield from my data.
- I can assess how well a procedure worked and suggest specific improvements.
- I can identify sources of experimental error and explain whether each raises or lowers the yield.
- I can report results with suitable precision and significant figures.
Words to know
precipitate
An insoluble solid that forms when two solutions are mixed.
filtrate
The liquid that passes through the filter paper.
gravity filtration
Separating a solid from a liquid by pouring the mixture through filter paper in a funnel.
precision
How finely a measurement is made (for example, to 0.01 g) and how closely repeated measurements agree.
accuracy
How close a result is to the accepted or theoretical value.
procedural improvement
A specific change to the method that would reduce a named source of error.
From paper to product
You can now predict exactly how much product a reaction should make. In this investigation you find out how much it does make, and why the two numbers differ. That gap between prediction and reality is where real chemistry, and real problem-solving, happens.
The investigation: making calcium carbonate
When solutions of calcium chloride and sodium carbonate are mixed, white calcium carbonate (the same compound as chalk and limestone) precipitates:
CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq)
Question (A1.1): What percentage yield of CaCO₃ can be obtained, and what limits it?
Hypothesis: If calcium chloride is the limiting reagent, the mass of dry CaCO₃ will be slightly less than the theoretical yield (about 90% to 100%), because some solid will be lost during filtering and transfer.
Outline of the procedure (done in a school lab):
- Weigh about 1.5 g of calcium chloride dihydrate, CaCl₂·2H₂O, and dissolve it in 50 mL of distilled water.
- Weigh about 1.5 g of sodium carbonate (an excess) and dissolve it in another 50 mL of distilled water.
- Pour one solution into the other and stir. A white precipitate forms.
- Weigh a dry filter paper. Filter the mixture. Rinse the beaker with small amounts of distilled water and pour the rinsings through the filter so no solid is left behind. Rinse the solid on the filter to wash away dissolved NaCl.
- Dry the filter paper and precipitate (overnight or in a drying oven). Weigh, dry again and reweigh to constant mass.
Safety (A1.4)
Wear safety goggles. Calcium chloride and sodium carbonate can irritate eyes and skin; check their WHMIS labels and safety data sheets. Wash hands after the lab. Dispose of the filtrate as your teacher directs. If a drying oven is used, handle hot items with tongs.
| Measurement | Value |
|---|
| mass of CaCl₂·2H₂O | 1.47 g |
| mass of Na₂CO₃ | 1.50 g |
| mass of dry filter paper | 0.912 g |
| mass of filter paper + dry precipitate (1st weighing) | 1.866 g |
| mass of filter paper + dry precipitate (2nd weighing) | 1.864 g |
| observations | white cloudy precipitate formed at once; filtrate clear and colourless |
Worked exampleExample: Analysing the sample data
Use the sample data to find the limiting reagent, theoretical yield, actual yield and percentage yield of CaCO₃.
- M(CaCl₂·2H₂O) = 40.08 + 2(35.45) + 2(18.02) = 147.02 g/mol. n = 1.47 ÷ 147.02 = 0.009999 mol.
- M(Na₂CO₃) = 105.99 g/mol. n = 1.50 ÷ 105.99 = 0.01415 mol.
- The mole ratio is 1 : 1, and 0.009999 < 0.01415, so CaCl₂ is limiting.
- Theoretical yield: n(CaCO₃) = 0.009999 mol × 100.09 g/mol = 1.001 g.
- Actual yield: 1.864 − 0.912 = 0.952 g.
- Percentage yield = 0.952 ÷ 1.001 × 100% = 95.1%.
Percentage yield = 95.1% (CaCl₂ limiting; theoretical 1.00 g; actual 0.952 g).
Assessing the procedure (A1.8)
A good analysis does three things:
- Names specific sources of error. Not "human error" but, for example, "a fine precipitate passed through the filter paper (the filtrate was slightly cloudy)".
- States the direction. Would this error make the yield too high or too low?
- Suggests an improvement that would fix it: "use finer filter paper" or "let the precipitate settle before filtering".
A 95.1% yield is lower than 100%, so the main errors were ones that lose product. Errors that add mass (a wet or impure precipitate) would push the yield up, and might hide some of the loss. That is why drying to constant mass and rinsing the precipitate matter.
Precision and significant figures (A1.13)
A centigram balance reads to 0.01 g; an analytical balance to 0.001 g. Your answer cannot be more precise than your measurements. In the sample, masses like 1.47 g (3 significant figures) limit the percentage yield to 3 significant figures: 95.1%, not 95.127%.
Small masses are hard to measure precisely. Losing 0.010 g of a 0.95 g product changes the yield by about 1%. Using larger amounts, or a more precise balance, reduces this effect.
Writing it up (A1.10, A1.11)
A complete lab report includes the question, hypothesis, materials, procedure, observations (qualitative and quantitative), calculations shown with units, a conclusion that answers the question, and an evaluation (sources of error and improvements). For example:
"The percentage yield of CaCO₃ was 95.1%, supporting the hypothesis that CaCl₂ would be limiting and that a small amount of product would be lost. The most likely source of error was precipitate left on the beaker walls, which lowers the yield; rinsing the beaker three times with distilled water would reduce this."
Just for funFilter paper has one of the most relaxed jobs in the lab: it just lets things go.
(Except the precipitate. It holds on to that.)
Hands-on activityHome investigation: how much CO₂ does baking soda release?
This is a safe kitchen version of a yield investigation. Reaction: NaHCO₃(s) + CH₃COOH(aq) → NaCH₃COO(aq) + H₂O(l) + CO₂(g). The CO₂ escapes, so the mass lost is the actual yield of CO₂.
You need: a kitchen scale (to 1 g, or better 0.1 g), baking soda, white vinegar (5% acetic acid), a large plastic cup, a small paper cup, a spoon. Work over the sink. Keep your face back from the cup; vinegar stings eyes.
- Predict: with 10.0 g of baking soda and excess vinegar, calculate the theoretical mass of CO₂. (10.0 g needs about 143 mL of vinegar, so use 250 mL to be sure vinegar is in excess.)
- Pour 250 mL of vinegar into the large cup. Put 10.0 g of baking soda in the small paper cup. Place both cups and the spoon on the scale together and record the total mass.
- Slowly tip the baking soda into the vinegar a little at a time (so it does not foam over), stirring gently. Put the empty paper cup and spoon back on the scale.
- When fizzing stops, wait 5 minutes, stir, then record the total mass again.
- Calculate: actual yield of CO₂ = mass before − mass after; percentage yield = actual ÷ theoretical × 100%.
- Evaluate: name one error that lowers your yield and one that raises it, and suggest an improvement.
What to look for: theoretical CO₂ = (10.0 ÷ 84.01) × 44.01 = 5.24 g. Your measured loss will probably be somewhat less than this, because some CO₂ stays dissolved in the liquid (yield too low). Spray or foam escaping the cup carries away liquid, which makes the mass loss, and the yield, too high. A scale that reads only to 1 g gives an uncertainty of about ±1 g in 5 g, about 20%, so the scale's precision is a major limitation. Improvements: use a 0.1 g scale, stir longer or warm the mixture slightly to drive out dissolved CO₂, and cover the cup loosely with a paper towel to catch spray.
The kitchen investigation. The whole system is weighed before and after; the lost mass is the CO₂ that escaped. (Sample readings shown: a 5 g loss.)Your turn
1In a repeat of the investigation, 2.21 g of CaCl₂·2H₂O is mixed with 2.00 g of Na₂CO₃, and 1.38 g of dry CaCO₃ is collected. Find the limiting reagent and the percentage yield.
n(CaCl₂·2H₂O) = 2.21 ÷ 147.02 = 0.01503 mol; n(Na₂CO₃) = 2.00 ÷ 105.99 = 0.01887 mol. 1 : 1, so CaCl₂ is limiting. Theoretical = 0.01503 × 100.09 = 1.505 g. Yield = 1.38 ÷ 1.505 × 100% = 91.7%.
2Why is the precipitate rinsed with distilled water while it is on the filter paper?
To wash away dissolved NaCl (and excess Na₂CO₃). If they dried on the precipitate, they would add mass and make the yield too high.
3A student's filtrate was cloudy. What does that suggest, and how does it affect the percentage yield?
Some fine precipitate passed through the filter paper, so less was collected: the yield is too low. Improvement: use finer filter paper, let the solid settle first, or re-filter the filtrate.
4Why do we use sodium carbonate in excess?
So that all of the calcium chloride reacts; CaCl₂ is then clearly the limiting reagent, and the theoretical yield can be calculated from it.
5Explain the difference between accuracy and precision using this lab.
Accuracy is how close the percentage yield is to 100% (the theoretical value). Precision is how finely the masses were measured (to 0.001 g) and how closely repeated trials or weighings agree.
6Write one sentence for the conclusion and one for the evaluation for a student who got 95.1%.
Sample conclusion: "The percentage yield of CaCO₃ was 95.1%, with CaCl₂ as the limiting reagent." Sample evaluation: "The yield was lower than 100% mainly because some precipitate stuck to the beaker; rinsing the beaker three times with distilled water would improve the result."
More practice: warm-up
1What is the precipitate in this investigation?
Calcium carbonate, CaCO₃ (white solid).
2What passes through the filter paper?
The filtrate: water with dissolved NaCl (and excess Na₂CO₃).
3Why must the filter paper be weighed before filtering?
So its mass can be subtracted to find the mass of precipitate alone.
4Name one error that makes the percentage yield too high.
Any one: precipitate still wet; dissolved impurities (NaCl) left in it; paper fibres or dust added.
5Name one error that makes the percentage yield too low.
Any one: precipitate passed through the filter; solid left in the beaker; spills during transfer.
More practice: on your own
1A student uses a balance that reads to 0.1 g. The product mass is 0.9 g. Why is a percentage yield of "95.13%" not justified?
The mass has only 1 significant figure (0.9 g), so the percentage yield can be given to about 1 significant figure at most (about 90%). The balance is not precise enough for more digits; a 0.001 g balance is needed.
2Explain why drying the precipitate to constant mass is important.
If water remains, the mass is too large and the yield is too high. When two weighings in a row agree, the water is gone.
3In the kitchen CO₂ investigation, a student measured a mass loss of 4.6 g from 10.0 g of baking soda. Find the percentage yield and give the most likely reason it is below 100%.
Theoretical = 5.24 g. Yield = 4.6 ÷ 5.24 × 100% = 88%. Most likely: some CO₂ stayed dissolved in the solution (or the reading was limited by the scale's precision).
4Suggest two procedural improvements that would make the calcium carbonate investigation more accurate.
Any two: rinse the beaker several times into the filter; use finer filter paper or let the solid settle; wash the precipitate with distilled water; dry to constant mass; use larger amounts so small losses matter less; do several trials and average.
5A student forgot to rinse the precipitate and got 103%. Explain the result, and say whether the reaction really produced more CaCO₃ than predicted.
Dissolved NaCl dried on the precipitate and added mass, so the measured mass was too large. The reaction did not make extra CaCO₃; matter cannot be created. The result reflects a measurement error, not more product.
Challenge
1Copper can be made by adding aluminium foil to copper(II) chloride solution: 2Al + 3CuCl₂ → 2AlCl₃ + 3Cu. A student uses 5.00 g of CuCl₂·2H₂O (M = 170.49 g/mol) and 1.00 g of Al foil. Find the limiting reagent and the theoretical yield of copper.
n(CuCl₂) = 5.00 ÷ 170.49 = 0.02933 mol. n(Al) = 1.00 ÷ 26.98 = 0.03706 mol. Al needed for all the CuCl₂ = 0.02933 × 2/3 = 0.01955 mol < 0.03706, so CuCl₂ is limiting. Theoretical Cu = 0.02933 mol × 63.55 = 1.86 g.
2In that experiment, the student reports 1.95 g of copper. Find the percentage yield and give two possible explanations.
1.95 ÷ 1.864 × 100% = 105%. Possible causes: the copper was still wet; pieces of unreacted aluminium foil were mixed in with the copper; or some copper became oxidized, adding oxygen's mass.
3Design an improvement to the kitchen CO₂ investigation that would make the result both more precise and more accurate. Explain how each change helps.
Sample: use a scale that reads to 0.1 g (more precise); use 20.0 g of baking soda with enough vinegar so the mass loss is larger relative to the scale's uncertainty (more precise); after fizzing stops, stir and wait longer or warm the cup slightly to drive out dissolved CO₂ (more accurate, raises a low yield); cover loosely with paper towel to stop spray escaping (more accurate, prevents a falsely high yield); do three trials and average.
Quiz
1. In the precipitation investigation, why is sodium carbonate used in excess?
2. Filter paper alone: 0.900 g. Paper + dry precipitate: 1.850 g. Theoretical yield: 1.000 g. What is the percentage yield?
3. Which error would make the percentage yield too HIGH?
4. Which statement is the best evaluation of a source of error?
5. A balance reads to 0.01 g. Which mass is recorded correctly?
6. In the kitchen CO₂ investigation, why does some CO₂ staying dissolved lower the measured yield?
Big idea: This is the inquiry required by D2.7: measure an actual yield, calculate the theoretical and percentage yields, assess the procedure, and suggest sources of error. It also practises many strand A skills: planning (A1.1), safety (A1.4), performing and recording (A1.5, A1.6), analysis of error (A1.8), conclusions and communication (A1.10, A1.11) and precision (A1.13).
Safety at home: the precipitation lab is for school. The baking soda and vinegar investigation is safe at home with simple precautions: work over a sink, keep faces back, and do not drink the mixture.
How to help: Do the kitchen investigation together and focus on the evaluation: "Which way would that error push your answer?" Encourage specific improvements, not "be more careful".
Common difficulties:
- Forgetting to subtract the mass of the filter paper or container.
- Using the excess reagent to calculate the theoretical yield.
- Writing "human error" without naming a cause or direction.
- Reporting more significant figures than the balance justifies.