Did you know that a limestone sample weighing 2.50 g may contain only about 2.00 g of calcium carbonate? If the sample is impure, the full mass on the balance is not automatically the mass of the chemical species reacting in your equation. For a sample known to contain only calcium carbonate and inert impurities, you can infer the CaCO₃ content from the amount of CO₂ formed when the carbonate reacts with acid, or from the amount of HCl actually consumed in a properly designed back titration.
This is one of the most useful skills for students searching for an H2 Chemistry tutor in Hougang, percentage purity calculation, calcium carbonate limestone with hydrochloric acid, volume of carbon dioxide, excess acid back titration, mole ratios, 9476 stoichiometry and JC1 chemistry worked examples.
The basic chemical equation is familiar:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l).
But a correct answer needs more than a balanced equation. The student must distinguish total impure sample mass from mass of actual CaCO₃, then identify whether the measured result represents carbon dioxide produced, hydrochloric acid added or hydrochloric acid remaining.
I like to teach this through two independent theoretical ways of analysing the same hypothetical sample. The first measures gas. The second uses excess acid and determines how much was left. Both should lead to the same percentage purity under the deliberately simplified assumptions; their experimental limitations are different.
At eduKate, the aim is to diagnose the first incorrect mole-concept assumption and move to changed, independent problems in our small groups of no more than three learners. Families travelling from Hougang can check current JC Chemistry lesson availability, physical venue and timetable directly.
Hougang Chemistry learning library · Check your SEAB H2 Chemistry syllabus
First define percentage purity correctly
Percentage purity by mass describes how much of a sample’s total mass is made up of the desired substance.
The formula is:
Percentage purity = (mass of desired pure substance / total sample mass) × 100%.
If an impure limestone sample has mass 2.50 g, and later analysis shows it contains 2.00 g of CaCO₃, then:
Percentage purity ≈ (2.00/2.50) × 100% = 80.0%.
This calculation only works if the pure substance’s mass is known or has been correctly inferred from the experimental data.
Do not put the mass of an entire gas product in the numerator. A product like CO₂ contains oxygen and carbon in a different chemical combination from the original CaCO₃, so its mass is not the original carbonate mass.
Likewise, do not divide the moles of HCl added directly by the sample mass. Those are different physical quantities, and the acid may have been in excess.
Read the balanced chemical equation as an atom ledger
The reaction is:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
From the stoichiometric coefficients:
- 1 mol CaCO₃ reacts with 2 mol HCl.
- 1 mol CaCO₃ produces 1 mol CO₂.
- 2 mol HCl consumed correspond to 1 mol CaCO₃.
These are the two ratios used in our worked analyses.
If a student obtains 0.0200 mol CO₂, the corresponding CaCO₃ amount is 0.0200 mol, not 0.0400 mol.
If an experiment establishes that 0.0400 mol HCl was consumed by the carbonate, the CaCO₃ amount is 0.0400/2 = 0.0200 mol.
The factor 2 belongs to the acid-to-carbonate ratio, not the carbonate-to-carbon-dioxide ratio.
This is why the half-minute spent reading coefficients can save several later marks.
Worked method A: use the carbon dioxide produced
Suppose a hypothetical impure limestone sample weighs:
m(sample) = 2.50 g.
It reacts completely with enough hydrochloric acid to consume all its CaCO₃.
Assume its impurities do not react with acid, do not produce gas and do not prevent the calcium carbonate from reacting. Assume all produced CO₂ is collected and corrected appropriately for its measurement conditions.
The problem states that the collected CO₂ amount corresponds to 0.0200 mol.
How much CaCO₃ was present?
From the 1:1 mole relationship:
n(CaCO₃) = n(CO₂) = 0.0200 mol.
Using a relative formula mass of approximately:
Mr(CaCO₃) = 40.1 + 12.0 + 3(16.0) = 100.1,
the pure carbonate mass is:
m(CaCO₃) = 0.0200 × 100.1 = 2.002 g.
To three significant figures:
m(CaCO₃) ≈ 2.00 g.
Then:
Percentage purity = (2.002/2.50) × 100% = 80.08% ≈ 80.1%.
That is the correct percentage purity under the stated assumptions.
Notice that the pure CaCO₃ mass is less than the original sample mass, as expected for a mixture containing some non-carbonate material.
If the gas amount is given as volume
Sometimes the question gives a volume of CO₂, rather than the number of moles.
Suppose the stated gas conditions allow a molar gas volume of 24.0 dm³ mol⁻¹. If CO₂ collected is 480 cm³, convert to dm³:
V = 480/1000 = 0.480 dm³.
Then:
n(CO₂) = 0.480/24.0 = 0.0200 mol.
The calculation returns the same 0.0200 mol CO₂ and therefore the same 80.1% purity for the 2.50 g limestone sample.
The molar gas volume depends on temperature and pressure. It is a common teaching approximation at specified room conditions, not an exact universal number.
If the actual exam supplies gas temperature and pressure instead, use the appropriate pV = nRT relationship and ensure the gas pressure corresponds to the species of interest.
Never combine volumes measured at one condition with a molar gas volume from another condition without checking the assumptions.
Why gas collection can underestimate purity
Imagine the sample genuinely contained enough CaCO₃ to generate 0.0200 mol CO₂, but the apparatus collected only 0.0180 mol because some CO₂ escaped or remained dissolved in the liquid.
Using the incomplete recovered amount:
n(CaCO₃,inferred) = 0.0180 mol.
Then:
m(CaCO₃,inferred) = 0.0180 × 100.1 ≈ 1.80 g.
The inferred purity becomes:
1.80/2.50 × 100% ≈ 72.1%.
That is lower than the original illustrative 80.1%.
The chemistry did not make the sample less pure. The measurement failed to account for all the CO₂ assumed by the analysis.
This is the kind of evaluation that a practical Chemistry question can ask: identify how a specific measurement limitation biases the calculated percentage.
A vague answer such as “human error” is less useful than naming the lost CO₂ and tracing its effect through the mole ratio.
What if another carbonate is present?
Real limestone can contain components beyond pure CaCO₃ and chemically inert impurities. For example, magnesium carbonate or other acid-reactive substances may be present in some natural mineral samples.
A carbonate impurity may also generate CO₂ when acid is added. In that case, total CO₂ collected is not uniquely attributable to CaCO₃.
If a student takes all gas moles as CaCO₃ moles, they may overestimate calcium carbonate content, because some gas came from a different carbonate.
This is an important limit on the method. The worked numerical answer is valid only because the hypothetical problem explicitly assumes the impurities are inert and do not generate CO₂.
A quantitative analysis of a real mineral may need additional compositional data or a more selective method. The same observation can have several chemical sources.
Worked method B: add known excess HCl and titrate what remains
Instead of collecting CO₂, another hypothetical experiment analyses the same type of limestone sample by a back titration.
We know the amount of hydrochloric acid added at the start. More is supplied than needed to react with the CaCO₃. After the carbonate reaction is complete, the amount of acid remaining is determined using a suitable standard base.
The conceptual equation is:
HCl initially added − HCl remaining = HCl consumed by the sample reaction.
Then the CaCO₃ moles are found from the 2:1 HCl:CaCO₃ stoichiometric ratio.
This method is especially useful when directly observing an endpoint between a solid and acid is inconvenient. But it still depends on correct assumptions about complete reaction, acid loss and other substances consuming acid.
Back titration is not simply a harder forward titration. It is a material-balance method that identifies the portion of a reagent consumed indirectly.
Set up our theoretical back-titration data
Suppose another sample from the same hypothetical uniform limestone material weighs:
2.50 g.
The sample is treated with:
100.0 cm³ of 0.600 mol dm⁻³ HCl.
We assume it contains only CaCO₃ and impurities inert to HCl, and that the carbonate reaction goes to completion.
After the reaction, the remaining solution is quantitatively transferred and made up to a final volume of 250.0 cm³, as specified by the hypothetical method.
A 25.0 cm³ aliquot of this final solution requires:
10.0 cm³ of 0.200 mol dm⁻³ NaOH for its stipulated neutralisation endpoint.
The problem asks for the original sample’s percentage purity.
This is a fictional data set designed to give the same result as the gas method. It is not a home reaction procedure or an actual lab observation.
Step 1: calculate the HCl added initially
Convert HCl volume to dm³:
100.0 cm³ = 0.1000 dm³.
Moles initially supplied:
n(HCl,initial) = 0.600 × 0.1000 = 0.0600 mol.
That is the total acid added, not the amount consumed by CaCO₃.
This distinction is essential because the question deliberately tells us that HCl was added in excess.
If a student treats 0.0600 mol as the amount consumed by the carbonate, they would calculate:
0.0600/2 = 0.0300 mol CaCO₃,
corresponding to about 3.00 g CaCO₃. That is impossible for a total sample weighing only 2.50 g under our assumptions.
The impossible purity above 100% is a valuable reasonableness check. It reveals that the student forgot to subtract the remaining HCl.
Step 2: calculate the acid in the 25.0 cm³ aliquot
The neutralisation between HCl and NaOH is:
HCl + NaOH → NaCl + H₂O.
Its mole ratio is 1:1.
Convert the NaOH titre:
10.0 cm³ = 0.0100 dm³.
Moles of NaOH used:
n(NaOH) = 0.200 × 0.0100 = 0.00200 mol.
Therefore the 25.0 cm³ aliquot contains:
n(HCl,aliquot) = 0.00200 mol.
The crucial interpretation is that the NaOH titre determines the acid remaining, not the amount of CaCO₃ or the acid originally added.
The 25.0 cm³ aliquot is only one tenth of the 250.0 cm³ total solution. We must account for the remaining nine tenths.
Step 3: find the HCl remaining in the total solution
The final solution has volume 250.0 cm³. We analysed 25.0 cm³.
The scale factor is:
250.0/25.0 = 10.0.
Therefore:
n(HCl,remaining total) = 0.00200 × 10.0 = 0.0200 mol.
This is the amount of hydrochloric acid present after the reaction, in the total analysed solution.
If a student uses only the aliquot value of 0.00200 mol as the entire remaining amount, they will subtract too little acid and dramatically overestimate the carbonate content.
That is not an error in acid chemistry; it is an aliquot scaling error. The repair is to make a diagram of the full volumetric flask and the smaller portion transferred for titration.
Step 4: find HCl actually consumed
We have:
Initial HCl = 0.0600 mol.
Remaining HCl = 0.0200 mol.
Subtract:
n(HCl,consumed) = 0.0600 − 0.0200 = 0.0400 mol.
That 0.0400 mol is the acid assumed to have reacted with CaCO₃ in our simplified sample.
Now use:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
One mole CaCO₃ consumes two moles HCl, so:
n(CaCO₃) = 0.0400/2 = 0.0200 mol.
The same CaCO₃ amount was independently inferred from the gas method.
The back titration has used what acid was consumed instead of what gas formed, but both rely on the same balanced equation.
Step 5: calculate mass and purity
Using Mr(CaCO₃) = 100.1:
m(CaCO₃) = 0.0200 × 100.1 = 2.002 g.
Now compare this to the original impure sample mass, 2.50 g:
Percentage purity = (2.002/2.50) × 100% ≈ 80.1%.
Our answer is:
Approximately 80.1% calcium carbonate by mass.
A useful final check is that the theoretical amount of CaCO₃ is 0.0200 mol, which could generate exactly 0.0200 mol CO₂.
Therefore the two independent numerical pathways agree under the same ideal assumptions.
This agreement does not prove that real measurements would necessarily match exactly. It shows that the hypothetical data were constructed to be chemically self-consistent.
Compare the two methods directly
| Question | Gas method | Back-titration method |
|---|---|---|
| Measured quantity | CO₂ amount or volume | Acid remaining after sample reaction |
| Key stoichiometry | 1 mol CaCO₃ → 1 mol CO₂ | 1 mol CaCO₃ consumes 2 mol HCl |
| Main first calculation | Convert gas to moles | Calculate HCl initially added |
| Important second calculation | Convert CO₂ moles to CaCO₃ moles | Scale remaining acid aliquot and subtract |
| Common error | Confuse gas mass with CaCO₃ mass | Use acid added instead of acid consumed |
| Example purity | 80.1% | 80.1% |
A student should be able to start either method from the measured quantity without attempting to force all questions into the same formula.
This is why mole concept and balanced reactions are more reliable than memorising a giant catalogue of numerical shortcuts.
Why a carbonate impurity and a basic impurity affect methods differently
Suppose the limestone contains another carbonate that also produces CO₂. The gas method may wrongly attribute some of its CO₂ to CaCO₃, potentially overestimating CaCO₃.
Suppose instead that the sample contains another acid-consuming substance that does not produce CO₂, such as an appropriate metal oxide impurity. A back titration could attribute too much acid consumption to CaCO₃, again potentially overestimating CaCO₃.
The two methods therefore have different chemical interference profiles.
A real laboratory might combine independent analyses to assess a complicated sample. But no calculation can recover a uniquely correct CaCO₃ purity if the question omits information needed to distinguish all interfering reactions.
The problem’s assumptions are not incidental. They are what make the inference from measured quantity to desired substance valid.
What if CO₂ dissolves in water?
CO₂ is soluble in water to some extent and can participate in aqueous acid–base equilibria. If the gas is collected by displacement of water, the measured gas volume may be less than the total CO₂ produced because some remains dissolved.
If the calculation assumes full gas collection, the CaCO₃ amount inferred from the measured gas is then biased downward.
A gas sample collected over water may also include water vapour, so using total pressure as though it were pure CO₂ pressure can introduce another bias. In a high-precision gas-law problem, the pressure of water vapour may need to be accounted for:
p(CO₂) = p(total gas) − p(H₂O vapour),
under the appropriate gas-mixture conditions.
In practical Chemistry, a question should specify what corrections or idealisations to use. Do not invent a vapour-pressure correction when the problem already gives the dry CO₂ amount, but do not ignore it when the question explicitly says wet gas was collected.
The same mass can produce different gas amounts at different purity
Imagine two hypothetical 2.50 g samples:
Sample A: 100% CaCO₃.
Sample B: 80% CaCO₃, with inert impurity.
For A:
n(CaCO₃) ≈ 2.50/100.1 ≈ 0.02498 mol.
It could theoretically generate approximately 0.02498 mol CO₂ if fully reacted.
For B:
m(CaCO₃) = 2.50 × 0.800 = 2.00 g, so:
n(CaCO₃) ≈ 2.00/100.1 ≈ 0.01998 mol.
It could produce approximately 0.01998 mol CO₂.
The difference in CO₂ is not caused by the total sample masses; those are equal. It comes from their different amounts of reacting CaCO₃.
That is why it is dangerous to assume the given sample mass is the mass of the desired compound.
Purity is not the same as percentage yield
Percentage purity describes the fraction of a sample made up of the target substance.
Percentage yield compares the actual amount of a reaction product obtained with the theoretical amount that should be possible from the limiting reactant under the stated stoichiometry.
A sample can be very pure but produce a poor isolated yield in a separate reaction due to incomplete recovery. A crude mixture can have low purity while a reaction earlier in the manufacturing process achieved high chemical conversion.
For limestone analysis, we are interested in what percentage of the original sample mass was CaCO₃.
That is different from asking how efficiently the laboratory collected CO₂. Gas recovery efficiency can affect our measurement of purity, but it is not the purity definition itself.
Keeping those quantities separate prevents a common mistake in data interpretation.
Why reaction completion must be checked
Our methods assumed every relevant CaCO₃ portion reacted completely.
If some carbonate remained unreacted because of limited contact, reaction conditions or an insufficient amount of acid, the produced gas or consumed acid would be lower than expected.
That could cause an underestimation of CaCO₃ purity under the simple inference.
Conversely, if an unintended side reaction consumed HCl, the back titration could overestimate carbonate content when the extra acid consumption was attributed to CaCO₃.
Before claiming a calculated purity is accurate, a chemist needs confidence that the measured quantity actually corresponds to the intended stoichiometric reaction.
This is why laboratory experimental design matters, not merely the arithmetic of the final answer.
A worked reversed question: sample mass from target purity
Suppose a limestone product is specified as 75.0% CaCO₃ by mass. How much of the impure material contains 0.0300 mol of CaCO₃?
First convert desired pure carbonate amount to mass:
m(CaCO₃) = 0.0300 × 100.1 = 3.003 g.
Let total impure sample mass be M.
Because purity is 75.0%:
3.003/M = 0.750.
Rearrange:
M = 3.003/0.750 = 4.004 g.
So the required impure sample mass is approximately:
4.00 g.
The fraction works in the opposite direction from the original analysis: to obtain a specified amount of pure material from a sample that is only 75% target compound, we need a larger total sample mass.
A student who multiplies by 0.750 instead of dividing gets a sample smaller than the required pure carbonate mass, which is physically inconsistent.
A practical reasonableness check
Whenever calculating percentage purity, ask:
- Is the inferred desired-substance mass non-negative?
- Does it exceed the original sample mass?
- Is the final percentage between 0% and 100% under the model?
- Is the measured gas or acid consumption consistent with the balanced mole ratio?
- Could other species produce the same experimental signal?
- Are sample mass, gas volume and solution concentration expressed in consistent units?
A result outside 0–100% is often a warning of a setup error or incompatible data and assumptions. It is not corrected by simply rounding it back into range.
One physically plausible result is not proof the method was correct, either. A calculation can yield 80% for the wrong chemical reason by coincidence. Verify the intermediate mole amounts.
The most common mistakes
| Student’s response | First error | Repair question |
|---|---|---|
| 2.50 g limestone means 2.50 g CaCO₃ | Ignored impurity | What fraction of the sample is the desired compound? |
| 0.0200 mol CO₂ means 0.0400 mol CaCO₃ | Used the wrong stoichiometric ratio | What is CaCO₃:CO₂ in the balanced reaction? |
| All added HCl reacted with carbonate | Ignored excess acid | How much HCl remained after reaction? |
| NaOH titre measures CaCO₃ directly | Misidentified the back-titration step | Which species was in the final aliquot? |
| Uses 25 cm³ aliquot moles as whole 250 cm³ flask amount | Forgot scaling | What fraction of the original solution did you titrate? |
| Every bubble means only CaCO₃ reacted | Ignored interfering carbonates or other gas-producing substances | Could another sample component produce CO₂? |
| Gas collected equals gas produced exactly | Ignored leakage, dissolution and wet-gas conditions | What did the apparatus actually measure? |
| Purity and yield are identical percentages | Confused composition and recovered product | Is the numerator sample content or product obtained? |
Five independent practice questions
- A 3.00 g impure CaCO₃ sample containing only inert impurities generates 0.0240 mol CO₂ on complete reaction. Use Mr(CaCO₃) = 100.0. Find its purity.
- A 2.00 g sample contains 0.0150 mol CaCO₃. Find the carbonate mass and percentage purity using Mr = 100.0.
- A carbonate sample reacts with 0.0500 mol HCl, and 0.0180 mol HCl remains. How many moles HCl reacted?
- Using Question 3, calculate the CaCO₃ moles consumed if only CaCO₃ reacts with HCl in a 1:2 molar ratio.
- Why would a magnesium carbonate impurity invalidate a simple “all CO₂ came from CaCO₃” assumption?
Answers
1. m(CaCO₃) = 0.0240 × 100.0 = 2.40 g; purity = 2.40/3.00 × 100 = 80.0%. 2. 0.0150 × 100.0 = 1.50 g; purity = 1.50/2.00 × 100 = 75.0%. 3. 0.0500 − 0.0180 = 0.0320 mol HCl. 4. 0.0320/2 = 0.0160 mol CaCO₃. 5. MgCO₃ can also release CO₂ with acid, so some gas may originate from the impurity rather than CaCO₃.
Question 5 is as important as the numerical answers. A student who can calculate 80% but cannot describe what the method measures is not yet ready to evaluate a real chemical analysis.
A one-week JC1–JC2 learning sequence
- Session 1: practise CaCO₃ + HCl mole ratios and identify which product contains carbon from the carbonate.
- Session 2: convert CO₂ gas moles or volume into CaCO₃ amount and mass.
- Session 3: calculate percentage purity from the inferred pure mass and original sample mass.
- Session 4: learn why back titration measures an excess reagent remaining rather than the reactant directly.
- Session 5: scale a 25.0 cm³ aliquot to the full volumetric solution and subtract acid moles.
- Session 6: compare gas loss, interfering impurities and acid-consuming side reactions.
- Session 7: solve a changed purity dataset independently and explain every calculation step aloud.
This is a better learning route than starting with a long formula sheet. The student gains a clear picture of where each mole quantity came from and when it is legitimate to identify it with CaCO₃.
What parents seeking H2 Chemistry tuition around Hougang can ask
Bring one wrongly answered percentage-purity or back-titration question and ask the tutor whether the student misread the excess reagent, omitted aliquot scaling, confused gas and carbonate moles, or used total sample mass as pure substance mass.
A three-student tutorial can use different datasets for each learner and compare the reasoning only after everyone has made a first attempt. Private lessons should also finish with an unfamiliar variant completed independently.
Confirm the tutor’s current H2 Chemistry 9476 knowledge, real location, schedule, class availability and travel needs. An educational article targeting Hougang searches is not proof that a physical Chemistry classroom operates there.
Why this matters outside school
Analytical chemists, geologists, environmental scientists and industrial materials engineers routinely distinguish total sample mass from the amount of a particular reactive component.
Real materials are mixtures. A sound chemical analysis needs a selective signal, justified reaction stoichiometry, calibrated measurements and explicit treatment of uncertainty.
Learning to see that chain of reasoning prepares JC students for more demanding practical data and later university science without promising any particular grade or admission outcome.
Frequently asked questions
How do I calculate limestone percentage purity?
Find the mass of CaCO₃ actually present from a justified chemical measurement, then divide by the total limestone sample mass and multiply by 100%.
Can CO₂ volume determine the amount of CaCO₃?
Yes under the appropriate conditions if all gas is collected and CaCO₃ is the only species producing that CO₂, using the 1:1 CaCO₃:CO₂ mole ratio.
Why is HCl added in excess for back titration?
The excess allows the carbonate to react; the remaining acid can then be measured by titration and subtracted from the known original amount.
Why do I multiply aliquot moles by a volume factor?
Because a measured aliquot contains only a known fraction of the total analysed solution.
What if limestone contains another carbonate?
The gas or acid-consumption signal may no longer be specific to CaCO₃. Further information or a more selective method may be necessary.
Is percentage purity the same as percentage yield?
No. Purity describes desired-substance content of a sample; yield compares product actually obtained with the theoretical amount possible from a reaction.
Continue learning
Back titration of antacids using HCl and NaOH · JC1 limiting reagents, yield and percentage purity · Gas collection over water and vapour pressure · Finding oxygen by difference in combustion analysis · Why concordant titres may still be inaccurate · Hougang Chemistry Tutor Library · eduKate Singapore H2 Chemistry enquiries.
The limestone and titration values are hypothetical teaching examples. Laboratory acids and gas-collection apparatus require properly trained supervision and approved safety procedures. Real mineral analyses require validation of the stated impurity assumptions.