How to Calculate Percentage Calcium Carbonate in a Mixture

Two samples of crushed limestone both fizz when hydrochloric acid is added. But one contains mostly calcium carbonate, while the other contains a large amount of quartz sand. The bubbles tell you that a carbonate is reacting. They don’t tell you how much calcium carbonate was originally present.

Suppose you add the same known excess of hydrochloric acid to both samples. Which one should leave more acid behind?

The sandier sample should leave more acid, assuming calcium carbonate is the only substance consuming it. Less calcium carbonate means less acid is used in the reaction.

To calculate the percentage of calcium carbonate, \(\ce{CaCO3}\), in a mixture, react a weighed sample with a known excess of hydrochloric acid, measure the acid left over using sodium hydroxide, and subtract to find the acid consumed. Use the reaction’s mole ratio to calculate the mass of calcium carbonate, then divide by the original mass of the mixture.

This is a back titration. The important distinction is between the acid you added, the acid remaining, and the acid actually consumed by calcium carbonate.

01Why measure the acid left over?

An obvious experiment would be to weigh some powdered limestone and add hydrochloric acid from a burette until the fizzing stops.

You might predict that the volume of acid added at that moment tells you exactly how much calcium carbonate was present.

Unfortunately, fizzing isn’t a reliable endpoint.

A large carbonate grain may react slowly because only its exposed surface contacts the acid. Bubbles can continue escaping after much of the carbonate has reacted. Fine grains may also become trapped beneath an insoluble layer of material.

The absence of visible bubbles does not prove that every calcium carbonate particle has reacted.

There is another problem. If the mixture contains other substances that react with acid, some hydrochloric acid may be consumed without producing carbon dioxide.

The revised experiment: Add excess acid, then measure what remains

Instead of trying to stop at exactly the right amount of acid, add more hydrochloric acid than the calcium carbonate should need.

The experimental arrangement works like this:

1. Weigh a representative, dry sample of the powdered mixture using a balance.
2. Transfer it to a conical flask and add a measured volume of hydrochloric acid of known concentration.
3. Swirl and allow sufficient time for the calcium carbonate to react completely. Carbon dioxide escapes as bubbles.
4. Allow the reaction mixture to settle. If necessary, filter off insoluble material, collecting all the acidic filtrate and rinses.
5. Titrate the remaining hydrochloric acid with sodium hydroxide of known concentration, using an appropriate endpoint.

A burette delivers the sodium hydroxide into the flask containing the acidic solution. The measured sodium hydroxide volume is called the titre.

The original direct-addition method depended on judging when a visible reaction had finished. The revised method replaces that uncertain judgement with a measured amount of acid remaining.

A positive titre of sodium hydroxide indicates that acid was left after the reaction, assuming the titre is measuring hydrochloric acid rather than another acidic species.

The numerical measurements throughout this article are illustrative, not records of actual experiments.

02The three acid quantities you must keep separate

Think of the hydrochloric acid as a measured budget. You know how much you started with and measure how much remains. The difference is what the sample used.

But the analogy has a limit: several different substances can consume the acid. The subtraction reveals the total acid consumed, not necessarily which mineral consumed it.

A common mistake is to treat the sodium hydroxide titre as a direct measurement of the acid consumed by calcium carbonate.

It’s the opposite. The sodium hydroxide measures the hydrochloric acid that survived the first reaction.

First reaction: Calcium carbonate consumes acid

Calcium carbonate reacts with hydrochloric acid:

\[
\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}
\]

One mole of calcium carbonate consumes two moles of hydrochloric acid.

The ratio is therefore:

\[
n(\ce{CaCO3}):n(\ce{HCl})=1:2
\]

Here, \(n\) represents amount of substance, measured in moles (mol).

The carbon dioxide is released as gas. Calcium chloride remains dissolved in the solution.

Second reaction: Sodium hydroxide measures the leftover acid

The remaining hydrochloric acid reacts with sodium hydroxide:

\[
\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}
\]

The mole ratio is \(1:1\).

That means the moles of sodium hydroxide used equal the moles of hydrochloric acid remaining, provided the entire remaining acid solution is titrated and no other substance significantly contributes to the endpoint.

We calculate moles using:

\[
n=cV
\]

Where:

• \(n\) is the amount in mol.
• \(c\) is the concentration in \(\mathrm{mol\,L^{-1}}\).
• \(V\) is the solution volume in litres (L).

Remember that \(1000\ \mathrm{mL}=1\ \mathrm{L}\).

If only a portion of the remaining solution is titrated, that portion is called an aliquot. You must scale its measured acid amount up to the full solution before subtracting.

Turn acid consumption into calcium carbonate percentage

Let:

• \(n_\mathrm{added}\) be the moles of HCl initially added.
• \(n_\mathrm{left}\) be the moles of HCl remaining.
• \(n_\mathrm{used}\) be the total moles of HCl consumed.
• \(n_\mathrm{other}\) be the moles of HCl consumed by substances other than calcium carbonate.

Then:

\[
\begin{aligned}
n_\mathrm{used}
&=n_\mathrm{added}-n_\mathrm{left}\\
n(\ce{CaCO3})
&=\frac{n_\mathrm{used}-n_\mathrm{other}}{2}
\end{aligned}
\]

For a mixture containing calcium carbonate and substances that do not react with hydrochloric acid, \(n_\mathrm{other}=0\).

To convert calcium carbonate moles into mass, use:

\[
m(\ce{CaCO3})=n(\ce{CaCO3})M
\]

Here, \(m\) is mass in grams and \(M\) is molar mass in \(\mathrm{g\,mol^{-1}}\).

For calcium carbonate:

\[
M(\ce{CaCO3})=100.1\ \mathrm{g\,mol^{-1}}
\]

Finally, percentage by mass is:

\[
p=\frac{m(\ce{CaCO3})}{m_\mathrm{sample}}\times100\%
\]

Here, \(p\) is the mass percentage of calcium carbonate and \(m_\mathrm{sample}\) is the original mass of the entire weighed mixture, including its impurities.

That denominator is important. A percentage composition describes how much of the original mixture was calcium carbonate, not how much of the acid solution was used.

04What if another ingredient also reacts with the acid?

Now consider a different powder containing three substances:

• Calcium carbonate, \(\ce{CaCO3}\).
• Magnesium oxide, \(\ce{MgO}\).
• Quartz, \(\ce{SiO2}\).

Under the stated dilute-acid conditions, quartz is essentially unreactive. Calcium carbonate produces carbon dioxide when it reacts with HCl.

But magnesium oxide also consumes acid:

\[
\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)}
\]

It does not produce carbon dioxide.

Would you be able to distinguish calcium carbonate from magnesium oxide using a back titration alone?

No. Both consume two moles of HCl per mole of substance.

The back titration tells you the combined acid consumption. It doesn’t identify the substances responsible.

If you assume all the acid was consumed by calcium carbonate, the calculated percentage will be too high.

05How to decide whether your percentage is trustworthy

The usual acid-subtraction method relies on several experimental assumptions.

The magnesium carbonate case is particularly interesting.

Like calcium carbonate, magnesium carbonate reacts with acid to produce carbon dioxide:

\[
\ce{MgCO3(s) + 2HCl(aq) -> MgCl2(aq) + CO2(g) + H2O(l)}
\]

Both carbonates consume the same number of moles of HCl and produce the same number of moles of carbon dioxide per mole of carbonate.

So even combining a back titration with a measurement of carbon dioxide might not distinguish calcium carbonate from magnesium carbonate.

You would need another independent measurement that responds differently to the two minerals.

Four important experimental checks

1. Was hydrochloric acid genuinely in excess?

If all the acid is consumed before the calcium carbonate finishes reacting, the experiment cannot determine the original calcium carbonate mass. It only measures how much reacted before the acid ran out.

A revised experiment needs a sufficiently large, known excess of acid.

2. Did the entire original sample react?

Large particles or a coating of unreactive material may slow the reaction. If some calcium carbonate remains unreacted, its mass will not be included in the calculated result.

The calculation assumes complete reaction, not merely that bubbling has slowed.

3. Was all the remaining HCl actually measured?

Losing acidic liquid during transfer, filtering, or heating can make the remaining acid appear smaller. The apparent acid consumption then increases, giving an overestimate of calcium carbonate.

Dissolved carbon dioxide can also affect a sodium hydroxide titration, depending on the endpoint. The method must control this interference rather than count carbonic acid as leftover HCl.

4. Does the denominator match the sample being analysed?

If a \(2.00\ \mathrm{g}\) damp mixture dries to \(1.80\ \mathrm{g}\), a percentage of the dry mixture is not the same as a percentage of the original damp mixture.

The numerator and denominator must refer to the same basis.

A result above \(100\%\) is a warning that a measurement, calculation, or assumption is wrong. But a result below \(100\%\) isn’t automatically trustworthy. Other acid-reactive compounds can produce a perfectly plausible but incorrect calcium carbonate percentage.

06Questions and solutions

The following questions increase in difficulty. The first checks the standard method. The later questions require you to decide what the measurements actually establish, including cases where the original experimental design cannot answer the question by itself.

All numerical scenarios are hypothetical.

Question 1

A \(1.000\ \mathrm{g}\) sample of powdered limestone contains calcium carbonate and unreactive sand.

It is reacted completely with \(40.00\ \mathrm{mL}\) of \(0.4000\ \mathrm{mol\,L^{-1}}\) HCl.

The remaining acid is neutralised by \(30.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) NaOH.

Calculate the percentage by mass of calcium carbonate in the limestone.

Solution 1

The limestone contains \(50.1\%\) calcium carbonate by mass.

The sodium hydroxide measures the leftover acid, so subtract it from the acid initially added.

First calculate the HCl added:

\[
\begin{aligned}
n_\mathrm{added}
&=0.4000\times0.04000\\
&=0.01600\ \mathrm{mol}
\end{aligned}
\]

Now calculate the HCl remaining using the \(1:1\) neutralisation ratio:

\[
\begin{aligned}
n_\mathrm{left}
&=0.2000\times0.03000\\
&=0.006000\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
n_\mathrm{used}
&=0.01600-0.006000\\
&=0.01000\ \mathrm{mol}
\end{aligned}
\]

Since calcium carbonate is the only acid-consuming substance:

\[
\begin{aligned}
n(\ce{CaCO3})
&=\frac{0.01000}{2}\\
&=0.005000\ \mathrm{mol}
\end{aligned}
\]

Convert to mass and percentage:

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.005000\times100.1\\
&=0.5005\ \mathrm{g}\\
p&=\frac{0.5005}{1.000}\times100\%\\
&=50.1\%
\end{aligned}
\]

Approximately half the original powder was calcium carbonate.

The tempting mistake is to use the NaOH titre as the acid consumed. That reverses what the second reaction measures.

Question 2

A \(2.000\ \mathrm{g}\) sample of damp limestone powder is dried to constant mass. The dry mass is \(1.840\ \mathrm{g}\).

A thoroughly mixed \(0.9200\ \mathrm{g}\) portion of the dried powder is reacted completely with \(60.00\ \mathrm{mL}\) of \(0.4000\ \mathrm{mol\,L^{-1}}\) HCl.

The entire acidic solution is transferred quantitatively to a \(200.0\ \mathrm{mL}\) volumetric flask and diluted to the mark.

A \(20.00\ \mathrm{mL}\) aliquot requires \(12.00\ \mathrm{mL}\) of \(0.1000\ \mathrm{mol\,L^{-1}}\) NaOH.

Assume drying removes only water and that calcium carbonate is the only acid-consuming solid.

Calculate:

(a) The percentage of calcium carbonate in the dry powder.

(b) The percentage of calcium carbonate in the original damp powder.

(c) Whether forgetting to scale the aliquot would produce an overestimate or underestimate of calcium carbonate.

Solution 2

The dry powder contains \(65.3\%\) calcium carbonate, while the original damp powder contains \(60.1\%\) by mass. Forgetting the aliquot correction would overestimate both percentages.

The central difficulty is that the titre represents only one-tenth of the remaining solution.

(a) Percentage in the dry powder

First calculate the initial HCl:

\[
\begin{aligned}
n_\mathrm{added}
&=0.4000\times0.06000\\
&=0.02400\ \mathrm{mol}
\end{aligned}
\]

The \(12.00\ \mathrm{mL}\) NaOH titre corresponds to:

\[
\begin{aligned}
n_\mathrm{aliquot}
&=0.1000\times0.01200\\
&=0.001200\ \mathrm{mol}
\end{aligned}
\]

The full solution is ten times the aliquot volume:

\[
\frac{200.0}{20.00}=10.00
\]

Therefore, the total remaining HCl is:

\[
\begin{aligned}
n_\mathrm{left}
&=0.001200\times10.00\\
&=0.01200\ \mathrm{mol}
\end{aligned}
\]

Now calculate the HCl consumed:

\[
\begin{aligned}
n_\mathrm{used}
&=0.02400-0.01200\\
&=0.01200\ \mathrm{mol}
\end{aligned}
\]

Use the carbonate reaction ratio:

\[
\begin{aligned}
n(\ce{CaCO3})
&=\frac{0.01200}{2}\\
&=0.006000\ \mathrm{mol}
\end{aligned}
\]

The calcium carbonate mass in the \(0.9200\ \mathrm{g}\) dry portion is:

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.006000\times100.1\\
&=0.6006\ \mathrm{g}
\end{aligned}
\]

Hence:

\[
\begin{aligned}
p_\mathrm{dry}
&=\frac{0.6006}{0.9200}\times100\%\\
&=65.3\%
\end{aligned}
\]

(b) Percentage in the original damp sample

The whole dry sample weighs \(1.840\ \mathrm{g}\), exactly twice the analysed dry portion.

Because the dried powder was thoroughly mixed, the estimated total calcium carbonate mass is:

\[
\begin{aligned}
m_\mathrm{total}
&=0.6006\times2\\
&=1.2012\ \mathrm{g}
\end{aligned}
\]

The original damp sample weighed \(2.000\ \mathrm{g}\), so:

\[
\begin{aligned}
p_\mathrm{damp}
&=\frac{1.2012}{2.000}\times100\%\\
&=60.1\%
\end{aligned}
\]

The damp percentage is lower because the original mass included water.

(c) Effect of forgetting the aliquot correction

Failing to multiply by ten would make the remaining HCl appear much smaller than it actually was.

That would make the calculated acid consumption too large and therefore overestimate the calcium carbonate content.

The deeper issue is matching the scale of the quantities. You cannot subtract the remaining acid in one-tenth of a solution from the acid originally added to the whole sample.

Question 3

A chemist is analysing a powdered mixture containing only calcium carbonate and unreactive sand.

In a first trial, a \(1.250\ \mathrm{g}\) sample receives \(30.00\ \mathrm{mL}\) of \(0.4000\ \mathrm{mol\,L^{-1}}\) HCl.

No excess HCl remains. Solid is still present, and adding a little more HCl causes bubbling to resume.

A second, equally representative \(1.250\ \mathrm{g}\) sample is tested with \(70.00\ \mathrm{mL}\) of the same HCl solution.

After the second reaction finishes, the remaining acid requires \(35.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) NaOH.

Determine:

(a) What the first trial establishes about the original percentage of calcium carbonate.

(b) The calcium carbonate percentage calculated from the second trial.

(c) Why the first trial cannot give the exact percentage, even though all the added HCl reacted.

Solution 3

The first trial establishes a lower bound of \(48.0\%\) calcium carbonate, while the second gives \(84.1\%\). The first trial cannot establish the exact percentage because hydrochloric acid ran out while calcium carbonate was still present.

(a) What the first trial establishes

The amount of HCl added was:

\[
\begin{aligned}
n(\ce{HCl})
&=0.4000\times0.03000\\
&=0.01200\ \mathrm{mol}
\end{aligned}
\]

All of it reacted. The amount of calcium carbonate that reacted was therefore:

\[
\begin{aligned}
n(\ce{CaCO3})
&=\frac{0.01200}{2}\\
&=0.006000\ \mathrm{mol}
\end{aligned}
\]

Its mass was:

\[
\begin{aligned}
m&=0.006000\times100.1\\
&=0.6006\ \mathrm{g}
\end{aligned}
\]

That represents:

\[
\begin{aligned}
p_\mathrm{reacted}
&=\frac{0.6006}{1.250}\times100\%\\
&=48.0\%
\end{aligned}
\]

But extra acid makes the remaining solid fizz, confirming that some calcium carbonate was still unreacted.

Therefore, the original calcium carbonate percentage must have been greater than \(48.0\%\). The measurement does not tell us exactly how much greater.

(b) Percentage from the second trial

The larger amount of HCl initially added was:

\[
\begin{aligned}
n_\mathrm{added}
&=0.4000\times0.07000\\
&=0.02800\ \mathrm{mol}
\end{aligned}
\]

The NaOH titre gives the acid remaining:

\[
\begin{aligned}
n_\mathrm{left}
&=0.2000\times0.03500\\
&=0.007000\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
n_\mathrm{used}
&=0.02800-0.007000\\
&=0.02100\ \mathrm{mol}
\end{aligned}
\]

Convert this into calcium carbonate:

\[
\begin{aligned}
n(\ce{CaCO3})
&=\frac{0.02100}{2}\\
&=0.01050\ \mathrm{mol}\\
m(\ce{CaCO3})
&=0.01050\times100.1\\
&=1.051\ \mathrm{g}
\end{aligned}
\]

Hence:

\[
\begin{aligned}
p&=\frac{1.051}{1.250}\times100\%\\
&=84.1\%
\end{aligned}
\]

(c) Why the first trial is insufficient

In the first trial, the maximum amount of calcium carbonate that could react was fixed by the available HCl.

A sample containing \(0.700\ \mathrm{g}\) of calcium carbonate and one containing \(1.000\ \mathrm{g}\) could both consume the same \(0.01200\ \mathrm{mol}\) of acid if acid ran out first.

Both would leave unreacted calcium carbonate.

So complete consumption of the acid does not imply complete consumption of the calcium carbonate.

The second trial changes the limiting condition: enough acid is supplied for all the carbonate to react, and measurable acid is still left over.

That is why its calculated percentage is meaningful under the stated assumptions.

Question 4

A \(1.400\ \mathrm{g}\) prepared mineral mixture contains calcium carbonate, magnesium oxide, and unreactive quartz.

The entire mixture reacts completely with \(50.00\ \mathrm{mL}\) of \(0.5000\ \mathrm{mol\,L^{-1}}\) HCl.

During the reaction, \(192.0\ \mathrm{mL}\) of dry carbon dioxide is collected without loss.

For the gas conditions used, take the molar volume to be \(24.0\ \mathrm{L\,mol^{-1}}\).

After the reaction, the remaining HCl requires \(27.00\ \mathrm{mL}\) of \(0.2000\ \mathrm{mol\,L^{-1}}\) NaOH.

Assume all reactions are complete and that calcium carbonate is the only substance producing carbon dioxide.

Calculate:

(a) The true percentage of calcium carbonate in the mixture.

(b) The mass of magnesium oxide.

(c) The percentage that would be reported if all the consumed HCl were incorrectly attributed to calcium carbonate.

Explain why the two percentages differ.

Use \(M(\ce{MgO})=40.3\ \mathrm{g\,mol^{-1}}\).

Solution 4

The true calcium carbonate content is \(57.2\%\), the mixture contains \(0.0725\ \mathrm{g}\) of magnesium oxide, and an acid-only calculation would incorrectly report \(70.1\%\) calcium carbonate.

The crucial distinction is that magnesium oxide consumes HCl but does not release CO2.

(a) True calcium carbonate percentage

The calcium carbonate reaction produces one mole of carbon dioxide for every mole of calcium carbonate consumed.

Convert the measured gas volume:

\[
192.0\ \mathrm{mL}=0.1920\ \mathrm{L}
\]

Using \(V=nV_m\), where \(V_m\) is molar gas volume:

\[
\begin{aligned}
n(\ce{CO2})
&=\frac{V}{V_m}\\
&=\frac{0.1920}{24.0}\\
&=0.00800\ \mathrm{mol}
\end{aligned}
\]

Because only calcium carbonate produces CO2:

\[
n(\ce{CaCO3})=0.00800\ \mathrm{mol}
\]

Therefore:

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.00800\times100.1\\
&=0.8008\ \mathrm{g}\\
p&=\frac{0.8008}{1.400}\times100\%\\
&=57.2\%
\end{aligned}
\]

(b) Magnesium oxide mass

First calculate the HCl added:

\[
\begin{aligned}
n_\mathrm{added}
&=0.5000\times0.05000\\
&=0.02500\ \mathrm{mol}
\end{aligned}
\]

The HCl remaining was:

\[
\begin{aligned}
n_\mathrm{left}
&=0.2000\times0.02700\\
&=0.005400\ \mathrm{mol}
\end{aligned}
\]

The total HCl consumed was:

\[
\begin{aligned}
n_\mathrm{used}
&=0.02500-0.005400\\
&=0.01960\ \mathrm{mol}
\end{aligned}
\]

The known calcium carbonate amount consumed:

\[
\begin{aligned}
n_{\ce{HCl,CaCO3}}
&=2(0.00800)\\
&=0.01600\ \mathrm{mol}
\end{aligned}
\]

The remaining acid consumption belongs to magnesium oxide:

\[
\begin{aligned}
n_{\ce{HCl,MgO}}
&=0.01960-0.01600\\
&=0.00360\ \mathrm{mol}
\end{aligned}
\]

Using the \(2:1\) HCl-to-MgO ratio:

\[
\begin{aligned}
n(\ce{MgO})
&=\frac{0.00360}{2}\\
&=0.00180\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
m(\ce{MgO})
&=0.00180\times40.3\\
&=0.0725\ \mathrm{g}
\end{aligned}
\]

(c) Incorrect percentage from acid consumption alone

If all \(0.01960\ \mathrm{mol}\) of consumed HCl were attributed to calcium carbonate:

\[
\begin{aligned}
n_\mathrm{apparent}
&=\frac{0.01960}{2}\\
&=0.00980\ \mathrm{mol}\\
m_\mathrm{apparent}
&=0.00980\times100.1\\
&=0.9810\ \mathrm{g}
\end{aligned}
\]

This would give:

\[
\begin{aligned}
p_\mathrm{apparent}
&=\frac{0.9810}{1.400}\times100\%\\
&=70.1\%
\end{aligned}
\]

That is an overestimate.

The two measurements reveal different information. The back titration measures total HCl consumption, while the gas measurement identifies the amount of calcium carbonate because MgO produces no carbon dioxide.

The gas measurement provides the additional constraint needed to separate the contributions.

This reasoning would fail if another carbonate, such as magnesium carbonate, were also present. That substance would consume HCl and release CO2, so the measured gas would no longer identify calcium carbonate uniquely.

Question 5

A \(1.800\ \mathrm{g}\) powdered mineral mixture contains calcium carbonate, magnesium carbonate, and quartz only.

The mixture is heated under carefully controlled conditions.

For this hypothetical experiment, assume:

• All magnesium carbonate decomposes into magnesium oxide and carbon dioxide.
• Calcium carbonate remains completely unchanged.
• Quartz remains unchanged.
• There is no moisture, no loss of solid particles, and all the mass decrease comes from escaping carbon dioxide.

After heating and cooling, the solid weighs \(1.668\ \mathrm{g}\).

The entire heated residue is then reacted completely with \(40.00\ \mathrm{mL}\) of \(0.6000\ \mathrm{mol\,L^{-1}}\) HCl.

The remaining acid requires \(25.00\ \mathrm{mL}\) of \(0.2400\ \mathrm{mol\,L^{-1}}\) NaOH.

A student argues:

“Heating removed the magnesium carbonate as carbon dioxide, so the heated sample must consume less hydrochloric acid than the original sample would have.”

Another student disagrees:

“If both samples consume the same amount of acid, then heating cannot have removed any carbonate.”

Determine:

(a) The percentage of calcium carbonate in the original mixture.

(b) The percentage of magnesium carbonate in the original mixture.

(c) Whether an identical, unheated portion would consume more HCl, less HCl, or the same amount of HCl as the heated residue.

(d) The moles of CO2 produced when the heated residue reacts with excess HCl, and whether two back titrations alone, one before heating and one after, could determine the calcium carbonate percentage.

Explain what is wrong with each student’s reasoning.

Use:

• \(M(\ce{CaCO3})=100.1\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{MgCO3})=84.3\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}}\)

Solution 5

The original mixture contains \(33.4\%\) calcium carbonate and \(14.1\%\) magnesium carbonate. An identical unheated portion would consume exactly the same amount of HCl as the heated residue, but the heated residue would produce only \(0.00600\ \mathrm{mol}\) of CO2 when reacted with acid.

Neither student’s conclusion is correct. The first overlooks the acid reactivity of magnesium oxide, while the second mistakes unchanged acid consumption for unchanged chemical composition.

(a) Percentage of calcium carbonate

The first important observation is the loss of mass during heating:

\[
\begin{aligned}
\Delta m
&=1.800-1.668\\
&=0.132\ \mathrm{g}
\end{aligned}
\]

The heating reaction is:

\[
\ce{MgCO3(s) -> MgO(s) + CO2(g)}
\]

Every mole of magnesium carbonate decomposed releases one mole of CO2.

Because the mass loss is entirely CO2:

\[
\begin{aligned}
n(\ce{CO2})
&=\frac{0.132}{44.0}\\
&=0.00300\ \mathrm{mol}
\end{aligned}
\]

Therefore:

\[
n(\ce{MgCO3})=0.00300\ \mathrm{mol}
\]

After heating, this magnesium carbonate has become \(0.00300\ \mathrm{mol}\) of magnesium oxide.

Now consider the back titration of the heated residue.

The HCl added was:

\[
\begin{aligned}
n_\mathrm{added}
&=0.6000\times0.04000\\
&=0.02400\ \mathrm{mol}
\end{aligned}
\]

The acid remaining was:

\[
\begin{aligned}
n_\mathrm{left}
&=0.2400\times0.02500\\
&=0.006000\ \mathrm{mol}
\end{aligned}
\]

So the heated residue consumed:

\[
\begin{aligned}
n_\mathrm{used}
&=0.02400-0.006000\\
&=0.01800\ \mathrm{mol}
\end{aligned}
\]

But this acid reacted with two substances: calcium carbonate and magnesium oxide.

The magnesium oxide consumes:

\[
\begin{aligned}
n_{\ce{HCl,MgO}}
&=2(0.00300)\\
&=0.00600\ \mathrm{mol}
\end{aligned}
\]

Therefore, the HCl consumed by calcium carbonate was:

\[
\begin{aligned}
n_{\ce{HCl,CaCO3}}
&=0.01800-0.00600\\
&=0.01200\ \mathrm{mol}
\end{aligned}
\]

Calculate the calcium carbonate moles:

\[
\begin{aligned}
n(\ce{CaCO3})
&=\frac{0.01200}{2}\\
&=0.006000\ \mathrm{mol}
\end{aligned}
\]

The mass of calcium carbonate was:

\[
\begin{aligned}
m(\ce{CaCO3})
&=0.006000\times100.1\\
&=0.6006\ \mathrm{g}
\end{aligned}
\]

Heating did not change the calcium carbonate, so this was also its original mass.

Use the original \(1.800\ \mathrm{g}\) sample as the denominator:

\[
\begin{aligned}
p(\ce{CaCO3})
&=\frac{0.6006}{1.800}\times100\%\\
&=33.4\%
\end{aligned}
\]

(b) Original percentage of magnesium carbonate

We already established that the original sample contained \(0.00300\ \mathrm{mol}\) of MgCO3.

Using its molar mass:

\[
\begin{aligned}
m(\ce{MgCO3})
&=0.00300\times84.3\\
&=0.2529\ \mathrm{g}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
p(\ce{MgCO3})
&=\frac{0.2529}{1.800}\times100\%\\
&=14.1\%
\end{aligned}
\]

As a consistency check, the remaining original mass must have been quartz:

\[
\begin{aligned}
m(\ce{SiO2})
&=1.800-0.6006-0.2529\\
&=0.9465\ \mathrm{g}
\end{aligned}
\]

All three original components have positive masses that add to \(1.800\ \mathrm{g}\).

(c) Would the unheated mixture consume more acid?

No. Under the specified assumptions, the unheated mixture and the heated residue consume the same \(0.01800\ \mathrm{mol}\) of HCl.

This is the conceptual hinge.

Before heating, magnesium carbonate reacts with acid:

\[
\ce{MgCO3(s) + 2HCl(aq) -> MgCl2(aq) + CO2(g) + H2O(l)}
\]

After heating, magnesium oxide reacts with acid:

\[
\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)}
\]

In both reactions, one mole of the magnesium-containing compound consumes two moles of HCl.

Heating removes carbon dioxide from magnesium carbonate, but leaves magnesium oxide behind.

The sample loses mass without losing the acid-neutralising capacity associated with those magnesium-containing formula units.

The first student assumed that magnesium carbonate disappearing meant its entire ability to consume acid disappeared.

But magnesium oxide is still present and reactive.

The second student correctly noticed that the acid consumption could remain unchanged, but incorrectly treated this as proof that no carbonate had decomposed.

Identical titration results do not necessarily mean identical compositions.

(d) Carbon dioxide from the heated residue

Only calcium carbonate remains capable of releasing carbon dioxide when the heated residue is treated with HCl.

The relevant reaction has a \(1:1\) ratio between calcium carbonate and carbon dioxide.

Therefore:

\[
\begin{aligned}
n(\ce{CO2})
&=n(\ce{CaCO3})\\
&=0.00600\ \mathrm{mol}
\end{aligned}
\]

An identical unheated sample would have produced:

\[
\begin{aligned}
n(\ce{CO2})
&=0.00600+0.00300\\
&=0.00900\ \mathrm{mol}
\end{aligned}
\]

The heated residue therefore produces less CO2 during its acid reaction, even though its acid consumption is unchanged.

Two back titrations alone would not distinguish the calcium carbonate from the magnesium carbonate in this experiment.

Before heating, both carbonates consume two moles of HCl per mole. After heating, calcium carbonate and magnesium oxide also consume two moles of HCl per mole.

The titrations provide the same combined acid-consumption information.

The heating mass loss supplies the independent measurement. It reveals the original magnesium carbonate amount, allowing its contribution to be separated from calcium carbonate.

This is why the original sample can have a precisely determined calcium carbonate percentage even though the acid-neutralising capacity does not change during heating.

07The next decision: What does your measurement actually identify?

For a mixture containing calcium carbonate and acid-inert material, a standard back titration is enough to calculate the calcium carbonate mass percentage.

But the calculation only works when the consumed HCl can be attributed to calcium carbonate, or when the contribution from other acid-reactive substances is independently known.

Before using the formula, decide what your measurement represents:

• A NaOH titre measures leftover acid.
• Subtracting from the acid added gives total acid consumption.
• The balanced equation converts acid consumption into calcium carbonate only when other acid consumers have been excluded or accounted for.
• The original sample mass determines the final percentage.

If you measure carbon dioxide instead, you are using a different chemical constraint. The relationship between carbonate moles, gas moles, and collected gas volume is developed in How to Calculate CO₂ Volume from a Carbonate Reaction.

The useful habit is to ask one question before calculating: does this experiment measure calcium carbonate specifically, or does it measure a property that several ingredients share?