How to Find the Percentage Purity of a Carbonate from Carbon Dioxide Produced

A 2.50 g sample of crushed limestone reacts with excess hydrochloric acid and produces 0.496 L of carbon dioxide. How much of the original sample was actually calcium carbonate, and how much was impurity?

If the gas was measured at 25.0°C and 100 kPa, the answer is 80.1% calcium carbonate by mass. The calculation works backwards from the carbon dioxide produced to the amount, and then the mass, of calcium carbonate in the original sample.

The method is:

Volume of CO₂ → moles of CO₂ → moles of CaCO₃ → mass of CaCO₃ → percentage purity.

Before using this method, you need to establish two things: the acid was in excess, and all the carbon dioxide produced was collected. Otherwise, the gas measurement might underestimate the amount of carbonate in the sample.

01How does the experiment measure carbonate purity?

Imagine two 2.50 g samples of solid.

• Sample A is 100% calcium carbonate.
• Sample B is 80% calcium carbonate, with the remaining 20% consisting of material that does not produce carbon dioxide.

Both samples have the same total mass. Which one should produce more carbon dioxide when treated with excess hydrochloric acid?

Sample A should produce more. It contains more calcium carbonate particles available to react.

This gives us a way to estimate purity: measure the carbon dioxide produced, work backwards to calculate the mass of calcium carbonate, and compare that mass with the original sample mass.

Setting up the experiment

Simply dropping limestone into an open beaker of acid makes the reaction visible, but it does not measure the gas accurately. The carbon dioxide escapes into the room, and counting bubbles is unreliable because bubbles vary in size and can merge.

A better design uses a gas syringe.

The arrangement is:

1. Accurately weigh the impure calcium carbonate sample.
2. Place the solid in a conical flask.
3. Place a measured amount of excess hydrochloric acid in a small separate container inside the flask.
4. Fit a stopper connected to a delivery tube and gas syringe.
5. Tip the flask to mix the acid and solid.
6. Allow the reaction to finish, then record the volume of carbon dioxide collected.

The acid and solid are kept separate until the apparatus is connected. This prevents gas escaping before measurement begins.

The gas syringe measures the volume of carbon dioxide produced. Its reading must be interpreted at the stated temperature and pressure.

The expected observation is straightforward: a sample containing less calcium carbonate should produce less carbon dioxide, provided the reaction conditions are identical.

However, this prediction has an important limitation. A smaller gas volume could also result from insufficient acid, incomplete reaction, or gas escaping from the apparatus. We need to control these possibilities before interpreting the result as purity.

02Start with the balanced chemical equation

Calcium carbonate reacts with hydrochloric acid according to:

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

Look carefully at the coefficients.

The equation tells us that:

• 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid.
• 1 mole of calcium carbonate produces 1 mole of carbon dioxide.
• 2 moles of hydrochloric acid produce 1 mole of carbon dioxide.

The most important relationship for calculating purity is:

\[ n(\ce{CaCO3}) : n(\ce{CO2}) = 1:1 \]

This is a mole ratio, not a mass ratio.

Think of the coefficients as a recipe. For every one mole of calcium carbonate consumed, the reaction produces one mole of carbon dioxide. However, one mole of calcium carbonate has a mass of 100.1 g, while one mole of carbon dioxide has a mass of only 44.0 g.

The number of moles matches. The masses do not.

This is why you must convert measurements into moles before applying the coefficients.

03The four steps for calculating percentage purity

The method is easiest when you keep the calculations in the correct order.

Step 1: Convert the carbon dioxide measurement into moles

If the question gives you a gas volume at 25°C and 100 kPa, you can use the molar volume:

\[ V_m=24.8\ \mathrm{L\,mol^{-1}} \]

The amount of carbon dioxide is:

\[ n(\ce{CO2})=\frac{V(\ce{CO2})}{V_m} \]

Here:

• \(n\) is the amount of substance, measured in mol.
• \(V\) is the gas volume, measured in L.
• \(V_m\) is the molar volume, measured in L mol⁻¹.

For other gas conditions, use the ideal gas equation:

\[ PV=nRT \]

where \(P\) is pressure, \(V\) is volume, \(n\) is amount, \(R\) is the gas constant, and \(T\) is absolute temperature in kelvin.

Step 2: Convert moles of CO₂ into moles of CaCO₃

Return to the balanced equation.

The mole ratio is 1:1, so:

\[ n(\ce{CaCO3})=n(\ce{CO2}) \]

This assumes that all the calcium carbonate has reacted and the acid was in excess.

Step 3: Convert moles of CaCO₃ into mass

Use the relationship:

\[ m=nM \]

where:

• \(m\) is mass in g;
• \(n\) is amount in mol;
• \(M\) is molar mass in g mol⁻¹.

The molar mass of calcium carbonate is:

\[ \begin{aligned} M(\ce{CaCO3})&=40.1+12.0+3(16.0)\\ &=100.1\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Therefore:

\[ m(\ce{CaCO3})=n(\ce{CaCO3})\times100.1 \]

Step 4: Calculate percentage purity

Now compare the mass of pure calcium carbonate with the original mass of the impure sample.

\[ \boxed{ \%\text{ purity}= \frac{m(\ce{CaCO3})} {m(\text{original sample})} \times100 } \]

The denominator is the mass you weighed before the reaction, not the mass of the solid remaining afterwards.

A useful way to organise the entire method is:

Let’s apply this method to an actual calculation.

04Worked examples

05The major misconception: Assuming all the carbonate reacted

There is one condition that can make an otherwise perfect calculation completely wrong.

The hydrochloric acid must be in excess.

Why?

Return to the balanced equation:

\[ \ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2} \]

Two moles of hydrochloric acid are required for every mole of calcium carbonate.

If the acid runs out first, some calcium carbonate remains unreacted. The measured carbon dioxide then tells you how much carbonate reacted, not how much carbonate was originally present.

Consider a contrasting example.

A 2.00 g sample of pure calcium carbonate is added to 20.0 mL of 0.500 mol L⁻¹ hydrochloric acid.

The amount of hydrochloric acid available is:

\[ \begin{aligned} n(\ce{HCl})&=cV\\ &=0.500\times0.0200\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The acid can produce a maximum of:

\[ n(\ce{CO2})=\frac{0.0100}{2} =0.00500\ \mathrm{mol} \]

At 25°C and 100 kPa, this corresponds to:

\[ \begin{aligned} V(\ce{CO2})&=0.00500\times24.8\\ &=0.124\ \mathrm{L} \end{aligned} \]

Now imagine someone measures 0.124 L of carbon dioxide and works backwards.

They obtain:

\[ \begin{aligned} n(\ce{CaCO3})&=0.00500\ \mathrm{mol}\\ m(\ce{CaCO3})&=0.00500\times100.1\\ &=0.5005\ \mathrm{g} \end{aligned} \]

They might conclude that the sample is:

\[ \frac{0.5005}{2.00}\times100=25.0\% \]

pure.

But we were told that the sample was actually 100% calcium carbonate!

The problem is that the acid only supplied enough reactant to consume approximately 0.5005 g of the original solid. The remaining calcium carbonate could not react.

The measured gas volume therefore underestimates the original carbonate content.

This gives you an important experimental design rule.

If the purity is unknown, use enough acid to react with the entire sample even if it were 100% calcium carbonate.

For a sample mass \(m\), the maximum amount of hydrochloric acid required is:

\[ n(\ce{HCl})_{\mathrm{required}} =2\left(\frac{m}{100.1}\right) \]

This ensures that acid availability does not restrict the carbon dioxide produced.

Other experimental limitations

Even with excess acid, a measured gas volume can be too small if:

• The apparatus leaks.
• Some carbon dioxide escapes before the apparatus is sealed.
• The reaction does not finish.
• Some carbon dioxide remains dissolved in the reaction mixture.
• The gas syringe sticks or is incorrectly read.

These errors generally make the calculated purity too low.

There is also a chemical limitation: carbon dioxide production does not automatically identify which carbonate produced it. If the sample contains both calcium carbonate and magnesium carbonate, both react with hydrochloric acid to produce carbon dioxide.

The calculation must account for the identity of the carbonate being measured.

If gas is collected over water, remember that the measured pressure includes water vapour. When using the ideal gas equation, subtract the water vapour pressure from the total pressure to obtain the carbon dioxide partial pressure.

06Questions and solutions

Use the following information unless a question states otherwise:

• \(M(\ce{CaCO3})=100.1\ \mathrm{g\,mol^{-1}}\)
• \(M(\ce{CO2})=44.0\ \mathrm{g\,mol^{-1}}\)
• \(V_m=24.8\ \mathrm{L\,mol^{-1}}\) at 25°C and 100 kPa.

Assume that gas measurements are accurate and all stated reactions proceed to completion, except where a question introduces an experimental limitation.

Question 1

A 1.25 g sample of impure calcium carbonate reacts with excess hydrochloric acid.

The reaction produces 0.440 g of carbon dioxide.

Calculate the percentage purity of the original sample.

Solution 1

The sample is 80.1% pure calcium carbonate by mass.

Start by converting the mass of carbon dioxide into moles.

\[ \begin{aligned} n(\ce{CO2})&=\frac{m}{M}\\ &=\frac{0.440}{44.0}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The balanced equation gives a 1:1 mole ratio between calcium carbonate and carbon dioxide.

Therefore:

\[ n(\ce{CaCO3})=0.0100\ \mathrm{mol} \]

Convert this amount into mass:

\[ \begin{aligned} m(\ce{CaCO3})&=nM\\ &=0.0100\times100.1\\ &=1.001\ \mathrm{g} \end{aligned} \]

Now calculate percentage purity:

\[ \begin{aligned} \%\text{ purity} &=\frac{1.001}{1.25}\times100\\ &=80.08\%\\ &\approx80.1\% \end{aligned} \]

The key is that the mass of carbon dioxide is not the mass of calcium carbonate. The mole ratio connects them, and the molar masses convert between their respective masses.

Question 2

A 1.50 g sample contains calcium carbonate and inert impurities only.

The sample is treated with 25.0 mL of 0.400 mol L⁻¹ hydrochloric acid. The reaction produces 0.124 L of carbon dioxide at 25°C and 100 kPa.

Assume that all the carbon dioxide produced is collected.

(a) Calculate the amount of calcium carbonate that reacted.

(b) Can the original percentage purity be determined uniquely?

(c) Determine the minimum and maximum possible percentage purity of the sample.

Solution 2

The original percentage purity cannot be determined uniquely. It must be between 33.4% and 100%.

(a) First, calculate the amount of carbon dioxide produced.

\[ \begin{aligned} n(\ce{CO2})&=\frac{0.124}{24.8}\\ &=0.00500\ \mathrm{mol} \end{aligned} \]

The mole ratio is 1:1, so:

\[ n(\ce{CaCO3})_{\mathrm{reacted}} =0.00500\ \mathrm{mol} \]

Convert this into mass:

\[ \begin{aligned} m(\ce{CaCO3})_{\mathrm{reacted}} &=0.00500\times100.1\\ &=0.5005\ \mathrm{g} \end{aligned} \]

At least 0.5005 g of calcium carbonate was present in the original sample.

(b) Now consider whether the acid was in excess.

Calculate the amount of hydrochloric acid supplied:

\[ \begin{aligned} n(\ce{HCl})&=cV\\ &=0.400\times0.0250\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

This amount of acid can react with:

\[ \frac{0.0100}{2}=0.00500\ \mathrm{mol} \]

of calcium carbonate.

That is exactly the amount that reacted.

The acid has been completely consumed. We cannot tell whether the original sample contained exactly 0.5005 g of calcium carbonate or whether additional calcium carbonate remained unreacted.

(c) The minimum possible purity occurs when the sample contained exactly the amount of calcium carbonate that reacted:

\[ \begin{aligned} \%\text{ purity}_{\min} &=\frac{0.5005}{1.50}\times100\\ &=33.4\% \end{aligned} \]

The maximum possible purity occurs if the entire sample was calcium carbonate:

\[ \%\text{ purity}_{\max} =\frac{1.50}{1.50}\times100 =100\% \]

Therefore:

\[ \boxed{33.4\%\leq\text{purity}\leq100\%} \]

This question demonstrates why measuring carbon dioxide is not sufficient by itself. You must also establish that the acid was in excess before treating the measured gas as a measure of the entire carbonate content.

Question 3

Three identical 1.50 g portions of an impure calcium carbonate powder are tested separately.

Different amounts of hydrochloric acid are used in each trial. The carbon dioxide volumes are measured at 25°C and 100 kPa.

Assume the powder has the same composition in each portion and contains only calcium carbonate and inert impurities.

(a) Identify the limiting reagent in each trial.

(b) Calculate the percentage purity of the powder.

(c) Explain why Trial A alone would not be sufficient to determine the original purity.

Solution 3

The powder is 66.7% pure calcium carbonate.

The important evidence is that increasing the acid amount from Trial B to Trial C does not increase the volume of carbon dioxide.

(a) Calculate the moles of carbon dioxide produced in each trial.

For Trial A:

\[ n(\ce{CO2})=\frac{0.124}{24.8} =0.00500\ \mathrm{mol} \]

The acid supplied was 0.0100 mol. It can produce at most:

\[ \frac{0.0100}{2}=0.00500\ \mathrm{mol\ CO2} \]

Therefore, hydrochloric acid is limiting in Trial A.

For Trial B:

\[ n(\ce{CO2})=\frac{0.248}{24.8} =0.0100\ \mathrm{mol} \]

The 0.0200 mol of HCl is exactly enough to produce 0.0100 mol of CO2.

The reactants are present in the stoichiometric ratio, so neither is left in excess.

For Trial C, the same 0.0100 mol of CO2 is produced, despite the larger amount of acid.

The 0.0300 mol of HCl could produce up to:

\[ \frac{0.0300}{2}=0.0150\ \mathrm{mol\ CO2} \]

However, only 0.0100 mol forms.

Therefore, calcium carbonate is limiting in Trial C, and hydrochloric acid is in excess.

(b) Trial C gives the amount of carbonate in the sample because the acid is in excess.

\[ n(\ce{CaCO3})=0.0100\ \mathrm{mol} \]

Convert to mass:

\[ \begin{aligned} m(\ce{CaCO3})&=0.0100\times100.1\\ &=1.001\ \mathrm{g} \end{aligned} \]

Calculate purity:

\[ \begin{aligned} \%\text{ purity} &=\frac{1.001}{1.50}\times100\\ &=66.7\% \end{aligned} \]

(c) Trial A is insufficient because hydrochloric acid is the limiting reagent.

It only tells us that 0.00500 mol of calcium carbonate reacted. There could have been additional calcium carbonate left in the flask.

Trial C removes this uncertainty because the acid is in excess. The amount of gas is now controlled by the amount of calcium carbonate in the sample.

The unchanged gas volume between Trials B and C is evidence that the reaction has reached its carbonate-limited plateau.

Question 4

A laboratory tests its gas collection apparatus using 1.00 g of pure calcium carbonate.

At 25°C and 100 kPa, the theoretical volume of carbon dioxide is 0.248 L. However, the apparatus collects only 0.223 L.

The laboratory therefore establishes a carbon dioxide recovery efficiency of 90.0%.

An unknown 2.00 g sample of impure calcium carbonate is then tested using the same apparatus and conditions. Excess hydrochloric acid is used, and the measured gas volume is 0.335 L.

Assume the same recovery efficiency applies to the unknown sample.

Calculate its corrected percentage purity.

Solution 4

After correcting for incomplete gas recovery, the sample is 75.1% pure calcium carbonate.

The tempting approach is to use the measured 0.335 L directly. However, the calibration experiment tells us that 10.0% of the carbon dioxide is not being collected.

We must correct the measurement before calculating purity.

Step 1: Determine the true carbon dioxide volume.

The apparatus collects 90.0% of the gas produced.

Therefore:

\[ \begin{aligned} V_{\mathrm{actual}} &=\frac{V_{\mathrm{measured}}}{0.900}\\ &=\frac{0.335}{0.900}\\ &=0.3722\ \mathrm{L} \end{aligned} \]

Step 2: Convert the corrected volume into moles.

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.3722}{24.8}\\ &=0.01501\ \mathrm{mol} \end{aligned} \]

Step 3: Use the mole ratio to find the amount of calcium carbonate.

\[ n(\ce{CaCO3})=0.01501\ \mathrm{mol} \]

Step 4: Convert the amount into mass.

\[ \begin{aligned} m(\ce{CaCO3}) &=0.01501\times100.1\\ &=1.502\ \mathrm{g} \end{aligned} \]

Step 5: Calculate percentage purity.

\[ \begin{aligned} \%\text{ purity} &=\frac{1.502}{2.00}\times100\\ &=75.1\% \end{aligned} \]

The corrected purity is 75.1%.

If you had ignored the calibration, you would have calculated:

\[ \begin{aligned} m(\ce{CaCO3}) &=\frac{0.335}{24.8}\times100.1\\ &=1.352\ \mathrm{g} \end{aligned} \]

This would give an apparent purity of 67.6%.

The difference is caused by gas that was produced but not collected.

This is an important distinction in experimental chemistry: the amount of product formed and the amount of product measured are not necessarily identical.

Question 5

A 2.000 g mineral sample contains three substances:

• Calcium carbonate, \(\ce{CaCO3}\).
• Magnesium carbonate, \(\ce{MgCO3}\).
• An inert, insoluble material that does not react with hydrochloric acid.

The sample is treated with excess hydrochloric acid.

Both carbonates react completely, and all the carbon dioxide is collected in a gas syringe. After the reaction, the mixture is filtered, washed, and dried. The remaining insoluble residue has a mass of 0.3246 g and consists entirely of the inert material.

The collected carbon dioxide volume is 0.4464 L at 25°C and 100 kPa.

Use:

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

\[ M(\ce{MgCO3})=84.3\ \mathrm{g\,mol^{-1}} \]

Calculate the percentage by mass of calcium carbonate in the original sample.

Solution 5

The original sample is 50.1% pure calcium carbonate by mass.

This is the most important conceptual distinction in the question: both calcium carbonate and magnesium carbonate produce carbon dioxide.

You cannot assume that every mole of carbon dioxide came from calcium carbonate.

We need two independent pieces of information to distinguish the two carbonates:

1. The total number of moles of carbonate, obtained from the carbon dioxide.
2. The total mass of carbonate, obtained by subtracting the inert residue from the original sample mass.

Step 1: Calculate the total moles of carbon dioxide.

\[ \begin{aligned} n(\ce{CO2}) &=\frac{0.4464}{24.8}\\ &=0.01800\ \mathrm{mol} \end{aligned} \]

Both carbonate reactions have a 1:1 mole ratio with carbon dioxide:

\[ \ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2} \]

\[ \ce{MgCO3 + 2HCl -> MgCl2 + H2O + CO2} \]

Therefore, the total moles of the two carbonates must add to 0.01800 mol.

Step 2: Calculate the total mass of carbonate in the original sample.

The original sample has a mass of 2.000 g.

The dried inert residue has a mass of 0.3246 g.

Therefore:

\[ \begin{aligned} m_{\mathrm{carbonates}} &=2.000-0.3246\\ &=1.6754\ \mathrm{g} \end{aligned} \]

The combined mass of calcium carbonate and magnesium carbonate is 1.6754 g.

Step 3: Set up simultaneous equations.

Let:

• \(x\) be the moles of calcium carbonate;
• \(y\) be the moles of magnesium carbonate.

From the total amount of carbon dioxide:

\[ x+y=0.01800 \]

From the total mass of the carbonates:

\[ 100.1x+84.3y=1.6754 \]

We now have two equations and two unknowns.

Step 4: Solve for the amount of calcium carbonate.

Rearrange the first equation:

\[ y=0.01800-x \]

Substitute into the mass equation:

\[ 100.1x+84.3(0.01800-x)=1.6754 \]

Expand:

\[ 100.1x+1.5174-84.3x=1.6754 \]

Collect like terms:

\[ 15.8x=0.1580 \]

Therefore:

\[ x=0.01000\ \mathrm{mol} \]

The original sample contained 0.01000 mol of calcium carbonate.

Step 5: Convert calcium carbonate into mass.

\[ \begin{aligned} m(\ce{CaCO3}) &=nM\\ &=0.01000\times100.1\\ &=1.001\ \mathrm{g} \end{aligned} \]

Step 6: Calculate percentage purity.

\[ \begin{aligned} \%\text{ purity} &=\frac{1.001}{2.000}\times100\\ &=50.05\%\\ &\approx50.1\% \end{aligned} \]

The original sample is therefore 50.1% calcium carbonate by mass.

We can verify the result using the magnesium carbonate amount:

\[ \begin{aligned} n(\ce{MgCO3})&=0.01800-0.01000\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

Its mass is:

\[ m(\ce{MgCO3})=0.00800\times84.3 =0.6744\ \mathrm{g} \]

The three original components therefore have masses:

The masses add back to the original sample mass, so the composition is consistent.

The tempting approach would have been to assign all 0.01800 mol of carbon dioxide to calcium carbonate. That would give 1.802 g of CaCO₃, which is greater than the total 1.6754 g mass of both carbonates combined.

The extra mass is impossible. The missing consideration is that magnesium carbonate also produces carbon dioxide, but has a different molar mass.

This is why knowing the identity of the reacting substances matters just as much as knowing the mole ratio.

07Where to go next

Once you can calculate purity from measured carbon dioxide, the complementary skill is predicting how much gas a known mass of carbonate should produce.

That requires working forwards through the same balanced equation and deciding whether the carbonate or the acid is the limiting reagent.

For that calculation, see How to Calculate CO₂ Volume from a Carbonate Reaction.

If an experiment measures the decrease in mass of an open reaction vessel instead of collecting gas in a syringe, the calculation follows a different measurement route. See How to Calculate Gas Mass Lost from an Open Reaction Vessel.