How to Solve Stoichiometry When Reactant Purity Is Below 100%

A 20.0 g sample of sodium carbonate is labelled 85.0% pure. It reacts with excess calcium chloride to produce calcium carbonate. How much calcium carbonate should form?

You might be tempted to use the full 20.0 g in your stoichiometry calculation. That would overestimate the product because 15% of the sample is not sodium carbonate.

The correct approach is to calculate the mass of pure reactant first, then perform the usual mass-to-moles-to-moles-to-mass calculation.

For an 85.0% pure sample:

\[ m(\text{pure reactant})=20.0\times0.850 =17.0\ \text{g} \]

Only 17.0 g of the sample contributes sodium carbonate to the reaction.

The remaining 3.00 g is impurity. If that impurity is chemically inert, it does not contribute to the amount of product formed.

This is the key adjustment when solving stoichiometry questions involving impure reactants. Everything else follows the usual mole-ratio method.

01Calculate the mass of pure reactant

Purity tells you what fraction of a sample consists of the desired substance.

For a solid sample, percentage purity is generally expressed as a percentage by mass.

The formula is:

\[ \%\text{ purity}= \frac{\text{mass of pure substance}} {\text{total sample mass}}\times100 \]

Rearranging:

\[ m(\text{pure substance})= m(\text{sample})\times \frac{\%\text{ purity}}{100} \]

The percentage must be divided by 100 before multiplying.

For example, consider three samples:

Notice that a 50.0 g sample that is 92.0% pure does not contain 50.0 g of the desired reactant. It contains 46.0 g.

The difference is the impurity.

\[ m(\text{impurity})= m(\text{sample})-m(\text{pure substance}) \]

For the first sample:

\[ m(\text{impurity})=20.0-17.0 =3.00\ \text{g} \]

This subtraction is a useful check that your purity correction makes sense.

What does this mean in an actual experiment?

Imagine a laboratory investigation using an impure sodium carbonate sample and calcium chloride solution.

The aim is to predict the mass of calcium carbonate precipitate formed.

The sample is weighed first, dissolved in water, and transferred into a beaker. Calcium chloride solution is then added in excess while the mixture is stirred.

A white solid forms. This is calcium carbonate.

What should the experiment show?

If the sodium carbonate sample is less pure, it contains fewer moles of sodium carbonate for the same total sample mass. Provided the calcium chloride is in excess and the reaction proceeds completely, less calcium carbonate should form.

There is an important experimental detail.

Simply evaporating the entire reaction mixture and weighing the residue would not give the mass of pure calcium carbonate. The residue would also contain sodium chloride, excess calcium chloride, and any soluble impurities.

Filtering the mixture, washing the precipitate, and drying it to constant mass gives a much more meaningful measurement of the calcium carbonate produced.

The apparatus and measurements here are illustrative, not results from a reported experiment.

This is the forward stoichiometry problem: the purity is already known, and we want to predict the product mass. It is different from an experiment that measures a product and works backwards to determine the original sample’s purity.

02Understand what the balanced equation tells you

Once you have the mass of pure reactant, you can use the balanced chemical equation.

For sodium carbonate reacting with calcium chloride:

\[ \ce{Na2CO3 + CaCl2 -> CaCO3 + 2NaCl} \]

The coefficients tell us the mole ratio:

\[ 1\ \text{mol Na}_2\text{CO}_3: 1\ \text{mol CaCO}_3 \]

Therefore, one mole of sodium carbonate produces one mole of calcium carbonate, assuming the reaction proceeds completely.

But be careful.

A 1:1 mole ratio does not mean a 1:1 mass ratio.

One mole of sodium carbonate has a mass of 106.0 g, while one mole of calcium carbonate has a mass of 100.1 g.

The coefficients count particles, or equivalently moles. They do not directly compare grams.

This is why you must convert the corrected reactant mass into moles before using the equation.

The complete calculation pathway

For an impure reactant, follow this sequence:

\[ \boxed{ \begin{aligned} &\text{Total sample mass}\\ &\downarrow\ \text{apply purity}\\ &\text{Pure reactant mass}\\ &\downarrow\ \text{divide by molar mass}\\ &\text{Moles of reactant}\\ &\downarrow\ \text{apply mole ratio}\\ &\text{Moles of product}\\ &\downarrow\ \text{multiply by molar mass}\\ &\text{Mass of product} \end{aligned} } \]

The important point is that purity correction happens before the mole calculation.

You should not apply the purity percentage to the final product mass as a second correction.

04What if the other reactant is also limited?

So far, we have assumed calcium chloride is in excess.

That means there is enough calcium chloride to react with all the available pure sodium carbonate.

But what happens if the question gives you a limited quantity of the second reactant?

You must identify the limiting reagent before calculating the product.

The limiting reagent is the reactant that runs out first. It determines the maximum amount of product that can form.

The purity correction still comes first, but it does not automatically mean the impure reactant controls the final product mass.

For example, consider the reaction:

\[ \ce{2Na3PO4 + 3CaCl2 -> Ca3(PO4)2 + 6NaCl} \]

This reaction has a 2:3:1 mole ratio between sodium phosphate, calcium chloride, and calcium phosphate.

Even if you know the exact amount of pure sodium phosphate, you cannot calculate the product until you check whether enough calcium chloride is available.

05Purity is not the same as percentage yield

These two ideas often appear in the same HSC Chemistry question, but they describe different things.

Consider the first worked example again.

The theoretical mass of calcium carbonate was 16.1 g.

Suppose the experiment only recovers 90.0% of this amount.

The actual mass recovered would be:

\[ \begin{aligned} m(\text{actual}) &=16.1\times0.900\\ &=14.4\ \text{g} \end{aligned} \]

The 85.0% purity determined how much sodium carbonate was available.

The 90.0% yield describes how much of the predicted calcium carbonate was actually recovered.

These percentages cannot be substituted for one another.

A sample can be 100% pure but produce a low percentage yield because the reaction is incomplete or product is lost during filtration.

Equally, a highly efficient reaction can start with an impure sample.

For a more detailed explanation, see How to Distinguish Actual, Theoretical and Percentage Yield.

There is also a related but different type of question where the product is measured first and the original reactant’s purity must be calculated. For example, How to Find the Percentage Purity of a Carbonate from Carbon Dioxide Produced works backwards from the measured gas to the amount of pure carbonate.

06Questions and solutions

The following questions progress from a straightforward purity correction to problems involving limiting reagents, experimental recovery, and chemically reactive impurities.

Use the following approximate atomic molar masses where needed:

Question 1

A 15.0 g sample of sodium carbonate is 80.0% pure by mass. The sample reacts with excess calcium chloride according to:

\[ \ce{Na2CO3 + CaCl2 -> CaCO3 + 2NaCl} \]

The impurities are inert.

Calculate the theoretical mass of calcium carbonate produced.

Solution 1

The theoretical mass of calcium carbonate is 11.3 g.

The first step is to correct the starting mass for purity. Using the full 15.0 g would incorrectly assume that the entire sample is sodium carbonate.

Step 1: Calculate the pure sodium carbonate mass.

\[ \begin{aligned} m(\ce{Na2CO3}) &=15.0\times0.800\\ &=12.0\ \text{g} \end{aligned} \]

Step 2: Convert to moles.

\[ \begin{aligned} n(\ce{Na2CO3}) &=\frac{12.0}{106.0}\\ &=0.1132\ \text{mol} \end{aligned} \]

Step 3: Apply the mole ratio.

The balanced equation gives a 1:1 ratio.

\[ n(\ce{CaCO3})=0.1132\ \text{mol} \]

Step 4: Convert into mass.

\[ \begin{aligned} m(\ce{CaCO3}) &=0.1132\times100.1\\ &=11.3\ \text{g} \end{aligned} \]

The result is the theoretical calcium carbonate mass, assuming complete reaction and no product loss.

Question 2

An impure iron sample has a total mass of 10.0 g and contains 88.0% iron by mass. The remaining material is chemically inert.

The sample reacts with 3.00 g of pure sulfur according to:

\[ \ce{Fe + S -> FeS} \]

Assume the reaction proceeds completely until the limiting reagent is consumed.

Calculate:

a. The limiting reagent.

b. The theoretical mass of iron(II) sulfide produced.

c. The mass of unreacted pure iron remaining.

Solution 2

Sulfur is the limiting reagent. The reaction produces 8.21 g of iron(II) sulfide, leaving 3.59 g of unreacted pure iron.

The tempting mistake is to use the full 10.0 g of iron when calculating its moles. Only 88.0% of that mass is actually iron.

Part a: Identify the limiting reagent.

Step 1: Calculate the mass of pure iron.

\[ \begin{aligned} m(\ce{Fe}) &=10.0\times0.880\\ &=8.80\ \text{g} \end{aligned} \]

Step 2: Convert both reactants into moles.

For iron:

\[ \begin{aligned} n(\ce{Fe}) &=\frac{8.80}{55.8}\\ &=0.1577\ \text{mol} \end{aligned} \]

For sulfur:

\[ \begin{aligned} n(\ce{S}) &=\frac{3.00}{32.1}\\ &=0.09346\ \text{mol} \end{aligned} \]

Step 3: Compare the mole amounts.

The equation has a 1:1 ratio.

Therefore, the reactant with fewer available moles is limiting.

Sulfur has 0.09346 mol, compared with 0.1577 mol of iron.

Sulfur is the limiting reagent.

Part b: Calculate the mass of iron(II) sulfide.

Step 4: Use the mole ratio.

\[ n(\ce{FeS})=0.09346\ \text{mol} \]

Step 5: Calculate the product mass.

The molar mass of iron(II) sulfide is:

\[ M(\ce{FeS})=55.8+32.1=87.9\ \text{g mol}^{-1} \]

Therefore:

\[ \begin{aligned} m(\ce{FeS}) &=0.09346\times87.9\\ &=8.21\ \text{g} \end{aligned} \]

Part c: Calculate the unreacted iron.

Step 6: Subtract the moles of iron consumed from the original amount.

The 1:1 ratio means 0.09346 mol of iron reacts.

\[ \begin{aligned} n(\ce{Fe})_{\text{remaining}} &=0.1577-0.09346\\ &=0.06425\ \text{mol} \end{aligned} \]

Convert this remaining amount into mass:

\[ \begin{aligned} m(\ce{Fe})_{\text{remaining}} &=0.06425\times55.8\\ &=3.59\ \text{g} \end{aligned} \]

The final results are:

• Limiting reagent: sulfur.
• Iron(II) sulfide produced: 8.21 g.
• Unreacted pure iron: 3.59 g.

The inert impurity is not included in the unreacted iron mass. It was never iron in the first place.

Question 3

An 18.0 g sample of magnesium carbonate is 80.0% pure by mass. The remaining material is inert and insoluble in water and hydrochloric acid.

The sample reacts with 250.0 mL of 0.500 mol/L hydrochloric acid.

The reaction is:

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

Assume the available limiting reagent is completely consumed, with no side reactions or loss of solid material.

Calculate:

a. The limiting reagent.

b. The theoretical mass of magnesium chloride produced.

c. The mass of unreacted magnesium carbonate remaining.

Solution 3

Hydrochloric acid is the limiting reagent. The reaction produces 5.96 g of magnesium chloride, leaving 9.13 g of unreacted magnesium carbonate.

The important point is that correcting for purity does not guarantee that all the pure magnesium carbonate will react. We must also check the available acid.

Step 1: Calculate the pure magnesium carbonate mass.

\[ \begin{aligned} m(\ce{MgCO3}) &=18.0\times0.800\\ &=14.4\ \text{g} \end{aligned} \]

Step 2: Convert magnesium carbonate into moles.

Its molar mass is:

\[ M(\ce{MgCO3})=24.3+12.0+3(16.0) =84.3\ \text{g mol}^{-1} \]

Therefore:

\[ \begin{aligned} n(\ce{MgCO3}) &=\frac{14.4}{84.3}\\ &=0.1708\ \text{mol} \end{aligned} \]

Step 3: Calculate the available hydrochloric acid.

Convert the volume into litres:

\[ 250.0\ \text{mL}=0.2500\ \text{L} \]

Using \(n=cV\):

\[ \begin{aligned} n(\ce{HCl}) &=0.500\times0.2500\\ &=0.1250\ \text{mol} \end{aligned} \]

Step 4: Identify the limiting reagent.

The equation requires:

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

One mole of magnesium carbonate needs two moles of hydrochloric acid.

To react with all 0.1708 mol of magnesium carbonate, the acid requirement would be:

\[ \begin{aligned} n(\ce{HCl})_{\text{required}} &=2\times0.1708\\ &=0.3416\ \text{mol} \end{aligned} \]

Only 0.1250 mol is available.

Therefore, hydrochloric acid is limiting.

Step 5: Calculate the magnesium chloride produced.

The mole ratio between hydrochloric acid and magnesium chloride is 2:1.

\[ \begin{aligned} n(\ce{MgCl2}) &=0.1250\times\frac{1}{2}\\ &=0.06250\ \text{mol} \end{aligned} \]

The molar mass of magnesium chloride is:

\[ M(\ce{MgCl2})=24.3+2(35.5) =95.3\ \text{g mol}^{-1} \]

Therefore:

\[ \begin{aligned} m(\ce{MgCl2}) &=0.06250\times95.3\\ &=5.96\ \text{g} \end{aligned} \]

Step 6: Calculate the remaining magnesium carbonate.

The 1:2 ratio means 0.1250 mol of hydrochloric acid consumes:

\[ n(\ce{MgCO3})_{\text{reacted}} =\frac{0.1250}{2} =0.06250\ \text{mol} \]

Subtract this from the original amount:

\[ \begin{aligned} n(\ce{MgCO3})_{\text{remaining}} &=0.1708-0.06250\\ &=0.1083\ \text{mol} \end{aligned} \]

Convert back to mass:

\[ \begin{aligned} m(\ce{MgCO3})_{\text{remaining}} &=0.1083\times84.3\\ &=9.13\ \text{g} \end{aligned} \]

The remaining magnesium carbonate has not disappeared. It simply could not react because the acid ran out.

The 3.60 g of inert impurity also remains as a separate solid.

This question combines three decisions: correct the sample for purity, identify the limiting reagent, and use the correct coefficient ratio.

Question 4

A 10.0 g sample of sodium carbonate reacts with excess calcium chloride.

After the reaction, filtration, washing, and drying, a student collects 6.80 g of calcium carbonate.

However, the student cannot establish whether the reaction went to completion or whether some precipitate was lost during filtration.

Can the percentage purity of the original sodium carbonate sample be determined uniquely from the recovered product mass?

Demonstrate your reasoning using two different combinations of purity and product recovery that could produce the same measured mass.

Solution 4

No. The original purity cannot be determined uniquely because the recovered product mass depends on both the starting purity and the overall product recovery.

A smaller recovered mass could mean that the original sample contained less sodium carbonate. It could also mean that some theoretical product was not formed or was lost during collection.

Let’s compare two possible explanations.

Model A: 72.0% purity and 100% product recovery

The pure sodium carbonate mass would be:

\[ 10.0\times0.720=7.20\ \text{g} \]

The moles of sodium carbonate would be:

\[ n(\ce{Na2CO3})=\frac{7.20}{106.0} =0.06792\ \text{mol} \]

The mole ratio between sodium carbonate and calcium carbonate is 1:1.

Therefore:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.06792\times100.1\\ &=6.80\ \text{g} \end{aligned} \]

If the reaction and collection were both complete, this sample would produce the measured 6.80 g.

Model B: 80.0% purity and 90.0% product recovery

The pure sodium carbonate mass would now be:

\[ 10.0\times0.800=8.00\ \text{g} \]

The theoretical calcium carbonate mass would be:

\[ \begin{aligned} m(\ce{CaCO3})_{\text{theoretical}} &=\frac{8.00}{106.0}\times100.1\\ &=7.55\ \text{g} \end{aligned} \]

If only 90.0% of this theoretical amount is recovered:

\[ \begin{aligned} m(\ce{CaCO3})_{\text{actual}} &=7.55\times0.900\\ &=6.80\ \text{g} \end{aligned} \]

Both models produce the same measured mass.

The tempting approach is to work backwards from the 6.80 g of product and assume it represents the theoretical amount.

That assumption is only valid if the reaction is complete and product recovery is quantitative.

Without independent information about reaction completion and product losses, purity and recovery cannot be separated.

This is a problem of insufficient information, not a problem with the stoichiometric equations.

Question 5

A 10.00 g sample is labelled as 80.0% pure sodium carbonate by mass.

The remaining 20.0% of the sample has not been identified.

The sample is dissolved and reacted with exactly 0.0800 mol of calcium chloride.

Consider two possible explanations for the impurity:

Model A: The remaining 20.0% is an inert, soluble substance that does not react with calcium chloride.

Model B: The remaining 20.0% is sodium hydrogencarbonate, \(\ce{NaHCO3}\), which also reacts with calcium chloride to produce calcium carbonate.

For Model B, use:

\[ \ce{2NaHCO3 + CaCl2 -> CaCO3 + 2NaCl + CO2 + H2O} \]

For both models, use the appropriate balanced equations and assume:

• Both reactions proceed until a limiting reactant is consumed.
• Carbon dioxide escapes from the solution.
• Calcium carbonate does not dissolve again.
• All calcium carbonate formed is collected.
• No other reactions occur.

Calculate the theoretical mass of calcium carbonate produced under each model.

Explain why the original 80.0% purity label does not necessarily determine one unique product mass.

Solution 5

The two models produce different theoretical masses of calcium carbonate:

• Model A: 7.55 g.
• Model B: 8.01 g.

The reason is that sodium hydrogencarbonate is not an inert impurity. It can also react to form calcium carbonate.

This changes the total amount of product that the sample can generate and, importantly, changes which reactant is limiting.

Model A: The impurity is inert.

Step 1: Calculate the mass of pure sodium carbonate.

\[ \begin{aligned} m(\ce{Na2CO3}) &=10.00\times0.800\\ &=8.00\ \text{g} \end{aligned} \]

Step 2: Convert sodium carbonate into moles.

\[ \begin{aligned} n(\ce{Na2CO3}) &=\frac{8.00}{106.0}\\ &=0.07547\ \text{mol} \end{aligned} \]

Step 3: Identify the limiting reagent.

The reaction is:

\[ \ce{Na2CO3 + CaCl2 -> CaCO3 + 2NaCl} \]

The mole ratio between sodium carbonate and calcium chloride is 1:1.

We have:

\[ n(\ce{Na2CO3})=0.07547\ \text{mol} \]

\[ n(\ce{CaCl2})=0.0800\ \text{mol} \]

There is more calcium chloride than sodium carbonate.

Therefore, sodium carbonate is limiting.

Step 4: Calculate the calcium carbonate mass.

The 1:1 ratio gives:

\[ n(\ce{CaCO3})=0.07547\ \text{mol} \]

Therefore:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.07547\times100.1\\ &=7.55\ \text{g} \end{aligned} \]

Model A produces 7.55 g of calcium carbonate.

Model B: The impurity is sodium hydrogencarbonate.

This is where the problem changes.

We still have 8.00 g of sodium carbonate, but the remaining 2.00 g is now another substance that reacts.

We must account for both reactions.

Step 1: Calculate the moles of sodium carbonate.

The sodium carbonate mass remains:

\[ m(\ce{Na2CO3})=8.00\ \text{g} \]

Therefore:

\[ n(\ce{Na2CO3})=\frac{8.00}{106.0} =0.07547\ \text{mol} \]

Step 2: Calculate the moles of sodium hydrogencarbonate.

The molar mass is:

\[ M(\ce{NaHCO3})=23.0+1.0+12.0+3(16.0) =84.0\ \text{g mol}^{-1} \]

Therefore:

\[ \begin{aligned} n(\ce{NaHCO3}) &=\frac{2.00}{84.0}\\ &=0.02381\ \text{mol} \end{aligned} \]

Step 3: Calculate the maximum calcium carbonate produced by each reactant.

From the sodium carbonate reaction:

\[ \ce{Na2CO3 + CaCl2 -> CaCO3 + 2NaCl} \]

The mole ratio is 1:1.

Therefore:

\[ n(\ce{CaCO3})=0.07547\ \text{mol} \]

From the sodium hydrogencarbonate reaction:

\[ \ce{2NaHCO3 + CaCl2 -> CaCO3 + 2NaCl + CO2 + H2O} \]

Two moles of sodium hydrogencarbonate produce one mole of calcium carbonate.

Therefore:

\[ \begin{aligned} n(\ce{CaCO3}) &=0.02381\times\frac{1}{2}\\ &=0.01190\ \text{mol} \end{aligned} \]

The total theoretical calcium carbonate capacity of the two reactants is:

\[ \begin{aligned} n(\ce{CaCO3})_{\text{total}} &=0.07547+0.01190\\ &=0.08738\ \text{mol} \end{aligned} \]

Step 4: Identify the limiting reagent.

The available calcium chloride is:

\[ n(\ce{CaCl2})=0.0800\ \text{mol} \]

The total carbonate reactants could produce 0.08738 mol of calcium carbonate.

However, the calcium chloride can only support 0.0800 mol of calcium carbonate because both reactions use one mole of calcium chloride per mole of calcium carbonate produced.

Therefore, calcium chloride is now the limiting reagent.

This is the crucial difference between the two models.

In Model A, sodium carbonate was limiting.

In Model B, the additional reactive impurity supplies enough carbonate to make calcium chloride limiting.

Step 5: Calculate the final calcium carbonate mass.

The calcium chloride-to-calcium carbonate ratio is 1:1 in both reactions.

Therefore:

\[ n(\ce{CaCO3})=0.0800\ \text{mol} \]

Convert this into mass:

\[ \begin{aligned} m(\ce{CaCO3}) &=0.0800\times100.1\\ &=8.01\ \text{g} \end{aligned} \]

Model B produces 8.01 g of calcium carbonate.

Why this final result matters

Both samples are described as 80.0% pure sodium carbonate.

However, the remaining 20.0% has a different chemical identity in each model.

In Model A, the impurity does not react, so it contributes nothing to the product.

In Model B, the impurity reacts and produces additional calcium carbonate. It also changes which reactant runs out first.

This exposes an important limitation of the usual purity correction:

You can only ignore the impurity if the question establishes that it is chemically unreactive under the reaction conditions.

A percentage purity tells you how much of the named substance is present. It does not automatically tell you what the remaining material does.

For ordinary HSC stoichiometry questions, the question will generally specify that impurities are inert or otherwise give enough information to account for them. When that assumption is not justified, you need to consider the chemistry of the impurity as well.

The reliable decision sequence is now clear: correct the reactant mass for purity, convert to moles, check the balanced equation, identify the limiting reagent if necessary, and calculate the theoretical product mass.

Once you have the theoretical amount, the next useful decision is whether the experimentally recovered product agrees with that prediction. That is where percentage yield, incomplete reactions, and product losses become important.