How to Find Product Mass When a Reactant Is Impure

A 13.0 g sample of zinc is labelled 80.0% pure. If you use the full 13.0 g in a stoichiometry calculation, you’ll predict more product than the sample can actually make. Why? Because 20% of the sample isn’t zinc at all.

Would this sample produce as much zinc chloride as 13.0 g of pure zinc? No. Only 10.4 g is zinc; the remaining 2.60 g is impurity.

To find the product mass when a reactant is impure, you need to:

1. Calculate the mass of pure reactant using its percentage purity.
2. Convert the pure reactant mass into moles.
3. Divide each reactant’s mole amount by its balanced equation coefficient to identify the limiting reactant.
4. Calculate the theoretical product mass and any leftover reactants.
5. Apply percentage yield if the question asks for the actual mass recovered.

The key is to exclude inert impurities from your stoichiometry calculations without accidentally excluding genuine reactants.

Throughout this explanation, assume purity is expressed as a mass percentage. Unless stated otherwise, impurities are inert, meaning they do not participate in the reaction.

01Correct the sample mass for purity

Percentage purity tells you what fraction of an impure sample is the substance you actually want to react.

The formula is:

\[ \text{Purity}= \frac{\text{Mass of pure substance}} {\text{Mass of impure sample}}\times100 \]

Rearranging gives the calculation you need:

\[ m_{\text{pure}}= m_{\text{sample}}\times \frac{\text{Purity}}{100} \]

For example, suppose you have 13.0 g of zinc that is 80.0% pure.

\[ \begin{aligned} m_{\text{pure}}&=13.0\times\frac{80.0}{100}\\ &=10.4\text{ g} \end{aligned} \]

Therefore:

• Pure zinc = 10.4 g
• Inert impurities = \(13.0-10.4=2.60\text{ g}\)

Only the 10.4 g of zinc contributes to the reaction.

Convert this pure mass into moles using:

\[ n=\frac{m}{M} \]

where:

• \(n\) = amount of substance, in mol
• \(m\) = mass of pure substance, in g
• \(M\) = molar mass, in g mol\(^{-1}\)

For zinc, \(M(\mathrm{Zn})=65.38\text{ g mol}^{-1}\).

\[ n(\mathrm{Zn})=\frac{10.4}{65.38} =0.159\text{ mol} \]

This is the amount of zinc available for the reaction.

An important distinction: purity and yield are not the same thing.

Purity corrects your starting material. Yield corrects your final product.

02Find the limiting reactant using amount divided by coefficient

Knowing the amount of pure reactant is only part of the calculation.

You also need to consider how much of the other reactants is available.

Consider the reaction between zinc and hydrochloric acid:

\[ \ce{Zn + 2HCl -> ZnCl2 + H2} \]

The balanced equation tells us that:

• 1 mol of zinc reacts with 2 mol of hydrochloric acid.
• 1 mol of zinc produces 1 mol of zinc chloride.
• 1 mol of zinc produces 1 mol of hydrogen gas.

The coefficients matter. You cannot simply compare the number of moles of each reactant and choose the smaller amount.

Instead, calculate:

\[ \boxed{\frac{n}{\text{balanced coefficient}}} \]

Do this for every reactant.

The reactant with the smallest value is the limiting reactant. It determines how much reaction can occur and, therefore, the maximum amount of product.

For example, suppose you have:

• 0.159 mol Zn
• 0.180 mol HCl

A common mistake is to compare the raw mole amounts and conclude that zinc is limiting because it has fewer moles.

That conclusion is wrong.

The reaction requires twice as many moles of HCl as Zn. Divide each amount by its coefficient:

Hydrochloric acid has the smaller value, so HCl is the limiting reactant.

Although there are more moles of HCl than zinc, there isn’t enough HCl to react with all the zinc.

If you need more practice with this comparison, see How to Compare Reactant Amounts and Find the Limiting Reactant.

03Calculate the theoretical product mass

Once you know the limiting reactant, use its amount divided by its coefficient to calculate the amount of product.

The general relationship is:

\[ n_{\text{product}}= \frac{n_{\text{limiting}}} {\nu_{\text{limiting}}} \times\nu_{\text{product}} \]

Here, \(\nu\) represents the coefficient in the balanced chemical equation.

Once you have the product amount in moles, convert it into mass:

\[ m_{\text{product}}=n_{\text{product}}M_{\text{product}} \]

This gives the theoretical yield: the maximum product mass possible if the limiting reactant is completely consumed and the reaction proceeds without side reactions or product losses.

What about the leftover reactants?

You should also calculate how much of each excess reactant remains.

The general formula is:

\[ n_{\text{left}}= n_{\text{initial}}- n_{\text{consumed}} \]

Using the limiting reactant to determine the amount consumed:

\[ n_{\text{left}}= n_{\text{initial}}- \left( \frac{n_{\text{limiting}}} {\nu_{\text{limiting}}} \times\nu_{\text{reactant}} \right) \]

Convert the leftover amount into mass using \(m=nM\), if the question asks for grams.

Remember that inert impurities are different from excess reactants:

• An excess reactant participates in the reaction but is not completely consumed.
• An inert impurity does not participate in the reaction at all.

Both can remain in the reaction mixture, but only the excess reactant belongs in your balanced equation.

For a separate worked explanation of calculating leftover reactant mass, see How to Calculate the Mass of Excess Reactant Left Over.

04Apply percentage yield at the end

Your theoretical product mass assumes perfect conversion and recovery.

Real reactions may produce less product because of incomplete reactions, side reactions, or losses during separation and purification.

Percentage yield measures the actual product recovered compared with the theoretical amount.

\[ \%\text{Yield}= \frac{m_{\text{actual}}} {m_{\text{theoretical}}}\times100 \]

If you know the percentage yield and need the actual product mass, rearrange:

\[ m_{\text{actual}}= m_{\text{theoretical}}\times \frac{\%\text{Yield}}{100} \]

For example, if the theoretical product mass is 20.0 g and the percentage yield is 85.0%:

\[ m_{\text{actual}}=20.0\times0.850 =17.0\text{ g} \]

Do not apply the percentage purity again at this stage. You have already accounted for the impurity when calculating the initial amount of pure reactant.

06How can you determine purity experimentally?

Sometimes a question gives you the percentage purity. In a laboratory investigation, however, you might need to determine it from measurements.

Consider an impure limestone sample containing calcium carbonate. How could you work out how much calcium carbonate is present?

One method is to heat the sample and measure the mass lost as carbon dioxide escapes.

The thermal decomposition reaction is:

\[ \ce{CaCO3(s) -> CaO(s) + CO2(g)} \]

One mole of calcium carbonate produces one mole of carbon dioxide.

The apparatus and the prediction

Place a known mass of powdered limestone in a pre-weighed crucible. The crucible sits on a clay triangle above a suitable burner.

As the sample is heated, carbon dioxide escapes through the open crucible. The remaining solid contains calcium oxide and any non-volatile impurities.

After heating, transfer the crucible to a desiccator to cool before weighing it. Repeat the heating, cooling, and weighing process until a constant mass is reached.

The prediction is straightforward:

If calcium carbonate is the only substance decomposing, the mass lost must equal the mass of carbon dioxide released.

Since the reaction has a 1:1 mole ratio between calcium carbonate and carbon dioxide, measuring the carbon dioxide allows us to calculate the original amount of calcium carbonate.

Why one heating is not enough

Suppose the first design heats the limestone once and immediately weighs the residue.

There is a problem: some calcium carbonate may not have decomposed completely.

The following illustrative readings show what happens when the sample is heated repeatedly.

The decreasing mass shows that further carbon dioxide was released during the second and third heating cycles.

Stopping after the first cycle would underestimate the original calcium carbonate content.

The revised method heats the sample repeatedly until consecutive cooled masses agree. Cooling in a desiccator also reduces the risk of calcium oxide absorbing moisture and carbon dioxide from the air before weighing.

This gives a more reliable estimate of the original calcium carbonate content.

Worked measurement: Determine the purity of limestone

A 10.00 g limestone sample is heated to constant mass. Its final residue mass is 6.48 g.

Calculate:

a) The mass of carbon dioxide released.

b) The percentage purity of the original limestone.

c) The theoretical mass of calcium oxide produced.

Step 1: Calculate the mass lost.

\[ m(\mathrm{CO_2})=10.00-6.48 \]

\[ m(\mathrm{CO_2})=3.52\text{ g} \]

Step 2: Convert carbon dioxide mass into moles.

The molar mass of carbon dioxide is 44.01 g mol\(^{-1}\).

\[ n(\mathrm{CO_2})=\frac{3.52}{44.01} =0.07998\text{ mol} \]

Step 3: Calculate the original amount of calcium carbonate.

From:

\[ \ce{CaCO3 -> CaO + CO2} \]

The mole ratio of calcium carbonate to carbon dioxide is 1:1.

Therefore:

\[ n(\mathrm{CaCO_3})=0.07998\text{ mol} \]

Convert this into mass using \(M(\mathrm{CaCO_3})=100.09\text{ g mol}^{-1}\).

\[ \begin{aligned} m(\mathrm{CaCO_3}) &=0.07998\times100.09\\ &=8.01\text{ g} \end{aligned} \]

Step 4: Calculate percentage purity.

\[ \begin{aligned} \%\text{Purity} &=\frac{8.01}{10.00}\times100\\ &=80.1\% \end{aligned} \]

The limestone is approximately 80.1% calcium carbonate by mass.

Step 5: Calculate the theoretical calcium oxide mass.

The mole ratio of calcium carbonate to calcium oxide is also 1:1.

\[ n(\mathrm{CaO})=0.07998\text{ mol} \]

Using \(M(\mathrm{CaO})=56.08\text{ g mol}^{-1}\):

\[ \begin{aligned} m(\mathrm{CaO}) &=0.07998\times56.08\\ &=4.49\text{ g} \end{aligned} \]

The theoretical calcium oxide mass is 4.49 g.

There is also a useful mass check. The original sample contained approximately 1.99 g of inert impurities. The final residue should therefore contain:

\[ 4.49+1.99=6.48\text{ g} \]

This agrees with the measured residue mass.

Limitations: This method assumes the impurities do not decompose or lose mass when heated. If other substances release gases, the measured mass loss cannot be attributed entirely to carbon dioxide. Solid loss through spitting and incomplete decomposition can also affect the result.

The experiment demonstrates an important idea: the total mass of an impure sample is not necessarily the mass of the substance participating in the reaction.

08Keep these quantities separate

When answering an HSC Chemistry question, make sure you know exactly which mass you are calculating.

A good final check is to ask yourself whether you have applied purity and yield at the correct stages.

Purity comes before the mole calculation. Limiting-reactant calculations come before product mass. Percentage yield comes last.

09Questions and solutions

The following questions progress from a straightforward purity calculation to problems involving competing reactants, reverse calculations, and an impurity that turns out to be chemically reactive.

Question 1

A 15.0 g sample of magnesium is 80.0% pure. The remaining material is inert and does not react with hydrochloric acid.

The magnesium reacts with excess hydrochloric acid according to:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

Calculate the theoretical mass of magnesium chloride produced.

Use:

• \(M(\mathrm{Mg})=24.31\text{ g mol}^{-1}\)
• \(M(\mathrm{MgCl_2})=95.21\text{ g mol}^{-1}\)

Solution 1

The theoretical mass of magnesium chloride is 47.0 g.

First, correct the magnesium sample mass for purity:

\[ \begin{aligned} m(\mathrm{Mg}) &=15.0\times\frac{80.0}{100}\\ &=12.0\text{ g} \end{aligned} \]

Convert this into moles:

\[ n(\mathrm{Mg})=\frac{12.0}{24.31} =0.4936\text{ mol} \]

The hydrochloric acid is in excess, so magnesium is the limiting reactant.

From the balanced equation, the mole ratio of magnesium to magnesium chloride is 1:1.

\[ n(\mathrm{MgCl_2})=0.4936\text{ mol} \]

Convert into mass:

\[ \begin{aligned} m(\mathrm{MgCl_2}) &=0.4936\times95.21\\ &=47.0\text{ g} \end{aligned} \]

The inert impurities contribute nothing to the magnesium chloride produced.

The tempting mistake is to calculate using all 15.0 g of the sample. That would count 3.00 g of material that contains no magnesium.

Question 2

A 25.0 g limestone sample contains 80.0% calcium carbonate by mass. The remaining material is inert.

The sample reacts with 200.0 mL of 1.50 mol L\(^{-1}\) hydrochloric acid according to:

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

Calculate:

a) The limiting reactant.

b) The theoretical mass of carbon dioxide produced.

c) The mass of calcium carbonate remaining after the reaction.

Use:

• \(M(\mathrm{CaCO_3})=100.09\text{ g mol}^{-1}\)
• \(M(\mathrm{CO_2})=44.01\text{ g mol}^{-1}\)

Solution 2

Hydrochloric acid is the limiting reactant. The theoretical carbon dioxide mass is 6.60 g, and 4.99 g of calcium carbonate remains unreacted.

Step 1: Calculate the pure calcium carbonate mass.

\[ m(\mathrm{CaCO_3})=25.0\times0.800 =20.0\text{ g} \]

Step 2: Convert both reactants into moles.

For calcium carbonate:

\[ n(\mathrm{CaCO_3})=\frac{20.0}{100.09} =0.1998\text{ mol} \]

For hydrochloric acid, convert the volume into litres:

\[ V=0.2000\text{ L} \]

\[ n(\mathrm{HCl})=1.50\times0.2000 =0.300\text{ mol} \]

Step 3: Identify the limiting reactant.

HCl has the smaller value and is therefore limiting.

Step 4: Calculate the theoretical carbon dioxide mass.

The mole ratio of HCl to CO2 is 2:1.

\[ n(\mathrm{CO_2})=\frac{0.300}{2} =0.150\text{ mol} \]

Therefore:

\[ \begin{aligned} m(\mathrm{CO_2}) &=0.150\times44.01\\ &=6.60\text{ g} \end{aligned} \]

Step 5: Calculate the leftover calcium carbonate.

The mole ratio of calcium carbonate to hydrochloric acid is 1:2.

\[ n(\mathrm{CaCO_3\ consumed}) =\frac{0.300}{2} =0.150\text{ mol} \]

Subtract this from the initial amount:

\[ \begin{aligned} n(\mathrm{CaCO_3\ left}) &=0.1998-0.150\\ &=0.0498\text{ mol} \end{aligned} \]

Convert to mass:

\[ \begin{aligned} m(\mathrm{CaCO_3\ left}) &=0.0498\times100.09\\ &=4.99\text{ g} \end{aligned} \]

The original inert impurity mass is also 5.00 g, but it must not be included in the 4.99 g of leftover calcium carbonate.

This question demonstrates why you must divide by the coefficients before deciding which reactant limits the reaction.

Question 3

A 24.0 g sample of aluminium contains an unknown percentage of inert impurities.

The aluminium reacts with excess chlorine according to:

\[ \ce{2Al + 3Cl2 -> 2AlCl3} \]

A dry, pure sample of aluminium chloride with a mass of 40.0 g is recovered. The reaction has a percentage yield of 80.0%.

Calculate the percentage purity of the original aluminium sample.

Use:

• \(M(\mathrm{Al})=26.98\text{ g mol}^{-1}\)
• \(M(\mathrm{AlCl_3})=133.33\text{ g mol}^{-1}\)

Solution 3

The original aluminium sample is 42.2% pure by mass.

The important step is to work backwards from the actual product mass to the theoretical product mass before calculating the original amount of aluminium.

Step 1: Calculate the theoretical aluminium chloride mass.

Rearrange the percentage yield formula:

\[ m_{\text{theoretical}}= \frac{m_{\text{actual}}} {\%\text{Yield}/100} \]

Substitute:

\[ m_{\text{theoretical}} =\frac{40.0}{0.800} =50.0\text{ g} \]

Step 2: Convert theoretical aluminium chloride mass into moles.

\[ n(\mathrm{AlCl_3})=\frac{50.0}{133.33} =0.3750\text{ mol} \]

Step 3: Calculate the amount of aluminium required.

From the balanced equation:

\[ \ce{2Al + 3Cl2 -> 2AlCl3} \]

The mole ratio of aluminium to aluminium chloride is 2:2, or 1:1.

Therefore:

\[ n(\mathrm{Al})=0.3750\text{ mol} \]

Step 4: Calculate the mass of pure aluminium.

\[ \begin{aligned} m(\mathrm{Al}) &=0.3750\times26.98\\ &=10.12\text{ g} \end{aligned} \]

Step 5: Calculate percentage purity.

The original sample weighed 24.0 g, but only 10.12 g was aluminium.

\[ \begin{aligned} \%\text{Purity} &=\frac{10.12}{24.0}\times100\\ &=42.2\% \end{aligned} \]

The sample contains 42.2% aluminium and 57.8% inert impurities by mass.

The main trap is confusing the 80.0% product yield with the purity of the original aluminium. They describe different stages of the process. You must undo the yield calculation before determining purity.

Question 4

A laboratory compares two batches of impure copper(II) oxide. Both are reduced using hydrogen gas according to:

\[ \ce{CuO + H2 -> Cu + H2O} \]

The experimental data are:

Calculate for each batch:

a) The limiting reactant.

b) The theoretical mass of copper produced.

c) The mass of excess reactant remaining.

d) The actual mass of copper recovered.

Which batch produces more copper?

Use:

• \(M(\mathrm{CuO})=79.55\text{ g mol}^{-1}\)
• \(M(\mathrm{Cu})=63.55\text{ g mol}^{-1}\)
• \(M(\mathrm{H_2})=2.016\text{ g mol}^{-1}\)

Solution 4

Batch B produces more recovered copper: 16.2 g compared with 13.5 g from Batch A.

The interesting part is that Batch A starts with more pure copper(II) oxide, but does not produce more copper. Its hydrogen supply is insufficient to react with all the copper(II) oxide.

Batch A

Step 1: Calculate the pure copper(II) oxide mass.

\[ m(\mathrm{CuO})=30.0\times0.800 =24.0\text{ g} \]

Convert into moles:

\[ n(\mathrm{CuO})=\frac{24.0}{79.55} =0.3017\text{ mol} \]

Hydrogen is already supplied in moles:

\[ n(\mathrm{H_2})=0.250\text{ mol} \]

Step 2: Identify the limiting reactant.

Both reactants have a coefficient of 1.

Hydrogen is limiting.

Step 3: Calculate the theoretical copper mass.

The mole ratio of hydrogen to copper is 1:1.

\[ n(\mathrm{Cu})=0.250\text{ mol} \]

\[ \begin{aligned} m(\mathrm{Cu}) &=0.250\times63.55\\ &=15.89\text{ g} \end{aligned} \]

Step 4: Calculate the leftover copper(II) oxide.

\[ n(\mathrm{CuO\ left}) =0.3017-0.250 =0.0517\text{ mol} \]

Convert into mass:

\[ \begin{aligned} m(\mathrm{CuO\ left}) &=0.0517\times79.55\\ &=4.11\text{ g} \end{aligned} \]

Hydrogen is completely consumed.

Step 5: Calculate the actual copper recovered.

The yield is 85.0%.

\[ \begin{aligned} m_{\text{actual}} &=15.89\times0.850\\ &=13.5\text{ g} \end{aligned} \]

Batch A recovers 13.5 g of copper.

Batch B

Step 6: Calculate the pure copper(II) oxide mass.

\[ m(\mathrm{CuO})=25.0\times0.900 =22.5\text{ g} \]

Convert into moles:

\[ n(\mathrm{CuO})=\frac{22.5}{79.55} =0.2828\text{ mol} \]

Hydrogen is supplied at:

\[ n(\mathrm{H_2})=0.400\text{ mol} \]

Step 7: Identify the limiting reactant.

Copper(II) oxide is limiting.

This is different from Batch A. Batch B has enough hydrogen to react with all its pure copper(II) oxide.

Step 8: Calculate the theoretical copper mass.

The mole ratio of copper(II) oxide to copper is 1:1.

\[ n(\mathrm{Cu})=0.2828\text{ mol} \]

\[ \begin{aligned} m(\mathrm{Cu}) &=0.2828\times63.55\\ &=17.97\text{ g} \end{aligned} \]

Step 9: Calculate the leftover hydrogen.

\[ \begin{aligned} n(\mathrm{H_2\ left}) &=0.400-0.2828\\ &=0.1172\text{ mol} \end{aligned} \]

Convert into mass:

\[ \begin{aligned} m(\mathrm{H_2\ left}) &=0.1172\times2.016\\ &=0.236\text{ g} \end{aligned} \]

No copper(II) oxide remains unreacted.

Step 10: Calculate the actual copper recovered.

\[ \begin{aligned} m_{\text{actual}} &=17.97\times0.900\\ &=16.2\text{ g} \end{aligned} \]

Batch B recovers 16.2 g of copper.

Final comparison

Batch B produces approximately 2.7 g more copper.

The tempting prediction is that Batch A should win because it starts with more pure copper(II) oxide. However, its hydrogen supply limits the reaction, while Batch B has enough hydrogen to consume all its copper(II) oxide and also achieves a higher percentage yield.

This is why you must consider purity, limiting reactants, and percentage yield together.

Question 5

A 12.0 g solid sample contains three substances:

• Magnesium metal, Mg.
• Magnesium oxide, MgO.
• An inert impurity that does not react with hydrochloric acid.

The sample reacts with 400.0 mL of 1.00 mol L\(^{-1}\) hydrochloric acid.

Two reactions occur:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

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

All the hydrochloric acid is consumed. The hydrogen gas produced is collected and has a volume of 2.40 L at 100 kPa and 25°C.

Calculate:

a) The mass of magnesium metal in the original sample.

b) The mass of magnesium oxide in the original sample.

c) The mass of inert impurity.

d) The theoretical mass of magnesium chloride produced.

e) The percentage yield if 17.1 g of dry, pure magnesium chloride is recovered.

Assume ideal gas behaviour, no gas escapes during collection, and hydrochloric acid reacts only with magnesium and magnesium oxide.

Use:

• \(M(\mathrm{Mg})=24.31\text{ g mol}^{-1}\)
• \(M(\mathrm{MgO})=40.31\text{ g mol}^{-1}\)
• \(M(\mathrm{MgCl_2})=95.21\text{ g mol}^{-1}\)
• \(R=8.314\text{ kPa L mol}^{-1}\text{ K}^{-1}\)

Solution 5

The original sample contains approximately:

• 2.35 g magnesium
• 4.16 g magnesium oxide
• 5.49 g inert impurity

The theoretical magnesium chloride mass is 19.0 g, giving a percentage yield of 89.8%.

This question has an important complication: magnesium oxide is not an inert impurity. It reacts with hydrochloric acid and produces the same salt as magnesium metal.

A calculation that treats everything other than magnesium as inert will therefore be incorrect.

Step 1: Calculate the initial amount of hydrochloric acid.

Convert the volume into litres:

\[ 400.0\text{ mL}=0.4000\text{ L} \]

Use \(n=cV\):

\[ \begin{aligned} n(\mathrm{HCl}) &=1.00\times0.4000\\ &=0.400\text{ mol} \end{aligned} \]

All 0.400 mol of hydrochloric acid is consumed.

Step 2: Calculate the amount of hydrogen gas collected.

Use the ideal gas equation:

\[ PV=nRT \]

Rearrange:

\[ n=\frac{PV}{RT} \]

Convert the temperature into kelvin:

\[ T=25+273=298\text{ K} \]

Substitute the supplied values:

\[ \begin{aligned} n(\mathrm{H_2}) &=\frac{100\times2.40} {8.314\times298}\\ &=0.0969\text{ mol} \end{aligned} \]

Step 3: Calculate the amount of magnesium metal.

Only the magnesium metal reaction produces hydrogen:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

The mole ratio of magnesium to hydrogen is 1:1.

Therefore:

\[ n(\mathrm{Mg})=0.0969\text{ mol} \]

Convert this into mass:

\[ \begin{aligned} m(\mathrm{Mg}) &=0.0969\times24.31\\ &=2.35\text{ g} \end{aligned} \]

The original sample contains 2.35 g of magnesium metal.

Step 4: Calculate the amount of magnesium oxide.

This is the important reasoning step.

Both reactions consume two moles of hydrochloric acid per mole of magnesium-containing reactant.

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

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

Therefore, the total amount of magnesium and magnesium oxide that reacted is:

\[ n(\mathrm{Mg})+n(\mathrm{MgO}) =\frac{n(\mathrm{HCl})}{2} \]

Substitute:

\[ n(\mathrm{Mg})+n(\mathrm{MgO}) =\frac{0.400}{2} =0.200\text{ mol} \]

We already know that magnesium metal contributes 0.0969 mol.

Therefore:

\[ \begin{aligned} n(\mathrm{MgO}) &=0.200-0.0969\\ &=0.1031\text{ mol} \end{aligned} \]

Convert this into mass:

\[ \begin{aligned} m(\mathrm{MgO}) &=0.1031\times40.31\\ &=4.16\text{ g} \end{aligned} \]

The original sample contains 4.16 g of magnesium oxide.

Step 5: Calculate the mass of inert impurity.

The original sample has a total mass of 12.0 g.

Subtract the masses of both reactive substances:

\[ \begin{aligned} m(\text{inert impurity}) &=12.0-2.35-4.16\\ &=5.49\text{ g} \end{aligned} \]

The mass balance is:

\[ 2.35+4.16+5.49=12.0\text{ g} \]

The result is consistent with the original sample mass.

Step 6: Calculate the theoretical magnesium chloride mass.

Here is the final important detail.

Both magnesium and magnesium oxide produce magnesium chloride.

From the first reaction:

\[ 1\text{ mol Mg}\rightarrow1\text{ mol MgCl}_2 \]

From the second reaction:

\[ 1\text{ mol MgO}\rightarrow1\text{ mol MgCl}_2 \]

Therefore, the total amount of magnesium chloride is:

\[ \begin{aligned} n(\mathrm{MgCl_2}) &=0.0969+0.1031\\ &=0.200\text{ mol} \end{aligned} \]

Convert into mass:

\[ \begin{aligned} m(\mathrm{MgCl_2}) &=0.200\times95.21\\ &=19.0\text{ g} \end{aligned} \]

The theoretical magnesium chloride mass is 19.0 g.

Step 7: Calculate the percentage yield.

The actual mass of dry magnesium chloride recovered is 17.1 g.

\[ \begin{aligned} \%\text{Yield} &=\frac{17.1}{19.042}\times100\\ &=89.8\% \end{aligned} \]

The percentage yield is 89.8%.

The conceptual trap is treating all material that is not magnesium metal as inert.

Magnesium oxide is chemically different from an inert impurity because it also reacts with hydrochloric acid and produces magnesium chloride. However, it does not produce hydrogen gas.

That means the hydrogen measurement tells us the amount of magnesium metal, while the acid consumption tells us the total amount of magnesium and magnesium oxide that reacted.

Combining those two pieces of evidence allows us to determine the composition of the original sample and the theoretical product mass.

10The final calculation checklist

When you encounter an impure reactant in an HSC Chemistry stoichiometry question, check your working in this order:

1. Balance the equation. Make sure every coefficient is correct before comparing reactants.
2. Correct for purity. Calculate the mass of the actual substance present, not the total sample mass.
3. Convert to moles. Use \(n=m/M\) for solids and \(n=cV\) for solutions.
4. Find the limiting reactant. Divide every initial reactant amount by its balanced coefficient. The smallest result identifies the limiting reactant.
5. Calculate theoretical product. Use the limiting reactant and the appropriate mole ratio.
6. Calculate leftovers. Subtract the amount consumed from the initial amount of each reactant.
7. Apply percentage yield. Only do this after calculating theoretical product mass.

The most important distinction is between an inert impurity and an excess reactant. An inert impurity never enters the reaction calculation. An excess reactant does enter the calculation, but some of it remains after the limiting reactant is consumed.

Once you can separate those two ideas, impure-reactant questions become much easier to organise. The same reasoning also prepares you for more demanding HSC questions that combine limiting reagents, excess reactants, percentage purity, and percentage yield in a single calculation.