Why a Product Can Weigh More Than One Reactant | HSC Chemistry

Mix 0.585 g of sodium chloride with excess silver nitrate, and you can theoretically produce 1.43 g of silver chloride. The product weighs more than twice as much as the sodium chloride you started with.

Has the reaction created mass?

No. The silver chloride contains atoms from two different reactants. Chlorine comes from sodium chloride, while silver comes from silver nitrate. The product’s mass includes contributions from both.

This is the key to understanding why a calculated product mass can exceed the mass of one reactant: a product can contain atoms supplied by other reactants, so comparing its mass with just one starting substance is not a complete mass balance.

Let’s follow the atoms through an actual type of chemical reaction, then build the calculation method you can use in HSC Chemistry questions.

01An experiment that reveals where the extra mass comes from

Consider a reaction between sodium chloride and silver nitrate.

Both substances dissolve in water. When their solutions are mixed, a white solid called silver chloride forms.

The balanced equation is:

\[ \ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)} \]

Notice that there are two products:

• Silver chloride, \(\ce{AgCl}\), which forms a solid precipitate.
• Sodium nitrate, \(\ce{NaNO3}\), which remains dissolved in the water.

The silver chloride is the substance we want to collect and weigh.

But where does each part of it come from?

Following the atoms through the apparatus

Imagine a gravimetric experiment, where the mass of a collected solid is used to investigate how much product has formed.

The apparatus has several parts:

• A balance measures the initial masses of the reactants and the final mass of the collected solid.
• Two beakers hold the separately dissolved reactants.
• A third beaker receives the combined solutions, where silver chloride forms.
• A filter funnel lined with pre-weighed filter paper separates the solid from the liquid.
• A conical flask below the funnel collects the liquid that passes through.
• The filter paper and solid are dried before their final mass is recorded.

The silver chloride remains on the filter paper. The dissolved sodium nitrate passes through into the flask.

What should the experiment show?

If the product contains silver atoms from silver nitrate as well as chlorine atoms from sodium chloride, its mass should be greater than the mass of the sodium chloride alone.

The ionic equation makes this especially clear:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

Silver ions combine with chloride ions to form the solid.

The sodium ions and nitrate ions do not disappear. They remain in solution and are present in the sodium nitrate.

The solid contains material from both reactants, even though only one of the reactants supplies the chloride.

What happens to the measured mass?

Consider the following illustrative theoretical data:

• Sodium chloride used: 0.585 g.
• Silver nitrate used: 2.00 g.
• Molar mass of sodium chloride: 58.5 g/mol.
• Molar mass of silver nitrate: 169.9 g/mol.
• Molar mass of silver chloride: 143.4 g/mol.

First, convert the reactant masses into moles.

For sodium chloride:

\[ n(\ce{NaCl})=\frac{0.585}{58.5} \]

\[ =0.0100\text{ mol} \]

For silver nitrate:

\[ n(\ce{AgNO3})=\frac{2.00}{169.9} \]

\[ =0.0118\text{ mol} \]

The balanced equation gives a 1:1 mole ratio between sodium chloride and silver nitrate.

Therefore, sodium chloride is the limiting reactant.

It can produce 0.0100 mol of silver chloride.

The theoretical product mass is:

\[ m(\ce{AgCl})=nM \]

\[ =0.0100\times143.4 \]

\[ =\boxed{1.43\text{ g}} \]

The calculated mass of silver chloride is greater than the original 0.585 g of sodium chloride.

That is entirely reasonable.

The silver chloride contains approximately:

The product gains silver from silver nitrate. Meanwhile, the sodium from sodium chloride remains in the solution as sodium nitrate.

The product is therefore not simply the original sodium chloride with a different formula. Its atoms have been rearranged.

Checking the complete mass balance

Now consider all the substances involved, rather than just the precipitate.

The balanced equation has a 1:1:1:1 mole ratio.

For every 0.0100 mol of sodium chloride consumed:

• 0.0100 mol of silver nitrate is consumed.
• 0.0100 mol of silver chloride forms.
• 0.0100 mol of sodium nitrate forms.

The mass balance is:

The total mass of the consumed reactants is:

\[ 0.585+1.699=2.284\text{ g} \]

The total mass of the products is:

\[ 1.434+0.850=2.284\text{ g} \]

Mass is conserved.

The original 2.00 g of silver nitrate included approximately 0.301 g that was not consumed. It remains in the solution, so it must also be accounted for when considering the entire mixture.

The solvent water is omitted from this solute-only calculation because it is not consumed in the reaction.

The important point is that the precipitate is only one of the products. Its mass does not have to equal the mass of either individual reactant.

Why the experimental design needs care

There is a measurement problem here.

Suppose the precipitate is collected and weighed immediately after filtration. The filter paper might still contain water and dissolved sodium nitrate.

Both would increase the measured mass, even though neither belongs in the theoretical mass of pure silver chloride.

A better design is to:

• Wash the precipitate to remove soluble substances.
• Dry it thoroughly.
• Cool it before weighing.
• Repeat the drying and weighing process until a constant mass is obtained.
• Subtract the original mass of the filter paper.

This produces a more reliable estimate of the mass of pure silver chloride.

There are still possible losses during transfer and filtration. The experiment also assumes the reaction proceeds as described and that the collected solid is pure.

A result greater than the theoretical mass is therefore a warning to investigate contamination, moisture, measurement errors, or incorrect assumptions. It is not evidence that the reaction created extra matter.

02The calculation method: mass to moles to moles to mass

The silver chloride experiment demonstrates the physical reason for the increase in product mass.

Now let’s turn that reasoning into a method you can apply to unfamiliar reaction questions.

The safest approach is:

\[ \boxed{ \text{Mass} \rightarrow \text{Moles} \rightarrow \text{Mole ratio} \rightarrow \text{Moles} \rightarrow \text{Mass} } \]

Step 1: Write a balanced chemical equation

The coefficients tell you the relative amounts, in moles, of each substance that reacts or forms.

For example:

\[ \ce{2Mg(s) + O2(g) -> 2MgO(s)} \]

This equation tells you that:

• 2 mol of magnesium react with 1 mol of oxygen gas.
• 2 mol of magnesium oxide form.

The mole ratio between magnesium and magnesium oxide is 2:2, which simplifies to 1:1.

Coefficients are mole ratios, not mass ratios.

Two moles of magnesium do not have the same mass as one mole of oxygen gas.

Their molar masses are:

Two moles of magnesium have a mass of 48.6 g.

One mole of oxygen gas has a mass of 32.0 g.

Together, they form two moles of magnesium oxide with a mass of 80.6 g.

\[ 48.6+32.0=80.6\text{ g} \]

The product weighs more than the magnesium because oxygen has been added.

Step 2: Convert the known reactant mass into moles

Use:

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

Where:

• \(n\) is the amount of substance, in moles (mol).
• \(m\) is the mass, in grams (g).
• \(M\) is the molar mass, in grams per mole (g/mol).

This conversion is essential because the balanced equation gives you mole ratios.

Step 3: Apply the coefficient ratio

Use the balanced equation to determine how many moles of product form.

The general relationship is:

\[ n(\text{product})= n(\text{reactant}) \times \frac{\text{product coefficient}} {\text{reactant coefficient}} \]

If more than one reactant has a specified mass, calculate the available moles of each and identify the limiting reactant first.

A useful decision rule is to divide each reactant’s available mole amount by its coefficient. The smallest result identifies the limiting reactant.

Step 4: Convert product moles into mass

Once you know the amount of product, use:

\[ m=nM \]

This gives the theoretical mass of product.

The molar mass of the product includes the mass of every atom in its formula. If those atoms came from different reactants, their masses are all included in the final answer.

05When can a product mass exceed a reactant mass?

It helps to compare two different types of reactions.

Case 1: Two reactants combine

Consider:

\[ \ce{2Mg + O2 -> 2MgO} \]

For every 2 mol of magnesium, 1 mol of oxygen gas is consumed.

The magnesium oxide contains atoms from both reactants.

Consequently, its mass is greater than the mass of magnesium alone.

Case 2: One reactant breaks into products

Now consider the thermal decomposition of calcium carbonate:

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

One mole of calcium carbonate has a mass of approximately 100.1 g.

It produces:

• 56.1 g of calcium oxide.
• 44.0 g of carbon dioxide.

Neither individual product is heavier than the original calcium carbonate.

The atoms from one reactant have simply been distributed between two products.

This is a useful contrast. A product can exceed the mass of an individual reactant when it incorporates atoms from other reactants. It cannot exceed the total mass of all consumed reactants in an ordinary chemical reaction when the complete system is accounted for.

Three checks to make before accepting your answer

Check 1: Have you used mole ratios rather than mass ratios?

Never apply the coefficients directly to reactant masses. Convert to moles first.

Check 2: Have you accounted for every reactant?

A product may contain atoms from multiple starting substances. If two reactants are supplied, check whether one is limiting.

Check 3: Have you compared the correct quantities?

Compare the total mass of products with the total mass of reactants consumed, not the mass of one product with one reactant.

For a more detailed method of checking your calculations, see How to Check Whether a Reaction-Mass Answer Conserves Atoms.

Also remember that theoretical product mass and experimentally collected product mass are different quantities. See How to Distinguish Actual, Theoretical and Percentage Yield.

06Questions and solutions

The following questions move from a straightforward mass calculation to more challenging problems involving limiting reactants, redox reactions, experimental errors, and the difference between reaction completion and product recovery.

Question 1

A sample of 0.240 g of carbon undergoes complete combustion in excess oxygen to produce carbon dioxide.

The reaction is:

\[ \ce{C + O2 -> CO2} \]

Calculate the theoretical mass of carbon dioxide produced.

Use the following molar masses:

• C = 12.0 g/mol.
• CO2 = 44.0 g/mol.

Solution 1

The theoretical mass of carbon dioxide is 0.880 g. This exceeds the original carbon mass because oxygen from the other reactant becomes part of the carbon dioxide.

Step 1: Calculate the moles of carbon.

\[ n(\ce{C})=\frac{0.240}{12.0} \]

\[ =0.0200\text{ mol} \]

Step 2: Apply the mole ratio.

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

Therefore:

\[ n(\ce{CO2})=0.0200\text{ mol} \]

Step 3: Calculate the product mass.

\[ m(\ce{CO2})=nM \]

\[ =0.0200\times44.0 \]

\[ =\boxed{0.880\text{ g}} \]

The mass difference is:

\[ 0.880-0.240=0.640\text{ g} \]

This 0.640 g is oxygen incorporated into the carbon dioxide.

The oxygen mass balances the difference between the carbon mass and the product mass.

Question 2

Aluminium reacts with chlorine gas to produce aluminium chloride.

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

A reaction mixture contains:

• 1.08 g of aluminium.
• 4.26 g of chlorine gas.

Assume both reactants are completely consumed.

(a) Determine the number of moles of each reactant.

(b) Determine whether either reactant is limiting.

(c) Calculate the theoretical mass of aluminium chloride produced.

(d) Explain how the product mass can exceed the starting mass of both reactants individually without violating conservation of mass.

Use:

• Al = 27.0 g/mol.
• Cl2 = 71.0 g/mol.
• AlCl3 = 133.5 g/mol.

Solution 2

The reaction produces 5.34 g of aluminium chloride. Neither reactant is limiting because the starting amounts are exactly in the required stoichiometric ratio.

(a) Calculate the reactant moles.

For aluminium:

\[ n(\ce{Al})=\frac{1.08}{27.0} \]

\[ =0.0400\text{ mol} \]

For chlorine gas:

\[ n(\ce{Cl2})=\frac{4.26}{71.0} \]

\[ =0.0600\text{ mol} \]

(b) Identify the limiting reactant.

The balanced equation requires:

\[ 2\text{ mol Al}:3\text{ mol Cl}_2 \]

For 0.0400 mol of aluminium, the required chlorine amount is:

\[ 0.0400\times\frac{3}{2} =0.0600\text{ mol} \]

Exactly 0.0600 mol of chlorine is available.

Therefore, neither reactant is in excess. Both are completely consumed.

(c) Calculate the aluminium chloride mass.

The mole ratio between aluminium and aluminium chloride is 2:2, or 1:1.

Therefore:

\[ n(\ce{AlCl3})=0.0400\text{ mol} \]

Using the molar mass:

\[ m(\ce{AlCl3})=0.0400\times133.5 \]

\[ =\boxed{5.34\text{ g}} \]

(d) Explain the mass increase.

The starting aluminium mass is 1.08 g, and the starting chlorine gas mass is 4.26 g.

The product mass is greater than either individual starting mass.

However:

\[ 1.08+4.26=5.34\text{ g} \]

The aluminium chloride contains aluminium and chlorine atoms from both reactants.

The complete mass of the reactants equals the complete mass of the product. There is no violation of conservation of mass.

Question 3

Iron can be extracted from iron(III) oxide using carbon monoxide.

The reaction is:

\[ \ce{Fe2O3 + 3CO -> 2Fe + 3CO2} \]

A reaction vessel initially contains:

• 16.0 g of pure iron(III) oxide.
• 5.60 g of carbon monoxide.

Assume the reaction proceeds completely until the limiting reactant is consumed, with no other reactions occurring.

(a) Determine the limiting reactant.

(b) Calculate the mass of iron produced.

(c) Calculate the mass of carbon dioxide produced.

(d) Explain why the carbon dioxide mass exceeds the original carbon monoxide mass.

(e) Calculate the mass of iron(III) oxide remaining and verify the mass balance.

Use:

• Fe = 55.8 g/mol.
• O = 16.0 g/mol.
• C = 12.0 g/mol.

Solution 3

Carbon monoxide is the limiting reactant. The reaction produces 7.44 g of iron and 8.80 g of carbon dioxide, while approximately 5.36 g of iron(III) oxide remains unreacted.

The important detail is that oxygen transfers from iron(III) oxide to carbon monoxide.

(a) Determine the limiting reactant.

First, calculate the molar masses:

\[ M(\ce{Fe2O3})=159.6\text{ g/mol} \]

\[ M(\ce{CO})=28.0\text{ g/mol} \]

Calculate the available moles.

For iron(III) oxide:

\[ n(\ce{Fe2O3})=\frac{16.0}{159.6} \]

\[ =0.1003\text{ mol} \]

For carbon monoxide:

\[ n(\ce{CO})=\frac{5.60}{28.0} \]

\[ =0.200\text{ mol} \]

The equation requires 3 mol of CO for every 1 mol of Fe2O3.

To consume all 0.1003 mol of iron(III) oxide, the required carbon monoxide would be:

\[ 0.1003\times3=0.3009\text{ mol} \]

Only 0.200 mol is available.

Therefore, carbon monoxide is the limiting reactant.

(b) Calculate the mass of iron produced.

The equation gives:

\[ 3\text{ mol CO}:2\text{ mol Fe} \]

Therefore:

\[ n(\ce{Fe})=0.200\times\frac{2}{3} \]

\[ =0.1333\text{ mol} \]

Convert to mass:

\[ m(\ce{Fe})=0.1333\times55.8 \]

\[ =\boxed{7.44\text{ g}} \]

(c) Calculate the mass of carbon dioxide produced.

The equation gives a 3:3 ratio between carbon monoxide and carbon dioxide.

Therefore:

\[ n(\ce{CO2})=0.200\text{ mol} \]

The molar mass of carbon dioxide is:

\[ M(\ce{CO2})=44.0\text{ g/mol} \]

Hence:

\[ m(\ce{CO2})=0.200\times44.0 \]

\[ =\boxed{8.80\text{ g}} \]

(d) Explain why carbon dioxide is heavier than carbon monoxide.

The initial carbon monoxide mass is 5.60 g.

The carbon dioxide mass is 8.80 g.

The difference is:

\[ 8.80-5.60=3.20\text{ g} \]

This extra 3.20 g is oxygen transferred from the iron(III) oxide.

Carbon monoxide contains one oxygen atom per molecule, while carbon dioxide contains two.

The additional oxygen comes from the iron oxide, which is being reduced to iron.

The carbon dioxide therefore contains atoms from two different reactants.

(e) Verify the mass balance.

First, calculate the amount of iron(III) oxide consumed.

The equation gives:

\[ n(\ce{Fe2O3})=0.200\times\frac{1}{3} \]

\[ =0.0667\text{ mol} \]

Its consumed mass is:

\[ m=0.0667\times159.6 \]

\[ =10.64\text{ g} \]

The remaining iron(III) oxide is:

\[ 16.0-10.64=\boxed{5.36\text{ g}} \]

Now compare the mass of the consumed reactants with the mass of the products.

Consumed reactants:

\[ 10.64+5.60=16.24\text{ g} \]

Products:

\[ 7.44+8.80=16.24\text{ g} \]

The reaction conserves mass.

The complete final contents also include the 5.36 g of unreacted iron(III) oxide.

This question demonstrates why you must account for every substance involved rather than comparing only the mass of one product with one reactant.

Question 4

A student performs the silver chloride precipitation experiment described earlier.

The starting quantities are:

• 0.585 g of sodium chloride.
• 2.00 g of silver nitrate.

After mixing, filtering and drying, the student collects 1.60 g of solid.

The student calculates:

\[ \text{Percentage yield} = \frac{1.60}{1.43}\times100 \]

\[ =112\% \]

The student concludes that the reaction has produced more silver chloride than theoretically possible.

Evaluate this conclusion.

Identify two possible experimental problems and suggest how the experimental design could be improved.

Solution 4

The student’s conclusion is incorrect. A measured yield of approximately 112% indicates that the collected solid is likely contaminated or that there is an error in the experimental data or assumptions.

The reaction cannot produce more pure silver chloride than the theoretical amount calculated from the limiting reactant.

Step 1: Calculate the theoretical yield.

The balanced equation is:

\[ \ce{AgNO3 + NaCl -> AgCl + NaNO3} \]

Calculate the moles of sodium chloride:

\[ n(\ce{NaCl})=\frac{0.585}{58.5} \]

\[ =0.0100\text{ mol} \]

Calculate the moles of silver nitrate:

\[ n(\ce{AgNO3})=\frac{2.00}{169.9} \]

\[ =0.0118\text{ mol} \]

The mole ratio is 1:1, so sodium chloride is limiting.

Therefore:

\[ n(\ce{AgCl})=0.0100\text{ mol} \]

The theoretical mass is:

\[ m(\ce{AgCl})=0.0100\times143.4 \]

\[ =1.434\text{ g} \]

Step 2: Calculate the apparent percentage yield.

\[ \text{Percentage yield} = \frac{\text{Actual yield}} {\text{Theoretical yield}}\times100 \]

\[ =\frac{1.60}{1.434}\times100 \]

\[ =\boxed{111.6\%} \]

This is approximately 112%.

However, the collected 1.60 g cannot represent pure silver chloride formed under the stated assumptions.

Step 3: Identify possible experimental problems.

Two likely explanations are:

• The precipitate was not washed properly. Dissolved sodium nitrate or excess silver nitrate may have remained on the filter paper and contributed to the measured mass.
• The precipitate was not completely dry. Retained water would increase the measured mass.

Other possibilities include contamination of the filter paper, incorrect weighing, or an error in the initial measurements.

Step 4: Improve the experimental design.

The student should:

• Wash the precipitate with distilled water.
• Dry it thoroughly.
• Cool it before weighing.
• Repeat drying and weighing until a constant mass is reached.
• Subtract the original filter paper mass.

The corrected measurement should approach the theoretical yield, although some product may be lost during filtration or transfer.

The key distinction is that an experimentally measured mass is not automatically the mass of pure product.

Question 5

A chemist mixes exactly 0.0100 mol of barium chloride with exactly 0.0100 mol of magnesium sulfate.

The reaction is:

\[ \ce{BaCl2 + MgSO4 -> BaSO4 + MgCl2} \]

The chemist filters the mixture and obtains 2.10 g of dry barium sulfate.

Use the following molar masses:

Answer the following:

(a) Calculate the initial mass of each reactant and the theoretical mass of each product.

(b) Demonstrate that the theoretical mass of barium sulfate exceeds the initial mass of either individual reactant.

(c) The collected barium sulfate mass is approximately 90.0% of the theoretical mass. Does this prove that only 90.0% of the reaction occurred?

(d) Construct two different explanations that are both consistent with the measured 2.10 g of barium sulfate:

• The reaction proceeds to only 90.0% completion, with all precipitate successfully collected.
• The reaction proceeds to 100% completion, but only 90.0% of the precipitate is collected.

For each explanation, calculate the amount of magnesium chloride formed.

(e) Suggest an additional measurement that could distinguish between the two explanations.

Solution 5

The theoretical mass of barium sulfate is 2.334 g, which exceeds the mass of either starting reactant individually. The collected mass of 2.10 g does not, by itself, reveal whether the reaction was incomplete or whether some precipitate was lost during collection.

This is an important distinction between chemical reaction completion and experimental product recovery.

(a) Calculate the initial reactant masses and theoretical product masses.

The initial amount of each reactant is 0.0100 mol.

For barium chloride:

\[ m(\ce{BaCl2})=0.0100\times208.3 \]

\[ =\boxed{2.083\text{ g}} \]

For magnesium sulfate:

\[ m(\ce{MgSO4})=0.0100\times120.4 \]

\[ =\boxed{1.204\text{ g}} \]

The balanced equation has a 1:1:1:1 mole ratio.

Therefore, both products form in amounts of 0.0100 mol.

For barium sulfate:

\[ m(\ce{BaSO4})=0.0100\times233.4 \]

\[ =\boxed{2.334\text{ g}} \]

For magnesium chloride:

\[ m(\ce{MgCl2})=0.0100\times95.3 \]

\[ =\boxed{0.953\text{ g}} \]

(b) Demonstrate that the product exceeds either individual reactant mass.

Compare the theoretical barium sulfate mass with the starting reactant masses:

\[ 2.334>2.083 \]

and

\[ 2.334>1.204 \]

Therefore, the barium sulfate is heavier than both the original barium chloride and the original magnesium sulfate individually.

This happens because the barium sulfate contains barium atoms from barium chloride and sulfate ions from magnesium sulfate.

The other atoms are accounted for in magnesium chloride.

Check the complete mass balance:

\[ \text{Total reactant mass} =2.083+1.204 \]

\[ =3.287\text{ g} \]

\[ \text{Total product mass} =2.334+0.953 \]

\[ =3.287\text{ g} \]

Mass is conserved.

(c) Calculate the apparent percentage yield.

The collected mass is 2.10 g.

The theoretical mass is 2.334 g.

\[ \text{Percentage yield} = \frac{2.10}{2.334}\times100 \]

\[ =\boxed{90.0\%} \]

This is the isolated percentage yield.

However, it does not necessarily mean that only 90.0% of the reactants reacted.

The measured mass tells us how much precipitate was collected. It does not directly tell us how much precipitate was originally formed.

(d) Compare the two possible explanations.

Explanation 1: Only 90.0% of the reaction occurs.

If only 90.0% of the initial reactants react, the amount of barium sulfate formed is:

\[ n(\ce{BaSO4})=0.0100\times0.900 \]

\[ =0.00900\text{ mol} \]

Its mass is:

\[ m=0.00900\times233.4 \]

\[ =2.101\text{ g} \]

This is approximately the 2.10 g collected.

Because the equation has a 1:1 ratio between barium sulfate and magnesium chloride:

\[ n(\ce{MgCl2})=0.00900\text{ mol} \]

Its mass is:

\[ m(\ce{MgCl2})=0.00900\times95.3 \]

\[ =\boxed{0.858\text{ g}} \]

The remaining unreacted reactants are:

\[ 0.0100-0.00900=0.00100\text{ mol} \]

For barium chloride:

\[ m=0.00100\times208.3 \]

\[ =0.208\text{ g} \]

For magnesium sulfate:

\[ m=0.00100\times120.4 \]

\[ =0.120\text{ g} \]

The remaining reactants stay in solution.

Explanation 2: The reaction is complete, but some precipitate is lost.

Now assume that all 0.0100 mol of each reactant reacts.

The theoretical amount of barium sulfate formed is still:

\[ 0.0100\text{ mol} \]

Its total mass is:

\[ 2.334\text{ g} \]

However, only 2.10 g is collected.

The missing mass is:

\[ 2.334-2.10=0.234\text{ g} \]

This 0.234 g may have remained on the original beaker, the filter funnel, or another part of the apparatus.

The full amount of magnesium chloride is still formed:

\[ n(\ce{MgCl2})=0.0100\text{ mol} \]

\[ m(\ce{MgCl2})=0.0100\times95.3 \]

\[ =\boxed{0.953\text{ g}} \]

The reaction is 100% complete, even though only 90.0% of the theoretical precipitate mass is collected.

(e) Suggest an additional measurement.

The chemist could quantitatively test the filtrate for unreacted barium ions or sulfate ions.

If the reaction stopped at 90.0% completion, approximately 0.00100 mol of each reactant would remain unreacted in solution.

If the reaction went to completion but some precipitate was lost during transfer, there should be no comparable quantity of unreacted starting ions remaining.

The filtrate analysis would therefore help distinguish incomplete reaction from product loss.

The important insight is that the same collected product mass can arise from two different chemical and experimental situations.

A percentage yield describes the amount of product isolated relative to the theoretical amount. It does not independently reveal exactly how much reactant reacted or how much product was lost during collection.

07The main idea to remember

A calculated product mass can exceed the mass of one reactant because the product may contain atoms from multiple reactants.

The balanced equation tells you the mole ratios, while molar masses tell you the mass of each substance.

For HSC Chemistry calculations, keep returning to this sequence:

\[ \boxed{ \text{Reactant mass} \rightarrow \text{Reactant moles} \rightarrow \text{Mole ratio} \rightarrow \text{Product moles} \rightarrow \text{Product mass} } \]

And when an answer seems surprisingly large, don’t immediately assume you’ve made a calculation error.

First ask whether you’ve accounted for every reactant that contributes atoms to the product.

The next useful skill is checking whether the limiting reactant, theoretical yield, and total atom balance all agree. Those checks will catch most errors before they reach your final answer.