What is the difference between percentage yield and percentage purity in Secondary 3 Chemistry? Percentage yield compares the amount of desired product actually obtained with the amount the balanced equation predicts could be obtained. Percentage purity compares the amount of the named pure substance with the total mass of a sample containing it. The two formulas resemble each other because both use percentages, but their denominators answer different questions. % yield = actual desired product ÷ theoretical desired product × 100; % purity = mass of pure substance ÷ total sample mass × 100.
Parents looking for a Secondary 3 Chemistry tutor in Hougang, percentage yield questions, percentage purity calculations, mole concept tuition near Kovan, Pure Chemistry G3 tuition or SEC Chemistry exam help often show me a striking kind of error: the child can calculate every number but chooses the wrong comparison. In one question, 8.0 g is a recovered product. In another, 8.0 g is the mass of a mixture. The arithmetic might be flawless while the meaning is mistaken. I want students to learn to ask what the mass belongs to before they touch a formula.
This guide builds a reliable percentage yield versus purity worked-example method, including theoretical yield from balanced equations, impure reactants, gas measurements, mass loss, why an apparent yield can exceed 100%, and how to tell a chemistry misunderstanding from a unit or reading error. It supports upper-secondary progression; students should match the depth of practice to their school’s G1, G2, G3, Pure or Combined Science course rather than assume every pathway is identical.
Hougang Secondary 3 Chemistry learning pathway · Hougang Chemistry tutor teaching-quality guide · 2027 SEAB G3 syllabuses
The answer in one minute
Before calculating anything, underline the noun after the number. Is it a product, a reactant, a mixture, or a pure component? Then identify the comparison the question wants.
- Yield question: How much of the target product did we actually obtain compared with its theoretical maximum?
- Purity question: How much of the weighed sample is actually the specified substance?
- Theoretical yield: How much desired product does a balanced equation predict from the limiting amount of reactant, assuming complete conversion to that product under the stated model?
- Actual yield: How much of the desired product was obtained and, where relevant, verified as the desired material.
- Impure sample: The weighed material contains something other than the substance whose amount we need.
- Final exam check: Is the denominator the predicted product amount, or is it the total sample mass?
If a student can choose the denominator for three unfamiliar situations without seeing an equation, the core idea is already becoming secure.
Two beakers that expose the problem
Imagine one labelled vessel contains an 8.0 g sample of material, of which 6.0 g is the desired chemical and 2.0 g is something else. This is a purity question: 6.0 ÷ 8.0 × 100 = 75% purity. We have learned about the *composition of the sample*.
Imagine a different experiment where a reaction was predicted to form 8.0 g of a product, but only 6.0 g of pure product was collected. This is a yield question: 6.0 ÷ 8.0 × 100 = 75% yield. We have learned about the *outcome relative to a theoretical reaction amount*.
The two numerical answers happen to be the same. The Chemistry is not. In the first situation, nothing tells us how efficiently the product was prepared. In the second, nothing tells us whether the starting reactant was pure unless the question gives that information. This is precisely why memorising a percentage formula without naming the quantities leads to so many marks lost.
I ask learners to finish this sentence: “The denominator represents …” If they cannot, I postpone the calculator and return to the story of the reaction.
Understanding percentage yield in plain language
A balanced chemical equation provides relationships between reactants and products. Convert the amount of an available reactant into moles, apply the equation’s coefficient ratio, and turn the predicted amount of desired product into the requested unit. That prediction is the theoretical yield for the specified reaction and assumptions.
The actual yield is the amount obtained in practice. It may be lower because the reaction does not go fully to completion, some reactant goes toward a competing reaction, product is lost during transfer or filtration, or the isolation process fails to collect everything. These are possible explanations, not a list to repeat automatically. You should connect the claimed cause with the actual experimental method.
Sometimes the reaction is complete but recovery is poor: for example, a portion of a solid product remains stuck to a vessel. Sometimes recovery is efficient but chemistry limits how much product forms. Both can give a yield below 100%, yet the repair a scientist would make is different.
The word “theoretical” does not mean imaginary. It means the maximum calculated under a stated chemical model, including the correct reactant amounts, equation and product. It is a reference point against which an observation can be evaluated.
Understanding percentage purity in plain language
A mass of sample is not always a mass of pure reacting substance. A 10.0 g sample of limestone, for example, need not be 10.0 g of calcium carbonate. It may contain substances that do not participate in the target reaction. If only 8.0 g of that sample is calcium carbonate, its purity with respect to CaCO₃ is 8.0/10.0 × 100 = 80%.
A purity percentage always needs a named target. Saying “this sample is 80% pure” is incomplete unless the question or context identifies the substance. A mixture can contain multiple components; purity is the proportion attributable to the specified one.
An important limitation is that a mass measurement alone seldom tells us the mass of the pure component. We may need reaction data, titration results, appropriate analytical testing or another defensible method to infer it. In school calculations, the question usually supplies a chemical reaction and enough information to identify the amount of the wanted substance. Do not assume every lost gram was an impurity unless the scenario justifies that model.
Worked example 1: simple percentage yield
A hypothetical reaction could produce 5.00 g of pure magnesium oxide from its limiting reactant. The measured amount of the desired, properly identified product is 4.00 g. The percentage yield is 4.00/5.00 × 100 = 80.0%.
The denominator is 5.00 g because it represents the theoretical product, not the starting reactant or total mixture. The numerator is 4.00 g because it represents the actual desired product, not a randomly weighed piece of apparatus.
We can immediately ask two useful follow-ups. Does 80% yield prove the reaction stopped when exactly 80% of reactant was used? No: incomplete conversion is one possibility, but loss during recovery is another. Does 80% yield mean the isolated solid is 80% pure? No: yield and purity describe different dimensions of the experiment.
That distinction is not an advanced extra. It is the scientific meaning of the answer.
Worked example 2: calculate theoretical yield from a reaction
Suppose 2.40 g of pure magnesium burns and forms magnesium oxide. Use Ar Mg = 24 and O = 16, and assume that magnesium is completely converted to MgO with sufficient oxygen available. The equation is 2Mg + O₂ → 2MgO.
Magnesium amount = 2.40/24 = 0.100 mol. The balanced equation gives one mole of MgO for each mole of Mg, so predicted MgO amount = 0.100 mol. Mr of MgO = 24 + 16 = 40. Theoretical yield of MgO = 0.100 × 40 = 4.00 g.
If the mass of the correctly identified pure product recovered is 3.20 g, the percentage yield is 3.20/4.00 × 100 = 80.0%. The original 2.40 g of Mg is not the denominator. The product contains oxygen acquired from the surroundings, so comparing 3.20 directly with 2.40 would not answer the yield question.
This makes a useful bridge from the earlier Hougang magnesium mass-gain explanation: compare the starting and product masses only after accounting for the atoms involved, and remember that the mass of the measured solid is not automatically the total mass of the whole reacting system.
Why balanced equations have to come before percentages
A student may memorise “actual over theoretical” but have no way to obtain the theoretical amount. The equation is the source of the product-to-reactant mole relationship. If the coefficients are wrong, the predicted amount is wrong. If the original formulae are wrong, the error travels all the way into Mr and product mass.
Consider 2H₂ + O₂ → 2H₂O. One mole of oxygen corresponds to two moles of water if sufficient hydrogen is supplied, not one mole. A mistaken 1:1 relationship would halve the theoretical water amount and could make a perfectly ordinary recovery look like more than 100% yield.
This is why a Chemistry tutor should inspect the first equation and the first conversion before criticising the final percentage. Often the visible error is downstream of a chemical model that the learner has not yet secured.
Worked example 3: simple percentage purity
A 12.0 g sample contains 9.0 g of pure calcium carbonate and 3.0 g of other material. Percentage purity with respect to calcium carbonate = 9.0/12.0 × 100 = 75.0%.
There is no theoretical product quantity in this problem. We did not predict how much a chemical reaction could make. We examined the composition of a sample. Even if both numerator and denominator are in grams, they are not playing the same roles as the two grams in a yield formula.
A neat way to see this is to draw a rectangle for the entire sample and shade three quarters of it to represent CaCO₃. The shaded fraction is the purity. Now draw a second rectangle representing theoretical product and a shorter bar for the product recovered. That pair represents yield. Similar percentages; different reference quantities.
Worked example 4: infer purity from carbon dioxide gas
Imagine a 10.0 g sample of material containing calcium carbonate is reacted under conditions where all its CaCO₃ reacts with excess acid. Assume, as the school problem states, that other components do not produce carbon dioxide and that the gas is collected without loss. The measured CO₂ is 1.76 g.
The acid–carbonate reaction gives one mole of CO₂ for each mole of CaCO₃: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Use Mr CO₂ = 44 and Mr CaCO₃ = 100 for the given school atomic masses. CO₂ amount = 1.76/44 = 0.0400 mol. Therefore the sample contained 0.0400 mol CaCO₃, corresponding to 0.0400 × 100 = 4.00 g CaCO₃.
Purity = 4.00/10.0 × 100 = 40.0% CaCO₃ by mass. The remaining 6.00 g is material other than CaCO₃ under the simplified assumptions. We do not know its identity from these numbers alone.
There is also a lesson about evidence. If another component releases CO₂ with acid, the conclusion is no longer uniquely supported. If some CO₂ dissolves or leaks, the measured 1.76 g may understate the amount produced. A good answer names the assumptions protecting the inference instead of pretending that any gas reading automatically proves a precise purity value.
One powerful comparison: 70% yield is not 70% purity
I sometimes give students two statements and ask them to say which is a reasonable conclusion: “Only 70% of the target mass was recovered” and “30% of the solid must therefore be impurity.” The first can describe a 70% yield relative to theory. The second does not follow.
You could isolate a perfectly pure product but recover only 70% of what the reaction could theoretically have supplied. You could also isolate nearly all of a crude material that contains significant impurities. These outcomes are independent enough that they need separate evidence.
A high percentage yield is not by itself proof of high purity. An apparently high crude-product mass may be due to trapped solvent, moisture, unreacted solids or contaminants. Conversely, an impressively pure crystal sample can have a disappointingly low yield because much of the dissolved substance was left in the mother liquor.
Chemistry becomes more interesting when the student stops treating every percentage as a grade. Each percentage has a defined job.
Worked example 5: the product is not completely pure
A reaction has a theoretical yield of 10.0 g of the desired pure substance. A student collects 8.0 g of crude solid, and a separate analysis shows that the collected solid is 75% pure with respect to the desired product. How much actual pure product was obtained?
Pure target product mass = 8.0 × 0.75 = 6.0 g. The yield of the pure target is 6.0/10.0 × 100 = 60%. Reporting 8.0/10.0 × 100 = 80% would compare the entire crude sample with a theoretical mass defined only for the wanted substance.
The lesson is subtle but important. Multiplying by the purity fraction is not an extra trick to memorise. It is a necessary adjustment so that numerator and denominator represent the same chemical species.
If the question explicitly defines “actual yield” in a different, simplified way, read its wording carefully and follow the stated assumptions. The general scientific standard is to keep unlike quantities from being silently compared.
What can make an apparent percentage yield exceed 100%?
Suppose the theoretical yield is 5.00 g and the recovered solid is recorded as 5.30 g. The arithmetic gives 106%. This cannot mean that the reaction produced 106% of the theoretical maximum of the pure target under the same complete model. Something about the calculation, sample or measurement needs investigation.
Possible reasons include residual water or solvent in the solid, impurities mixed with the product, contamination, weighing the container incorrectly, wrong stoichiometry, incorrect relative masses or an assumption about the starting reactant that does not match the actual sample. In unusual multi-reaction systems the simple stated model may not describe the chemistry, but that calls for a revised model rather than a celebration of impossible efficiency.
Ask which explanation fits the actual experiment. A wet precipitate can be heavier than the dry target. A dirty weighing boat can affect mass. A gas-volume error cannot always be used to explain why a recovered dry solid mass increased. We should move from generic “human error” toward directional reasoning grounded in the apparatus.
In an exam, a result above 100% is a clue to interrogate the data and assumptions—not a reason to force the arithmetic to 100%.
Does incomplete reaction always explain a low yield?
No. An actual yield lower than theory can be affected by incomplete conversion, competing reactions and isolation losses, but the experiment determines which possibilities make sense. If a solid product is filtered, some may remain in the filtrate, adhere to glassware or be lost when transferring. If a gas is collected, leaks or dissolution may cause an apparently lower collected amount even when more gas was produced.
There is a difference between chemical production and measurement or recovery. The student should distinguish “not produced” from “produced but not collected” and “collected but not correctly measured.” That separation makes a much stronger practical-evaluation answer.
A strong learner can explain why an experimental improvement targets one of those causes. Improving sealing addresses loss of gas. Drying a wet solid addresses an overestimated crude mass, not a low reaction conversion. Using excess of the appropriate reagent can address a shortage of that reagent, but may be irrelevant if the true problem is transfer loss.
Yield starts with the limiting reactant
If two reactants are supplied in specified quantities, the theoretical amount may depend on which one runs out first. The smallest raw mass is not necessarily the limiting reactant. Convert reactant quantities into moles, compare available amounts with their stoichiometric coefficients, and identify the reactant that can support the least progress through the balanced equation.
Once the limiting reactant is known, calculate theoretical target product from that reactant. Only then apply actual/theoretical × 100. You cannot get a meaningful yield from a guessed denominator.
For a full method and longer worked comparisons, continue to How to Find the Limiting Reactant: Hougang Chemistry. This article owns the difference between efficiency and composition; that companion guide owns the limiting-reagent decision.
Check the units before using either formula
Both numerator and denominator must use matching units and refer to comparable material. If actual product is given in grams while the theoretical amount is given in moles, you need the relevant molar mass. If the sample mass is in kilograms and the pure component mass is in grams, convert before dividing. A unitless percentage should emerge from a division of like quantities.
A student who writes 0.080 mol divided by 4.00 g has not built a valid yield ratio. It is possible for a calculator to output a number, but the result has units of mol/g and is not the requested percentage. This is a lovely place where the unit can become a built-in second marker.
It also helps to write the noun with the unit: 4.00 g MgO theoretical, 3.20 g MgO actual. That habit prevents a learner from accidentally substituting 2.40 g magnesium in place of the product denominator.
Forward and reverse purity calculations
The easy direction is pure mass ÷ sample mass. The reverse direction can be more useful. If 80.0% of a 25.0 g sample is the named substance, its mass is 0.800 × 25.0 = 20.0 g. If you require 8.00 g of pure substance from a material that is 80.0% pure, the required sample mass is 8.00/0.800 = 10.0 g.
These two calculations go in different directions. The first multiplies by the purity fraction because only part of the sample is useful. The second divides because you must obtain a larger sample to supply the desired pure amount. Students who memorise “percentage always means multiply” will get the second wrong.
I would ask a child to estimate before calculating. If a sample is only 80% pure, should you need more or less than 8.00 g of it to provide 8.00 g of the wanted substance? More. The expected direction makes the algebra easier to judge.
Forward and reverse yield calculations
If theory predicts 12.0 g and actual yield is 75.0%, recovered desired product mass = 0.750 × 12.0 = 9.00 g. If actual pure product is 9.00 g and percentage yield was 75.0%, the theoretical yield must be 9.00/0.750 = 12.0 g.
Again, estimation protects the method. If the yield is below 100%, the theoretical amount should be at least as large as the recovered pure target amount under the model. If your reverse calculation gives a smaller theoretical mass, check whether you multiplied rather than divided.
The ability to reverse a formula is a good indication that a student understands what the fraction represents. It is more durable than memorising two separate pieces of algebra.
A combined sample problem with purity and yield
A hypothetical reactant sample has mass 12.0 g and is 75.0% pure in a compound that produces the desired product in a 1:1 mole ratio. The pure reactant mass is 12.0 × 0.750 = 9.00 g. Suppose its molar mass is 90 g mol⁻¹. The amount of active reactant is 9.00/90 = 0.100 mol.
Under the given 1:1 reaction, the theoretical amount of product is 0.100 mol. If the product has molar mass 50 g mol⁻¹, its theoretical mass is 5.00 g. If 4.00 g of verified pure product is recovered, percentage yield = 4.00/5.00 × 100 = 80.0%.
Two different percentages have entered the same calculation. The 75.0% purity corrects the starting material before stoichiometry. The 80.0% yield compares the final desired product with what the active reactant could theoretically produce. Swapping the percentages without checking their reference objects is not a valid procedure, even if multiplication happens to commute numerically in a simplified sequence.
This is the kind of question where a careful tutor should train nouns and units before training speed.
Why purity can matter before stoichiometry
Suppose a 20.0 g sample is said to contain 60% by mass of a reacting compound. If you treat all 20.0 g as the compound, you overestimate the available amount of reacting substance. The active mass is only 0.60 × 20.0 = 12.0 g under the stated simplification.
The unreactive remainder does not suddenly acquire the compound’s molar mass. Treating the full sample mass as chemically active is a modelling error, not just an arithmetical one. It can distort the theoretical product and every subsequent yield calculation.
A reliable student draws a small box around the starting sample and labels its pure fraction. Only then does the moles calculation begin. The visual boundary makes it harder to forget which mass is appropriate for the chemical equation.
Purity is not identical to concentration
When a question gives a solution concentration such as 0.200 mol dm⁻³, it is describing amount of a solute per volume of solution, not directly the percentage purity of a weighed solid. Concentration and purity can interact in preparation problems, but they are not synonyms.
If 25.0 cm³ of a 0.200 mol dm⁻³ solution is involved, the amount of solute is 0.200 × 0.0250 = 0.00500 mol, using litres/dm³ consistently. That is a solution-quantity calculation. Without further information, you cannot infer what percentage by mass of an original impure solid it represents.
This distinction becomes increasingly important in titration and quantitative practical questions. Students should not use every percentage-like relationship interchangeably simply because the topic contains numbers.
Read practical evidence cautiously
A practical report might contain masses before and after heating, dry-product masses, changes in gas volume or repeated titre values. Those measurements are observations. Concluding the sample’s purity or the reaction’s yield requires a chemical model connecting those observations to the specific substance.
If a solid loses mass on heating, did water evaporate, a gas escape from a decomposition reaction, or some material get spilt? If a collected precipitate is heavy, is it dry, is the filter paper included, and has washing removed soluble contaminants? A correct formula cannot repair an unsupported assumption about what was weighed.
I would teach students to underline the observation in one colour and the inference in another. Then ask what assumption carries the inference. This discipline is useful long after any one school examination.
Diagnostic: three wrong answers, three different repairs
Student A uses mass of the starting reactant as the theoretical product mass. They need stoichiometry and the meaning of the product’s molar mass. Student B divides the mass of pure limestone by the mass of carbon dioxide measured. They need the distinction between composition and reaction outputs. Student C correctly computes 105% yield and writes “very efficient” without noticing the contradiction. They need a physical plausibility and experimental-quality check.
Those are not interchangeable mistakes. Give Student A a molecule-to-molecule ratio problem. Give Student B a shaded sample diagram followed by a gas-inference question. Give Student C a dry-versus-wet product comparison. Then ask all three to try a new mixed problem without hints.
This is the teaching advantage a deliberately managed three-student group can offer: common scientific foundations, individual diagnostic work, and independent verification of improvement. It should be evaluated by those practices, not by a promise that a small class automatically raises marks.
A 15-minute home practice set
- 1. Theoretical yield 15.0 g; pure product obtained 12.0 g. Find percentage yield.
- 2. A 30.0 g sample contains 21.0 g of a named compound. Find its percentage purity.
- 3. A 25.0 g material is 64.0% pure. Find the mass of its active compound.
- 4. A desired 12.0 g of pure compound must be supplied from an 80.0% pure batch. What mass of the batch is required?
- 5. A reaction has theoretical yield 8.00 g. The recovered crude solid is 6.00 g at 90.0% purity. Find the yield of pure desired product.
- 6. The theoretical product is 4.00 g, but the supposedly pure solid weighs 4.40 g. Is 110% yield a satisfactory chemical conclusion? Explain one plausible cause.
- 7. A 10.0 g sample gives enough data to infer 6.00 g of pure reactive compound. Which percentage can be calculated directly, yield or purity?
- 8. A reaction gave only 70% of the theoretical desired product. Can you conclude that the remaining 30% of the collected solid is impurity?
- 9. An impure sample contains 80.0% of a reactant, and 15.0 g of the sample is used. What mass should enter the mole calculation for that reactant?
- 10. A child knows the equation for % yield but consistently gets the wrong answer. Name two chemical decisions worth checking before criticising their arithmetic.
Have the student write a noun beside every number: sample, pure component, theoretical product, actual pure product. That small habit can be more effective than another page of unexplained answers.
Answers and what each reveals
- 1. 12.0/15.0 × 100 = 80.0% yield.
- 2. 21.0/30.0 × 100 = 70.0% purity.
- 3. 25.0 × 0.640 = 16.0 g active compound.
- 4. 12.0/0.800 = 15.0 g impure material.
- 5. Pure product mass = 6.00 × 0.900 = 5.40 g; yield = 5.40/8.00 × 100 = 67.5%.
- 6. No. The 110% figure is an apparent mass ratio, not a valid pure-product yield under the stated model. The solid may be wet or contain contamination; other calculation or weighing errors are also possible.
- 7. Purity = 6.00/10.0 × 100 = 60.0%; no theoretical product has been supplied.
- 8. No. Unrecovered product is not necessarily impurity in the solid that was collected.
- 9. 15.0 × 0.800 = 12.0 g of active reactant.
- 10. Check the balanced equation and limiting-reactant/theoretical-yield route; also check whether the measured mass is of the correct pure product.
A strong revision session returns to these same ideas with altered quantities, not memorised numerical answers. Change the purity to 72%, change the reagent amounts, or supply a molar volume. Ask whether the method changes and why.
A seven-day pathway from confusion to control
Day 1: compare a yield story and a purity story without using numbers. Day 2: practise the two simple percentage formulas. Day 3: calculate a theoretical yield from a balanced equation. Day 4: infer pure reactant mass from an impure sample before calculating moles. Day 5: combine product purity with theoretical yield. Day 6: evaluate experimental results over 100% and propose directional improvements. Day 7: mix all question types with no headings and require a brief denominator explanation before every calculation.
Each session can be short. Spacing matters because learning a method during a lesson is not the same as retrieving it after several days. Include one cold question the following week. If the child cannot begin, help reconstruct the model rather than simply reread the solved page.
Parents can watch for progress in how the child speaks: “This is the mass of the sample, not the compound”; “I need to convert to moles first”; “The denominator is theoretical product.” These sentences are evidence of chemical classification, not just greater calculator confidence.
Study choices for G1, G2 and G3 students
Secondary 1 and 2 students often learn mass, volume, mixtures and basic evidence within integrated Science before meeting full upper-secondary calculations. In Secondary 3, subject combinations and levels become important. G3 Pure Chemistry is not interchangeable with every G2 Combined Science route or every school’s Secondary 3 scheme of work.
The official 2027 Singapore–Cambridge SEC G3 Chemistry K324 syllabus includes calculation of percentage yield and percentage purity under Section 4.2(h). It also includes stoichiometric reacting masses, gas volumes and possible limiting-reactant calculations under Section 4.2(f). This gives a clear national reference for G3 Pure Chemistry students in the 2027 cohort; the year in which an individual school teaches the skill may differ.
A tutor should confirm the student’s current examination year, subject code and school materials before choosing the quantity and difficulty of practice. For the broader curriculum routes, use the Hougang Chemistry reference library.
How a parent should assess a Chemistry tutor for this topic
Ask the tutor to show a diagnostic where the student must identify the numerator and denominator in two different stories. Ask whether feedback is followed by a new independent problem, whether a genuine practical error is distinguished from a calculation error, and whether the student’s own marked work sets the teaching priority.
Avoid judging a programme from claims about instant grade improvement or vague promises of examination secrets. A useful Chemistry tutorial changes what a student can do independently. It should leave a trail: initial mistake, explanation, changed question, later retrieval and a more reliable answer.
For families in Hougang, Kovan and nearby neighbourhoods, verify the actual teaching location and timetable. A page about Hougang Chemistry does not by itself guarantee that the lesson venue is in Hougang. The Hougang Secondary 3 tutor guide is a useful starting point for evaluating teaching quality.
Questions Hougang parents and Secondary 3 students ask
Is a percentage yield of 100% always a perfect experiment?
It indicates that the measured actual desired product amount matches the theoretical amount under the assumptions. It does not, by itself, prove every measurement was flawless or that no other contaminants were present. Check the evidence used to define the actual pure product.
Is a percentage purity above 100% possible?
Not for the fraction of a specified substance within the same accurately weighed sample. Such a result points to a bad measurement, incorrect inference, wrong basis or calculation problem.
What if the question gives moles instead of grams?
A yield ratio can compare like amounts in moles when both refer to the same target product; many school questions use masses. For purity by mass, follow the mass basis specified and convert when necessary.
Does 80% pure mean 20% did not react?
No. The remaining 20% describes material other than the target compound in the starting sample. Whether it reacted depends on its identity and conditions, which require separate evidence.
Can a product be both high yield and low purity?
Yes. A large mass of crude product may be collected while significant contaminating material is present. The pure-product yield should be evaluated with the purity-adjusted amount when that information is available.
What should I study before this chapter?
Stable chemical formulae, relative atomic masses, moles, balanced equations, stoichiometric ratios and clear units. For a related calculation bridge, read Empirical Formula from Percentage Composition.
Why does the practical question ask for an improvement?
Because an answer about why measured results differ from predictions should connect to a specific experimental mechanism: loss, contamination, incomplete reaction, instrument precision or another defensible source of uncertainty.
The thought that makes percentages reliable
There are two kinds of “how much” here. One is how much of this sample is the substance I want? The other is how much of the product did the reaction give compared with its chemical prediction? Purity answers the first. Yield answers the second.
When I see a student label those two questions correctly before calculating, I know we are making progress. The formulas are short. The judgement is the real Chemistry.
Further reading: SEC 2027 G3 Chemistry K324 official syllabus · Gas-syringe graphs: rate versus final amount · Hougang Chemistry Tutor · Hougang Chemistry tuition guide · Parent–student consultation.