Quick answer: A sublimation mass ledger begins with net feed mass, adds the deposited fraction and residue, and subtracts that recovered total from the feed. Check the tare, the percentage denominator and the precision before claiming anything about loss or efficiency.
This G3 SEC Mathematics guide extends the physical separation decision in Volume 0857 into reversible percentages, equations, bounds, multiple trials and real-world feasibility. Its original examples are designed to practise K310 mathematical reasoning and should not be mistaken for released exam questions.
At a glance: define each mass; subtract containers; calculate total recovered; identify the unaccounted difference; distinguish recovery from purity; reverse percentages when needed; compare bounds and repeated trials; state only supported conclusions.
First identify what the numbers measure
A practical record gives masses for a starting mixture, a deposited fraction and a residue. The mathematical task begins before subtraction. Identify which readings are gross masses that include containers and which are net masses of material. Ask whether a reading is a starting quantity, an amount recovered at a particular location or a reference value used to calculate a percentage. These categories cannot be exchanged just because every entry has the unit grams.
Suppose a mixture starts at 12.00 g and the experiment retrieves 4.30 g at the cool collecting surface and 7.40 g in the lower vessel. The measured total recovered is 11.70 g, so the difference from the feed is 0.30 g. That difference is 2.5% of the starting mass. It is not the purity of either fraction and it does not identify which substance was uncollected. It is simply a measure of incomplete accounting against the starting measurement.
The science explanation belongs to Volume 0857. This Mathematics volume owns the arithmetic and logic required to check the mass account, compare repeated runs, handle uncertainty and rule out claims that exceed the evidence. The goal is not to decorate a Chemistry experiment with calculations. It is to demonstrate the mathematical habits used in contextual questions.
Two clear paragraphs beat one large equation
Write a mass ledger with three lines. The first line is feed mass. The second is recovered mass: deposited fraction plus residue. The third is the difference between feed and recovery. With the illustrative figures, those lines read 12.00 g; 4.30 + 7.40 = 11.70 g; and 12.00 − 11.70 = 0.30 g.
Then write an interpretation in words: ‘The two weighed fractions total 0.30 g less than the original sample.’ A mathematically correct subtraction is only the beginning. If the question asks whether material was destroyed, the mass ledger alone does not justify that conclusion. If the question asks which component was lost, the data also do not answer it. Do not make the ledger more ambitious than the recorded variables permit.
K310, the 2027 G3 Mathematics syllabus, assesses standard techniques, problem solving in varied contexts and mathematical communication. Its Paper 1 and Paper 2 each last 2 hours 15 minutes and carry 90 marks and 50% of the subject. Both require all questions to be answered. The final Paper 2 question focuses on applying mathematics in a real-world scenario, though it need not be a science setting. The worked questions here are original training cases, not official specimens.
Before subtracting, remove the container
Imagine that a small dish weighs 18.24 g when empty and 22.69 g when it contains recovered crystals. The material mass is 22.69 − 18.24 = 4.45 g. If a learner records 22.69 g as product mass, the entire subsequent mass account becomes false. The correct first operation is a tare subtraction, not a percentage.
Now suppose a second dish has an empty mass of 24.80 g and a loaded mass of 31.05 g. Its material contains 6.25 g. These recovered components total 10.70 g. If the original mixture was 11.00 g, the unaccounted difference is 0.30 g. Writing those intermediate values makes it possible for someone else to check the model.
A sensible solution labels columns ‘container alone’, ‘container plus material’ and ‘net material’. The extra words prevent a common mistake: assuming two numbers refer to the same physical quantity. Mathematical precision starts with a clearly defined variable and continues through the arithmetic. It is not achieved by writing more decimal places at the end.
Percentage of feed: which denominator?
The fraction of the original sample recovered is recovered mass divided by initial mass. In a 20.0 g starting sample with 18.5 g recovered in total, this is 18.5 ÷ 20.0 × 100% = 92.5%. The unaccounted share is 7.5%. These two percentages sum to 100% because they refer to a partition of the initial measured mass.
The percentage of the recovered product located at the cool surface is a different question. If 6.0 g is at the surface and 12.5 g is residue, the surface share of recovered mass is 6.0 ÷ 18.5 × 100%, approximately 32.4%. The same 6.0 g as a percentage of initial feed is 6.0 ÷ 20.0 × 100% = 30.0%. Both are valid calculations for different questions. Only one answers each prompt.
Before using the percentage key, complete the sentence: ‘[Quantity] as a percentage of [reference total].’ The reference becomes your denominator. A good mathematical explanation declares it in plain language, because many data questions are lost through a perfectly executed calculation with the wrong base.
Recovery is not the same as purity
Suppose a reference method establishes that 5.0 g of the desired substance was initially present. A recovery operation collects 4.4 g of material believed to contain the substance. If the collected material were entirely the desired substance, the apparent recovery is 4.4 ÷ 5.0 × 100% = 88%. But ‘apparent’ matters when the composition of the recovered material is unverified.
A pure sample could have low recovery because some of the target was left behind or escaped the defined collecting zone. A high mass recovery might still conceal a contaminant. A fraction can weigh more than the reference target amount because it contains something else, because weighing was incorrect or because the reference assumption is wrong. A result above 100% is a diagnostic flag, not proof of miraculous production of matter.
This is a problem of identifying the quantity being estimated. Recovery concerns how much target was obtained relative to the possible target amount. Purity concerns the target proportion in what was collected. The two measurements have different denominators and may need different evidence. Volume 0850 owns crystal-batch accounting; here the owner is the specific two-location sublimation ledger.
The missing mass percentage and its meaning
Use an example with 15.00 g of feed, 5.70 g deposited fraction and 8.85 g residue. Recovery totals 14.55 g. The difference is 0.45 g. Its percentage of feed is 0.45 ÷ 15.00 × 100 = 3.0%. A student who computes 0.45 ÷ 14.55 has answered a different question: missing mass relative to material recovered, approximately 3.09%.
A precise sentence is: ‘The unaccounted mass is 0.45 g, or 3.0% of the original measured feed.’ Notice what the sentence does not say. It does not claim that 3.0% chemically reacted. It does not assign all missing material to the sublimable component. It does not declare a purity of 97%. Each of those statements would require facts not present in this ledger.
If a data table omits the starting mass but gives the exact recovered mass and the unaccounted difference, you can reconstruct feed by addition. If it omits two related values, there may be multiple solutions. The habit of counting unknowns and independent equations avoids producing a single attractive answer from insufficient evidence.
Reverse problems: recovering the original feed
Suppose a student reports 11.40 g recovered and says that this represents 95% of the original starting mass. Under the stated model, starting mass = 11.40 ÷ 0.95 = 12.00 g. The missing amount is therefore 0.60 g. Writing ‘11.40 × 0.95’ would give a smaller answer and cannot represent a starting amount when recovered mass is only 95% of it.
Check the direction of the calculation before deciding which operation to use. If a quantity is a stated fraction of a larger total, divide by the fraction to reconstruct the total. A quick reasonableness test confirms that the original feed must exceed the recovered mass when recovery is below 100%.
Now change the information. Suppose ‘recovery is 95%’ refers only to the target substance rather than the entire feed. It would be wrong to divide the total two-fraction recovery by 0.95 to find the feed. The percentage’s meaning must be defined first. Reverse percentages are powerful only when the reference is explicit.
Interval data instead of falsely exact readings
Measurements may be presented to a stated decimal place. A mass recorded as 4.3 g to the nearest 0.1 g represents a value from 4.25 g up to, but not including, 4.35 g under the ordinary rounding convention. A feed recorded as 12.0 g to the nearest 0.1 g lies from 11.95 g up to 12.05 g. These bounds become relevant if the question asks for extreme possible totals.
For two positive quantities, the minimum sum comes from the two lower bounds and the maximum possible sum approaches the sum of the two upper bounds. For a difference, the smallest possible feed-minus-recovery uses the smallest feed and largest possible recovery, while the greatest uses the largest feed and smallest recovery. This is why a point estimate and a bounded estimate have different roles.
Do not introduce rounding intervals into every straightforward mass calculation. Use them when the question provides precision information and asks for bounds, reliability, or feasible extremes. Label strict versus inclusive endpoints only when it matters to the mathematical answer. The aim is to represent uncertainty faithfully, not to turn a simple three-number ledger into an unnecessary algebra exercise.
A full bound example
Suppose the feed is 12.0 g to the nearest 0.1 g, and the two recovered fractions are 4.3 g and 7.4 g, each to the nearest 0.1 g. Feed lies in [11.95, 12.05), deposit lies in [4.25, 4.35), and residue lies in [7.35, 7.45). The recovered sum lies in [11.60, 11.80). These are the interval statements.
The unaccounted difference, defined as feed minus recovered sum, can be greater than 11.95 − 11.80 = 0.15 g and less than 12.05 − 11.60 = 0.45 g; exact endpoint inclusion follows from the rounding intervals. A simple answer can say the missing mass is approximately 0.30 g, with a possible range approaching 0.15–0.45 g under the stated rounding model. It is not safe to call the discrepancy definitely 0.30 g to unlimited precision.
The interpretation requires care. All feasible differences in this example are positive, so the rounded data support an unaccounted amount rather than full equality within those reported precisions. This does not identify the physical loss mechanism; it shows how securely the numerical conclusion survives the rounding uncertainty.
A zero difference may still conceal uncertainty
Suppose a student rounds the feed to 10.0 g and the recovered sum is also displayed as 10.0 g. They conclude that the process had absolutely no losses. That conclusion may be too strong if measurements were rounded and separate components were weighed with limited precision. Two reported totals can match even though their underlying values differ slightly.
The correct wording is that the reported masses balance at the stated precision. This is not the same as proving perfect physical recovery. It is also possible that systematic errors cancel. A balance that has not been tared, a damp weighing vessel or material adhering to an unmeasured part of the apparatus can distort a result despite an attractive zero in the final line.
This example helps with mathematical communication. An equality sign should represent exact equality in the model or arithmetic, not a promise about the physical world. Use an approximation sign when you report a rounded calculated result. Distinguishing the two keeps a real-world model honest.
What if the total appears to be too large?
A trial begins with 10.00 g and reports 4.80 g at the collecting surface plus 5.45 g in the residue. The recovered total is 10.25 g, which exceeds the feed by 0.25 g. The record cannot be accepted as a simple closed mass partition without investigating. It might include contamination, incorrectly subtracted container mass, wet material, an earlier entry error or another explanation supported by new observations.
Do not force the ledger to show a positive ‘loss’ by subtracting in whichever direction gives a pleasant result. Write the signed discrepancy if that is helpful: feed − recovered = −0.25 g. Then explain that the negative sign indicates excess measured recovery under this convention. Mathematically sound solutions preserve the sign until the interpretation stage.
The next step is diagnostic rather than cosmetic. Check which entries were tare-corrected, whether the containers were dry, whether the same units were used and whether any material was added unintentionally. A learner should be able to identify the impossible or suspicious part of a model without pretending to know its cause.
Working with ratios instead of percentages
A question may ask for the ratio of collected fraction to residue rather than a percentage of feed. If deposit is 4.5 g and residue is 7.5 g, the ratio is 4.5:7.5, simplified to 3:5. That ratio describes the two recovered fractions, not the composition of the original material unless full recovery and correct identities have been established.
Suppose the ledger also has 0.4 g of unaccounted material. You should not use 3:5 to conclude that the feed originally contained exactly three parts target to five parts other substance. The unaccounted amount might change those shares, and the deposit could contain contamination. The simplification is correct arithmetic but a potentially false physical story.
The same principle appears in recipes, cost allocations and sample compositions. Ratios answer what the compared quantities are, whereas percentages answer what share one quantity represents of a specified whole. To earn marks for interpretation, carry the labels into the simplified answer.
An unknown fraction and an equation
Let x be the mass deposited at the collector. The original feed is 18.0 g, the residue is 11.7 g, and the unaccounted difference is 0.6 g. The mass account is 18.0 = x + 11.7 + 0.6. Solve to obtain x = 5.7 g. Checking gives 5.7 + 11.7 + 0.6 = 18.0.
This is ordinary algebra applied to a material ledger. The equation follows from the way the variables have been defined. It does not come from a new physical law. The question may instead define an ‘excess’ when recovered mass is greater than feed, in which case the sign convention must change accordingly.
Students sometimes form 18 = 11.7 + 0.6 − x because they memorised a formula with no diagram. Avoid that by drawing a partition of the feed: recovered at source, recovered at collector and unaccounted within the measurement model. The visual organisation is valuable even if the problem is a single equation.
When a problem has insufficient data
A sample starts at 15 g and ends with 9 g of residue. The task asks for the mass of pure target recovered by sublimation. The difference of 6 g does not necessarily answer it. The 6 g represents material not in the residue, but it might include collected deposit, escaped vapour or measurement discrepancy. If the question supplies a complete-recovery assumption and identities, then 6 g could be justified. Without that assumption, the answer is underdetermined.
A sophisticated mathematical response names what is missing. For example: ‘The mass of the deposit cannot be determined uniquely because the unaccounted fraction is not known.’ You may express a relationship such as deposit mass + unaccounted mass = 6 g, assuming the feed and residue values are comparable. That relationship contains more useful information than a fabricated single answer.
Examinations often test whether the learner can distinguish a known quantity from a possible quantity. Good reasoning sometimes produces a range or equation rather than a point estimate. This is a sign of control, not indecision.
A bound from a non-negative missing fraction
Take the 15 g feed and 9 g residue again. If the problem states that any unaccounted mass is non-negative, the deposited fraction cannot exceed 6 g under the simple partition. The lower bound may be 0 g if there is no evidence of positive collection. Thus 0 ≤ deposit ≤ 6 g is a possible mathematical bound under this model, even though the exact amount is unknown.
Now suppose the collector is directly measured to contain at least 4 g of material under the given accuracy conditions. The feasible interval narrows. If the deposit is from 4 to 6 g, the unaccounted fraction is from 2 to 0 g in the opposite direction. This is why constraints can reduce uncertainty without completely eliminating it.
Write the assumptions. A bound can be valid for a simplified model but invalid if material was added externally or the residue was wet. Mathematical answers for contextual problems should identify which physical quantities are being treated as non-negative and which conservation relationship is assumed.
Three trial runs and a fair comparison
Two groups compare reported recovery rates. Group A has runs of 91%, 93% and 92%. Group B has runs of 85%, 99% and 92%. Both means equal 92%. Yet the spread differs greatly: Group A varies by 2 percentage points, Group B by 14. The mean alone does not explain the consistency of the apparatus or process.
A meaningful statement is: ‘Both groups have the same mean recorded recovery, but Group A shows a smaller range across these runs.’ Do not claim Group A is necessarily more accurate, because the true recovery and possible systematic bias are unknown. Consistency and accuracy are different concepts. If both groups share a calibration error, tight clustering may still produce the wrong result.
This style of statistical explanation is relevant to K310’s objectives of interpreting data and justifying conclusions. You can calculate an average quickly; the more demanding work is to decide what the summary can tell you and what it cannot. Always check whether runs used comparable feed masses, apparatus and rules.
Normalising runs of unequal size
A group collects 1.8 g of target material from a 5.0 g feed in one trial and 3.9 g from a 10.0 g feed in another. Comparing absolute deposit masses alone would favour the larger trial, but the feed masses differ. The proportions are 36% and 39% of feed respectively. The second still has a larger measured proportion, but the gap is only three percentage points.
This does not necessarily mean the second apparatus is better. The starting mixture’s composition might differ, so deposited fraction as a percentage of total feed is not automatically a measure of target recovery. If the target content is independently known, use target recovered divided by original target amount instead. The denominator should match the question’s defined comparison.
Normalisation is a useful habit across science and Mathematics. Convert to a common basis, but only if that basis measures the property you intend to compare. An incorrect denominator makes a standardised number look authoritative while concealing the underlying mismatch.
The difference between percentage points and percentage change
Suppose recorded recovery rises from 80% to 90%. The increase is 10 percentage points. Relative to the original 80%, it is a percentage increase of (90 − 80) ÷ 80 × 100% = 12.5%. Both statements describe the same change from different perspectives. They are not interchangeable.
When a chart or promotional claim states that a method ‘improved by 10%’, ask whether the writer means 10 percentage points or a 10% relative increase. If the earlier recovery was 50%, a rise to 60% would be a 10-point gain but a 20% relative increase. The distinction matters whenever proportions are used to compare experimental methods.
The learner’s task is to read the denominator beneath the English words. This is also a useful bridge to the next volume, where a reader must distinguish a narrator’s or writer’s claim from the evidence supplied. Mathematical clarity becomes part of clear English.
A real-world decision with constraints
Imagine a school demonstration project has two proposed models. Model A recovers 8.8 g per 10.0 g of reference target, with a maximum permitted unaccounted mass of 1.5 g. Model B recovers 9.1 g per 10.0 g but has an extra preparation cost, and the budget permits only Model A. Neither can be declared the overall winner without reading the task’s stated decision criteria.
If the objective is to achieve at least 90% target recovery and stay within budget, Model A fails the recovery threshold at 88%, while Model B might fail the budget. The correct result may be that neither meets both requirements. This possibility is often missed when students are trained to choose one option at all costs.
Write a small feasibility table in your working: recovery criterion, budget criterion, evidence-quality criterion. Mark whether each option meets each condition. Then state the decision and reason. The final real-world question on K310 Paper 2 rewards selection, modelling and interpretation; it does not require every case to have an attractive approved option.
Constructing a defensible conclusion
A conclusion should name the quantity, the comparison base and the limitation. Instead of ‘The method is best because 94%’, write ‘In these three trials, the method recorded a mean recovery of 94% of the independently measured starting target amount; the evidence does not yet establish purity or performance under different conditions.’ The second sentence has more information but makes a smaller, better-supported claim.
Mathematical communication includes correct notation, units, approximate values and explanations tied to the context. Under the K310 syllabus, essential working matters. Students should ordinarily give non-exact numerical answers to three significant figures, or angles to one decimal place, unless the question states a different level of accuracy. In a laboratory table, respect the stated measurement precision where it governs the task rather than manufacturing decimal places.
A calculation has three parts: define, compute and interpret. When your solution is difficult to read, put those parts on separate lines. This makes marks easier to award and errors easier to repair.
An error clinic: eight mistakes and repairs
Mistake one: using the loaded dish mass as product mass. Repair: subtract the empty container. Mistake two: calling missing mass ‘pure impurity’. Repair: use ‘unaccounted difference’ unless composition has been established. Mistake three: dividing deposited mass by recovered total when the question asks for share of feed. Repair: write the denominator in words. Mistake four: multiplying by a recovery percentage when reconstructing a larger starting amount. Repair: reverse by division.
Mistake five: rounding intermediate values too early. Repair: retain adequate working precision and round at the end as instructed. Mistake six: dropping the negative sign when recovered mass exceeds feed. Repair: preserve the defined discrepancy sign and interpret it. Mistake seven: claiming a high average proves consistency. Repair: examine spread as well as centre. Mistake eight: forcing an exact answer where only a bound is justified. Repair: record the unknowns and constraints.
In each case, the correction changes the reasoning, not just the final digit. The purpose of an error ledger is to see the earliest step that failed and practise a new problem of the same logical type. See the Secondary 4 Mathematics past-paper forensics owner for this wider revision approach.
Practice set: arithmetic and interpretation
Question 1. A feed of 14.0 g produces 5.2 g deposit and 8.5 g residue. Find the unaccounted amount. Answer: 14.0 − 13.7 = 0.3 g. Question 2. Express that amount as a percentage of feed. Answer: 0.3 ÷ 14.0 × 100 ≈ 2.14%. Question 3. What percentage of feed was recovered in total? Answer: 13.7 ÷ 14.0 × 100 ≈ 97.9%. These answers follow the three-significant-figure convention when no different instruction is given.
Question 4. A dish alone weighs 25.68 g and with deposit weighs 29.93 g. Find the net deposit. Answer: 4.25 g. Question 5. 9.6 g is 80% of the original target mass. Find that target mass. Answer: 9.6 ÷ 0.8 = 12.0 g. Question 6. If total recovered is 10.2 g from a feed of 10.0 g, what is feed minus recovered? Answer: −0.2 g, indicating measured excess under that convention.
Question 7. A deposit is 6 g and a residue is 9 g. Give their ratio in simplest terms. Answer: 2:3. Question 8. A 15 g feed has 9 g residue and an unknown deposit. Assuming no external matter is introduced and an unaccounted amount cannot be negative, what is the greatest deposit? Answer: 6 g.
Practice set: modelling and reasoning
Question 9. Group A has recoveries of 91%, 92% and 93%; Group B has 85%, 92% and 99%. Compare the means and ranges. Answer: both means are 92%; the ranges are 2 and 14 percentage points respectively. Question 10. Do those figures prove Group A more accurate? Answer: no, because accuracy needs comparison to an appropriate true or reference value.
Question 11. Feed mass is 12.0 g and recovered masses are 4.3 g and 7.4 g, each rounded to the nearest 0.1 g. Can you assert that the unaccounted amount is exactly 0.30 g? Answer: no; 0.30 g is the point estimate, with a broader interval under the rounding model. Question 12. If a product has a high recovery percentage, can you deduce it is pure? Answer: no; purity is a different measurement.
Question 13. A result rises from 70% to 84%. State the percentage-point increase and the relative percentage increase. Answer: 14 percentage points and (14 ÷ 70) × 100 = 20%. Question 14. A method is cheaper but below the required recovery threshold. Must it be selected? Answer: not if the task requires both budget and recovery constraints. Question 15. Two distinct unknown masses remain but the problem gives one total. Can both be found uniquely? Answer: not without another independent relationship or constraint.
The two-minute calculation check
Before you finish a real-world Maths question, run a small check. First, inspect units. All masses must be comparable; grams and kilograms cannot be added without conversion. Second, inspect signs. A negative feed-minus-recovery difference means the measured recovery exceeds the feed. Third, inspect the base for each percentage. Say it aloud: ‘out of what?’ Fourth, inspect magnitude. A recovered fraction cannot be greater than the whole feed in a simple no-addition model without triggering further investigation.
Fifth, inspect the assumption that makes your equation possible. Sixth, check whether the question asks for an exact quantity, an estimate, a bound or a judgement. Seventh, retain enough working for another reader to reproduce the answer. Finally, write the concluding sentence using the actual nouns from the context instead of ‘it is 94%’ with no reference.
The next guide will shift from numerical proof to narrative control in G3 English Paper 1. The shared theme is evidence: a number needs a defined denominator; a story needs a credible cause, consequence and point of view.
Further practice and official references
To strengthen this skill, take three ordinary percentage questions and rewrite each as a material ledger. Then take a mass ledger and construct a version with one missing value. Work out whether it is solvable or underdetermined. A final extension is to round each reading differently and investigate whether the sign of the discrepancy is still certain. These tasks develop reasoning far beyond a single familiar example.
The official 2027 G3 Mathematics syllabus K310 sets the papers, accuracy expectations and assessment objectives. The SEC overview explains the new qualification beginning in 2027. For fuller mathematical notation, review the Secondary 4 notation owner. For transition from PSLE Science to G3 experimental reasoning, return to the transition guide.
The central habit is transferable: numbers answer the question that their definitions permit, not the stronger question a reader wishes they could answer. A good calculator result and a good interpretation are two different achievements. Practise both.
Read in sequence: Volume 0857: chemistry of the separation. Then continue to the next English volume for narrative and audience reasoning.
