SCIENCE · INVESTIGATE AND EXPLAIN

Solve a composition balance and reject impossible mixtures

Use two independent mass equations to infer a mixture fraction, then check feasibility and sensitivity.

Learning outcome: At total 20 g compare element percentages 40, 50, 60 and 70. Inspect algebraic masses and feasibility before changing total mass. Propagate the±.5-percentage-point composition bound.

Starting recap

Use algebra and dimensional consistency; distinguish a measured reading, a model estimate and the evidence needed to accept the model.

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Try

A20 g sample with 50% E fits 10 g of 40%-E component and 10 g of 60%-E component. An observed 70% E cannot fit either nonnegative mixture.

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Predict

Can adding more of a 40%-E component to a 60%-E component produce 70% E overall?

Your prediction and reason

Experiment

At total 20 g compare element percentages 40, 50, 60 and 70. Inspect algebraic masses and feasibility before changing total mass. Propagate the±.5-percentage-point composition bound.

Solve and challenge a two-component mass balance

Worked teaching model · synthetic values · Two stipulated nonreacting components contain 40% and 60% E by mass. Solve for mass fraction of the 60% component. Observed fractions outside 40–60% reject this two-component model; negative inferred masses are algebraic diagnostics, not physical outputs. Element fraction has an assigned±0.5 percentage-point bound.

Total sample mass: 20 g Observed mass percent of element E: 50 %

Reset keeps your prediction, captures and practice history.

Conditional calculations; not physical measurements
Quantity	Value	Unit
Algebraic mass fraction of 60% component	0.5	fraction
Lower fraction from element bound	0.475	fraction
Upper fraction from element bound	0.525	fraction
Algebraic mass of 60% component	10	g
Algebraic mass of 40% component	10	g
Feasible nominal two-component model	1	1=yes,0=no

The composition balance identifies a fraction only inside the stipulated two-component model.

m 60+m 40=M and .60m 60+.40m 40=fM imply x=(f−.40)/.20. Out-of-range solutions reject assumptions, calibration or component completeness; do not clip them and silently discard discrepancy.

Predict before changing controls, retain contrasting captures, and explain a changed case. Displayed digits are computational precision, not an uncertainty claim. Flags check stipulated numerical compatibility, not identity, reasoning quality or practical proficiency.

The composition balance identifies a fraction only inside the stipulated two-component model.

m 60+m 40=M and .60m 60+.40m 40=fM imply x=(f−.40)/.20. Out-of-range solutions reject assumptions, calibration or component completeness; do not clip them and silently discard discrepancy.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Specify both conserved quantities

A total mass measurement and an elemental mass fraction constrain different aspects of a mixture. Our two fictional components contain 40% and 60% of element E by mass. Let their masses be m40 and m60, and total mass M. Conservation gives m40 + m60 = M. Element accounting gives 0.40m40 + 0.60m60 = fM, where f is the measured overall E fraction.

Derive the inverse solution

Substitute m40 = M − m60 into the element equation. The result is 0.40M + 0.20m60 = fM. Thus x = m60/M = (f − 0.40)/0.20. This calculation is dimensionally a fraction divided by a fraction. The remaining component fraction is 1 − x. A physically feasible two-component mixture requires both fractions between zero and one.

Work a feasible and an impossible case

At M = 20 g and f = 0.50, x = 0.50, so each component mass is 10 g. At f = 0.70, x = 1.50, giving algebraic masses 30 and −10 g. The negative value is not a new kind of matter or a slider display to clip. It demonstrates that the nominal observation cannot satisfy the stipulated component model with nonnegative masses.

Propagate composition sensitivity

The assigned elemental-fraction bound is ±0.005, equivalent to ±0.5 percentage point. Because x depends linearly on f with slope 1/0.20 = 5, the fraction bound is ±0.025. At f = 0.50, x lies in [0.475,0.525]. The total mass scales component masses, not these fractions. Capture two total masses with the same f and verify this distinction.

Diagnose rather than erase a contradiction

An out-of-range observation may indicate an omitted component, mistaken component composition, contamination, a calibration problem or a violated nonreaction assumption. The balance equations alone do not identify which explanation is correct. Propose a separate measurement that would discriminate at least two explanations. Repeating the same biased composition method may only reproduce the contradiction.

Transfer: an ill-conditioned contrast

Imagine component E fractions 0.49 and 0.51 instead. The inversion divides by 0.02, making the same ±0.005 input bound expand to ±0.25 in mixture fraction. Derive this result and explain why nearly indistinguishable components produce a poorly determined composition. More displayed decimal places cannot recover information that the measurement contrast does not contain.

Source and evidence boundary

These explanations, calculations and diagrams are original. Conceptual references checked 30 October 2026: OpenStax dimensional calculations, NIST uncertainty framework. The linked textbook is a reference, not copied lesson text or imagery. All displayed datasets are synthetic teaching packets. Saved numerical agreement or prose is not a performed practical, independent reasoning review or laboratory certification. No handling, heating or chemical preparation procedure is requested.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why clipping a negative mass hides a failed model and why a fitted fraction does not prove the component list.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Carry named entities through mass and mole conversions Next: Infer a mass-loss formula without confusing assumptions with observations

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