Diagnose product yield and impure isolates

A heavy isolate is not necessarily more product

Suppose a reaction could produce at most 10.0 g of a named target, but the collected material weighs 12.0 g. Before congratulating the process, ask what was actually weighed. The isolate may include water, retained solution, another solid or an incorrect tare. The starting amount or target formula may also have been recorded incorrectly. A large reading is a reason to investigate the evidence, not evidence that atoms were created.

The OpenStax treatment of reaction yields distinguishes the theoretical amount predicted by stoichiometry from the amount obtained. Percent yield compares actual target with theoretical target. Here we add an explicit composition check: total isolated mass is not automatically target mass. All scenarios and balance records in this lesson are original synthetic teaching evidence, not laboratory measurements.

Establish the denominator before weighing

Our assigned feed contains 0.100 mol calcium ions and 0.150 mol carbonate ions. The target reaction is:

Ca²⁺ + CO₃²⁻ → CaCO₃

Each mole of calcium ions requires one mole of carbonate ions and can form one mole of calcium carbonate. Calcium therefore limits target formation to 0.100 mol. Using the rounded teaching molar mass 100 g/mol, theoretical target mass is 0.100 × 100 = 10.0 g. Carbonate is in excess by 0.050 mol relative to complete target formation. Charge is balanced: +2 and −2 sum to zero, matching the neutral solid.

This net ionic equation omits spectator ions. The 10.0 g target is not the mass of all initial solutions. Water, spectator ions and excess carbonate still belong to a complete process balance. The widget does not model solubility equilibrium, precipitation kinetics or how the solid was collected. It uses the stoichiometric ceiling as a reference. This is an analysis exercise, not a physical mixing or drying procedure.

Separate three ratios

Let m be collected isolate mass in grams, p its assigned target mass fraction between zero and one, and T theoretical target mass in grams. The calculated target in the isolate is mp. Then:

• Apparent uncorrected yield = 100m/T percent, which assumes the entire isolate is target.
• Corrected yield = 100mp/T percent, which uses the assigned target composition.
• Purity by mass = 100p percent, whose denominator is isolate mass rather than theoretical target.

For 10.0 g isolate at 80% target, p = 0.80. Target mass is 8.0 g, other material is 2.0 g, apparent yield is 100%, and corrected yield is 80%. The extra material has not been identified merely by doing the subtraction. It could be more than one substance.

Now increase isolate mass to 12.0 g while keeping p = 0.80. The target is 9.6 g; other material is 2.4 g. Apparent yield is 120%, but corrected yield is 96%, compatible with the stated ceiling. Set composition to 100%. The resulting 12.0 g target exceeds the available 10.0 g theoretical target. The interface flags inconsistent inputs and preserves the reported 120%; silently clipping it to 100% would conceal the contradiction.

At zero isolate mass, calculated collected target is zero. The assigned mass fraction remains a hypothetical parameter; no purity has been measured in an empty sample. A reported zero collection also does not establish that no target formed elsewhere in the apparatus.

Read a balance record before choosing a claim

The following fictional record belongs to one collected isolate, repeatedly weighed in the same container. Each displayed reading has an assigned bound of ±0.01 g. These are not independent reaction replicates. The endpoint readings are two successive observations of the same sample.

Your action: subtract the tare from each contents reading before expanding the answer. Which part of a claim that “the process produced 12 g of pure target” is contradicted or unsupported? Which further evidence would you request?

Check the balance-record analysis

The net masses are 12.00, 10.35, 10.01 and 10.00 g. The last two differ by 0.01 g, consistent with near-stability at the stated reading bounds. That supports a claim about observed mass stability, not chemical purity or complete reaction. If a separate, valid analysis assigns the final isolate 80% target by mass, its nominal target mass is 8.00 g and corrected yield is 80%. The mass fraction must apply to the final sample; an earlier wet-sample composition cannot simply be reused after drying.

Drying-related mass change does not create or destroy atoms. Material has left the weighed boundary, or a measurement assumption has failed. Without further evidence, do not identify every lost gram as water: loss of volatile material, sample loss or chemical change are competing possibilities. Stable mass likewise does not establish purity; a stable contaminant can remain.

Carry uncertainty through the subtraction

For the final contents reading 42.40 ± 0.01 g and tare 32.40 ± 0.01 g, a conservative net-mass interval is [9.98, 10.02] g. Find the lower endpoint by subtracting the largest allowed tare from the smallest allowed gross reading, and reverse those choices for the upper endpoint. This worst-case interval is not a statistical confidence interval. Correlated balance errors may behave differently; no correlation model is supplied.

If a separate analysis bounds the target fraction between 0.78 and 0.82, the corresponding target-mass interval is [9.98 × 0.78, 10.02 × 0.82] = [7.7844, 8.2164] g. Holding the theoretical 10.0 g fixed as a teaching assumption gives a corrected-yield interval of 77.844% to 82.164%. The extra digits show arithmetic, not warranted experimental precision. In a real report, uncertainty in the feed, identity and molar mass assumptions must also be addressed. The sliders show nominal scenarios; they do not calculate these intervals.

Explain low yield without inventing its cause

An 8.0 g collected target is 2.0 g below the theoretical target. That difference is an unrecovered target-equivalent, not necessarily 2.0 g of target discarded. Reactants may remain unconverted; competing products may form; target may remain dissolved or be lost during transfer. A feed conversion measurement asks how much reactant reacted. A yield measurement asks how much desired target was obtained on the stated basis. They answer different questions.

Propose discriminating evidence: analyze remaining reactant to investigate incomplete conversion, examine dissolved target in the filtrate to investigate collection, or identify additional products to investigate selectivity. A single mass reading cannot distinguish these explanations. Do not remove inconvenient readings merely to improve the reported yield.

Transfer: Another process has a theoretical target of 5.00 g. An isolate weighs 4.50 g and contains 90% target by mass. Calculate apparent yield, corrected yield and other material. Explain why the difference from 5.00 g does not establish a transfer loss.

Check the yield transfer

Apparent yield is 90%. Target mass is 4.50 × 0.90 = 4.05 g, so corrected yield is 81%; other material is 0.45 g. The 0.95 g unrecovered target-equivalent does not identify a mechanism. A matched analysis of remaining reactant, other products and collection streams would help distinguish causes.