2.4Models and Applications

Percent-composition and empirical-formula problems are proportional word models start to finish — given masses or percentages, build a mole ratio, then scale it — the same translate-data-into-a-proportion skill Models and Applications drills on rate and mixture problems.

The lead-in-tap-water worked example explicitly rearranges ppb = mass solute/mass solution x 10^9 to mass solute = ppb x mass solution/10^9 in symbols ('Rearranging the equation defining the ppb unit and substituting the given quantities yields') before any of the three given numbers (15 ppb, 300 mL, 1.00 g/mL) are substituted — the identical solve-a-formula-for-a-specified-variable step already used for molarity and dilution, now applied to the ppb definition.

The section opens by explicitly modeling stoichiometric factors on a pancake recipe scaled by a ratio, then reuses that same 'multiply by the ratio' move for every mol-to-mol and mass-to-mass conversion — the identical translate-a-word-problem-into-a-proportion skill Models and Applications teaches.

Identifying the limiting reactant means comparing a provided-amount ratio (e.g. 1.33:1) to the stoichiometric ratio from the equation (1.5:1) — a direct proportional-reasoning comparison, the same skeleton as a rate or mixture model problem.

KE_avg = (1/2)Mu_rms^2 and KE_avg = (3/2)RT are set equal and solved for u_rms — u_rms = sqrt(3RT/M) — entirely in the section's own exposition, in symbols, before the worked example ever introduces a number; by the time the student reaches 'Calculation of u_rms' they are handed the already-rearranged formula and only ever substitute into it. The literal-equation isolate-a-variable step is performed by the text rather than required of the reader, but the resulting formula is exactly a solve-for-a-specified-variable result, and the section leans on the reader being able to follow that derivation to trust the formula rather than merely memorize it.

q = c×m×ΔT is a literal formula in four quantities; the ‘Determining Other Quantities’ example isolates c from 6640 J = c×(348 g)×(21.2 °C), the same rearrange-and-solve-for-a-letter move as any formula application.

Calorimetry problems chain q = c×m×ΔT for two substances into one equation, e.g. (c×m×(Tf−Ti))rebar = −(c×m×ΔT)water, then isolate the one unknown temperature buried inside a product on one side — a multi-term literal equation, not just a single formula.

Capillary rise h = 2T cos θ/(rρg) is a five-quantity formula; the Check Your Learning problem gives h and solves back for the tube radius, the same isolate-one-variable-among-several skill as any formula rearrangement, just with more factors crowding the denominator.

The Bragg equation nλ = 2d sin θ is rearranged to solve for d given n, λ, θ (or for θ given d and λ) — another literal-equation isolate-a-variable exercise, here with a trig factor riding along instead of a purely algebraic one.

Finding the molarity, molality, and percent by mass of 0.94 g H₂ dissolved in 215 g Pd (solution density 10.8 g/cm³) means feeding the same mass and density data through three different concentration-formula rearrangements — the same model-and-solve-for-a-variable skill repeated with a different target variable each time, where reusing the volume computed for molarity as though it were also the solvent mass for molality mixes up the two definitions.

Determining a molar mass from a freezing-point depression chains three formula rearrangements — solve ΔTf = Kf·m for m, convert m and solvent mass to moles, then divide mass by moles — the multi-step ‘solve a formula, feed the result into the next formula’ workflow rather than a single substitution.

‘Will Ice Spontaneously Melt?’ substitutes into ΔSuniv = ΔSsys + qsurr/T at two different temperatures and compares the resulting signs — literal-formula evaluation where the payoff is a sign, not a magnitude.

ΔG° = ΔH° − TΔS° mixes units — ΔH° arrives in kJ while TΔS° comes out in J — so evaluating this literal equation correctly requires inserting a ×(1 kJ/1000 J) conversion mid-formula, a step the worked examples show explicitly and that is easy to drop.

Setting up change-in-concentration terms like Δ[H2] = +3x and Δ[NH3] = −2x from a mole ratio is building a linear model in one unknown x from a word-problem context, a skill the chemistry assumes rather than teaches.

'Rearrangement of the Kw expression shows that [OH-] is inversely proportional to [H3O+]' — the text performs the isolate-the-variable step in symbols, [OH-] = Kw/[H3O+], as its own stated move, and only after that substitutes 1.0e-14 and 2.0e-6 to divide. This is the same solve-a-formula-for-a-specified-variable skill used for molarity and dilution in chapter 6, reused on the water ion-product expression.

Faraday's-law problems (Q = It, n = Q/F, t = Q/I) are multi-step dimensional-analysis word problems built on rearranging a simple product/quotient formula — an assumed literal-equation and unit-conversion skill the chemistry exercises rather than introduces.

Setting the fast pre-equilibrium step's forward and reverse rates equal, k1[NO][Cl2] = k-1[NOCl2], and solving for the intermediate, [NOCl2] = (k1/k-1)[NO][Cl2], is done entirely in symbols — no numbers appear anywhere in this derivation — exactly the solve-a-formula-for-a-specified-variable move, here used to eliminate an unmeasurable reaction intermediate rather than to find a named physical quantity from given data.

Converting ΔH = −350 kJ mol⁻¹ for CaO + H₂O to kJ per gram, then scaling to a 1-ton batch of slaked lime, chains a divide-by-molar-mass step to a multiply-by-mass-ratio step — the same two-stage unit-rate model as a work-rate or mixture problem, where forgetting to convert the ton to grams before applying the per-gram rate is the standing slip.

Faraday's-law electrolysis problems (‘100,000 A through a melt for 1.00 h, 85% yield — what mass of metal forms?’) are rate × time = amount word problems, structurally identical to the work-rate and mixture problems algebra §2.4 trains — charge Q = It stands in for distance = rate × time, and stoichiometry then converts charge to moles of metal.

A silicon hydride at 306 torr, 26 °C, in 57.0 mL with mass 0.0861 g requires solving PV = nRT for n and then M = mass/n — the same substitute-then-solve-for-a-variable skeleton as any rate-and-quantity word problem, where using torr and °C directly instead of converting to atm and kelvin first throws the molar mass off by a large factor.

Finding the mass of CaH₂ needed to fill a 4.5 L balloon at 20 °C and 0.8 atm chains PV = nRT (solve for moles H₂) to the 2:1 mole ratio from the balanced equation to a mass — three linked model-and-solve steps performed in sequence, the same layered word-problem skill as a multi-step mixture question, where inverting the mole ratio halves or doubles the answer.

Deciding whether 3.0 g H₂ or 3.0 g N₂ limits ammonia production means converting both to moles and comparing the provided ratio against the 3:1 stoichiometric ratio, and the titration exercise (25.00 mL CsOH neutralized by 35.27 mL of 0.1062 M HNO₃) is the same M₁V₁ = M₂V₂ proportional model — both are the mixture/rate-problem skeleton applied to new reagents.

Finding the NaOH volume that neutralizes the acid from 2.00 g of PCl₃, and separately finding the mass of phosphorus needed for 1.00×10⁴ kg of phosphoric acid at 98.85% yield, are both multi-step model-and-solve problems chaining a mole ratio to a concentration or a yield percentage — treating percent yield as a multiplier instead of a divisor when working backward from product to reactant is the recurring trap.

Finding the volume of 0.250 M H₂SO₄ that neutralizes a solution made from 5.00 g CaCO₃ chains a mass-to-mole conversion through the reaction's stoichiometric ratio to M = mol/V solved for volume — the same multi-step model-and-solve skeleton as any concentration word problem, where dividing by the molarity instead of multiplying by its reciprocal inverts the answer.

Finding the grams of Epsom salts (MgSO₄·7H₂O) that form from 5.0 kg of magnesium chains a mass-to-mole conversion through a 1:1 stoichiometric ratio to a final mass in the hydrate's larger molar mass, landing on an answer reported as 5.1×10⁴ g — losing track of the power of ten during the kg-to-g conversion is the standing error in this shape of problem.

Recovering the grams of NaCl in 25 mL of 0.16 M physiological saline is a direct M = mol/V rearrangement multiplied by molar mass — the same solve-for-the-other-variable move as any concentration word problem, where treating 25 mL as 25 L is the unit-conversion slip this shape of question invites.

Recovering XeF₆'s empirical formula from 81 mL of Xe at STP and a titration of the resulting HF (68.43 mL of 0.3172 M NaOH) chains three separate model-and-solve steps — gas volume to moles, titration volume to moles, then a mole ratio — into one problem, the same layered word-problem skill as a multi-stage mixture question.

Titrating a 2.5000 g iron-ore sample with 19.17 mL of 0.0100 M Na₂Cr₂O₇ to find percent iron, and finding percent chloride from a 3.03707 g AgCl precipitate out of a 2.5624 g sample, both chain a volume-or-mass measurement through a mole ratio to a percent-by-mass — the same model-and-solve-then-convert-to-percent skeleton as any stoichiometry word problem.

Producing exactly 1000 kg of MTBE at 100% yield and working backward to the volume of methanol (density 0.7915 g/mL) needed chains a stoichiometric mass ratio through d = m/v solved for volume, landing on 4.593×10² L — the same multi-step model-and-solve-for-a-variable skeleton as any density or mixture word problem, and this is the section a teammate zeroed out for being ordinary stoichiometry, which is exactly why it repeats here.

Computing percent yield for 13.0 g of ethyl acetate obtained from 10.0 g of acetic acid requires finding the theoretical yield through the stoichiometric mole ratio first, then dividing actual by theoretical — treating the 13.0 g as though it were already the theoretical yield, instead of computing it from the limiting reagent, is the trap this shape of question sets.