Advanced Stoichiometry
- Solve combined limiting-reactant and percent-yield problems in a single chain
- Use gravimetric analysis — the mass of a precipitate — to find the amount of an unknown
- Work back-titration and multi-step solution-stoichiometry problems while tracking units throughout
Chaining limiting reactant into percent yield
Hard exam problems stack the ideas from earlier lessons. First find the limiting reactant by converting each reactant to moles and seeing which yields the least product — that amount is the theoretical yield. Then compare it against the actual yield to get percent yield. The discipline is to finish the limiting-reactant analysis completely (all the way to grams of product) before touching the percent-yield step, so the theoretical value you divide by is trustworthy.
Gravimetric analysis: weighing a precipitate
Gravimetric analysis measures an unknown by converting it quantitatively into an insoluble solid, filtering, drying, and weighing that precipitate. The mass of precipitate → moles of precipitate → (via the net ionic mole ratio) → moles of the target ion → mass or concentration of the unknown. A chloride sample, for example, is precipitated with excess Ag⁺ as AgCl (Ag⁺ + Cl⁻ → AgCl), and the AgCl mass reveals exactly how much chloride was present.
Back-titration: measuring what is left over
When an analyte is a solid, slow to react, or has no clean endpoint, a back-titration helps: add a known excess of reagent so it fully consumes the analyte, then titrate the leftover reagent with a second standard solution. The reagent that reacted with the analyte equals initial reagent − leftover reagent. Applying the mole ratio to that difference gives the analyte. Keep every quantity in moles and label each one, because the answer is a subtraction of two amounts that are easy to swap.
For 2 Na + Cl₂ → 2 NaCl you combine 4.60 g Na (23.0 g·mol⁻¹) with 10.65 g Cl₂ (71.0 g·mol⁻¹) and collect 9.36 g NaCl (58.5 g·mol⁻¹). Find the limiting reactant, the theoretical yield, and the percent yield.
- 1.Moles of each reactant: Na = 4.60 g ÷ 23.0 g·mol⁻¹ = 0.200 mol; Cl₂ = 10.65 g ÷ 71.0 g·mol⁻¹ = 0.150 mol.
- 2.Test against the 2 Na : 1 Cl₂ ratio. To consume 0.200 mol Na you need 0.200 ÷ 2 = 0.100 mol Cl₂; you have 0.150 mol, so Cl₂ is in excess and Na is limiting.
- 3.Theoretical yield from Na: 0.200 mol Na × (2 mol NaCl ÷ 2 mol Na) = 0.200 mol NaCl → 0.200 mol × 58.5 g·mol⁻¹ = 11.7 g NaCl.
- 4.Percent yield = (actual ÷ theoretical) × 100% = (9.36 g ÷ 11.7 g) × 100% = 80.0%.
A solution of pure NaCl is treated with excess AgNO₃, and the AgCl precipitate is filtered, dried, and found to weigh 1.433 g (AgCl = 143.3 g·mol⁻¹). What mass of NaCl (58.5 g·mol⁻¹) was in the original solution?
- 1.Moles of precipitate: 1.433 g ÷ 143.3 g·mol⁻¹ = 0.0100 mol AgCl.
- 2.Net ionic reaction Ag⁺ + Cl⁻ → AgCl(s) is 1:1, so moles of Cl⁻ = 0.0100 mol.
- 3.Each NaCl supplies one Cl⁻, so moles of NaCl = 0.0100 mol.
- 4.Convert to mass: 0.0100 mol × 58.5 g·mol⁻¹ = 0.585 g NaCl.
Track units on every line and convert volumes to liters before multiplying by molarity. A back-titration answer is a difference of moles — subtracting mL from mol, or forgetting that the leftover reagent is what the second titration measures, flips the result completely.
A sample containing Ba²⁺ is treated with excess Na₂SO₄, precipitating 0.466 g of BaSO₄ (233 g·mol⁻¹) via Ba²⁺ + SO₄²⁻ → BaSO₄(s). How many moles of Ba²⁺ were in the sample?
An impure CaCO₃ tablet is dissolved in 50.0 mL of 0.200 M HCl (an excess). Titrating the leftover acid takes 20.0 mL of 0.100 M NaOH. How many moles of HCl reacted with the CaCO₃?
Multi-step AP problems reward a clear moles ledger: write moles of every species, label each with its source, and apply the balanced mole ratio at each junction. Do all conversions to grams or molarity only after the moles are settled — that keeps limiting-reactant, gravimetric, and back-titration problems on the same reliable track.
Answer the 2 checkpoints as you read.
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