10.2 Stoichiometry: Conservation of Matter/Energy & Reactant/Product Calculations

Stoichiometry: Conservation of Matter/Energy & Reactant/Product Calculations

Quick Answer: Stoichiometry is the arithmetic of chemical reactions. The PA-CAT Bulletin of Information (rev. 20240815) lists stoichiometry under Chemistry Table 5 and even provides a sample item built around acetylation of p-aminophenol by acetic anhydride — with a hydrolysis side reaction. The rule that anchors everything is the law of conservation of mass: a balanced equation accounts for every atom.

The Mole and Molar Mass

One mole = 6.022 × 10²³ particles (Avogadro's number). Molar mass (g/mol) converts grams to moles: n = mass ÷ M. For CO₂, M = 12.01 + 2(16.00) = 44.01 g/mol, so 88.02 g CO₂ = 2.00 mol = 1.20 × 10²⁴ molecules.

Mole Ratios from Balanced Equations

Coefficients are mole ratios. In 2 H₂ + O₂ → 2 H₂O, 2 mol H₂ reacts with 1 mol O₂ to give 2 mol H₂O. The general procedure:

1. Balance the equation.
2. Convert given mass → moles (÷ M).
3. Use the mole ratio (coefficient target ÷ coefficient given).
4. Convert moles target → mass (× M).

Limiting Reactant, Theoretical Yield, Percent Yield

The limiting reactant is consumed first and caps the product. The theoretical yield is the maximum product possible from the limiting reactant. Percent yield = (actual yield ÷ theoretical yield) × 100%.

To find the limiting reactant, compute product moles from each reactant separately; the smaller result identifies the limiting reactant. Excess reactant remains unreacted.

The PA-CAT Worked Example: Acetylation of p-Aminophenol

The Bulletin's stoichiometry sample uses the synthesis of acetaminophen (paracetamol) from p-aminophenol and acetic anhydride:

Main reaction (acetylation):

HO–C₆H₄–NH₂ + (CH₃CO)₂O → HO–C₆H₄–NHCOCH₃ + CH₃COOH

p-aminophenol + acetic anhydride → acetaminophen + acetic acid

Side reaction (hydrolysis):

(CH₃CO)₂O + H₂O → 2 CH₃COOH

Acetic anhydride that does not acetylate the amine is hydrolyzed by water, producing extra acetic acid. The Bulletin reports 29.47 g of acetaminophen isolated and 13.35 g of acetic acid recovered, and asks how much acetic anhydride was consumed by acetylation versus how much underwent hydrolysis. This is a mass-balance problem with a competing side reaction, not a percent-yield problem — both answers are expressed in grams of acetic anhydride.

Step-by-step (molar masses: p-aminophenol 109.13 g/mol; acetic anhydride 102.09 g/mol; acetaminophen 151.16 g/mol; acetic acid 60.05 g/mol):

1. Moles acetaminophen isolated = 29.47 g ÷ 151.16 g/mol = 0.195 mol.
2. By the 1:1 acetylation stoichiometry, moles of acetic anhydride consumed by acetylation = 0.195 mol → mass = 0.195 × 102.09 = 19.91 g.
3. Acetic acid co-produced by acetylation (1:1 with acetaminophen) = 0.195 mol × 60.05 g/mol = 11.71 g.
4. Acetic acid from hydrolysis = total recovered − acetylation co-product = 13.35 − 11.71 = 1.64 g.
5. Hydrolysis stoichiometry is 1 mol anhydride → 2 mol acetic acid. Moles acetic acid from hydrolysis = 1.64 ÷ 60.05 = 0.0273 mol → anhydride consumed by hydrolysis = 0.0273 ÷ 2 = 0.0137 mol → mass = 0.0137 × 102.09 = 1.39 g.
6. Answer: acetylation consumed 19.91 g of acetic anhydride and hydrolysis consumed 1.39 g of acetic anhydride (Bulletin answer choice B).

The lesson: when a side reaction competes for a reagent, account for the reagent each pathway consumes. The product mass (acetaminophen) fixes the acetylation branch; the surplus acetic acid (recovered minus the acetylation co-product) fixes the hydrolysis branch. The PA-CAT may ask which mass corresponds to which pathway, so label every arrow with the species it refers to — 19.91 g and 1.39 g are both grams of acetic anhydride, not of acetaminophen or acetic acid.

flowchart TD
    A[Acetic anhydride charged] --> B{Fate of anhydride}
    B -->|Acetylation: 0.195 mol| C[Acetaminophen 29.47 g isolated + 11.71 g acetic acid]
    B -->|Hydrolysis: 0.0137 mol| D[1.64 g acetic acid]
    C --> E[Anhydride consumed = 19.91 g]
    D --> F[Anhydride consumed = 1.39 g]

Conservation of Energy

Reactions also conserve energy. The enthalpy change ΔH of a balanced reaction is fixed regardless of pathway (Hess's law, covered in 10.3). Stoichiometry extends to heat: if 2 mol H₂ + O₂ → 2 H₂O releases 572 kJ, then 1 mol H₂ releases 286 kJ, and combusting 4 g H₂ (2 mol) releases 572 kJ. Coefficients scale both mass and energy.

Quick Checklist

• Is the equation balanced? (Count atoms each side.)
• Is the given in mass, moles, or particles? Convert to moles first.
• Which reactant is limiting? Compute product from each.
• Does the problem mention a side reaction? Account for the reagent it consumes.
• Percent yield uses actual ÷ theoretical, never actual ÷ given.

Worked Limiting-Reactant Mass-to-Mass Calculation

A common PA-CAT stoichiometry item gives two reactant masses and asks for the product mass. Use the four-step mole-ratio pipeline: mass → moles → mole ratio → mass.

Problem: 25.0 g of Fe₂O₃ reacts with 15.0 g of Al via the thermite reaction. What mass of Fe is produced, and which reactant is limiting?

Balanced equation: Fe₂O₃ + 2 Al → 2 Fe + Al₂O₃

Step 1 — Convert each reactant to moles. Moles Fe₂O₃ = 25.0 g ÷ 159.69 g/mol = 0.1566 mol. Moles Al = 15.0 g ÷ 26.98 g/mol = 0.5560 mol.

Step 2 — Use mole ratios to find product moles from each reactant. From Fe₂O₃: 0.1566 × (2 mol Fe / 1 mol Fe₂O₃) = 0.3132 mol Fe. From Al: 0.5560 × (2 mol Fe / 2 mol Al) = 0.5560 mol Fe.

Step 3 — Identify the limiting reactant. The smaller product amount is 0.3132 mol Fe, so Fe₂O₃ is limiting and Al is in excess. All downstream stoichiometry uses the limiting reactant's 0.3132 mol Fe.

Step 4 — Convert product moles to mass. Mass Fe = 0.3132 mol × 55.85 g/mol = 17.5 g Fe.

Excess check: Al consumed = 0.1566 mol Fe₂O₃ × (2 mol Al / 1) = 0.3132 mol = 8.45 g. Excess Al = 15.0 − 8.45 = 6.55 g unreacted.

The limiting reactant is always the one giving the smaller product amount; never average or add the two product estimates. This pipeline works for any balanced reaction on the exam.