6.5 Stoichiometry & Chemical Equations

Stoichiometry & Chemical Equations

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction — the arithmetic that turns a balanced equation into predictions about mass, moles, and yield. It is one of the most heavily calculation-based topics on the MCAT, and because no calculator is allowed, exam numbers are chosen to work out with clean mental math, a habit worth practicing now.

Molecular weight (also called molar mass) is the sum of the atomic weights of every atom in a molecule's formula, expressed in grams per mole (g/mol). For ionic compounds, the equivalent term is formula weight, since ionic solids exist as extended lattices rather than discrete molecules.

The empirical formula gives the simplest whole-number ratio of atoms in a compound (for example, CH₂O). The molecular formula gives the actual number of atoms in one molecule (for example, C₆H₁₂O₆ for glucose) and is always a whole-number multiple of the empirical formula:

Molecular formula = empirical formula × n, where n = (molecular weight) ÷ (empirical formula weight)

Two compounds can share an empirical formula while being entirely different substances — formaldehyde (CH₂O, molecular weight ≈30 g/mol) and glucose (C₆H₁₂O₆, molecular weight ≈180 g/mol, n = 6) both reduce to the same 1:2:1 carbon:hydrogen:oxygen ratio.

Metric Units and Percent Composition

Chemistry calculations run on the metric system: mass in grams (g) or kilograms (kg), volume in liters (L) or milliliters (mL), with standard prefixes (kilo- = 1000×, milli- = 1/1000×, micro- = 1/1,000,000×) converting between scales. Solution concentration is typically expressed as molarity (mol/L). Always convert to consistent units — grams and moles, not grams and millimoles — before applying any stoichiometric ratio.

Percent composition by mass describes what fraction of a compound's total mass comes from one element:

Percent composition = (mass of element in 1 mole of compound ÷ molar mass of compound) × 100%

For water (H₂O, molar mass ≈18 g/mol), hydrogen contributes 2 g of the 18 g total molar mass, so percent H ≈ (2 ÷ 18) × 100% ≈ 11.1%, and percent O ≈ (16 ÷ 18) × 100% ≈ 88.9%.

The Mole Concept, Avogadro's Number, and Density

A mole is a counting unit, exactly like "dozen," except it counts 6.022 × 10²³ particles — Avogadro's number (NA). One mole of any substance contains NA atoms, molecules, or ions, and its mass in grams equals that substance's molar mass. The mole is the bridge between the microscopic formula (atoms and molecules) and the macroscopic quantities (grams) actually measured on a laboratory balance.

Converting between mass and moles uses molar mass as the conversion factor: moles = mass (g) ÷ molar mass (g/mol), and mass (g) = moles × molar mass (g/mol).

Density is defined as mass per unit volume, ρ = m/V, typically reported in g/mL or g/cm³ for solids and liquids and g/L for gases. Density links a substance's mass to its volume and is often the missing piece needed to find moles when only a volume is given — for example, when a liquid reagent is measured out with a graduated cylinder rather than weighed directly.

Oxidation Numbers, Oxidizing/Reducing Agents, and Disproportionation

An oxidation number (oxidation state) is a bookkeeping charge assigned to an atom as if every bond in the species were fully ionic. Key rules: an element in its natural, uncombined form has an oxidation number of 0; a monatomic ion's oxidation number equals its charge; oxygen is usually −2 (except −1 in peroxides and +2 in OF₂); hydrogen is usually +1 (except −1 in metal hydrides); and the oxidation numbers within a species must sum to that species' overall charge.

An oxidizing agent is the species that is itself reduced (gains electrons, oxidation number decreases) while causing another species to be oxidized. A reducing agent is the species that is itself oxidized (loses electrons, oxidation number increases) while causing another species to be reduced.

Disproportionation is a special redox reaction in which a single element, starting in a single oxidation state, is simultaneously oxidized and reduced. The classic MCAT example is hydrogen peroxide decomposition:

2 H₂O₂ (O = −1) → 2 H₂O (O = −2) + O₂ (O = 0)

Oxygen starts at −1 in peroxide, then splits: some oxygen atoms are reduced to −2 (ending up in water) while others are oxidized to 0 (ending up in O₂ gas) — one starting element, one reaction, two different directions.

Describing Reactions with Chemical Equations

A balanced chemical equation obeys conservation of mass: the same number of atoms of each element must appear on both sides, achieved by adjusting whole-number coefficients in front of formulas — never subscripts within a formula, which would change the compound's identity rather than balance the equation. State symbols — (s), (l), (g), (aq) — indicate physical state, and reaction conditions such as a catalyst or added heat are often written above or below the reaction arrow.

Balancing redox equations typically uses the half-reaction method: split the overall equation into an oxidation half-reaction and a reduction half-reaction, balance atoms other than oxygen and hydrogen first, balance oxygen and hydrogen (using H₂O and H⁺ in acidic solution), balance charge by adding electrons to each half-reaction, then scale each half-reaction so the electrons lost in oxidation equal the electrons gained in reduction before combining the two halves and canceling any species that appear on both sides.

Worked Example: Limiting Reactant and Theoretical Yield

Question: Nitrogen and hydrogen gas react to form ammonia: N₂(g) + 3 H₂(g) → 2 NH₃(g). A flask contains 14 g of N₂ and 6 g of H₂. Which reactant is limiting, and what is the theoretical yield of NH₃ in grams?

Solution:

1. Convert each given mass to moles using molar mass (N₂ ≈ 28 g/mol, H₂ ≈ 2 g/mol):
   • moles N₂ = 14 g ÷ 28 g/mol = 0.5 mol
   • moles H₂ = 6 g ÷ 2 g/mol = 3 mol
2. Compare the available mole ratio to the required 1:3 (N₂:H₂) ratio from the balanced equation. For 0.5 mol N₂, the reaction needs 0.5 × 3 = 1.5 mol H₂ — but 3 mol H₂ is available, twice what's actually needed. N₂ is the limiting reactant; H₂ is in excess, with 1.5 mol of the original 3 mol left unreacted.
3. Calculate theoretical yield from the limiting reactant using the N₂-to-NH₃ mole ratio (1:2): 0.5 mol N₂ × (2 mol NH₃ ÷ 1 mol N₂) = 1.0 mol NH₃.
4. Convert moles of product to grams using NH₃'s molar mass (≈17 g/mol): 1.0 mol × 17 g/mol = 17 g NH₃, the theoretical yield — the maximum mass of product obtainable if the reaction proceeds to completion with no losses.

Worked Example in a Biological Context: Glucose and Oxygen

The same limiting-reactant logic governs metabolic stoichiometry. Complete aerobic oxidation of glucose is written:

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O

(This is the overall stoichiometry of cellular respiration's carbon skeleton; real cells capture free energy as ATP rather than releasing it entirely as heat, but the atom balance is identical to combustion.)

Question: A closed culture flask initially contains 0.10 mol glucose and 0.30 mol O₂. Assuming the reaction can proceed to completion as written, which reactant is limiting, and what is the theoretical yield of CO₂ in moles?

Solution:

1. Required ratio from the balanced equation: 1 mol glucose needs 6 mol O₂.
2. For 0.10 mol glucose, O₂ required = 0.10 × 6 = 0.60 mol. Only 0.30 mol O₂ is present — half of what complete oxidation of all glucose would need — so O₂ is limiting and leftover glucose remains.
3. Theoretical CO₂ yield is set by O₂ using the 6 O₂ : 6 CO₂ (1:1) ratio: 0.30 mol O₂ produces 0.30 mol CO₂.
4. Glucose consumed = 0.30 mol O₂ × (1 mol glucose ÷ 6 mol O₂) = 0.05 mol; unreacted glucose = 0.10 − 0.05 = 0.05 mol.

This pattern appears constantly on Chem/Phys passages: drug–receptor binding with a scarce ligand, titration of an amino acid with strong base, enzyme assays run under substrate-limiting conditions, and hemoglobin oxygen-binding stoichiometry when oxygen is scarce. Always convert to moles (or millimoles), compare to the balanced ratio, and compute product only from the limiting species. Percent yield in a lab or industrial context is then (actual mass of product ÷ theoretical mass) × 100%, never exceeding 100% without experimental error or an incorrect theoretical calculation.

Bridge to empirical formulas in biochemistry: many carbohydrates share the empirical formula CH₂O ("hydrates of carbon"). Distinguishing glucose (C₆H₁₂O₆, n = 6) from ribose (C₅H₁₀O₅, n = 5) or a triose (C₃H₆O₃, n = 3) requires molecular weight or structural information — empirical formula alone is never enough, a point already illustrated by the CH₂O / 180 g·mol⁻¹ → C₆H₁₂O₆ calculation.

Common MCAT Traps

• Comparing masses instead of moles to find the limiting reactant — the reactant with the smaller mass is not necessarily the limiting one; always convert to moles first and compare against the balanced-equation ratio.
• Forgetting that theoretical yield is calculated only from the limiting reactant, never from the reactant present in excess.
• Balancing equations by changing subscripts instead of coefficients, which silently turns one compound into a different compound rather than balancing the equation.
• Missing a disproportionation reaction because the reactants "look like" an ordinary redox pair — check whether the same element shows up in more than one oxidation state among the products before assuming a simple oxidizer/reducer relationship.