Stoichiometric Calculations

Learning Objective 4.5.A Explain changes in the amounts of reactants and products based on the balanced reaction equation for a chemical process.

Quick Notes

Mole ratios = coefficients. In a balanced equation the coefficients are exact mole ratios that link reactants and products.
Stoichiometry calculations are based on mole conversions.
- From mass: n = m / M
- Solutions: n = M × V (L)
- Gases: PV = nRT
- Particles: N = n × N_A
Limiting reagent: reactant that will be 'used up' first in a reaction and determines the final amount of product formed.
Yields: theoretical yield is the maximum amount of product that can be formed (based on limiting reagent) and the percent yield = (actual yield ÷ theoretical yield) × 100%
Gas stoichiometry: at the same Temperature & Pressure, volume ratios of gases are the same as their molar ratios
Solution stoichiometry, For titrations: find moles of known concentration solution (M × V), apply the mole ratio, then solve for unknown M or V.
Units - always balance first; convert °C → K , use L and atm with the matching R and report sig figs sensibly.

Full Notes

Balanced Equations & Mole Ratios

A balanced chemical equation shows the smallest whole-number ratio of reacting species. Those coefficients are moles, so they act as exact conversion factors between substances.

Example:2 H₂ + O₂ → 2 H₂O
gives the ratios 2 mol H₂ : 1 mol O₂ : 2 mol H₂O.

Mass–Mass, Mass–Volume, and Particle Calculations

Convert to moles using molar mass (or n = M × V for solutions; or n = PV/RT for gases).
Apply the mole ratio from the balanced equation to switch substances.
Convert from moles to requested units: grams, liters (gas), liters/molarity (solution), or particles (via Avogadro’s number).

Matt’s Exam Tip

Things to look out for

Forgetting to balance the equation (wrong ratios).
Using grams directly in ratios – always convert to moles first.
Ignoring the limiting reagent when both reactants are provided.
In gas problems, mixing units (use K, atm, and the matching R; convert °C→K).
In solution problems, remember volume in liters for molarity.

Worked Example 1 — Mass → Mass

Water from Oxygen

Question: Using 2 H₂ + O₂ → 2 H₂O, what mass of water forms from 5.00 g O₂ (excess H₂)?

Moles O₂: 5.00 g ÷ 32.00 g·mol⁻¹ = 0.1563 mol.
Mole ratio to H₂O: 0.1563 mol O₂ × (2 mol H₂O / 1 mol O₂) = 0.3126 mol H₂O.
Mass H₂O: 0.3126 mol × 18.02 g·mol⁻¹ = 5.63 g.

Answer: 5.63 g H₂O.

Worked Example 2 — Limiting Reagent & Percent Yield

Ammonia Synthesis

Given: N₂ + 3 H₂ → 2 NH₃. Mix 25.0 g N₂ and 10.0 g H₂. Find the limiting reagent and theoretical mass of NH₃.

n(N₂) = 25.0 g ÷ 28.02 = 0.893 mol;
n(H₂) = 10.0 g ÷ 2.016 = 4.96 mol.
Needed H₂ for 0.893 mol N₂: 3×0.893 = 2.68 mol (available 4.96 mol) → N₂ is limiting.
n(NH₃) = 0.893 × (2/1) = 1.786 mol → mass = 1.786 × 17.03 = 30.4 g.

Theoretical yield: 30.4 g NH₃. If the actual yield were 27.5 g, percent yield = 27.5/30.4 × 100% = 90.5%.

Gas Stoichiometry

At a single temperature and pressure, gas volumes are proportional to moles, so you can use the equation’s coefficients as volume ratios. Otherwise use PV = nRT.

Same Templerature & Pressure example: For N₂ + 3 H₂ → 2 NH₃, reacting 12.0 L H₂ needs 4.0 L N₂ and makes 8.0 L NH₃.
General case: Find moles via n = PV/RT, apply the ratio, then convert back to volume if needed.

Solution Stoichiometry

Use molarity to connect volume and moles: n = M × V (V in liters).

Acid–Base Titration

Question: What is the concentration of NaOH if 25.00 mL of 0.1000 M HCl neutralizes 20.00 mL of NaOH? (Reaction: HCl + NaOH → NaCl + H₂O)

n(HCl) = 0.1000 mol·L⁻¹ × 0.02500 L = 0.002500 mol.
1:1 ratio → n(NaOH) at equivalence = 0.002500 mol.
M(NaOH) = n/V = 0.002500 ÷ 0.02000 = 0.125 M.

Answer: 0.125 M NaOH.