Stoichiometry & Percent Yield — Full Lecture Notes

Any question that involves a chemical reaction (reactants → products) and asks for amounts produced or required is a stoichiometry problem.
- Tests rarely label it “stoichiometry”; you must infer it.
Core indicator: you are given a mass, volume, particles, or yield for one substance and asked to find the amount of another.
Percent-yield problems add an extra comparison step between the actual yield (measured in the lab) and the theoretical yield (calculated from stoichiometry).
- Formula: \%\,\text{yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100\%

Write & balance the chemical equation.
- Balancing guarantees conservation of mass ("you cannot create or destroy matter").
- Correct ionic charges must be used when writing formulas.
Convert the given quantity to moles.
- Current toolbox: grams → moles via molar mass.
- Future chapters: other pathways (liters → moles using \frac{V}{22.4\,L} for gases at STP, particles → moles using Avogadro’s number, etc.)
Identify the limiting reactant (LR) if more than one reactant is given.
- Limiting reactant determines the maximum amount of product (sandwich analogy below).
Use the mole ratio (coefficients) to move from moles of LR → moles of desired product.
Convert final moles to the requested unit (grams, liters, particles, etc.).
If percent yield is asked for, compare the calculated theoretical yield with the provided actual yield using the equation above.

Equation: n=\frac{m}{M} where n = moles, m = mass (g), M = molar mass ((g\,mol^{-1})).
Carry full calculator precision internally; round only the final answer to the correct significant figures.
- Example in transcript: student kept 3 sig figs for 3.00 g, but more sig figs internally for intermediate molar-mass values.

Analogy: making sandwiches that require 1 slice ham + 2 slices bread.
- If you still have 32 slices ham but no bread, production stops: bread is the LR.
In chemistry, once one reactant is consumed, the reaction ceases even if other reactants remain in excess.
Practical test-tip: calculate the moles of product each reactant could form; the smaller value identifies the LR.

Students practiced classifying reactions (single vs. double displacement).
- Example discussed was double displacement.
General double-displacement pattern: AB + CD \to AD + CB.
Correct product formulas require swapping partners and ensuring ionic charge neutrality.

Charges determine subscripts inside each compound; coefficients outside compounds balance the number of atoms.
Example walked through in class:
- Unbalanced (ionic formulas first): \text{Pb(NO}3)2 + \text{FeCl}3 \to \text{PbCl}2 + \text{Fe(NO}3)3
- Balanced: 3\text{Pb(NO}3)2 + 2\text{FeCl}3 \to 3\text{PbCl}2 + 2\text{Fe(NO}3)3
- Coefficient 3 in front of Pb(NO$3$)$2$ means 6 nitrates total on the left, matching 6 nitrates on the right.
Diatomic elements reminder: H2,N2,O2,F2,Cl2,Br2,I_2 appear with subscript 2 when uncombined.

Stoichiometric factor: \frac{\text{coefficient of desired}}{\text{coefficient of given}}.
- Shown on board: \frac{2~\text{mol AgCl}}{2~\text{mol AgNO}_3} for a reaction whose balanced equation contains 2 in front of each.
Even if coefficients are identical, you must still write and use them; they often cancel but they define the ratio.

A student used 1.243 g Pb(NO$3$)$2$.
- Molar mass mis-entered as 255.197 g mol$^{-1}$ (correct ≈ 331.2 g mol$^{-1}$; error traced to forgetting “2 NO$_3$”).
Another value cited: divisor 2.06 9 (?) g mol$^{-1}$—illustrates importance of re-checking calculator inputs.
Interim mole value heard: 0.048707 mol (mole counts differ if molar mass off).

Verify every empirical formula (charges & subscripts).
Balance the equation—count each atom on both sides.
Compute molar masses carefully; include parentheses multipliers (e.g., $\text{Pb(NO}3)2$ → 1 Pb + 2(N + 3 O)).
Keep units with each number to catch placement errors (g, mol, g mol$^{-1}$).
For LR, always divide theoretical moles of product by coefficient if comparing directly.
Only round final answers; keep guard-digits all the way.

Balanced equations embody the Law of Conservation of Mass; if your work seems “impossible to balance,” re-check formulas.
Percent yield quantifies efficiency; values >100 % signal calculation or experimental errors.
Mastery flows from repetition: convert→compare→ratio→convert back—every problem follows that skeleton.

Industrial chemists rely on limiting-reactant & percent-yield analysis to minimize waste and cost.
Environmental impact: maximizing yield reduces by-products and resource consumption.