Chemistry

5% of exam

Stoichiometry and equilibrium, acids/bases and pH, oxidation-reduction and corrosion, and gas laws.

3 concepts

Exam areas

A. Oxidation and reduction

Oxidation-Reduction and Corrosion

Oxidation states, balanced half-reactions, cell EMF from standard potentials, and the galvanic-series logic behind corrosion and cathodic protection.

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B. Acids and bases

Acids, Bases, and pH

pH and pOH, the strong-vs-weak distinction through Ka and Kb, buffer design with Henderson-Hasselbalch, and titration stoichiometry.

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C. Chemical reactions

Stoichiometry and Chemical Equilibrium

The mole, balanced equations, limiting reagent and yield, plus the equilibrium constant and Le Chatelier's principle that govern how far a reaction goes.

Stoichiometry is the grammar of chemistry: a balanced equation is a molar recipe, and once you can read it in moles you can answer almost any 'how much' question on the FE Other Disciplines exam. The two skills the exam rewards are finding which reactant runs out first (the limiting reagent), and knowing how far a reaction actually proceeds (equilibrium). Both rest on the mole, the bridge between the grams you weigh and the molecules that react. The handbook's Chemistry and Biology section supplies Avogadro's number, molar volume, and the equilibrium-constant expression, but the procedure — balance, convert to moles, find the limiter, then check equilibrium — is yours to drill. This is the most reused chemistry skill on the exam, so it pays to make it automatic.

The mole and molar mass

A mole is Avogadro's number,

6.022\times10^{23}

, of particles, and its practical power is the molar mass: the mass in grams of one mole equals the formula's atomic-weight sum. Converting a measured mass to moles,

n=m/M

, is the first move in nearly every problem, because balanced equations speak in moles, not grams. For gases at standard conditions (

0^\circ\text{C}

1\,\text{atm}

) one mole occupies

22.4\,\text{L}

, which lets you turn a gas volume straight into moles without weighing anything.

n = \dfrac{m}{M}, \qquad V_{\text{molar}} = 22.4\;\text{L/mol at STP}

Balancing chemical equations

A balanced equation conserves every element and the total charge — atoms are rearranged, never created or destroyed. You balance by adjusting stoichiometric coefficients (never subscripts, which would change the substance) until each element's count matches on both sides. Those coefficients are exactly the mole ratios you carry through the rest of the problem. For ammonia synthesis the balanced form below says one mole of nitrogen demands three of hydrogen and yields two of ammonia — the ratio that drives every downstream calculation.

\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}

Limiting reagent

When reactants are not supplied in the exact stoichiometric ratio, one runs out first and caps the product; that one is the limiting reagent, and the rest are in excess. The reliable test is to divide each reactant's available moles by its coefficient — the smallest quotient identifies the limiter. Everything else (product formed, excess remaining) is then computed from the limiting reagent alone. Skipping this step and basing the answer on the wrong reactant is the single most common stoichiometry error on the exam.

\text{limiting reagent} = \min_i\!\left(\dfrac{n_i}{\nu_i}\right)

Theoretical and percent yield

The theoretical yield is the maximum product the limiting reagent can give, computed by mole ratio and converted back to grams. Real reactions fall short because of side reactions, incomplete conversion, or losses, so the percent yield compares what you actually got to that maximum. A yield above 100% is physically impossible and signals an arithmetic slip or impure product — a useful self-check on exam day.

\%\,\text{yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}}\times100\%

The equilibrium constant

Most reactions are reversible and settle at a dynamic equilibrium where forward and reverse rates match. The equilibrium constant

K_{eq}

is the ratio of product activities to reactant activities, each raised to its stoichiometric coefficient; pure solids and liquids take an activity of 1 and drop out. A large

K_{eq}

means equilibrium favors products, a small one favors reactants. For gases you can write

K_p

in partial pressures or

K_c

in concentrations, related by

K_p=K_c(RT)^{\Delta n}

where

\Delta n

is the change in moles of gas. Comparing the reaction quotient

Q

K_{eq}

tells you which way a non-equilibrium mixture will shift.

K_{eq} = \dfrac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}} \quad\text{for}\quad a\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}

Le Chatelier's principle

Le Chatelier's principle predicts the direction of an equilibrium shift without any calculation: when a system at equilibrium is disturbed, it responds so as to partially offset the disturbance. Add a reactant and it shifts toward products; remove product and it shifts to make more. Increase pressure (decrease volume) and it shifts toward the side with fewer gas moles. Raise temperature and it shifts in the endothermic direction, consuming the added heat — which means an exothermic reaction's

K_{eq}

falls as temperature rises. A catalyst speeds both directions equally and so does not move the equilibrium position, only the time to reach it.

Bioconversion: stoichiometry with cells

The same conservation bookkeeping extends to biological reactions, a flavor the FE Other Disciplines exam draws from environmental and biological practice. In aerobic biodegradation a substrate such as glucose is oxidized while microbes build new cell mass, written with an empirical cell formula like

\text{CH}_{1.8}\text{O}_{0.5}\text{N}_{0.2}

and an ammonia nitrogen source. The coefficients still come from balancing C, H, O, and N across the equation; the cell-formation coefficient is a theoretical maximum yield, trimmed in practice by a yield factor. The respiratory quotient (RQ), the ratio of

\text{CO}_2

produced to

\text{O}_2

consumed, ties the balance to measurable gas exchange.

\mathrm{C_6H_{12}O_6 + aO_2 + bNH_3 \rightarrow c\,CH_{1.8}O_{0.5}N_{0.2} + dCO_2 + eH_2O}

Exam strategy

Build the same reflex every time: balance the equation, convert all given masses or volumes to moles, then divide each reactant's moles by its coefficient to find the limiter before computing anything else. Carry the limiting reagent through to product moles, then back to grams only at the very end. For equilibrium questions, decide first whether you need a number (set up an ICE table and solve for the unknown extent) or just a direction (apply Le Chatelier and move on). Use the impossible-yield (

>100\%

) and the

Q

-versus-

K

comparisons as built-in sanity checks, and keep significant figures honest to three at the finish.

Key equations

Moles from mass

n = \dfrac{m}{M}

$n$ = moles, $m$ = mass (g), $M$ = molar mass (g/mol). The first conversion in nearly every stoichiometry problem.

Avogadro's number

1\;\text{mol} = 6.022\times10^{23}\;\text{particles}

Bridges molecule counts and moles. Rarely needed numerically but defines the mole.

Molar volume at STP

V = 22.4\;\text{L/mol}\quad(0^\circ\text{C},\,1\,\text{atm})

Converts an ideal-gas volume to moles at standard conditions; $359\,\text{ft}^3$ /lb-mol in USCS.

Molarity

M = \dfrac{n_{\text{solute}}}{V_{\text{solution, L}}}

Concentration in mol/L. Used to convert solution volumes to moles in reaction and equilibrium work.

Limiting-reagent test

\text{limiter} = \min_i\!\left(\dfrac{n_i}{\nu_i}\right)

Divide each reactant's moles $n_i$ by its coefficient $\nu_i$ ; the smallest quotient is the limiting reagent that sets the yield.

Percent yield

\%\,\text{yield} = \dfrac{m_{\text{actual}}}{m_{\text{theoretical}}}\times100\%

Actual product relative to the limiting-reagent maximum. A value above 100% flags an error or impurity.

Equilibrium constant

K_{eq} = \dfrac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}

Products over reactants, each to its coefficient; pure solids/liquids are taken as activity 1. Large $K_{eq}$ favors products.

Reaction quotient

Q = \dfrac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}[\mathrm{B}]^{b}}\;\text{(any instant)}

Same form as $K_{eq}$ but at non-equilibrium concentrations. $Q<K$ shifts forward, $Q>K$ shifts reverse.

Kp–Kc relation

K_p = K_c\,(RT)^{\Delta n}

$\Delta n$ = moles of gaseous product minus reactant. Use $R=0.08206\,\text{L·atm/(mol·K)}$ with pressures in atm.

Solubility product

K_{sp} = [\mathrm{A^{n+}}]^{m}[\mathrm{B^{m-}}]^{n}

Equilibrium constant for a slightly soluble salt $\text{A}_m\text{B}_n$ dissolving; predicts precipitation when $Q>K_{sp}$ .

Worked examples

Limiting reagent and percent yield in ammonia synthesis

Problem. In the reaction $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$ , $28.0\,\text{g}$ of $\text{N}_2$ reacts with $10.0\,\text{g}$ of $\text{H}_2$ . Identify the limiting reagent, find the theoretical mass of ammonia, and the percent yield if $29.0\,\text{g}$ of $\text{NH}_3$ is actually collected. Use $M_{N_2}=28.02$ , $M_{H_2}=2.016$ , $M_{NH_3}=17.03\,\text{g/mol}$ .

Solution. Convert to moles:

n_{N_2}=28.0/28.02=1.00\,\text{mol}

;

n_{H_2}=10.0/2.016=4.96\,\text{mol}

. Apply the limiting test:

n_{N_2}/1=1.00

versus

n_{H_2}/3=1.65

. The smaller is nitrogen, so

\text{N}_2

is limiting. Theoretical ammonia:

1.00\,\text{mol N}_2\times\dfrac{2\,\text{mol NH}_3}{1\,\text{mol N}_2}=2.00\,\text{mol}

, and

2.00\times17.03=34.06\,\text{g}

\%\,\text{yield}=\frac{29.0}{34.06}\times100\%=\mathbf{85.1\%}.

Sanity:

\text{H}_2

used

=3(1.00)=3.00\,\text{mol}

, leaving

1.96\,\text{mol}

(

3.95\,\text{g}

) of

\text{H}_2

in excess — consistent with nitrogen running out first, and the yield is safely under 100%.

Equilibrium concentrations from an ICE table

Problem. For $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ , $K_c=50.0$ at the reaction temperature. A $1.00\,\text{L}$ vessel is charged with $1.00\,\text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ and no HI. Find the equilibrium concentration of HI.

Solution. ICE:

[\text{H}_2]=[\text{I}_2]=1.00-x

[\text{HI}]=2x

K_c=\frac{(2x)^2}{(1.00-x)^2}=50.0.

The form is a perfect square, so take roots:

\dfrac{2x}{1.00-x}=\sqrt{50.0}=7.071

. Solve:

2x=7.071-7.071x\Rightarrow 9.071x=7.071\Rightarrow x=0.779

. Thus

[\text{HI}]=2(0.779)=\mathbf{1.56\;M}

, with

[\text{H}_2]=[\text{I}_2]=0.221\,\text{M}

. Check:

(1.56)^2/(0.221)^2=2.43/0.0488=49.8

, matching

K_c=50.0

within rounding (carrying the unrounded

x=0.7795

gives

(1.559)^2/(0.2205)^2=50.0

exactly).

Le Chatelier direction

Problem. The exothermic synthesis $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ is at equilibrium. State the direction the equilibrium shifts when (a) the total pressure is increased by compression, and (b) the temperature is raised.

Solution. (a) The left side has

1+3=4

moles of gas, the right side has

2

. Increasing pressure shifts toward fewer gas moles, so the equilibrium shifts toward

\text{NH}_3

(the products). (b) The forward reaction is exothermic, so heat is effectively a product. Adding heat shifts the equilibrium in the endothermic (reverse) direction, toward

\text{N}_2

and

\text{H}_2

, and

K_{eq}

decreases. Sanity: these are exactly why industrial ammonia synthesis uses high pressure but only moderate temperature (with a catalyst to reach equilibrium fast despite the cooler operation).

Common pitfalls

•Computing yield or product from the reactant given in the largest amount instead of the limiting reagent. Always run the $n_i/\nu_i$ test first.
•Balancing by changing subscripts. Only coefficients may change — altering a subscript changes the chemical identity of the species.
•Forgetting to convert grams to moles before using mole ratios. Stoichiometric coefficients are mole ratios, never mass ratios.
•Including pure solids or liquids in $K_{eq}$ . Their activity is 1, so they are omitted; only gases and dissolved species appear.
•Using $K_p=K_c$ when moles of gas change. Apply $K_p=K_c(RT)^{\Delta n}$ whenever $\Delta n\neq 0$ , with $R=0.08206\,\text{L·atm/(mol·K)}$ .
•Thinking a catalyst shifts equilibrium or raises yield. A catalyst speeds both directions equally — it changes the rate, not the equilibrium position or $K_{eq}$ .
•Reporting a percent yield above 100%. This is physically impossible and means a math slip or that the product was impure or wet — recheck the theoretical value.

References

NCEES FE Reference Handbook — Chemistry and Biology (Definitions, Equilibrium Constant of a Chemical Reaction, Solubility Product, Bioconversion)
NCEES FE Reference Handbook — Chemistry and Biology (Periodic Table of Elements) — Atomic weights for molar-mass calculations.
NCEES FE Reference Handbook — Thermodynamics — Heats of reaction and the temperature dependence of equilibrium.