Science Reference

Chemistry Equations & Formulas

Complete reference with LaTeX syntax and MATLAB commands for every formula
How this document works

Each formula is shown three ways: (1) rendered equation, (2) the LaTeX code to reproduce it in Obsidian or Overleaf, (3) MATLAB code to compute it numerically. MATLAB's Symbolic Math Toolbox commands are included where symbolic manipulation is useful.

1. Fundamental Constants & Unit Conversions
ConstantSymbolValueMATLAB
Avogadro's number$N_A$$6.022 \times 10^{23}$ mol$^{-1}$NA = 6.022e23;
Gas constant$R$$8.314$ J/(mol·K)R = 8.314;
Gas constant (atm)$R$$0.08206$ L·atm/(mol·K)R_atm = 0.08206;
Boltzmann constant$k_B$$1.381 \times 10^{-23}$ J/KkB = 1.381e-23;
Planck's constant$h$$6.626 \times 10^{-34}$ J·sh = 6.626e-34;
Speed of light$c$$2.998 \times 10^8$ m/sc = 2.998e8;
Faraday constant$F$$96{,}485$ C/molF = 96485;
Elementary charge$e$$1.602 \times 10^{-19}$ Ce_charge = 1.602e-19;
Electron mass$m_e$$9.109 \times 10^{-31}$ kgme = 9.109e-31;
Rydberg constant$R_H$$1.097 \times 10^7$ m$^{-1}$RH = 1.097e7;
Standard pressure—$1$ atm $= 101.325$ kPa $= 760$ mmHgP_std = 101325;
Standard temperature—$0°$C $= 273.15$ KT_std = 273.15;
Molar volume (STP)$V_m$$22.414$ L/molVm = 22.414;
Temperature Conversions
$$T_K = T_C + 273.15 \qquad T_C = \frac{5}{9}(T_F - 32) \qquad T_F = \frac{9}{5}T_C + 32$$

LaTeX: T_K = T_C + 273.15

T_C = 25;
T_K = T_C + 273.15;         % 298.15 K
T_F = (9/5)*T_C + 32;       % 77 °F
Pressure Conversions
$$1 \text{ atm} = 101.325 \text{ kPa} = 760 \text{ mmHg} = 14.696 \text{ psi} = 1.01325 \text{ bar}$$
P_atm = 2.5;
P_kPa = P_atm * 101.325;       % 253.3 kPa
P_mmHg = P_atm * 760;          % 1900 mmHg
P_psi = P_atm * 14.696;        % 36.74 psi
2. Atomic Structure & Quantum Numbers
Photon Energy, Frequency & Wavelength
$$E = h\nu = \frac{hc}{\lambda} \qquad c = \lambda\nu$$

LaTeX: E = h\nu = \frac{hc}{\lambda}

h = 6.626e-34; c = 2.998e8;
lambda = 550e-9;                 % 550 nm (green light)
nu = c / lambda;                 % 5.45e14 Hz
E = h * nu;                     % 3.61e-19 J per photon
E_eV = E / 1.602e-19;           % 2.25 eV
de Broglie Wavelength
$$\lambda = \frac{h}{mv} = \frac{h}{p}$$

Every particle has wave-like behavior. $m$ = mass, $v$ = velocity, $p$ = momentum.

h = 6.626e-34; me = 9.109e-31; v = 1e6;  % electron at 10^6 m/s
lambda = h / (me * v);           % 7.27e-10 m = 0.727 nm
Rydberg Equation — Hydrogen Emission Lines
$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \qquad n_2 > n_1$$

$n_1 = 1$: Lyman (UV). $n_1 = 2$: Balmer (visible). $n_1 = 3$: Paschen (IR).

RH = 1.097e7;
n1 = 2; n2 = 3;                 % Balmer series, H-alpha
inv_lambda = RH*(1/n1^2 - 1/n2^2);
lambda = 1/inv_lambda;           % 6.56e-7 m = 656 nm (red)
Bohr Model — Hydrogen Energy Levels
$$E_n = -\frac{13.6 \text{ eV}}{n^2} = -\frac{2.18 \times 10^{-18} \text{ J}}{n^2}$$
$$\Delta E = E_{\text{final}} - E_{\text{initial}} = -2.18 \times 10^{-18}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$
n_i = 3; n_f = 1;
dE = -2.18e-18 * (1/n_f^2 - 1/n_i^2);    % -1.94e-18 J (emitted)
Heisenberg Uncertainty Principle
$$\Delta x \cdot \Delta p \geq \frac{h}{4\pi} \qquad \text{or equivalently} \qquad \Delta x \cdot m\Delta v \geq \frac{\hbar}{2}$$

LaTeX: \Delta x \cdot \Delta p \geq \frac{h}{4\pi}

Quantum Numbers
NameSymbolValuesDetermines
Principal$n$$1, 2, 3, \ldots$Energy level / shell
Angular momentum$\ell$$0, 1, \ldots, n-1$Subshell shape (s, p, d, f)
Magnetic$m_\ell$$-\ell, \ldots, 0, \ldots, +\ell$Orbital orientation
Spin$m_s$$+\frac{1}{2}, -\frac{1}{2}$Electron spin

Max electrons per shell: $2n^2$. Max per subshell: $2(2\ell + 1)$.

n = 3;
max_electrons_shell = 2*n^2;         % 18
for ell = 0:n-1
    max_sub = 2*(2*ell+1);
    fprintf('l=%d: %d electrons\n', ell, max_sub)
end
3. Stoichiometry & Molar Relationships
Moles, Mass & Molar Mass
$$n = \frac{m}{M} \qquad m = nM \qquad N = n \cdot N_A$$

$n$ = moles, $m$ = mass (g), $M$ = molar mass (g/mol), $N$ = number of particles.

m = 36;                          % grams of water
M_H2O = 2*1.008 + 16.00;        % 18.016 g/mol
n = m / M_H2O;                   % 1.998 mol
N = n * 6.022e23;                % 1.203e24 molecules
Molarity (Molar Concentration)
$$M = \frac{n}{V} = \frac{\text{moles of solute}}{\text{liters of solution}}$$

LaTeX: M = \frac{n}{V} (note: $M$ for molarity, not to be confused with molar mass)

n_solute = 0.5;                  % 0.5 mol NaCl
V_liters = 0.25;                 % 250 mL = 0.25 L
Molarity = n_solute / V_liters;  % 2.0 M
Dilution
$$M_1 V_1 = M_2 V_2$$
M1 = 12; V1_needed = [];         % 12 M stock HCl
M2 = 1.5; V2 = 0.500;           % want 500 mL of 1.5 M
V1 = M2*V2/M1;                   % 0.0625 L = 62.5 mL of stock
Percent Composition
$$\%\text{ element} = \frac{n \times A_{\text{element}}}{M_{\text{compound}}} \times 100\%$$
% Percent oxygen in H2O
n_O = 1; A_O = 16.00; M_H2O = 18.016;
pct_O = (n_O * A_O / M_H2O) * 100;    % 88.8%
Limiting Reagent & Theoretical Yield
$$\text{moles product} = \text{moles limiting reagent} \times \frac{\text{coeff. product}}{\text{coeff. reagent}}$$
$$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$$
% 2H₂ + O₂ → 2H₂O
n_H2 = 3; n_O2 = 1;
% H2 needs O2 in 2:1 ratio → need 1.5 mol O2 but have 1
% O2 is limiting
n_H2O_theor = n_O2 * (2/1);        % 2 mol H2O
actual = 30;                         % 30 g actual yield
theoretical = n_H2O_theor * 18.016;  % 36.03 g
pct_yield = actual/theoretical*100;   % 83.3%
Empirical & Molecular Formulas
$$\text{molecular formula} = (\text{empirical formula})_n \qquad n = \frac{M_{\text{molecular}}}{M_{\text{empirical}}}$$
% Compound is 40.0% C, 6.7% H, 53.3% O. M = 180 g/mol
m_C = 40.0; m_H = 6.7; m_O = 53.3;
mol_C = m_C/12.01; mol_H = m_H/1.008; mol_O = m_O/16.00;
min_mol = min([mol_C mol_H mol_O]);
ratio = [mol_C mol_H mol_O] / min_mol;  % [1 2 1] → CH₂O
M_emp = 12.01 + 2*1.008 + 16.00;        % 30.03
n = round(180 / M_emp);                  % 6
% Molecular formula: C₆H₁₂O₆ (glucose)
4. Gas Laws
Ideal Gas Law
$$PV = nRT$$

$P$ in atm, $V$ in L, $n$ in mol, $R = 0.08206$ L·atm/(mol·K), $T$ in K. Or: $P$ in Pa, $V$ in m³, $R = 8.314$ J/(mol·K).

n = 2; T = 300; R = 0.08206;     % 2 mol, 300 K
P = 1;                            % 1 atm
V = n*R*T/P;                      % 49.24 L
Combined Gas Law
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
P1 = 1; V1 = 10; T1 = 300;
P2 = 2; T2 = 450;
V2 = P1*V1*T2 / (T1*P2);         % 7.5 L
Individual Gas Laws
LawEquationHeld Constant
Boyle's$P_1 V_1 = P_2 V_2$$T$, $n$
Charles's$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$P$, $n$
Gay-Lussac's$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$V$, $n$
Avogadro's$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$P$, $T$
Dalton's Law of Partial Pressures
$$P_{\text{total}} = P_1 + P_2 + \cdots + P_n \qquad P_i = x_i P_{\text{total}} \qquad x_i = \frac{n_i}{n_{\text{total}}}$$

$x_i$ = mole fraction of gas $i$.

n = [0.5 0.3 0.2];              % moles of 3 gases
x = n / sum(n);                  % mole fractions
P_total = 2;                     % 2 atm total
P_partial = x * P_total;         % [1.0, 0.6, 0.4] atm
Graham's Law of Effusion
$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$

Lighter gases effuse faster. $r$ = rate, $M$ = molar mass.

M_H2 = 2.016; M_O2 = 32.00;
ratio = sqrt(M_O2/M_H2);        % 3.99 → H₂ effuses ~4× faster
van der Waals Equation (Real Gases)
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

$a$ corrects for intermolecular attractions. $b$ corrects for molecular volume.

LaTeX: \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

% CO₂: a = 3.59 L²atm/mol², b = 0.0427 L/mol
a = 3.59; b = 0.0427; n = 1; R = 0.08206; T = 500;
% Solve for V numerically using fzero
f = @(V) (1 + a*n^2/V^2)*(V - n*b) - n*R*T;  % simplified
% V_ideal = nRT/P as starting guess
Root-Mean-Square Speed
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \qquad v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \qquad v_{\text{mp}} = \sqrt{\frac{2RT}{M}}$$

$M$ must be in kg/mol (not g/mol). $R = 8.314$ J/(mol·K).

R = 8.314; T = 300; M = 0.032;    % O₂ in kg/mol
v_rms = sqrt(3*R*T/M);             % 483.6 m/s
v_avg = sqrt(8*R*T/(pi*M));        % 445.6 m/s
v_mp = sqrt(2*R*T/M);              % 394.8 m/s
Kinetic Energy of Gas Molecules
$$KE_{\text{avg}} = \frac{3}{2}k_B T = \frac{3}{2}\frac{RT}{N_A}$$
kB = 1.381e-23; T = 300;
KE = 1.5 * kB * T;                  % 6.21e-21 J per molecule
5. Thermodynamics
First Law of Thermodynamics
$$\Delta U = q + w \qquad w = -P\Delta V \quad \text{(at constant pressure)}$$

$\Delta U$ = internal energy change, $q$ = heat, $w$ = work. Sign convention: $q > 0$ = heat absorbed, $w > 0$ = work done on system.

Enthalpy
$$\Delta H = q_p \quad \text{(at constant pressure)} \qquad q = mc\Delta T = nC\Delta T$$

$c$ = specific heat (J/(g·K)), $C$ = molar heat capacity (J/(mol·K)).

m = 100; c = 4.184; dT = 25;        % 100g water, heated 25°C
q = m * c * dT;                       % 10,460 J = 10.46 kJ
Hess's Law
$$\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$$
% CH₄ + 2O₂ → CO₂ + 2H₂O
dHf_CO2 = -393.5; dHf_H2O = -285.8;   % kJ/mol
dHf_CH4 = -74.8; dHf_O2 = 0;
dH_rxn = (1*dHf_CO2 + 2*dHf_H2O) - (1*dHf_CH4 + 2*dHf_O2);
% -890.3 kJ/mol (exothermic)
Entropy
$$\Delta S_{\text{rxn}} = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$$
$$\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}$$
Gibbs Free Energy
$$\Delta G = \Delta H - T\Delta S$$
$$\Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})$$

$\Delta G < 0$: spontaneous. $\Delta G > 0$: non-spontaneous. $\Delta G = 0$: equilibrium.

dH = -890.3e3; dS = -242.2;  T = 298;     % J and J/K
dG = dH - T*dS;                             % -817,985 J (spontaneous)
Gibbs Free Energy & Equilibrium
$$\Delta G^\circ = -RT\ln K \qquad \Leftrightarrow \qquad K = e^{-\Delta G^\circ / RT}$$
$$\Delta G = \Delta G^\circ + RT\ln Q$$

$Q$ = reaction quotient, $K$ = equilibrium constant. At equilibrium: $Q = K$, $\Delta G = 0$.

R = 8.314; T = 298;
dG0 = -33.0e3;                       % -33.0 kJ/mol
K = exp(-dG0/(R*T));                  % 6.48e5
Calorimetry — Bomb Calorimeter
$$q_{\text{rxn}} = -(q_{\text{water}} + q_{\text{cal}}) = -(mc\Delta T + C_{\text{cal}}\Delta T)$$
m_water = 500; c = 4.184; C_cal = 850;   % J/°C
dT = 3.2;
q_rxn = -(m_water*c*dT + C_cal*dT);      % -9,414 J
Bond Energy Method
$$\Delta H_{\text{rxn}} \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})$$

Breaking bonds requires energy (positive). Forming bonds releases energy (negative).

6. Chemical Kinetics
Rate Law
$$\text{rate} = k[A]^m[B]^n$$

$k$ = rate constant, $m$ and $n$ = reaction orders (determined experimentally, not from coefficients).

Integrated Rate Laws
OrderIntegrated LawLinear PlotHalf-Life
Zero$[A] = [A]_0 - kt$$[A]$ vs. $t$$t_{1/2} = \frac{[A]_0}{2k}$
First$\ln[A] = \ln[A]_0 - kt$$\ln[A]$ vs. $t$$t_{1/2} = \frac{\ln 2}{k}$
Second$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$\frac{1}{[A]}$ vs. $t$$t_{1/2} = \frac{1}{k[A]_0}$
 
% First-order decay
A0 = 0.5; k = 0.03;              % M, s⁻¹
t = linspace(0, 200, 500);
A = A0 * exp(-k*t);              % [A] over time
t_half = log(2)/k;                % 23.1 s
Arrhenius Equation
$$k = A e^{-E_a/RT} \qquad \ln k = \ln A - \frac{E_a}{RT}$$

Two-temperature form (find $E_a$ from two data points):

$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
k1 = 0.005; T1 = 300; k2 = 0.08; T2 = 350; R = 8.314;
Ea = R * log(k2/k1) / (1/T1 - 1/T2);   % 52,660 J/mol = 52.7 kJ/mol

% Predict k at new temperature
T3 = 400;
k3 = k1 * exp((Ea/R)*(1/T1 - 1/T3));   % 0.765 s⁻¹
Reaction Half-Life — Radioactive Decay
$$N(t) = N_0 e^{-\lambda t} \qquad \lambda = \frac{\ln 2}{t_{1/2}} \qquad t = \frac{1}{\lambda}\ln\frac{N_0}{N}$$
t_half = 5730;                    % C-14 half-life (years)
lambda = log(2)/t_half;
N0 = 1000; t = 11460;            % 2 half-lives
N = N0 * exp(-lambda*t);          % 250 (as expected: 1/4 remaining)
7. Chemical Equilibrium
Equilibrium Constant Expressions
$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \qquad K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$

For $aA + bB \rightleftharpoons cC + dD$. Pure solids and liquids are omitted.

Relationship Between $K_p$ and $K_c$
$$K_p = K_c(RT)^{\Delta n} \qquad \Delta n = \sum n_{\text{products}} - \sum n_{\text{reactants}} \;\text{(gas moles only)}$$
Kc = 0.042; R = 0.08206; T = 500;
dn = 2 - 3;                          % example: 3 mol gas → 2 mol gas
Kp = Kc * (R*T)^dn;                  % 1.02e-3
Reaction Quotient $Q$ vs. $K$
$$Q < K \;\Rightarrow\; \text{forward (→)} \qquad Q = K \;\Rightarrow\; \text{equilibrium} \qquad Q > K \;\Rightarrow\; \text{reverse (←)}$$
van't Hoff Equation — Temperature Dependence of $K$
$$\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
K1 = 1.5e5; T1 = 298; T2 = 350; dH = -92.2e3; R = 8.314;
K2 = K1 * exp((-dH/R)*(1/T2 - 1/T1));   % 5.96e3 (K decreases for exothermic)
8. Acids, Bases & pH
pH, pOH & pK
$$\text{pH} = -\log_{10}[\text{H}^+] \qquad \text{pOH} = -\log_{10}[\text{OH}^-] \qquad \text{pH} + \text{pOH} = 14 \;\text{(at 25°C)}$$
$$[\text{H}^+] = 10^{-\text{pH}} \qquad K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$$
H = 3.5e-4;                         % [H+] in mol/L
pH = -log10(H);                      % 3.46
pOH = 14 - pH;                       % 10.54
OH = 10^(-pOH);                      % 2.88e-11 M
Weak Acid Equilibrium
$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \qquad \text{p}K_a = -\log K_a$$

For weak acid $HA$ at concentration $C$ (assuming $x \ll C$):

$$[\text{H}^+] \approx \sqrt{K_a \cdot C} \qquad \text{pH} \approx \frac{1}{2}(\text{p}K_a - \log C)$$
Ka = 1.8e-5; C = 0.1;               % acetic acid, 0.1 M
H = sqrt(Ka * C);                    % 1.34e-3 M
pH = -log10(H);                      % 2.87
pct_ionization = (H/C)*100;          % 1.34%
Weak Base Equilibrium
$$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \qquad K_a \cdot K_b = K_w$$
Kb = 1.8e-5; C = 0.15;              % ammonia, 0.15 M
OH = sqrt(Kb * C);
pOH = -log10(OH);
pH = 14 - pOH;                       % 11.21
Henderson–Hasselbalch Equation (Buffer pH)
$$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$$

For a base buffer: $\text{pOH} = \text{p}K_b + \log\frac{[\text{BH}^+]}{[\text{B}]}$

pKa = 4.74;                          % acetic acid
conc_A = 0.20; conc_HA = 0.15;      % conjugate base, acid
pH = pKa + log10(conc_A/conc_HA);    % 4.87
Polyprotic Acid
$$K_{a1} \gg K_{a2} \gg K_{a3}$$

Each successive ionization is weaker. For $\text{H}_3\text{PO}_4$: $K_{a1} = 7.5 \times 10^{-3}$, $K_{a2} = 6.2 \times 10^{-8}$, $K_{a3} = 4.8 \times 10^{-13}$. The first ionization dominates the pH.

Titration — Equivalence Point
$$n_{\text{acid}} = n_{\text{base}} \qquad M_a V_a = M_b V_b \quad \text{(monoprotic)}$$
Ma = 0.1; Va = 25e-3;                % 25 mL of 0.1 M acid
Mb = 0.15;
Vb = Ma*Va/Mb;                        % 16.67 mL base to equivalence
9. Solubility & Precipitation
Solubility Product
$$K_{sp} = [\text{cation}]^m[\text{anion}]^n \qquad \text{for } M_m A_n \rightleftharpoons mM^{n+} + nA^{m-}$$
% AgCl ⇌ Ag⁺ + Cl⁻,  Ksp = 1.8e-10
Ksp = 1.8e-10;
s = sqrt(Ksp);                         % 1.34e-5 M (molar solubility)
Molar Solubility — Multi-Ion Salt
$$\text{For } A_2B_3: \quad K_{sp} = (2s)^2(3s)^3 = 108s^5$$
% Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻,  Ksp = 2.07e-33
Ksp = 2.07e-33;
% Ksp = (3s)^3 * (2s)^2 = 108 s^5
s = (Ksp/108)^(1/5);                   % 1.14e-7 M
Common Ion Effect
$$\text{If } [\text{common ion}] = C: \qquad K_{sp} = s(s + C) \approx sC \;\text{ when } C \gg s$$
% AgCl in 0.1 M NaCl solution
Ksp = 1.8e-10; C_Cl = 0.1;
s = Ksp / C_Cl;                        % 1.8e-9 M (much less soluble!)
Precipitation Criterion
$$Q_{sp} > K_{sp} \;\Rightarrow\; \text{precipitate forms} \qquad Q_{sp} \leq K_{sp} \;\Rightarrow\; \text{no precipitate}$$
10. Solutions & Colligative Properties
Concentration Units
UnitFormulaMATLAB
Molarity ($M$)$\frac{\text{mol solute}}{\text{L solution}}$M = n/V;
Molality ($m$)$\frac{\text{mol solute}}{\text{kg solvent}}$m = n/kg_solvent;
Mole fraction$x_i = \frac{n_i}{\sum n}$x = n_i/sum(n);
Mass percent$\frac{m_{\text{solute}}}{m_{\text{total}}} \times 100$pct = m_s/m_t*100;
ppm$\frac{m_{\text{solute}}}{m_{\text{total}}} \times 10^6$ppm = m_s/m_t*1e6;
Raoult's Law (Vapor Pressure Lowering)
$$P_{\text{solution}} = x_{\text{solvent}} \cdot P^\circ_{\text{solvent}} \qquad \Delta P = x_{\text{solute}} \cdot P^\circ_{\text{solvent}}$$
P0 = 23.8;                        % vapor pressure of pure water at 25°C (mmHg)
x_solvent = 0.98;
P_soln = x_solvent * P0;          % 23.3 mmHg
Boiling Point Elevation & Freezing Point Depression
$$\Delta T_b = iK_b m \qquad \Delta T_f = iK_f m$$

$i$ = van't Hoff factor (1 for non-electrolytes, 2 for NaCl, 3 for CaCl₂, etc.). $K_b$, $K_f$ = ebullioscopic/cryoscopic constants. $m$ = molality.

% NaCl in water (i=2)
i = 2; Kf = 1.86; Kb = 0.512;       % water constants (°C/m)
m = 0.5;                              % 0.5 molal
dTf = i*Kf*m;                         % 1.86 °C depression
dTb = i*Kb*m;                         % 0.512 °C elevation
fp = 0 - dTf;                         % -1.86 °C
bp = 100 + dTb;                       % 100.51 °C
Osmotic Pressure
$$\Pi = iMRT$$

$\Pi$ = osmotic pressure (atm), $M$ = molarity, $R = 0.08206$ L·atm/(mol·K).

i = 1; M_conc = 0.2; R = 0.08206; T = 298;
Pi = i*M_conc*R*T;                     % 4.89 atm
Henry's Law (Gas Solubility)
$$C = k_H P_{\text{gas}}$$

$C$ = dissolved gas concentration, $k_H$ = Henry's law constant, $P_{\text{gas}}$ = partial pressure.

kH = 3.4e-2;                      % O₂ in water, mol/(L·atm)
P_O2 = 0.21;                      % atm in air
C = kH * P_O2;                    % 7.14e-3 M dissolved O₂
11. Electrochemistry
Standard Cell Potential
$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$

Cathode = reduction (gains electrons). Anode = oxidation (loses electrons). $E^\circ_{\text{cell}} > 0$ means spontaneous.

E_cathode = 0.34;                    % Cu²⁺/Cu
E_anode = -0.76;                      % Zn²⁺/Zn
E_cell = E_cathode - E_anode;         % 1.10 V
Nernst Equation
$$E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{0.0257}{n}\ln Q \;\text{(at 25°C)}$$

Often written with $\log_{10}$: $E = E^\circ - \frac{0.0592}{n}\log Q$

E0 = 1.10; n = 2; T = 298;
R = 8.314; F_const = 96485;
Q = 0.01;                             % reaction quotient
E = E0 - (R*T/(n*F_const))*log(Q);   % 1.16 V
Gibbs Free Energy & Cell Potential
$$\Delta G^\circ = -nFE^\circ \qquad \Delta G = -nFE$$
n = 2; F_const = 96485; E0 = 1.10;
dG0 = -n * F_const * E0;              % -212,267 J = -212.3 kJ
Equilibrium Constant from Cell Potential
$$\ln K = \frac{nFE^\circ}{RT} \qquad K = e^{nFE^\circ/RT}$$
n = 2; F_const = 96485; E0 = 1.10; R = 8.314; T = 298;
K = exp(n*F_const*E0/(R*T));           % 1.69e37
Faraday's Laws of Electrolysis
$$m = \frac{MIt}{nF} \qquad Q = It$$

$m$ = mass deposited (g), $M$ = molar mass, $I$ = current (A), $t$ = time (s), $n$ = electrons transferred.

M_Cu = 63.55; I = 5; t = 3600;       % 5 A for 1 hour
n = 2; F_const = 96485;               % Cu²⁺ + 2e⁻ → Cu
m = M_Cu*I*t/(n*F_const);             % 5.93 g copper deposited
12. Nuclear Chemistry
Radioactive Decay Law
$$N(t) = N_0 e^{-\lambda t} \qquad \lambda = \frac{\ln 2}{t_{1/2}} \qquad A = \lambda N$$

$A$ = activity (disintegrations/sec = Bq), $\lambda$ = decay constant.

t_half = 5730*365.25*24*3600;         % C-14 half-life in seconds
lambda = log(2)/t_half;
N0 = 1e12;                            % initial atoms
t = 5730*365.25*24*3600;              % 1 half-life
N = N0 * exp(-lambda*t);              % 5e11 (half remaining)
A = lambda * N;                        % activity in Bq
Mass–Energy Equivalence
$$E = \Delta m \cdot c^2$$

$\Delta m$ = mass defect (kg), $c = 2.998 \times 10^8$ m/s. Often $\Delta m$ in amu: $1 \text{ amu} = 931.5$ MeV.

dm_amu = 0.0304;                       % mass defect for He-4
E_MeV = dm_amu * 931.5;               % 28.3 MeV binding energy
Binding Energy per Nucleon
$$\frac{E_b}{A} = \frac{\Delta m \cdot 931.5 \text{ MeV}}{A}$$

$A$ = mass number (protons + neutrons). Peak stability: Fe-56 at 8.79 MeV/nucleon.

dm = 0.0304; A = 4;                    % He-4
BE_per_nucleon = dm*931.5/A;           % 7.07 MeV/nucleon
Types of Radioactive Decay
TypeParticleSymbol$\Delta Z$$\Delta A$
AlphaHe-4 nucleus${}^4_2\text{He}$ or $\alpha$$-2$$-4$
Beta minusElectron${}^0_{-1}e$ or $\beta^-$$+1$$0$
Beta plusPositron${}^0_{+1}e$ or $\beta^+$$-1$$0$
GammaPhoton$\gamma$$0$$0$
Electron captureInner electronEC$-1$$0$

LaTeX for isotope notation: {}^{A}_{Z}\text{X} renders as ${}^{A}_{Z}\text{X}$, e.g., {}^{238}_{92}\text{U} → ${}^{238}_{92}\text{U}$

Carbon-14 Dating
$$t = \frac{t_{1/2}}{\ln 2}\ln\frac{A_0}{A} = \frac{1}{\lambda}\ln\frac{N_0}{N}$$
t_half = 5730;                         % years
lambda = log(2)/t_half;
A0 = 15.3;                             % modern activity (dpm/g)
A = 9.8;                               % sample activity
t = (1/lambda)*log(A0/A);              % 3,690 years old
Nuclear Reaction Energy (Q-value)
$$Q = (\sum m_{\text{reactants}} - \sum m_{\text{products}}) \times 931.5 \text{ MeV}$$

$Q > 0$: exothermic (releases energy). $Q < 0$: endothermic (requires energy input).

Cross-reference: Chemistry ↔ Algebra

Many of these formulas directly use the algebra you've been studying:

Logarithms: pH, pK, Arrhenius equation, Nernst equation, radioactive decay dating, decibels

Exponentials: First-order kinetics, radioactive decay, Arrhenius temperature dependence, Boltzmann distribution

Quadratics: Weak acid/base equilibrium (ICE tables), multi-ion solubility products

Systems of equations: Hess's Law (combining reactions), partial pressure calculations, buffer problems

Rational expressions: Rate laws, equilibrium expressions, van der Waals equation