Master Review Sheet

M = coefficient, exactly one nonzero digit left of the decimal; n = integer exponent (positive for numbers above 1, negative for numbers below 1)

Write any very large or very small number this way, and let the calculator carry the powers of ten while you handle the sig figs.

Try it: 0.00456 = 4.56 × 10⁻³ (decimal moved 3 places right, so n = −3)

K = Kelvin temperature (K); °C = Celsius temperature; 273.15 is exact and never limits sig figs

Convert to kelvins before any gas law; Kelvin is the scale whose zero is absolute zero, so no reading on it is ever negative.

Try it: 25 °C + 273.15 = 298.15 K

°F = Fahrenheit reading; °C = Celsius reading; 1.8 and 32 are exact defined numbers

Order of operations is the whole game: the 32 is added last going to °F and subtracted first coming back, and it never gets multiplied by 1.8.

Try it: 35.0 °C → 1.8(35.0) + 32 = 63.0 + 32 = 95.0 °F

ΔT = T_final − T_initial, a gap between two readings, not a reading

This is why q = mcΔT comes out identical whether the endpoint readings were in °C or K.

Try it: A rise from 20 °C to 45 °C is ΔT = 25 C°, which is also 25 K

d = density (g⁄mL or g⁄cm³ for solids and liquids, g⁄L for gases); mass in g; volume in mL

Reach for density whenever a problem hands you mass and asks for volume, or the reverse; it rides the same units-cancel rail as any conversion factor.

Try it: A 323 g sample occupying 53.0 mL: d = 323 g ⁄ 53.0 mL = 6.09 g⁄mL

range = highest − lowest trial (same units as the data); average = sum of trials ÷ number of trials

This is the precision verdict: it uses only the trials themselves, and the true value never enters it.

Try it: Trials 51.42, 51.38, 51.44, 51.40 g: range 0.06 g, average 51.41 g, so 0.06 ⁄ 51.41 × 100 = 0.12%, precise

average = mean of the trials; true value = the accepted or known value

This is the accuracy verdict; agreement between trials cannot expose an error they all share, so a miscalibrated balance passes precision and fails here.

Try it: Average 51.41 g against a 50.00 g standard: (51.41 − 50.00) ⁄ 50.00 × 100 = 2.8%, not accurate

fractional abundance = percent abundance ÷ 100; isotope masses in amu (1 amu = exactly 1⁄12 the mass of a carbon-12 atom)

The periodic table prints this weighted average, which is why it shows decimals; the result must land between the lightest and heaviest isotope, closest to the most abundant one.

Try it: Copper: (0.6917 × 62.9296) + (0.3083 × 64.9278) = 43.53 + 20.02 = 63.55 amu

atomic masses from the periodic table (g⁄mol); subscripts count the atoms of each element

The periodic-table number works for one atom in amu and one mole in grams, so molar mass is the equality that lets you count atoms by weighing.

Try it: Al(OH)₃: 26.98 + 3(16.00) + 3(1.008) = 78.00 g⁄mol

molar mass in g⁄mol; Avogadro's number = 6.022 × 10²³ particles⁄mol

This is the master solution map of the course; every mole problem is some walk along this rail with each factor written so the unwanted unit cancels.

Try it: 10.5 g S × (1 mol ⁄ 32.07 g) × (6.022 × 10²³ atoms ⁄ 1 mol) = 1.97 × 10²³ atoms S

mass of element = (number of atoms) × (atomic mass); molar mass = whole-formula total (g⁄mol)

Percent composition is fixed for a pure compound at any sample size, and the 100% total is both your check and a shortcut for the last element.

Try it: K₂S (110.27 g⁄mol): %K = 78.20 ⁄ 110.27 × 100 = 70.92%, leaving 29.08% S

empirical formula = smallest whole-number atom ratio; molecular formula = empirical formula × n; n must be a whole number, and n × empirical mass must rebuild the given molar mass

Percents alone only reach the empirical formula, since every multiple of it shares the same percent composition; the measured molar mass is what picks out the multiple.

Try it: 40.0% C, 6.7% H, 53.3% O, molar mass 180.16 g⁄mol: moles 3.33 : 6.65 : 3.33 → CH₂O (30.03 g⁄mol); n = 180.16 ⁄ 30.03 = 6 → C₆H₁₂O₆

M = molarity (mol⁄L); mol solute; L of finished solution (not solvent added); convert mL to L first

Molarity converts between solution volume and moles of solute in either direction; for an ion's concentration, scale by the formula subscript after dissociation.

Try it: 0.20 mol KCl dissolved to make 250.0 mL: 0.20 mol ⁄ 0.2500 L = 0.80 M

both masses in the same unit (they cancel); mass solution = solute + solvent; 1 ppm = 1000 ppb; 1% = 10,000 ppm

Reach for ppm and ppb when a percent would read as a string of zeros: trace solutes like fluoride, dissolved minerals, and pollutants.

Try it: Fluoride at 0.70 ppm in a 2.0 L pitcher: 0.70 mg⁄L × 2.0 L = 1.4 mg F⁻

a, b, c, d = balanced-equation coefficients; they count moles, never grams

The heart of every stoichiometry problem is this mole-to-mole step, and the ratios come only from a balanced equation, so balance first, always.

Try it: N₂ + 3 H₂ → 2 NH₃: 4.2 mol H₂ × (2 mol NH₃ ⁄ 3 mol H₂) = 2.8 mol NH₃

molar masses from the formulas (g⁄mol); mole ratio from the balanced coefficients

You cannot jump straight from grams of A to grams of B; only the mole ratio connects two substances, and it only speaks moles.

Try it: 3 Si + 2 Cr₂O₃ → 3 SiO₂ + 4 Cr: 59.4 g Si × (1 mol ⁄ 28.09 g) × (4 mol Cr ⁄ 3 mol Si) × (52.00 g ⁄ 1 mol) = 147 g Cr

compare amounts of product, never masses or moles of reactant; excess remaining = starting amount − amount consumed (via the mole ratio from the limiting reactant)

One reactant runs out first and the reaction stops there; picking the limiting reactant by smaller mass or fewer moles is the classic trap, because coefficients matter.

Try it: 2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe: 4.60 mol Al gives 2.30 mol Al₂O₃, 3.76 mol Fe₂O₃ gives 3.76 mol; smaller wins, Al is limiting

actual yield = measured in the lab, stated in the problem; theoretical yield = computed by stoichiometry from the given amounts (from the limiting reactant if two are given)

Reactions come up short like cookie recipes do; a percent above 100 means the fraction is upside down or the product came out wet.

Try it: Theoretical 22.0 g CO₂, collected 18.5 g: 18.5 ⁄ 22.0 × 100 = 84.1%

molarity from the solution label (mol⁄L); mole ratio from the balanced equation; molar mass from the formula

Molarity connects volume and moles of one substance; only the balanced equation connects two different substances, so never skip the middle link.

Try it: BaCl₂ + 2 AgNO₃ → 2 AgCl + Ba(NO₃)₂: 1.500 L × (0.400 mol ⁄ 1 L) × (2 mol AgCl ⁄ 1 mol BaCl₂) × (143.3 g ⁄ 1 mol) = 172 g AgCl

[H₃O⁺] = hydronium ion concentration (M); pH is unitless

The log scale compresses concentrations spanning many powers of ten into a number you can read off a meter; lower pH means more acidic.

Try it: [H₃O⁺] = 1 × 10⁻³ M gives pH 3, acidic

q = heat (J); m = mass (g); c = specific heat (J⁄(g·°C)); ΔT = T_final − T_initial (°C)

Rearrange the same equation for q, m, c, or ΔT; a cooling sample has negative ΔT, so q comes out negative, meaning heat was released.

Try it: Warming 200. g of water by 10.0 °C: q = (200. g)(4.184 J⁄(g·°C))(10.0 °C) = 8370 J ≈ 8.37 kJ

Water's unusually large specific heat is why it dominates every calorimetry balance and why the final temperature lands so close to the water's start.

Try it: Gold's c is 0.129 J⁄(g·°C): the same heat warms gold about 32 times more than water

ΔE = change in internal energy (J); q positive when heat flows into the system; w positive when work is done on the system (compression), negative when the system does work (expansion)

Add signed values, never magnitudes; a gas that absorbs heat and then spends some of it pushing on the surroundings keeps only the difference.

Try it: A gas absorbs 125 J and does 78 J of work: ΔE = (+125) + (−78) = +47 J

each m and c stay with their own body; T_f = shared final temperature, always between the two starting temperatures

T final is dragged toward whichever body carries the larger m times c, which is nearly always the water. A negative ΔT is not an error: it marks the body that released heat. A T final outside the starting pair means the minus sign in q_A = −q_B was dropped.

Try it: 100 g Fe (c = 0.449) at 90 °C into 150 g water at 20 °C: 44.9 × (T_f − 90) = −627.6 × (T_f − 20), so T_f = (44.9 × 90 + 627.6 × 20) ⁄ 672.5 = 24.7 °C. Iron's ΔT = −65.3 °C; the water's ΔT = +4.7 °C.

each given thermochemical equation carries its own ΔH (kJ); shared species on opposite sides cancel in the sum

Enthalpy is a state function, so a reaction no calorimeter can isolate still has a computable ΔH: fix the equations first and the ΔH values follow.

Try it: C + O₂ → CO₂ (−393.5 kJ) plus the reverse of CO + ½ O₂ → CO₂ (+283.0 kJ) gives C + ½ O₂ → CO: ΔH = −393.5 + 283.0 = −110.5 kJ

n = balanced-equation coefficients; ΔH°f values from the table (kJ⁄mol), row chosen by physical state

Products minus reactants, coefficients multiplied in, elements counted as zero: the table method is Hess's law, pre-packaged.

Try it: CH₄ + 2 O₂ → CO₂ + 2 H₂O(l): [−393.5 + 2(−285.8)] − [−74.8 + 0] = −890.3 kJ

P and V in any units, matched on both sides; T and n held constant

Squeeze a gas into a smaller volume and its particles hit the walls more often; finish with the sense check that pressure up means volume down.

Try it: 745 mmHg in 65.0 L moved to 25.0 L: P₂ = (745 × 65.0) ⁄ 25.0 = 1940 mmHg

V in any unit, matched on both sides; T always in kelvins

Convert Celsius to Kelvin before anything else; the proportionality runs through absolute zero, so Celsius numbers give nonsense.

Try it: 3.0 L at −15 °C (258 K) warmed to 27 °C (300 K): V₂ = 3.0 × (300 ⁄ 258) = 3.5 L

P in any unit, matched on both sides; T always in kelvins

This is why aerosol cans warn against incineration: heating a sealed, rigid can drives the pressure up dangerously.

Try it: A sealed can at 2.0 atm and 300. K heated to 450. K: P₂ = 2.0 × (450 ⁄ 300) = 3.0 atm

P and V in any units, matched on both sides; T always in kelvins; n fixed

One equation covers every before-and-after problem; list the knowns as (P₁, V₁, T₁) and (P₂, V₂, T₂), find the one unknown, sense-check the direction.

Try it: 125 mL at STP taken to 65 °C and 320. torr: V₂ = 125 × (760 ⁄ 320.) × (338 ⁄ 273) = 368 mL

P = pressure (atm); V = volume (L); n = amount of gas (mol); R = 0.08206 L·atm⁄(mol·K); T = temperature (K, always)

One equation relates all four properties of a single gas sample; leaving temperature in Celsius is the single most common gas-law error.

Try it: n = PV ⁄ RT = (1.00 atm × 22.4 L) ⁄ (0.08206 × 273 K) = 1.00 mol, the molar volume recovered

M = molar mass (g⁄mol); m = sample mass (g); R = 0.08206 L·atm⁄(mol·K); P in atm, V in L, T in K

Weigh a bulb of unknown gas, read P, V, and T, and the identity is one algebra step away; the units must come out g⁄mol or the setup is wrong.

Try it: 0.210 g in 0.333 L at 0.568 atm and 296 K: M = (0.210 × 0.08206 × 296) ⁄ (0.568 × 0.333) = 27.0 g⁄mol

P_T = total pressure; each partial pressure in the same unit as the total; mole fraction × P_T carves out one gas's slice

Ideal gas particles ignore each other, so pressures simply add; always check that the partial pressures sum back to the stated total.

Try it: 8.24 mol CH₄ of 8.78 mol total at 1.37 atm: P(CH₄) = (8.24 ⁄ 8.78) × 1.37 = 1.29 atm

P_total = barometer reading once the water levels are matched; P_H₂O = vapor pressure of water at the water's temperature, from the table

Dry the pressure first, then feed it to PV = nRT; putting the barometer reading straight in overcounts the gas.

Try it: Barometer 762 torr, water at 24 °C (vapor pressure 22.4 torr): P_gas = 762 − 22.4 = 739.6 torr

22.4 L⁄mol valid only at 0.00 °C and 1 atm; mole ratio from the balanced equation

A balanced equation speaks in moles but a gas is measured as a volume; this link is how airbag chemistry turns grams of solid into liters of gas.

Try it: 2 NaN₃ → 2 Na + 3 N₂: 130. g NaN₃ × (1 mol ⁄ 65.02 g) × (3 mol N₂ ⁄ 2 mol NaN₃) × (22.4 L ⁄ 1 mol) = 67.2 L N₂

a = attraction constant (L²·atm⁄mol²), b = excluded volume per mole (L⁄mol), both tabulated per gas; P in atm, V in L, n in mol, T in K, R = 0.08206 L·atm⁄(mol·K)

Real gases stray from PV = nRT at high pressure and low temperature; solved for P, the a term lowers the pressure and the V − n·b term raises it, so the real pressure can land on either side of nRT⁄V depending on which correction wins.

Try it: 1.00 mol CO₂ (a = 3.640, b = 0.04267) in 1.00 L at 300. K: P = (1.00 × 0.08206 × 300.) ⁄ (1.00 − 0.04267) − 3.640 × 1.00² ⁄ 1.00² = 25.7 − 3.64 = 22.1 atm

ΔH = heat of fusion (melting) or vaporization (boiling); J⁄g pairs with grams, kJ⁄mol pairs with moles; freezing and condensation release the same amounts (negative q)

Melting and boiling spend energy on rearranging molecules instead of warming them, which is why ice sits at 0 °C until the last crystal is gone.

Try it: Melting 25.0 g of ice at 0 °C: 25.0 g × (335 J ⁄ 1 g) = 8380 J

M₁, V₁ = concentrated (before); M₂, V₂ = dilute (after); any volume unit, as long as both sides match; V₂ is the final total volume

Dilution only, never across a chemical reaction: adding solvent spreads the same moles through more volume, and this equation is that sentence in algebra.

Try it: Volume of 12 M HCl for 500.0 mL of 0.10 M: V₁ = (0.10 × 500.0) ⁄ 12 = 4.2 mL

m = molality (mol⁄kg, read "molal"); kg of solvent only

Molality counts moles against mass, so warming the solution changes nothing; it is the concentration unit colligative properties will demand.

Try it: 18.0 g glucose (180.16 g⁄mol) in 250. g water: 0.0999 mol ⁄ 0.250 kg = 0.400 m

X = mole fraction (unitless); total moles = every component counted, solvent included

When one component's fraction is known, the other follows by subtraction from 1, no extra work.

Try it: 62.0 g glycol (0.999 mol) with 180. g water (9.99 mol): X(glycol) = 0.999 ⁄ 10.99 = 0.0909

i = van't Hoff factor, particles per formula unit (glucose 1; NaCl 2; CaCl₂ 3); m = molality (mol solute ⁄ kg solvent); K from the solvent

Only the number of dissolved particles matters, never their identity, which is why salt earns its keep on icy roads; converting the given concentration to molality is the first move.

Try it: 1.20 m NaCl (i = 2): ΔT_f = 2 × 1.86 × 1.20 = 4.46 °C, so the solution freezes at −4.46 °C

M of base and mL of base = the titrant (known); mL of acid = the sample; ratio = base units per acid unit from the balanced equation; each molarity pairs with its own solution's volume

At the equivalence point (the indicator flip) the moles of OH⁻ added equal the moles of H⁺ available, so the base volume is the measurement; write the balanced equation first, the ratio lives there and nowhere else.

Try it: 20.00 mL of H₂SO₄ takes 32.00 mL of 0.250 M NaOH: (0.250 × 32.00) ⁄ (2 × 20.00) = 0.200 M H₂SO₄

This is the counting unit that lets chemists count atoms by weighing, like a grocer counting fruit.

Try it: 2.00 mol × (6.022 × 10²³ particles ⁄ 1 mol) = 1.20 × 10²⁴ particles

This is the R for PV = nRT and every molar-mass-from-gas problem; the units it carries dictate the units everything else must arrive in.

This form appears in kinetic-molecular theory, where average kinetic energy tracks Kelvin temperature alone, regardless of the gas's identity.

You cannot state a gas volume without its temperature and pressure, so STP is the agreed reference condition, and the fine print behind 22.4 L⁄mol.

Treat 22.4 L⁄mol as a conversion factor between liters of gas and moles, valid only at STP; anywhere else, use PV = nRT.

Try it: 0.250 mol O₂ × (22.4 L ⁄ 1 mol) = 5.60 L at STP