General Chemistry: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Stoichiometry

Moles
n = m/M
Concentration
c = n/V
Mole ratio
use the coefficients of the balanced equation
Limiting reactant
the one that runs out first
Percent yield
actual / theoretical × 100%

Method

  1. Write and balance the chemical equation.
  2. Convert the given mass to moles: n = m/M.
  3. Use the coefficients as the mole ratio to find the moles of the substance you want.
  4. Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.

Molarity & dilution

Molarity
c = n/V (mol/L)
Moles
n = m/M
Dilution
c1V1 = c2V2
Mixing
c = (c1V1 + c2V2)/(V1 + V2)

Method

  1. Convert mass to moles with n = m/M.
  2. Convert the volume to litres.
  3. Molarity c = n/V in mol/L.
  4. For a dilution, the moles stay the same: c₁V₁ = c₂V₂.

Ideal gas law

Ideal gas law
pV = nRT
Gas constant
R = 8.314 J/(mol·K)
Combined gas law
p1V1/T1 = p2V2/T2
Moles
n = m/M
Temperature
T[K] = T[°C] + 273.15

Calorimetry & enthalpy

Heat
q = m·c·ΔT
Water
c = 4.18 J/(g·°C)
Calorimeter
heat lost = heat gained
Enthalpy of reaction
ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants)

Chemical equilibrium

Equilibrium constant
Kc = [C]c[D]d/([A]a[B]b)
ICE table
Initial, Change (×coefficient), Equilibrium
Quadratic
x = (−b + √(b² − 4ac))/(2a)

Method

  1. Write Kc with products over reactants, each raised to the power of its coefficient.
  2. Set up an ICE table: Initial, Change (in steps of x times the coefficients), Equilibrium.
  3. Put the equilibrium expressions into Kc and solve for x.
  4. Check whether x is small enough for an approximation; otherwise solve the quadratic.

pH & buffers

pH
pH = −log[H⁺]
pOH
pOH = −log[OH⁻], pH + pOH = 14
Weak acid
[H⁺] ≈ √(Ka·c)
Buffer
pH = pKa + log([A⁻]/[HA])
Dilution
c1V1 = c2V2

Method

  1. Decide whether the acid or base is strong (fully dissociated) or weak.
  2. Strong acid: [H⁺] equals its concentration. Strong base: find [OH⁻] first.
  3. Weak acid: solve Ka = x²/(c − x) for x = [H⁺].
  4. pH = −log[H⁺], and pH + pOH = 14 at 25 °C.

Reaction kinetics

First order
ln(CA/CA0) = −kt
Half-life (first order)
t½ = ln 2/k
Second order
1/CA = 1/CA0 + kt
Arrhenius
k = A·e−Ea/(RT)
Two temperatures
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

2Practice problems

1.StoichiometryMedium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

4 g of H2 and 24 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
m(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 8 g of Na reacts completely (the other reactants are in excess) and 15.25 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 15.25 g
Y = %
3.Molarity & dilutionMedium
1.5 mol/LV1​0.3 mol/L250 mL

You need 250 mL of 0.3 mol/L solution and have a 1.5 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 1.5 mol/Lc2 = 0.3 mol/LV2 = 250 mL
Vwater = mL
4.Molarity & dilutionHard
1.5 mol/L150 mL0.75 mol/L50 mLc = ?200 mL

150 mL of a 1.5 mol/L solution is mixed with 50 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1.5 mol/LV1 = 150 mLc2 = 0.75 mol/LV2 = 50 mL
c = mol/L
5.Ideal gas lawMedium
450 g of O₂V = 50 LT = 75 °CP = ?

A tank of 50 L contains 450 g of O₂ at 75 °C. Treat it as an ideal gas and find the pressure P.

m = 450 gM = 32 g/molV = 50 LT = 75 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 400 kPaT = 160 °Cρ = ?

Find the density ρ of Ar at 400 kPa and 160 °C, treating it as an ideal gas.

M = 39.95 g/molP = 400 kPaT = 160 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Zincm = 200 gT = 190 °Cc = 0.388 J/(g·°C)Waterm = 250 gT = 23 °Cc = 4.18 J/(g·°C)

A 200 g block of zinc (c = 0.388 J/(g·°C)) at 190 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 23 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.388 J/(g·°C)Tmetal = 190 °Cmwater = 250 gTwater = 23 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H2CH3​OH + 3O₂reactants2CO₂ + 4H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction 2CH3OH + 3O₂ → 2CO₂ + 4H₂O(l).

ΔH°f(CH3OH) = −238.7 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
ΔH° = kJ
9.Chemical equilibriumMedium
H2​I2​HII0.40.40C−x−x+2xE0.4 − x0.4 − x2x

0.4 mol/L of H₂ and 0.4 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 25. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 25
[HI] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.100C−x+x+xE0.1 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.1 mol/LKc = 0.1
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.2 mol/L solution of formic acid (Ka = 0.00018).

c = 0.2 mol/LKa = 0.00018
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.5 mol/L hypochlorous acid and 0.15 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.

[HA] = 0.5[A⁻] = 0.15Ka = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
t (min)C (mol/L)10 minC = ?

A reaction with rate law −rA = k·CA² has k = 0.8 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 10 min.

k = 0.8 L/(mol·min)CA0 = 1 mol/Lt = 10 min
CA = mol/L
14.Reaction kineticsHard
Tk35 °C0.0565 °C?

A reaction has a rate constant of 0.05 min⁻¹ at 35 °C and an activation energy of 60 kJ/mol. Find the rate constant k2 at 65 °C.

k1 = 0.05 min⁻¹T1 = 35 °CT2 = 65 °CEa = 60 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹

3Solutions

1.Stoichiometry

Answer: m = 27.03 g

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Moles of each reactant

n(H2) = 4/2.016 = 1.984 mol
n(O2) = 24/32 = 0.75 mol

Limiting reactant

1.984/2 = 0.9921, 0.75/1 = 0.75 → O2

Mass of product

n(H2O) = 0.75·2 = 1.5 mol
m = 1.5·18.02 = 27.03 g

Answer

m = 27.03 g
2.Stoichiometry

Answer: Y = 74.99 %

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Theoretical yield

n(Na) = 8/22.99 = 0.348 mol
m = 0.348·2/2·58.44 = 20.34 g

Percent yield

Y = 15.25/20.34·100 = 74.99%

Answer

Y = 74.99%
3.Molarity & dilution

Answer: Vwater = 200 mL

Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.

Volume of stock solution

V1 = c2·V2/c1 = 0.3·250/1.5 = 50 mL

Water to add

Vwater = 250 − 50 = 200 mL

Answer

Vwater = 200 mL
4.Molarity & dilution

Answer: c = 1.313 mol/L

Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.

Total amount

n = 1.5·150 + 0.75·50 = 262.5 mmol

Concentration of the mixture

c = 262.5/200 = 1.313 mol/L

Answer

c = 1.313 mol/L
5.Ideal gas law

Answer: P = 814.1 kPa

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = m/M = 450/32 = 14.06 mol
T = 75 + 273.15 = 348.1 K

Pressure

P = nRT/V = 14.06·8.314·348.1/0.05 = 814.1 kPa

Answer

P = 814.1 kPa
6.Ideal gas law

Answer: ρ = 4.437 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 160 + 273.15 = 433.1 K

Density

ρ = 400·39.95/(8.314·433.1) = 4.437 kg/m³

Answer

ρ = 4.437 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 34.54 °C

Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.

Heat balance

200·0.388·(190 − Tf) = 250·4.18·(Tf − 23)

Solve for the final temperature

Tf = (77.6·190 + 1045·23)/(77.6 + 1045) = 34.54 °C

Answer

Tf = 34.54 °C
8.Calorimetry & enthalpy

Answer: ΔH° = −1453 kJ

Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.

Enthalpy of reaction

ΔH° = [2·(−393.5) + 4·(−285.8)] − [2·(−238.7) + 3·0] = −1453 kJ

Answer

ΔH° = −1453 kJ

Negative: the reaction gives off heat (exothermic).

9.Chemical equilibrium

Answer: [HI] = 0.5714 mol/L

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

25 = (2x)²/(0.4 − x)²

Take the square root

5 = 2x/(0.4 − x) → x = 0.4·5/(2 + 5) = 0.2857 mol/L

Equilibrium concentration

[HI] = 2x = 0.5714 mol/L

Answer

[HI] = 0.5714 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.0618 mol/L

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

0.1 = x²/(0.1 − x) → x² + 0.1·x − 0.01 = 0

Solve the quadratic

x = (−0.1 + √(0.1² + 4·0.01))/2 = 0.0618 mol/L

Equilibrium concentration

[Cl₂] = x = 0.0618 mol/L

Answer

[Cl2] = 0.0618 mol/L
11.pH & buffers

Answer: pH = 2.228

pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).

Equilibrium

HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.

Ka = x²/(0.2 − x) → x = 0.005911 mol/L

pH

pH = −log(0.005911) = 2.228

Answer

pH = 2.228
12.pH & buffers

Answer: pH = 7

pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).

pKa

pKa = 7.523

Henderson–Hasselbalch

pH = 7.523 + log(0.15/0.5) = 7

Answer

pH = 7
13.Reaction kinetics

Answer: CA = 0.1111 mol/L

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Integrated second-order law

1/CA = 1/CA0 + kt = 1 + 0.8·10 = 9 L/mol
CA = 1/9 = 0.1111 mol/L

Answer

CA = 0.1111 mol/L
14.Reaction kinetics

Answer: k2 = 0.3993 min⁻¹

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Absolute temperatures

T1 = 308.1 K, T2 = 338.1 K

Arrhenius ratio

ln(k2/k1) = −(60000/8.314)·(1/338.1 − 1/308.1) = 2.078
k2 = 0.05·e2.078 = 0.3993 min⁻¹

Answer

k2 = 0.3993 min⁻¹

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