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(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

8 g of C3H8 and 64 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
m(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 16 g of H2 reacts completely (the other reactants are in excess) and 134.4 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 134.4 g
Y = %
3.Molarity & dilutionMedium
2 mol/LV1​ = ?0.3 mol/L200 mL

You need 200 mL of 0.3 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 2 mol/Lc2 = 0.3 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
0.5 mol/L100 mL2.5 mol/L250 mLc = ?350 mL

100 mL of a 0.5 mol/L solution is mixed with 250 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.5 mol/LV1 = 100 mLc2 = 2.5 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
30 g of O₂V = 50 LT = 50 °CP = ?

A tank of 50 L contains 30 g of O₂ at 50 °C. Treat it as an ideal gas and find the pressure P.

m = 30 gM = 32 g/molV = 50 LT = 50 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
He, M = 4.003 g/molP = 700 kPaT = 190 °Cρ = ?

Find the density ρ of He at 700 kPa and 190 °C, treating it as an ideal gas.

M = 4.003 g/molP = 700 kPaT = 190 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Zincm = 100 gT = 90 °Cc = 0.388 J/(g·°C)Waterm = 250 gT = 21 °Cc = 4.18 J/(g·°C)

A 100 g block of zinc (c = 0.388 J/(g·°C)) at 90 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 21 °C in an insulated cup. Find the final temperature Tf.

mmetal = 100 gcmetal = 0.388 J/(g·°C)Tmetal = 90 °Cmwater = 250 gTwater = 21 °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​HII0.50.50C−x−x+2xE0.5 − x0.5 − x2x

0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 36. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 36
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.400C−x+x+xE0.4 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.4 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.05 mol/L solution of hydrofluoric acid (Ka = 0.00068).

c = 0.05 mol/LKa = 0.00068
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A 1.00 L buffer contains 0.3 mol acetic acid and 0.2 mol of its conjugate base (Ka = 1.8 × 10⁻⁵). 0.02 mol of NaOH is added. Find the new pH pH.

n(HA) = 0.3 moln(A⁻) = 0.2 moln(OH⁻) = 0.02 molKa = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
t (min)C (mol/L)5 minC = ?

A reaction with rate law −rA = k·CA² has k = 0.5 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 5 min.

k = 0.5 L/(mol·min)CA0 = 1 mol/Lt = 5 min
CA = mol/L
14.Reaction kineticsHard
1/Tln kT = 25 °Cln 0.05ln A = ?Ea​ = 100 kJ/mol

A first-order reaction has a rate constant of 0.05 s⁻¹ at 25 °C and an activation energy of 100 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.05 s⁻¹T = 25 °CEa = 100 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹

3Solutions

1.Stoichiometry

Answer: m = 23.95 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(C3H8) = 8/44.1 = 0.1814 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.1814/1 = 0.1814, 2/5 = 0.4 → C3H8

Mass of product

n(CO2) = 0.1814·3 = 0.5442 mol
m = 0.5442·44.01 = 23.95 g

Answer

m = 23.95 g
2.Stoichiometry

Answer: Y = 94 %

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(H2) = 16/2.016 = 7.937 mol
m = 7.937·2/2·18.02 = 143 g

Percent yield

Y = 134.4/143·100 = 94%

Answer

Y = 94%
3.Molarity & dilution

Answer: V1 = 30 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 = 0.3·200/2 = 30 mL

Answer

V1 = 30 mL
4.Molarity & dilution

Answer: c = 1.929 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 = 0.5·100 + 2.5·250 = 675 mmol

Concentration of the mixture

c = 675/350 = 1.929 mol/L

Answer

c = 1.929 mol/L
5.Ideal gas law

Answer: P = 50.38 kPa

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

Moles

n = m/M = 30/32 = 0.9375 mol
T = 50 + 273.15 = 323.1 K

Pressure

P = nRT/V = 0.9375·8.314·323.1/0.05 = 50.38 kPa

Answer

P = 50.38 kPa
6.Ideal gas law

Answer: ρ = 0.7277 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K

Density

ρ = 700·4.003/(8.314·463.1) = 0.7277 kg/m³

Answer

ρ = 0.7277 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 23.47 °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

100·0.388·(90 − Tf) = 250·4.18·(Tf − 21)

Solve for the final temperature

Tf = (38.8·90 + 1045·21)/(38.8 + 1045) = 23.47 °C

Answer

Tf = 23.47 °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: [H2] = 0.125 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

36 = (2x)²/(0.5 − x)²

Take the square root

6 = 2x/(0.5 − x) → x = 0.5·6/(2 + 6) = 0.375 mol/L

Equilibrium concentration

[H₂] = 0.5 − x = 0.125 mol/L

Answer

[H2] = 0.125 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.1433 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

0.08 = x²/(0.4 − x) → x² + 0.08·x − 0.032 = 0

Solve the quadratic

x = (−0.08 + √(0.08² + 4·0.032))/2 = 0.1433 mol/L

Equilibrium concentration

[Cl₂] = x = 0.1433 mol/L

Answer

[Cl2] = 0.1433 mol/L
11.pH & buffers

Answer: pH = 2.26

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/(0.05 − x) → x = 0.005501 mol/L

pH

pH = −log(0.005501) = 2.26

Answer

pH = 2.26
12.pH & buffers

Answer: pH = 4.64

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]).

Neutralisation

n(HA) = 0.3 − 0.02 = 0.28 mol
n(A⁻) = 0.2 + 0.02 = 0.22 mol

Henderson–Hasselbalch

pH = 4.745 + log(0.22/0.28) = 4.64

Answer

pH = 4.64
13.Reaction kinetics

Answer: CA = 0.2857 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 + 0.5·5 = 3.5 L/mol
CA = 1/3.5 = 0.2857 mol/L

Answer

CA = 0.2857 mol/L
14.Reaction kinetics

Answer: A = 1.656 × 10¹⁶ s⁻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 temperature

T = 25 + 273.15 = 298.1 K

Solve Arrhenius for A

Ea/(RT) = 100000/(8.314·298.1) = 40.34
A = k·e40.34 = 0.05·3.313 × 10¹⁷ = 1.656 × 10¹⁶ s⁻¹

Answer

A = 1.656 × 10¹⁶ s⁻¹

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