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(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

16 g of N2 and 32 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
2.StoichiometryHard
m(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

In the reaction N2 + 3H2 → 2NH3, 8 g of N2 reacts completely (the other reactants are in excess) and 7.29 g of NH3 is collected. Find the percent yield Y.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 7.29 g
Y = %
3.Molarity & dilutionMedium
2.5 mol/LV1​0.2 mol/L100 mL

You need 100 mL of 0.2 mol/L solution and have a 2.5 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 2.5 mol/Lc2 = 0.2 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
1 mol/L50 mL1.2 mol/L250 mLc = ?300 mL

50 mL of a 1 mol/L solution is mixed with 250 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1 mol/LV1 = 50 mLc2 = 1.2 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
470 g of CO₂P = 500 kPaT = 70 °CV = ?

What volume V does 470 g of CO₂ occupy at 500 kPa and 70 °C? Treat it as an ideal gas.

m = 470 gM = 44.01 g/molP = 500 kPaT = 70 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 350 kPaT = 200 °Cρ = ?

Find the density ρ of CH₄ at 350 kPa and 200 °C, treating it as an ideal gas.

M = 16.04 g/molP = 350 kPaT = 200 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Zincm = 50 gT = 140 °Cc = 0.388 J/(g·°C)Waterm = 200 gT = 22 °Cc = 4.18 J/(g·°C)

A 50 g block of zinc (c = 0.388 J/(g·°C)) at 140 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 22 °C in an insulated cup. Find the final temperature Tf.

mmetal = 50 gcmetal = 0.388 J/(g·°C)Tmetal = 140 °Cmwater = 200 gTwater = 22 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H10 g C3​H8​C3​H8​ + 5O₂reactants3CO₂ + 4H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 10 g of C3H8 (M = 44.1 g/mol) burns completely: C3H8 + 5O₂ → 3CO₂ + 4H₂O(l). Give q as a positive number.

ΔH°f(C3H8) = −103.8 kJ/molΔH°f(CO₂) = −393.5 kJ/molΔH°f(H₂O, l) = −285.8 kJ/molΔH°f(O₂) = 0
q = kJ
9.Chemical equilibriumMedium
H2​I2​HII0.20.20C−x−x+2xE0.2 − x0.2 − x2x

0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 4
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I100C−x+x+xE1 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 1 mol/LKc = 0.04
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.25 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.25 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.2 mol/L acetic acid and 0.5 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.

[HA] = 0.2[A⁻] = 0.5Ka = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
t (min)C (mol/L)30 minC = ?

A reaction with rate law −rA = k·CA² has k = 0.4 L/(mol·min) and starts at 1.5 mol/L. Find the concentration CA after 30 min.

k = 0.4 L/(mol·min)CA0 = 1.5 mol/Lt = 30 min
CA = mol/L
14.Reaction kineticsHard
1/Tln kT = 50 °Cln 0.1ln A = ?Ea​ = 90 kJ/mol

A first-order reaction has a rate constant of 0.1 s⁻¹ at 50 °C and an activation energy of 90 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.1 s⁻¹T = 50 °CEa = 90 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹

3Solutions

1.Stoichiometry

Answer: m = 19.45 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(N2) = 16/28.02 = 0.571 mol
n(H2) = 32/2.016 = 15.87 mol

Limiting reactant

0.571/1 = 0.571, 15.87/3 = 5.291 → N2

Mass of product

n(NH3) = 0.571·2 = 1.142 mol
m = 1.142·17.03 = 19.45 g

Answer

m = 19.45 g
2.Stoichiometry

Answer: Y = 74.97 %

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(N2) = 8/28.02 = 0.2855 mol
m = 0.2855·2/1·17.03 = 9.724 g

Percent yield

Y = 7.29/9.724·100 = 74.97%

Answer

Y = 74.97%
3.Molarity & dilution

Answer: Vwater = 92 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 = 0.2·100/2.5 = 8 mL

Water to add

Vwater = 100 − 8 = 92 mL

Answer

Vwater = 92 mL
4.Molarity & dilution

Answer: c = 1.167 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 = 1·50 + 1.2·250 = 350 mmol

Concentration of the mixture

c = 350/300 = 1.167 mol/L

Answer

c = 1.167 mol/L
5.Ideal gas law

Answer: V = 60.94 L

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

Moles

n = 470/44.01 = 10.68 mol
T = 70 + 273.15 = 343.1 K

Volume

V = nRT/P = 10.68·8.314·343.1/500 = 60.94 L

Answer

V = 60.94 L
6.Ideal gas law

Answer: ρ = 1.427 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K

Density

ρ = 350·16.04/(8.314·473.1) = 1.427 kg/m³

Answer

ρ = 1.427 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 24.68 °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

50·0.388·(140 − Tf) = 200·4.18·(Tf − 22)

Solve for the final temperature

Tf = (19.4·140 + 836·22)/(19.4 + 836) = 24.68 °C

Answer

Tf = 24.68 °C
8.Calorimetry & enthalpy

Answer: q = 503.4 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° = [3·(−393.5) + 4·(−285.8)] − [1·(−103.8) + 5·0] = −2220 kJ

Heat for the given mass

n = 10/44.1 = 0.2268 mol
q = n·|ΔH°/1| = 0.2268·2220 = 503.4 kJ

Answer

q = 503.4 kJ
9.Chemical equilibrium

Answer: [H2] = 0.1 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

4 = (2x)²/(0.2 − x)²

Take the square root

2 = 2x/(0.2 − x) → x = 0.2·2/(2 + 2) = 0.1 mol/L

Equilibrium concentration

[H₂] = 0.2 − x = 0.1 mol/L

Answer

[H2] = 0.1 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.181 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

0.04 = x²/(1 − x) → x² + 0.04·x − 0.04 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.04))/2 = 0.181 mol/L

Equilibrium concentration

[Cl₂] = x = 0.181 mol/L

Answer

[Cl2] = 0.181 mol/L
11.pH & buffers

Answer: pH = 2.405

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/(0.25 − x) → x = 0.003937 mol/L

pH

pH = −log(0.003937) = 2.405

Answer

pH = 2.405
12.pH & buffers

Answer: pH = 5.143

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 = 4.745

Henderson–Hasselbalch

pH = 4.745 + log(0.5/0.2) = 5.143

Answer

pH = 5.143
13.Reaction kinetics

Answer: CA = 0.07895 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 = 0.6667 + 0.4·30 = 12.67 L/mol
CA = 1/12.67 = 0.07895 mol/L

Answer

CA = 0.07895 mol/L
14.Reaction kinetics

Answer: A = 3.534 × 10¹³ 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 = 50 + 273.15 = 323.1 K

Solve Arrhenius for A

Ea/(RT) = 90000/(8.314·323.1) = 33.5
A = k·e33.5 = 0.1·3.534 × 10¹⁴ = 3.534 × 10¹³ s⁻¹

Answer

A = 3.534 × 10¹³ s⁻¹

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