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​)

10 g of N2 and 64 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(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 6 g of Fe2O3 reacts completely (the other reactants are in excess) and 4.51 g of CO2 is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 4.51 g
Y = %
3.Molarity & dilutionMedium
2 mol/LV1​ = ?0.1 mol/L200 mL

You need 200 mL of 0.1 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 2 mol/Lc2 = 0.1 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
0.4 mol/L300 mL0.6 mol/L200 mLc = ?500 mL

300 mL of a 0.4 mol/L solution is mixed with 200 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.4 mol/LV1 = 300 mLc2 = 0.6 mol/LV2 = 200 mL
c = mol/L
5.Ideal gas lawMedium
140 g of HeV = 30 LT = 10 °CP = ?

A tank of 30 L contains 140 g of He at 10 °C. Treat it as an ideal gas and find the pressure P.

m = 140 gM = 4.003 g/molV = 30 LT = 10 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
He, M = 4.003 g/molP = 950 kPaT = 20 °Cρ = ?

Find the density ρ of He at 950 kPa and 20 °C, treating it as an ideal gas.

M = 4.003 g/molP = 950 kPaT = 20 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Copperm = 75 gT = 80 °Cc = 0.385 J/(g·°C)Waterm = 250 gT = 19 °Cc = 4.18 J/(g·°C)

A 75 g block of copper (c = 0.385 J/(g·°C)) at 80 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 19 °C in an insulated cup. Find the final temperature Tf.

mmetal = 75 gcmetal = 0.385 J/(g·°C)Tmetal = 80 °Cmwater = 250 gTwater = 19 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H2C2​H6​ + 7O₂reactants4CO₂ + 6H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(l).

ΔH°f(C2H6) = −84.7 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​HII0.10.10C−x−x+2xE0.1 − x0.1 − x2x

0.1 mol/L of H₂ and 0.1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 9. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 9
[HI] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.200C−x+x+xE0.2 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.2 mol/LKc = 0.04
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.01 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.01 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A 1.00 L buffer contains 0.25 mol acetic acid and 0.25 mol of its conjugate base (Ka = 1.8 × 10⁻⁵). 0.02 mol of NaOH is added. Find the new pH pH.

n(HA) = 0.25 moln(A⁻) = 0.25 moln(OH⁻) = 0.02 molKa = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?75%

A first-order reaction in a batch reactor has k = 0.2 min⁻¹. How long does it take to reach 75% conversion? Find t.

k = 0.2 min⁻¹X = 75 %
t = min
14.Reaction kineticsHard
Tk35 °C0.00545 °C?

A reaction has a rate constant of 0.005 min⁻¹ at 35 °C and an activation energy of 60 kJ/mol. Find the rate constant k2 at 45 °C.

k1 = 0.005 min⁻¹T1 = 35 °CT2 = 45 °CEa = 60 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹

3Solutions

1.Stoichiometry

Answer: m = 12.16 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) = 10/28.02 = 0.3569 mol
n(H2) = 64/2.016 = 31.75 mol

Limiting reactant

0.3569/1 = 0.3569, 31.75/3 = 10.58 → N2

Mass of product

n(NH3) = 0.3569·2 = 0.7138 mol
m = 0.7138·17.03 = 12.16 g

Answer

m = 12.16 g
2.Stoichiometry

Answer: Y = 90.91 %

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(Fe2O3) = 6/159.69 = 0.03757 mol
m = 0.03757·3/1·44.01 = 4.961 g

Percent yield

Y = 4.51/4.961·100 = 90.91%

Answer

Y = 90.91%
3.Molarity & dilution

Answer: V1 = 10 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 = 0.1·200/2 = 10 mL

Answer

V1 = 10 mL
4.Molarity & dilution

Answer: c = 0.48 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 = 0.4·300 + 0.6·200 = 240 mmol

Concentration of the mixture

c = 240/500 = 0.48 mol/L

Answer

c = 0.48 mol/L
5.Ideal gas law

Answer: P = 2744 kPa

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

Moles

n = m/M = 140/4.003 = 34.97 mol
T = 10 + 273.15 = 283.1 K

Pressure

P = nRT/V = 34.97·8.314·283.1/0.03 = 2744 kPa

Answer

P = 2744 kPa
6.Ideal gas law

Answer: ρ = 1.56 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 20 + 273.15 = 293.1 K

Density

ρ = 950·4.003/(8.314·293.1) = 1.56 kg/m³

Answer

ρ = 1.56 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 20.64 °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

75·0.385·(80 − Tf) = 250·4.18·(Tf − 19)

Solve for the final temperature

Tf = (28.88·80 + 1045·19)/(28.88 + 1045) = 20.64 °C

Answer

Tf = 20.64 °C
8.Calorimetry & enthalpy

Answer: ΔH° = −3119 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° = [4·(−393.5) + 6·(−285.8)] − [2·(−84.7) + 7·0] = −3119 kJ

Answer

ΔH° = −3119 kJ

Negative: the reaction gives off heat (exothermic).

9.Chemical equilibrium

Answer: [HI] = 0.12 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

9 = (2x)²/(0.1 − x)²

Take the square root

3 = 2x/(0.1 − x) → x = 0.1·3/(2 + 3) = 0.06 mol/L

Equilibrium concentration

[HI] = 2x = 0.12 mol/L

Answer

[HI] = 0.12 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.07165 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/(0.2 − x) → x² + 0.04·x − 0.008 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.008))/2 = 0.07165 mol/L

Equilibrium concentration

[Cl₂] = x = 0.07165 mol/L

Answer

[Cl2] = 0.07165 mol/L
11.pH & buffers

Answer: pH = 3.118

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/(0.01 − x) → x = 0.0007629 mol/L

pH

pH = −log(0.0007629) = 3.118

Answer

pH = 3.118
12.pH & buffers

Answer: pH = 4.814

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) = 0.25 − 0.02 = 0.23 mol
n(A⁻) = 0.25 + 0.02 = 0.27 mol

Henderson–Hasselbalch

pH = 4.745 + log(0.27/0.23) = 4.814

Answer

pH = 4.814
13.Reaction kinetics

Answer: t = 6.931 min

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).

Conversion and concentration

C/C0 = 1 − X = 0.25

Solve for the time

t = −ln(1 − X)/k = −ln(0.25)/0.2 = 6.931 min

Answer

t = 6.931 min
14.Reaction kinetics

Answer: k2 = 0.01044 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 = 318.1 K

Arrhenius ratio

ln(k2/k1) = −(60000/8.314)·(1/318.1 − 1/308.1) = 0.7361
k2 = 0.005·e0.7361 = 0.01044 min⁻¹

Answer

k2 = 0.01044 min⁻¹

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