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(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

8 g of Mg and 16 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
2.StoichiometryHard
m(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 6 g of CH4 reacts completely (the other reactants are in excess) and 12.35 g of CO2 is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 12.35 g
Y = %
3.Molarity & dilutionMedium
1.5 mol/LV1​ = ?0.1 mol/L500 mL

You need 500 mL of 0.1 mol/L solution and have a 1.5 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 1.5 mol/Lc2 = 0.1 mol/LV2 = 500 mL
V1 = mL
4.Molarity & dilutionHard
0.4 mol/L100 mL2.5 mol/L400 mLc = ?500 mL

100 mL of a 0.4 mol/L solution is mixed with 400 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.4 mol/LV1 = 100 mLc2 = 2.5 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
400 g of O₂P = 100 kPaT = 35 °CV = ?

What volume V does 400 g of O₂ occupy at 100 kPa and 35 °C? Treat it as an ideal gas.

m = 400 gM = 32 g/molP = 100 kPaT = 35 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 700 kPaT = 140 °Cρ = ?

Find the density ρ of O₂ at 700 kPa and 140 °C, treating it as an ideal gas.

M = 32 g/molP = 700 kPaT = 140 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Copperm = 200 gT = 110 °Cc = 0.385 J/(g·°C)Waterm = 150 gT = 24 °Cc = 4.18 J/(g·°C)

A 200 g block of copper (c = 0.385 J/(g·°C)) at 110 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.385 J/(g·°C)Tmetal = 110 °Cmwater = 150 gTwater = 24 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
HC6​H12​O6​ + 6O₂reactants6CO₂ + 6H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l).

ΔH°f(C6H12O6) = −1274 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​HII110C−x−x+2xE1 − x1 − x2x

1 mol/L of H₂ and 1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 50. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 1 mol/LKc = 50
[H2] = 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 0.05 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.05 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.5 mol/L acetic acid and 0.1 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.

[HA] = 0.5[A⁻] = 0.1Ka = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?60%

A first-order reaction in a batch reactor has k = 0.3 min⁻¹. How long does it take to reach 60% conversion? Find t.

k = 0.3 min⁻¹X = 60 %
t = min
14.Reaction kineticsHard
Tk30 °C0.0560 °C?

A reaction has a rate constant of 0.05 min⁻¹ at 30 °C and an activation energy of 55 kJ/mol. Find the rate constant k2 at 60 °C.

k1 = 0.05 min⁻¹T1 = 30 °CT2 = 60 °CEa = 55 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹

3Solutions

1.Stoichiometry

Answer: m = 13.26 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(Mg) = 8/24.31 = 0.3291 mol
n(O2) = 16/32 = 0.5 mol

Limiting reactant

0.3291/2 = 0.1645, 0.5/1 = 0.5 → Mg

Mass of product

n(MgO) = 0.1645·2 = 0.3291 mol
m = 0.3291·40.3 = 13.26 g

Answer

m = 13.26 g
2.Stoichiometry

Answer: Y = 75.02 %

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(CH4) = 6/16.04 = 0.3741 mol
m = 0.3741·1/1·44.01 = 16.46 g

Percent yield

Y = 12.35/16.46·100 = 75.02%

Answer

Y = 75.02%
3.Molarity & dilution

Answer: V1 = 33.33 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 = 0.1·500/1.5 = 33.33 mL

Answer

V1 = 33.33 mL
4.Molarity & dilution

Answer: c = 2.08 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 = 0.4·100 + 2.5·400 = 1040 mmol

Concentration of the mixture

c = 1040/500 = 2.08 mol/L

Answer

c = 2.08 mol/L
5.Ideal gas law

Answer: V = 320.2 L

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

Moles

n = 400/32 = 12.5 mol
T = 35 + 273.15 = 308.1 K

Volume

V = nRT/P = 12.5·8.314·308.1/100 = 320.2 L

Answer

V = 320.2 L
6.Ideal gas law

Answer: ρ = 6.521 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K

Density

ρ = 700·32/(8.314·413.1) = 6.521 kg/m³

Answer

ρ = 6.521 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 33.41 °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

200·0.385·(110 − Tf) = 150·4.18·(Tf − 24)

Solve for the final temperature

Tf = (77·110 + 627·24)/(77 + 627) = 33.41 °C

Answer

Tf = 33.41 °C
8.Calorimetry & enthalpy

Answer: ΔH° = −2802 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° = [6·(−393.5) + 6·(−285.8)] − [1·(−1274) + 6·0] = −2802 kJ

Answer

ΔH° = −2802 kJ

Negative: the reaction gives off heat (exothermic).

9.Chemical equilibrium

Answer: [H2] = 0.2205 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

50 = (2x)²/(1 − x)²

Take the square root

7.071 = 2x/(1 − x) → x = 1·7.071/(2 + 7.071) = 0.7795 mol/L

Equilibrium concentration

[H₂] = 1 − x = 0.2205 mol/L

Answer

[H2] = 0.2205 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 = 2.759

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 = 0.001744 mol/L

pH

pH = −log(0.001744) = 2.759

Answer

pH = 2.759
12.pH & buffers

Answer: pH = 4.046

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(0.1/0.5) = 4.046

Answer

pH = 4.046
13.Reaction kinetics

Answer: t = 3.054 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 = 0.4

Solve for the time

t = −ln(1 − X)/k = −ln(0.4)/0.3 = 3.054 min

Answer

t = 3.054 min
14.Reaction kinetics

Answer: k2 = 0.3568 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 = 303.1 K, T2 = 333.1 K

Arrhenius ratio

ln(k2/k1) = −(55000/8.314)·(1/333.1 − 1/303.1) = 1.965
k2 = 0.05·e1.965 = 0.3568 min⁻¹

Answer

k2 = 0.3568 min⁻¹

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