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(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

4 g of Al and 20 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?

M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
2.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 30 g of CaCO3 reacts completely (the other reactants are in excess) and 9.23 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 9.23 g
Y = %
3.Molarity & dilutionMedium
5 mol/LV1​0.1 mol/L100 mL

You need 100 mL of 0.1 mol/L solution and have a 5 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 5 mol/Lc2 = 0.1 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
0.8 mol/L300 mL0.3 mol/L400 mLc = ?700 mL

300 mL of a 0.8 mol/L solution is mixed with 400 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.8 mol/LV1 = 300 mLc2 = 0.3 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
60 g of CH₄P = 350 kPaT = 45 °CV = ?

What volume V does 60 g of CH₄ occupy at 350 kPa and 45 °C? Treat it as an ideal gas.

m = 60 gM = 16.04 g/molP = 350 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
O₂, M = 32 g/molP = 300 kPaT = 70 °Cρ = ?

Find the density ρ of O₂ at 300 kPa and 70 °C, treating it as an ideal gas.

M = 32 g/molP = 300 kPaT = 70 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Ironm = 75 gT = 130 °Cc = 0.449 J/(g·°C)Waterm = 200 gT = 24 °Cc = 4.18 J/(g·°C)

A 75 g block of iron (c = 0.449 J/(g·°C)) at 130 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.

mmetal = 75 gcmetal = 0.449 J/(g·°C)Tmetal = 130 °Cmwater = 200 gTwater = 24 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H10 g C6​H12​O6​C6​H12​O6​ + 6O₂reactants6CO₂ + 6H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 10 g of C6H12O6 (M = 180.16 g/mol) burns completely: C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l). Give q as a positive number.

ΔH°f(C6H12O6) = −1274 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​HII0.250.250C−x−x+2xE0.25 − x0.25 − x2x

0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 9. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 9
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.800C−x+x+xE0.8 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.8 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.25 mol/L solution of lactic acid (Ka = 0.00014).

c = 0.25 mol/LKa = 0.00014
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.5 mol/L lactic acid and 0.5 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.

[HA] = 0.5[A⁻] = 0.5Ka = 0.00014
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?80%

A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.8 L/(mol·min) and starts at 1.5 mol/L. How long does it take to reach 80% conversion? Find t.

k = 0.8 L/(mol·min)CA0 = 1.5 mol/LX = 80 %
t = min
14.Reaction kineticsHard
Tk35 °C0.155 °C0.851

The rate constant of a reaction is 0.1 min⁻¹ at 35 °C and 0.851 min⁻¹ at 55 °C. Find the activation energy Ea.

k1 = 0.1 min⁻¹T1 = 35 °Ck2 = 0.851 min⁻¹T2 = 55 °CR = 8.314 J/(mol·K)
Ea = kJ/mol

3Solutions

1.Stoichiometry

Answer: m = 19.77 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(Al) = 4/26.98 = 0.1483 mol
n(Cl2) = 20/70.9 = 0.2821 mol

Limiting reactant

0.1483/2 = 0.07413, 0.2821/3 = 0.09403 → Al

Mass of product

n(AlCl3) = 0.07413·2 = 0.1483 mol
m = 0.1483·133.33 = 19.77 g

Answer

m = 19.77 g
2.Stoichiometry

Answer: Y = 69.97 %

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(CaCO3) = 30/100.09 = 0.2997 mol
m = 0.2997·1/1·44.01 = 13.19 g

Percent yield

Y = 9.23/13.19·100 = 69.97%

Answer

Y = 69.97%
3.Molarity & dilution

Answer: Vwater = 98 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 = 0.1·100/5 = 2 mL

Water to add

Vwater = 100 − 2 = 98 mL

Answer

Vwater = 98 mL
4.Molarity & dilution

Answer: c = 0.5143 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 = 0.8·300 + 0.3·400 = 360 mmol

Concentration of the mixture

c = 360/700 = 0.5143 mol/L

Answer

c = 0.5143 mol/L
5.Ideal gas law

Answer: V = 28.27 L

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

Moles

n = 60/16.04 = 3.741 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 3.741·8.314·318.1/350 = 28.27 L

Answer

V = 28.27 L
6.Ideal gas law

Answer: ρ = 3.365 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 70 + 273.15 = 343.1 K

Density

ρ = 300·32/(8.314·343.1) = 3.365 kg/m³

Answer

ρ = 3.365 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 28.1 °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

75·0.449·(130 − Tf) = 200·4.18·(Tf − 24)

Solve for the final temperature

Tf = (33.68·130 + 836·24)/(33.68 + 836) = 28.1 °C

Answer

Tf = 28.1 °C
8.Calorimetry & enthalpy

Answer: q = 155.5 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

Heat for the given mass

n = 10/180.16 = 0.05551 mol
q = n·|ΔH°/1| = 0.05551·2802 = 155.5 kJ

Answer

q = 155.5 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

9 = (2x)²/(0.25 − x)²

Take the square root

3 = 2x/(0.25 − x) → x = 0.25·3/(2 + 3) = 0.15 mol/L

Equilibrium concentration

[H₂] = 0.25 − x = 0.1 mol/L

Answer

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

Answer: [Cl2] = 0.2161 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/(0.8 − x) → x² + 0.08·x − 0.064 = 0

Solve the quadratic

x = (−0.08 + √(0.08² + 4·0.064))/2 = 0.2161 mol/L

Equilibrium concentration

[Cl₂] = x = 0.2161 mol/L

Answer

[Cl2] = 0.2161 mol/L
11.pH & buffers

Answer: pH = 2.233

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 = 0.005846 mol/L

pH

pH = −log(0.005846) = 2.233

Answer

pH = 2.233
12.pH & buffers

Answer: pH = 3.854

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 = 3.854

Henderson–Hasselbalch

pH = 3.854 + log(0.5/0.5) = 3.854

Answer

pH = 3.854
13.Reaction kinetics

Answer: t = 3.333 min

Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).

Concentration at the end

CA = 1.5·(1 − 0.8) = 0.3 mol/L

Integrated second-order law

t = (1/CA − 1/CA0)/k = (3.333 − 0.6667)/0.8 = 3.333 min

Answer

t = 3.333 min
14.Reaction kinetics

Answer: Ea = 90.01 kJ/mol

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 = 328.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.141/(-1.978e-4) = 90008 J/mol
Ea = 90.01 kJ/mol

Answer

Ea = 90.01 kJ/mol

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