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​)

6 g of C3H8 and 80 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(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 10 g of CaCO3 reacts completely (the other reactants are in excess) and 3.87 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 3.87 g
Y = %
3.Molarity & dilutionMedium
4 mol/LV1​0.25 mol/L100 mL

You need 100 mL of 0.25 mol/L solution and have a 4 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 4 mol/Lc2 = 0.25 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
0.25 mol/L200 mL0.05 mol/L50 mLc = ?250 mL

200 mL of a 0.25 mol/L solution is mixed with 50 mL of a 0.05 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.25 mol/LV1 = 200 mLc2 = 0.05 mol/LV2 = 50 mL
c = mol/L
5.Ideal gas lawMedium
320 g of CO₂P = 650 kPaT = 75 °CV = ?

What volume V does 320 g of CO₂ occupy at 650 kPa and 75 °C? Treat it as an ideal gas.

m = 320 gM = 44.01 g/molP = 650 kPaT = 75 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 250 kPaT = 100 °Cρ = ?

Find the density ρ of CO₂ at 250 kPa and 100 °C, treating it as an ideal gas.

M = 44.01 g/molP = 250 kPaT = 100 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Zincm = 150 gT = 90 °Cc = 0.388 J/(g·°C)Waterm = 250 gT = 21 °Cc = 4.18 J/(g·°C)

A 150 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 = 150 gcmetal = 0.388 J/(g·°C)Tmetal = 90 °Cmwater = 250 gTwater = 21 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H50 g C3​H8​C3​H8​ + 5O₂reactants3CO₂ + 4H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 50 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​HII0.50.50C−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 = 25. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 25
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I1.500C−x+x+xE1.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 1.5 mol/LKc = 0.5
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.01 mol/L solution of formic acid (Ka = 0.00018).

c = 0.01 mol/LKa = 0.00018
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.5 mol/L hypochlorous acid and 0.2 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.

[HA] = 0.5[A⁻] = 0.2Ka = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?20%

A zero-order reaction has k = 0.05 mol/(L·min) and starts at 2 mol/L. How long does it take to reach 20% conversion? Find t.

k = 0.05 mol/(L·min)CA0 = 2 mol/LX = 20 %
t = min
14.Reaction kineticsHard
Tk40 °C0.160 °C0.564

The rate constant of a reaction is 0.1 min⁻¹ at 40 °C and 0.564 min⁻¹ at 60 °C. Find the activation energy Ea.

k1 = 0.1 min⁻¹T1 = 40 °Ck2 = 0.564 min⁻¹T2 = 60 °CR = 8.314 J/(mol·K)
Ea = kJ/mol

3Solutions

1.Stoichiometry

Answer: m = 17.96 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) = 6/44.1 = 0.1361 mol
n(O2) = 80/32 = 2.5 mol

Limiting reactant

0.1361/1 = 0.1361, 2.5/5 = 0.5 → C3H8

Mass of product

n(CO2) = 0.1361·3 = 0.4082 mol
m = 0.4082·44.01 = 17.96 g

Answer

m = 17.96 g
2.Stoichiometry

Answer: Y = 88.01 %

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) = 10/100.09 = 0.09991 mol
m = 0.09991·1/1·44.01 = 4.397 g

Percent yield

Y = 3.87/4.397·100 = 88.01%

Answer

Y = 88.01%
3.Molarity & dilution

Answer: Vwater = 93.75 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 = 0.25·100/4 = 6.25 mL

Water to add

Vwater = 100 − 6.25 = 93.75 mL

Answer

Vwater = 93.75 mL
4.Molarity & dilution

Answer: c = 0.21 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 = 0.25·200 + 0.05·50 = 52.5 mmol

Concentration of the mixture

c = 52.5/250 = 0.21 mol/L

Answer

c = 0.21 mol/L
5.Ideal gas law

Answer: V = 32.38 L

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

Moles

n = 320/44.01 = 7.271 mol
T = 75 + 273.15 = 348.1 K

Volume

V = nRT/P = 7.271·8.314·348.1/650 = 32.38 L

Answer

V = 32.38 L
6.Ideal gas law

Answer: ρ = 3.546 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 100 + 273.15 = 373.1 K

Density

ρ = 250·44.01/(8.314·373.1) = 3.546 kg/m³

Answer

ρ = 3.546 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 24.64 °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

150·0.388·(90 − Tf) = 250·4.18·(Tf − 21)

Solve for the final temperature

Tf = (58.2·90 + 1045·21)/(58.2 + 1045) = 24.64 °C

Answer

Tf = 24.64 °C
8.Calorimetry & enthalpy

Answer: q = 2517 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 = 50/44.1 = 1.134 mol
q = n·|ΔH°/1| = 1.134·2220 = 2517 kJ

Answer

q = 2517 kJ
9.Chemical equilibrium

Answer: [H2] = 0.1429 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

25 = (2x)²/(0.5 − x)²

Take the square root

5 = 2x/(0.5 − x) → x = 0.5·5/(2 + 5) = 0.3571 mol/L

Equilibrium concentration

[H₂] = 0.5 − x = 0.1429 mol/L

Answer

[H2] = 0.1429 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.6514 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

0.5 = x²/(1.5 − x) → x² + 0.5·x − 0.75 = 0

Solve the quadratic

x = (−0.5 + √(0.5² + 4·0.75))/2 = 0.6514 mol/L

Equilibrium concentration

[Cl₂] = x = 0.6514 mol/L

Answer

[Cl2] = 0.6514 mol/L
11.pH & buffers

Answer: pH = 2.901

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 = 0.001255 mol/L

pH

pH = −log(0.001255) = 2.901

Answer

pH = 2.901
12.pH & buffers

Answer: pH = 7.125

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 = 7.523

Henderson–Hasselbalch

pH = 7.523 + log(0.2/0.5) = 7.125

Answer

pH = 7.125
13.Reaction kinetics

Answer: t = 8 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).

Amount converted

CA0·X = 2·0.2 = 0.4 mol/L

Time

t = CA0·X/k = 0.4/0.05 = 8 min

Answer

t = 8 min
14.Reaction kinetics

Answer: Ea = 75.02 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 = 313.1 K, T2 = 333.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.73/(-1.917e-4) = 75022 J/mol
Ea = 75.02 kJ/mol

Answer

Ea = 75.02 kJ/mol

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