General Chemistry: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

30 g of C3H8 and 16 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
m(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 12 g of Al reacts completely (the other reactants are in excess) and 52.19 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 52.19 g
Y = %
3.Molarity & dilutionMedium
2 mol/LV1​0.25 mol/L500 mL

You need 500 mL of 0.25 mol/L solution and have a 2 mol/L stock solution. How much water Vwater do you add to the stock solution?

c1 = 2 mol/Lc2 = 0.25 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
1 mol/L200 mL0.3 mol/L500 mLc = ?700 mL

200 mL of a 1 mol/L solution is mixed with 500 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1 mol/LV1 = 200 mLc2 = 0.3 mol/LV2 = 500 mL
c = mol/L
5.Ideal gas lawMedium
60 g of CO₂P = 700 kPaT = 85 °CV = ?

What volume V does 60 g of CO₂ occupy at 700 kPa and 85 °C? Treat it as an ideal gas.

m = 60 gM = 44.01 g/molP = 700 kPaT = 85 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 300 kPa, 2 L, 35 °C to 750 kPa and 90 °C. Find the new volume V2.

P1 = 300 kPaV1 = 2 LT1 = 35 °CP2 = 750 kPaT2 = 90 °C
V2 = L
7.Calorimetry & enthalpyMedium
Tf​ = ?Ironm = 50 gT = 160 °Cc = 0.449 J/(g·°C)Waterm = 150 gT = 16 °Cc = 4.18 J/(g·°C)

A 50 g block of iron (c = 0.449 J/(g·°C)) at 160 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.

mmetal = 50 gcmetal = 0.449 J/(g·°C)Tmetal = 160 °Cmwater = 150 gTwater = 16 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
HC2​H5​OH + 3O₂reactants2CO₂ + 3H₂O(l)productsΔH° = ?

Use the standard enthalpies of formation to find ΔH° for the reaction C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l).

ΔH°f(C2H5OH) = −277.7 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​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 = 16. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 1 mol/LKc = 16
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I1.500C−x+x+xE1.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 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 = 0.1
[Cl2] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.5 mol/L solution of formic acid (Ka = 0.00018).

c = 0.5 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)C (mol/L)10 minC = ?

A reaction with rate law −rA = k·CA² has k = 0.2 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 10 min.

k = 0.2 L/(mol·min)CA0 = 1 mol/Lt = 10 min
CA = mol/L
14.Reaction kineticsHard
1/Tln kT = 30 °Cln 0.05ln A = ?Ea​ = 75 kJ/mol

A first-order reaction has a rate constant of 0.05 s⁻¹ at 30 °C and an activation energy of 75 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.05 s⁻¹T = 30 °CEa = 75 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹

3Solutions

1.Stoichiometry

Answer: m = 7.208 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) = 30/44.1 = 0.6803 mol
n(O2) = 16/32 = 0.5 mol

Limiting reactant

0.6803/1 = 0.6803, 0.5/5 = 0.1 → O2

Mass of product

n(H2O) = 0.1·4 = 0.4 mol
m = 0.4·18.02 = 7.208 g

Answer

m = 7.208 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(Al) = 12/26.98 = 0.4448 mol
m = 0.4448·2/2·133.33 = 59.3 g

Percent yield

Y = 52.19/59.3·100 = 88.01%

Answer

Y = 88.01%
3.Molarity & dilution

Answer: Vwater = 437.5 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 = 0.25·500/2 = 62.5 mL

Water to add

Vwater = 500 − 62.5 = 437.5 mL

Answer

Vwater = 437.5 mL
4.Molarity & dilution

Answer: c = 0.5 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 = 1·200 + 0.3·500 = 350 mmol

Concentration of the mixture

c = 350/700 = 0.5 mol/L

Answer

c = 0.5 mol/L
5.Ideal gas law

Answer: V = 5.799 L

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

Moles

n = 60/44.01 = 1.363 mol
T = 85 + 273.15 = 358.1 K

Volume

V = nRT/P = 1.363·8.314·358.1/700 = 5.799 L

Answer

V = 5.799 L
6.Ideal gas law

Answer: V2 = 0.9428 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 308.1 K, T2 = 363.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(300/750)·(363.1/308.1) = 0.9428 L

Answer

V2 = 0.9428 L
7.Calorimetry & enthalpy

Answer: Tf = 20.98 °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

50·0.449·(160 − Tf) = 150·4.18·(Tf − 16)

Solve for the final temperature

Tf = (22.45·160 + 627·16)/(22.45 + 627) = 20.98 °C

Answer

Tf = 20.98 °C
8.Calorimetry & enthalpy

Answer: ΔH° = −1367 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° = [2·(−393.5) + 3·(−285.8)] − [1·(−277.7) + 3·0] = −1367 kJ

Answer

ΔH° = −1367 kJ

Negative: the reaction gives off heat (exothermic).

9.Chemical equilibrium

Answer: [H2] = 0.3333 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

16 = (2x)²/(1 − x)²

Take the square root

4 = 2x/(1 − x) → x = 1·4/(2 + 4) = 0.6667 mol/L

Equilibrium concentration

[H₂] = 1 − x = 0.3333 mol/L

Answer

[H2] = 0.3333 mol/L
10.Chemical equilibrium

Answer: [Cl2] = 0.3405 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/(1.5 − x) → x² + 0.1·x − 0.15 = 0

Solve the quadratic

x = (−0.1 + √(0.1² + 4·0.15))/2 = 0.3405 mol/L

Equilibrium concentration

[Cl₂] = x = 0.3405 mol/L

Answer

[Cl2] = 0.3405 mol/L
11.pH & buffers

Answer: pH = 2.027

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/(0.5 − x) → x = 0.009397 mol/L

pH

pH = −log(0.009397) = 2.027

Answer

pH = 2.027
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: CA = 0.3333 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 = 1 + 0.2·10 = 3 L/mol
CA = 1/3 = 0.3333 mol/L

Answer

CA = 0.3333 mol/L
14.Reaction kinetics

Answer: A = 4.192 × 10¹¹ 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 = 30 + 273.15 = 303.1 K

Solve Arrhenius for A

Ea/(RT) = 75000/(8.314·303.1) = 29.76
A = k·e29.76 = 0.05·8.384 × 10¹² = 4.192 × 10¹¹ s⁻¹

Answer

A = 4.192 × 10¹¹ s⁻¹

ExamProblems · examproblems.com/courses/general-chemistry