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(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

10 g of Mg and 40 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(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

In the reaction 2Na + Cl2 → 2NaCl, 8 g of Na reacts completely (the other reactants are in excess) and 18.51 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 18.51 g
Y = %
3.Molarity & dilutionMedium
2 mol/LV1​0.25 mol/L100 mL

You need 100 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 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
1 mol/L300 mL0.05 mol/L250 mLc = ?550 mL

300 mL of a 1 mol/L solution is mixed with 250 mL of a 0.05 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1 mol/LV1 = 300 mLc2 = 0.05 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
280 g of CO₂P = 550 kPaT = 35 °CV = ?

What volume V does 280 g of CO₂ occupy at 550 kPa and 35 °C? Treat it as an ideal gas.

m = 280 gM = 44.01 g/molP = 550 kPaT = 35 °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​ = ?Copperm = 250 gT = 150 °Cc = 0.385 J/(g·°C)Waterm = 250 gT = 16 °Cc = 4.18 J/(g·°C)

A 250 g block of copper (c = 0.385 J/(g·°C)) at 150 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.

mmetal = 250 gcmetal = 0.385 J/(g·°C)Tmetal = 150 °Cmwater = 250 gTwater = 16 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H25 g C3​H8​C3​H8​ + 5O₂reactants3CO₂ + 4H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 25 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​HII0.10.10C−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 = 64. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 64
[HI] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.500C−x+x+xE0.5 − xxx

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

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

Find the pH pH of a 0.25 mol/L solution of hydrofluoric acid (Ka = 0.00068).

c = 0.25 mol/LKa = 0.00068
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A 1.00 L buffer contains 0.1 mol benzoic acid and 0.1 mol of its conjugate base (Ka = 6.3 × 10⁻⁵). 0.05 mol of NaOH is added. Find the new pH pH.

n(HA) = 0.1 moln(A⁻) = 0.1 moln(OH⁻) = 0.05 molKa = 6.3 × 10⁻⁵
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?90%

A first-order reaction in a batch reactor has k = 0.1 min⁻¹. How long does it take to reach 90% conversion? Find t.

k = 0.1 min⁻¹X = 90 %
t = min
14.Reaction kineticsHard
Tk35 °C0.0155 °C?

A reaction has a rate constant of 0.01 min⁻¹ at 35 °C and an activation energy of 90 kJ/mol. Find the rate constant k2 at 55 °C.

k1 = 0.01 min⁻¹T1 = 35 °CT2 = 55 °CEa = 90 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹

3Solutions

1.Stoichiometry

Answer: m = 16.58 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) = 10/24.31 = 0.4114 mol
n(O2) = 40/32 = 1.25 mol

Limiting reactant

0.4114/2 = 0.2057, 1.25/1 = 1.25 → Mg

Mass of product

n(MgO) = 0.2057·2 = 0.4114 mol
m = 0.4114·40.3 = 16.58 g

Answer

m = 16.58 g
2.Stoichiometry

Answer: Y = 91.02 %

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(Na) = 8/22.99 = 0.348 mol
m = 0.348·2/2·58.44 = 20.34 g

Percent yield

Y = 18.51/20.34·100 = 91.02%

Answer

Y = 91.02%
3.Molarity & dilution

Answer: Vwater = 87.5 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 = 0.25·100/2 = 12.5 mL

Water to add

Vwater = 100 − 12.5 = 87.5 mL

Answer

Vwater = 87.5 mL
4.Molarity & dilution

Answer: c = 0.5682 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 = 1·300 + 0.05·250 = 312.5 mmol

Concentration of the mixture

c = 312.5/550 = 0.5682 mol/L

Answer

c = 0.5682 mol/L
5.Ideal gas law

Answer: V = 29.64 L

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

Moles

n = 280/44.01 = 6.362 mol
T = 35 + 273.15 = 308.1 K

Volume

V = nRT/P = 6.362·8.314·308.1/550 = 29.64 L

Answer

V = 29.64 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 = 27.3 °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

250·0.385·(150 − Tf) = 250·4.18·(Tf − 16)

Solve for the final temperature

Tf = (96.25·150 + 1045·16)/(96.25 + 1045) = 27.3 °C

Answer

Tf = 27.3 °C
8.Calorimetry & enthalpy

Answer: q = 1258 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 = 25/44.1 = 0.5669 mol
q = n·|ΔH°/1| = 0.5669·2220 = 1258 kJ

Answer

q = 1258 kJ
9.Chemical equilibrium

Answer: [HI] = 0.16 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

64 = (2x)²/(0.1 − x)²

Take the square root

8 = 2x/(0.1 − x) → x = 0.1·8/(2 + 8) = 0.08 mol/L

Equilibrium concentration

[HI] = 2x = 0.16 mol/L

Answer

[HI] = 0.16 mol/L
10.Chemical equilibrium

Answer: [PCl5] = 0.191 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/(0.5 − x) → x² + 0.5·x − 0.25 = 0

Solve the quadratic

x = (−0.5 + √(0.5² + 4·0.25))/2 = 0.309 mol/L

Equilibrium concentration

[PCl₅] = 0.5 − 0.309 = 0.191 mol/L

Answer

[PCl5] = 0.191 mol/L
11.pH & buffers

Answer: pH = 1.896

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 = 0.0127 mol/L

pH

pH = −log(0.0127) = 1.896

Answer

pH = 1.896
12.pH & buffers

Answer: pH = 4.678

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) = 0.1 − 0.05 = 0.05 mol
n(A⁻) = 0.1 + 0.05 = 0.15 mol

Henderson–Hasselbalch

pH = 4.201 + log(0.15/0.05) = 4.678

Answer

pH = 4.678
13.Reaction kinetics

Answer: t = 23.03 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 = 0.1

Solve for the time

t = −ln(1 − X)/k = −ln(0.1)/0.1 = 23.03 min

Answer

t = 23.03 min
14.Reaction kinetics

Answer: k2 = 0.08508 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 = 328.1 K

Arrhenius ratio

ln(k2/k1) = −(90000/8.314)·(1/328.1 − 1/308.1) = 2.141
k2 = 0.01·e2.141 = 0.08508 min⁻¹

Answer

k2 = 0.08508 min⁻¹

ExamProblems · examproblems.com/courses/general-chemistry