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(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

4 g of C3H8 and 64 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(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 8 g of C6H12O6 reacts completely (the other reactants are in excess) and 3.48 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 3.48 g
Y = %
3.Molarity & dilutionMedium
3 mol/LV1​ = ?0.4 mol/L250 mL

You need 250 mL of 0.4 mol/L solution and have a 3 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 3 mol/Lc2 = 0.4 mol/LV2 = 250 mL
V1 = mL
4.Molarity & dilutionHard
2 mol/L150 mL0.75 mol/L100 mLc = ?250 mL

150 mL of a 2 mol/L solution is mixed with 100 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 2 mol/LV1 = 150 mLc2 = 0.75 mol/LV2 = 100 mL
c = mol/L
5.Ideal gas lawMedium
160 g of CO₂P = 450 kPaT = 20 °CV = ?

What volume V does 160 g of CO₂ occupy at 450 kPa and 20 °C? Treat it as an ideal gas.

m = 160 gM = 44.01 g/molP = 450 kPaT = 20 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CH₄, M = 16.04 g/molP = 150 kPaT = 0 °Cρ = ?

Find the density ρ of CH₄ at 150 kPa and 0 °C, treating it as an ideal gas.

M = 16.04 g/molP = 150 kPaT = 0 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Ironm = 200 gT = 200 °Cc = 0.449 J/(g·°C)Waterm = 400 gT = 18 °Cc = 4.18 J/(g·°C)

A 200 g block of iron (c = 0.449 J/(g·°C)) at 200 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 18 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.449 J/(g·°C)Tmetal = 200 °Cmwater = 400 gTwater = 18 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H100 g C2​H6​2C2​H6​ + 7O₂reactants4CO₂ + 6H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 100 g of C2H6 (M = 30.07 g/mol) burns completely: 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(l). Give q as a positive number.

ΔH°f(C2H6) = −84.7 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 = 36. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 36
[HI] = 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 [PCl5].

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

Find the pH pH of a 0.5 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.5 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A 1.00 L buffer contains 0.3 mol hydrofluoric acid and 0.1 mol of its conjugate base (Ka = 0.00068). 0.05 mol of NaOH is added. Find the new pH pH.

n(HA) = 0.3 moln(A⁻) = 0.1 moln(OH⁻) = 0.05 molKa = 0.00068
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?75%

A first-order reaction in a batch reactor has k = 0.08 min⁻¹. How long does it take to reach 75% conversion? Find t.

k = 0.08 min⁻¹X = 75 %
t = min
14.Reaction kineticsHard
Tk40 °C0.00260 °C0.0201

The rate constant of a reaction is 0.002 min⁻¹ at 40 °C and 0.0201 min⁻¹ at 60 °C. Find the activation energy Ea.

k1 = 0.002 min⁻¹T1 = 40 °Ck2 = 0.0201 min⁻¹T2 = 60 °CR = 8.314 J/(mol·K)
Ea = kJ/mol

3Solutions

1.Stoichiometry

Answer: m = 6.538 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) = 4/44.1 = 0.0907 mol
n(O2) = 64/32 = 2 mol

Limiting reactant

0.0907/1 = 0.0907, 2/5 = 0.4 → C3H8

Mass of product

n(H2O) = 0.0907·4 = 0.3628 mol
m = 0.3628·18.02 = 6.538 g

Answer

m = 6.538 g
2.Stoichiometry

Answer: Y = 85.05 %

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(C6H12O6) = 8/180.16 = 0.0444 mol
m = 0.0444·2/1·46.07 = 4.091 g

Percent yield

Y = 3.48/4.091·100 = 85.05%

Answer

Y = 85.05%
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 = 0.4·250/3 = 33.33 mL

Answer

V1 = 33.33 mL
4.Molarity & dilution

Answer: c = 1.5 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 = 2·150 + 0.75·100 = 375 mmol

Concentration of the mixture

c = 375/250 = 1.5 mol/L

Answer

c = 1.5 mol/L
5.Ideal gas law

Answer: V = 19.69 L

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

Moles

n = 160/44.01 = 3.636 mol
T = 20 + 273.15 = 293.1 K

Volume

V = nRT/P = 3.636·8.314·293.1/450 = 19.69 L

Answer

V = 19.69 L
6.Ideal gas law

Answer: ρ = 1.059 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K

Density

ρ = 150·16.04/(8.314·273.1) = 1.059 kg/m³

Answer

ρ = 1.059 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 27.28 °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

200·0.449·(200 − Tf) = 400·4.18·(Tf − 18)

Solve for the final temperature

Tf = (89.8·200 + 1672·18)/(89.8 + 1672) = 27.28 °C

Answer

Tf = 27.28 °C
8.Calorimetry & enthalpy

Answer: q = 5187 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° = [4·(−393.5) + 6·(−285.8)] − [2·(−84.7) + 7·0] = −3119 kJ

Heat for the given mass

n = 100/30.07 = 3.326 mol
q = n·|ΔH°/2| = 3.326·1560 = 5187 kJ

Answer

q = 5187 kJ
9.Chemical equilibrium

Answer: [HI] = 0.75 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

36 = (2x)²/(0.5 − x)²

Take the square root

6 = 2x/(0.5 − x) → x = 0.5·6/(2 + 6) = 0.375 mol/L

Equilibrium concentration

[HI] = 2x = 0.75 mol/L

Answer

[HI] = 0.75 mol/L
10.Chemical equilibrium

Answer: [PCl5] = 0.8486 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

[PCl₅] = 1.5 − 0.6514 = 0.8486 mol/L

Answer

[PCl5] = 0.8486 mol/L
11.pH & buffers

Answer: pH = 2.253

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 = 0.005581 mol/L

pH

pH = −log(0.005581) = 2.253

Answer

pH = 2.253
12.pH & buffers

Answer: pH = 2.946

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) = 0.3 − 0.05 = 0.25 mol
n(A⁻) = 0.1 + 0.05 = 0.15 mol

Henderson–Hasselbalch

pH = 3.167 + log(0.15/0.25) = 2.946

Answer

pH = 2.946
13.Reaction kinetics

Answer: t = 17.33 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 = 0.25

Solve for the time

t = −ln(1 − X)/k = −ln(0.25)/0.08 = 17.33 min

Answer

t = 17.33 min
14.Reaction kinetics

Answer: Ea = 100.1 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) = −8.314·2.308/(-1.917e-4) = 100075 J/mol
Ea = 100.1 kJ/mol

Answer

Ea = 100.1 kJ/mol

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