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(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

30 g of N2 and 8 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
2.StoichiometryHard
m(CaCO3​)n(CaCO3​)n(CaO)m(CaO)÷ M(CaCO3​)× 1/1× M(CaO)

In the reaction CaCO3 → CaO + CO2, 20 g of CaCO3 reacts completely (the other reactants are in excess) and 9.19 g of CaO is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 9.19 g
Y = %
3.Molarity & dilutionMedium
6 mol/LV1​ = ?0.5 mol/L100 mL

You need 100 mL of 0.5 mol/L solution and have a 6 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 6 mol/Lc2 = 0.5 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
1.5 mol/L300 mL0.3 mol/L100 mLc = ?400 mL

300 mL of a 1.5 mol/L solution is mixed with 100 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1.5 mol/LV1 = 300 mLc2 = 0.3 mol/LV2 = 100 mL
c = mol/L
5.Ideal gas lawMedium
270 g of O₂V = 10 LT = 0 °CP = ?

A tank of 10 L contains 270 g of O₂ at 0 °C. Treat it as an ideal gas and find the pressure P.

m = 270 gM = 32 g/molV = 10 LT = 0 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 250 kPa, 10 L, 25 °C to 700 kPa and 140 °C. Find the new volume V2.

P1 = 250 kPaV1 = 10 LT1 = 25 °CP2 = 700 kPaT2 = 140 °C
V2 = L
7.Calorimetry & enthalpyMedium
Tf​ = ?Leadm = 200 gT = 130 °Cc = 0.129 J/(g·°C)Waterm = 250 gT = 25 °Cc = 4.18 J/(g·°C)

A 200 g block of lead (c = 0.129 J/(g·°C)) at 130 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 25 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.129 J/(g·°C)Tmetal = 130 °Cmwater = 250 gTwater = 25 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H100 g CH4​CH4​ + 2O₂reactantsCO₂ + 2H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 100 g of CH4 (M = 16.04 g/mol) burns completely: CH4 + 2O₂ → CO₂ + 2H₂O(l). Give q as a positive number.

ΔH°f(CH4) = −74.8 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​HII0.250.250C−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​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 lactic acid (Ka = 0.00014).

c = 0.5 mol/LKa = 0.00014
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.2 mol/L benzoic acid and 0.3 mol/L of its sodium salt. Ka = 6.3 × 10⁻⁵. Find the pH pH of the buffer.

[HA] = 0.2[A⁻] = 0.3Ka = 6.3 × 10⁻⁵
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?90%

A first-order reaction in a batch reactor has k = 0.08 min⁻¹. How long does it take to reach 90% conversion? Find t.

k = 0.08 min⁻¹X = 90 %
t = min
14.Reaction kineticsHard
Tk30 °C0.0140 °C?

A reaction has a rate constant of 0.01 min⁻¹ at 30 °C and an activation energy of 50 kJ/mol. Find the rate constant k2 at 40 °C.

k1 = 0.01 min⁻¹T1 = 30 °CT2 = 40 °CEa = 50 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹

3Solutions

1.Stoichiometry

Answer: m = 36.47 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(N2) = 30/28.02 = 1.071 mol
n(H2) = 8/2.016 = 3.968 mol

Limiting reactant

1.071/1 = 1.071, 3.968/3 = 1.323 → N2

Mass of product

n(NH3) = 1.071·2 = 2.141 mol
m = 2.141·17.03 = 36.47 g

Answer

m = 36.47 g
2.Stoichiometry

Answer: Y = 82.01 %

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) = 20/100.09 = 0.1998 mol
m = 0.1998·1/1·56.08 = 11.21 g

Percent yield

Y = 9.19/11.21·100 = 82.01%

Answer

Y = 82.01%
3.Molarity & dilution

Answer: V1 = 8.333 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 = 0.5·100/6 = 8.333 mL

Answer

V1 = 8.333 mL
4.Molarity & dilution

Answer: c = 1.2 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 = 1.5·300 + 0.3·100 = 480 mmol

Concentration of the mixture

c = 480/400 = 1.2 mol/L

Answer

c = 1.2 mol/L
5.Ideal gas law

Answer: P = 1916 kPa

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

Moles

n = m/M = 270/32 = 8.438 mol
T = 0 + 273.15 = 273.1 K

Pressure

P = nRT/V = 8.438·8.314·273.1/0.01 = 1916 kPa

Answer

P = 1916 kPa
6.Ideal gas law

Answer: V2 = 4.949 L

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

Absolute temperatures

T1 = 298.1 K, T2 = 413.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 10·(250/700)·(413.1/298.1) = 4.949 L

Answer

V2 = 4.949 L
7.Calorimetry & enthalpy

Answer: Tf = 27.53 °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

200·0.129·(130 − Tf) = 250·4.18·(Tf − 25)

Solve for the final temperature

Tf = (25.8·130 + 1045·25)/(25.8 + 1045) = 27.53 °C

Answer

Tf = 27.53 °C
8.Calorimetry & enthalpy

Answer: q = 5550 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° = [1·(−393.5) + 2·(−285.8)] − [1·(−74.8) + 2·0] = −890.3 kJ

Heat for the given mass

n = 100/16.04 = 6.234 mol
q = n·|ΔH°/1| = 6.234·890.3 = 5550 kJ

Answer

q = 5550 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] = 0.3405 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 = 2.081

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 = 0.008297 mol/L

pH

pH = −log(0.008297) = 2.081

Answer

pH = 2.081
12.pH & buffers

Answer: pH = 4.377

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 = 4.201

Henderson–Hasselbalch

pH = 4.201 + log(0.3/0.2) = 4.377

Answer

pH = 4.377
13.Reaction kinetics

Answer: t = 28.78 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)/0.08 = 28.78 min

Answer

t = 28.78 min
14.Reaction kinetics

Answer: k2 = 0.01884 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 = 303.1 K, T2 = 313.1 K

Arrhenius ratio

ln(k2/k1) = −(50000/8.314)·(1/313.1 − 1/303.1) = 0.6335
k2 = 0.01·e0.6335 = 0.01884 min⁻¹

Answer

k2 = 0.01884 min⁻¹

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