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)× 2/ν× M(H2​O)

30 g of H2 and 48 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 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 9.53 g of CaO is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 9.53 g
Y = %
3.Molarity & dilutionMedium
4 mol/LV1​ = ?0.25 mol/L500 mL

You need 500 mL of 0.25 mol/L solution and have a 4 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 4 mol/Lc2 = 0.25 mol/LV2 = 500 mL
V1 = mL
4.Molarity & dilutionHard
1 mol/L300 mL0.75 mol/L100 mLc = ?400 mL

300 mL of a 1 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 = 1 mol/LV1 = 300 mLc2 = 0.75 mol/LV2 = 100 mL
c = mol/L
5.Ideal gas lawMedium
380 g of CO₂V = 20 LT = 80 °CP = ?

A tank of 20 L contains 380 g of CO₂ at 80 °C. Treat it as an ideal gas and find the pressure P.

m = 380 gM = 44.01 g/molV = 20 LT = 80 °CR = 8.314 J/(mol·K)
P = kPa
6.Ideal gas lawHard
State 1State 2

A fixed amount of O₂ goes from 350 kPa, 2 L, 30 °C to 350 kPa and 240 °C. Find the new volume V2.

P1 = 350 kPaV1 = 2 LT1 = 30 °CP2 = 350 kPaT2 = 240 °C
V2 = L
7.Calorimetry & enthalpyMedium
Tf​ = ?Aluminiumm = 200 gT = 180 °Cc = 0.897 J/(g·°C)Waterm = 300 gT = 22 °Cc = 4.18 J/(g·°C)

A 200 g block of aluminium (c = 0.897 J/(g·°C)) at 180 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 22 °C in an insulated cup. Find the final temperature Tf.

mmetal = 200 gcmetal = 0.897 J/(g·°C)Tmetal = 180 °Cmwater = 300 gTwater = 22 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H10 g C3​H8​C3​H8​ + 5O₂reactants3CO₂ + 4H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 10 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​HII0.250.250C−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 = 25. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 25
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.800C−x+x+xE0.8 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].

[PCl₅]₀ = 0.8 mol/LKc = 0.5
[PCl5] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.1 mol/L solution of hydrofluoric acid (Ka = 0.00068).

c = 0.1 mol/LKa = 0.00068
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.4 mol/L formic acid and 0.1 mol/L of its sodium salt. Ka = 0.00018. Find the pH pH of the buffer.

[HA] = 0.4[A⁻] = 0.1Ka = 0.00018
pH =
13.Reaction kineticsMedium
t (min)C (mol/L)20 minC = ?

A reaction with rate law −rA = k·CA² has k = 1 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 20 min.

k = 1 L/(mol·min)CA0 = 0.5 mol/Lt = 20 min
CA = mol/L
14.Reaction kineticsHard
1/Tln kT = 30 °Cln 0.005ln A = ?Ea​ = 40 kJ/mol

A first-order reaction has a rate constant of 0.005 s⁻¹ at 30 °C and an activation energy of 40 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.005 s⁻¹T = 30 °CEa = 40 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹

3Solutions

1.Stoichiometry

Answer: m = 54.06 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(H2) = 30/2.016 = 14.88 mol
n(O2) = 48/32 = 1.5 mol

Limiting reactant

14.88/2 = 7.44, 1.5/1 = 1.5 → O2

Mass of product

n(H2O) = 1.5·2 = 3 mol
m = 3·18.02 = 54.06 g

Answer

m = 54.06 g
2.Stoichiometry

Answer: Y = 85.04 %

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 = 9.53/11.21·100 = 85.04%

Answer

Y = 85.04%
3.Molarity & dilution

Answer: V1 = 31.25 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 = 0.25·500/4 = 31.25 mL

Answer

V1 = 31.25 mL
4.Molarity & dilution

Answer: c = 0.9375 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 + 0.75·100 = 375 mmol

Concentration of the mixture

c = 375/400 = 0.9375 mol/L

Answer

c = 0.9375 mol/L
5.Ideal gas law

Answer: P = 1268 kPa

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

Moles

n = m/M = 380/44.01 = 8.634 mol
T = 80 + 273.15 = 353.1 K

Pressure

P = nRT/V = 8.634·8.314·353.1/0.02 = 1268 kPa

Answer

P = 1268 kPa
6.Ideal gas law

Answer: V2 = 3.385 L

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

Absolute temperatures

T1 = 303.1 K, T2 = 513.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(350/350)·(513.1/303.1) = 3.385 L

Answer

V2 = 3.385 L
7.Calorimetry & enthalpy

Answer: Tf = 41.77 °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

200·0.897·(180 − Tf) = 300·4.18·(Tf − 22)

Solve for the final temperature

Tf = (179.4·180 + 1254·22)/(179.4 + 1254) = 41.77 °C

Answer

Tf = 41.77 °C
8.Calorimetry & enthalpy

Answer: q = 503.4 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 = 10/44.1 = 0.2268 mol
q = n·|ΔH°/1| = 0.2268·2220 = 503.4 kJ

Answer

q = 503.4 kJ
9.Chemical equilibrium

Answer: [H2] = 0.07143 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/(0.25 − x)²

Take the square root

5 = 2x/(0.25 − x) → x = 0.25·5/(2 + 5) = 0.1786 mol/L

Equilibrium concentration

[H₂] = 0.25 − x = 0.07143 mol/L

Answer

[H2] = 0.07143 mol/L
10.Chemical equilibrium

Answer: [PCl5] = 0.3699 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/(0.8 − x) → x² + 0.5·x − 0.4 = 0

Solve the quadratic

x = (−0.5 + √(0.5² + 4·0.4))/2 = 0.4301 mol/L

Equilibrium concentration

[PCl₅] = 0.8 − 0.4301 = 0.3699 mol/L

Answer

[PCl5] = 0.3699 mol/L
11.pH & buffers

Answer: pH = 2.102

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/(0.1 − x) → x = 0.007913 mol/L

pH

pH = −log(0.007913) = 2.102

Answer

pH = 2.102
12.pH & buffers

Answer: pH = 3.143

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 = 3.745

Henderson–Hasselbalch

pH = 3.745 + log(0.1/0.4) = 3.143

Answer

pH = 3.143
13.Reaction kinetics

Answer: CA = 0.04545 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 = 2 + 1·20 = 22 L/mol
CA = 1/22 = 0.04545 mol/L

Answer

CA = 0.04545 mol/L
14.Reaction kinetics

Answer: A = 39036 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) = 40000/(8.314·303.1) = 15.87
A = k·e15.87 = 0.005·7.807 × 10⁶ = 39036 s⁻¹

Answer

A = 39036 s⁻¹

ExamProblems · examproblems.com/courses/general-chemistry