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(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

6 g of Mg and 16 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, 6 g of Na reacts completely (the other reactants are in excess) and 8.39 g of NaCl is collected. Find the percent yield Y.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 8.39 g
Y = %
3.Molarity & dilutionMedium
2.5 mol/LV1​ = ?0.4 mol/L100 mL

You need 100 mL of 0.4 mol/L solution and have a 2.5 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 2.5 mol/Lc2 = 0.4 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
1.5 mol/L200 mL0.6 mol/L50 mLc = ?250 mL

200 mL of a 1.5 mol/L solution is mixed with 50 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 1.5 mol/LV1 = 200 mLc2 = 0.6 mol/LV2 = 50 mL
c = mol/L
5.Ideal gas lawMedium
490 g of CH₄P = 700 kPaT = 45 °CV = ?

What volume V does 490 g of CH₄ occupy at 700 kPa and 45 °C? Treat it as an ideal gas.

m = 490 gM = 16.04 g/molP = 700 kPaT = 45 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
Ar, M = 39.95 g/molP = 900 kPaT = 140 °Cρ = ?

Find the density ρ of Ar at 900 kPa and 140 °C, treating it as an ideal gas.

M = 39.95 g/molP = 900 kPaT = 140 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Zincm = 75 gT = 110 °Cc = 0.388 J/(g·°C)Waterm = 300 gT = 23 °Cc = 4.18 J/(g·°C)

A 75 g block of zinc (c = 0.388 J/(g·°C)) at 110 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 23 °C in an insulated cup. Find the final temperature Tf.

mmetal = 75 gcmetal = 0.388 J/(g·°C)Tmetal = 110 °Cmwater = 300 gTwater = 23 °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​HII0.40.40C−x−x+2xE0.4 − x0.4 − x2x

0.4 mol/L of H₂ and 0.4 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 50. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 50
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.500C−x+x+xE0.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 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 = 0.04
[PCl5] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.2 mol/L solution of formic acid (Ka = 0.00018).

c = 0.2 mol/LKa = 0.00018
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A 1.00 L buffer contains 0.2 mol hydrofluoric acid and 0.25 mol of its conjugate base (Ka = 0.00068). 0.02 mol of NaOH is added. Find the new pH pH.

n(HA) = 0.2 moln(A⁻) = 0.25 moln(OH⁻) = 0.02 molKa = 0.00068
pH =
13.Reaction kineticsMedium
t (min)X (%)t = ?90%

A first-order reaction in a batch reactor has k = 0.05 min⁻¹. How long does it take to reach 90% conversion? Find t.

k = 0.05 min⁻¹X = 90 %
t = min
14.Reaction kineticsHard
Tk20 °C0.150 °C0.983

The rate constant of a reaction is 0.1 min⁻¹ at 20 °C and 0.983 min⁻¹ at 50 °C. Find the activation energy Ea.

k1 = 0.1 min⁻¹T1 = 20 °Ck2 = 0.983 min⁻¹T2 = 50 °CR = 8.314 J/(mol·K)
Ea = kJ/mol

3Solutions

1.Stoichiometry

Answer: m = 9.947 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) = 6/24.31 = 0.2468 mol
n(O2) = 16/32 = 0.5 mol

Limiting reactant

0.2468/2 = 0.1234, 0.5/1 = 0.5 → Mg

Mass of product

n(MgO) = 0.1234·2 = 0.2468 mol
m = 0.2468·40.3 = 9.947 g

Answer

m = 9.947 g
2.Stoichiometry

Answer: Y = 55.01 %

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(Na) = 6/22.99 = 0.261 mol
m = 0.261·2/2·58.44 = 15.25 g

Percent yield

Y = 8.39/15.25·100 = 55.01%

Answer

Y = 55.01%
3.Molarity & dilution

Answer: V1 = 16 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 = 0.4·100/2.5 = 16 mL

Answer

V1 = 16 mL
4.Molarity & dilution

Answer: c = 1.32 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 = 1.5·200 + 0.6·50 = 330 mmol

Concentration of the mixture

c = 330/250 = 1.32 mol/L

Answer

c = 1.32 mol/L
5.Ideal gas law

Answer: V = 115.4 L

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

Moles

n = 490/16.04 = 30.55 mol
T = 45 + 273.15 = 318.1 K

Volume

V = nRT/P = 30.55·8.314·318.1/700 = 115.4 L

Answer

V = 115.4 L
6.Ideal gas law

Answer: ρ = 10.47 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K

Density

ρ = 900·39.95/(8.314·413.1) = 10.47 kg/m³

Answer

ρ = 10.47 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 24.97 °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

75·0.388·(110 − Tf) = 300·4.18·(Tf − 23)

Solve for the final temperature

Tf = (29.1·110 + 1254·23)/(29.1 + 1254) = 24.97 °C

Answer

Tf = 24.97 °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] = 0.08819 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

50 = (2x)²/(0.4 − x)²

Take the square root

7.071 = 2x/(0.4 − x) → x = 0.4·7.071/(2 + 7.071) = 0.3118 mol/L

Equilibrium concentration

[H₂] = 0.4 − x = 0.08819 mol/L

Answer

[H2] = 0.08819 mol/L
10.Chemical equilibrium

Answer: [PCl5] = 0.3772 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

0.04 = x²/(0.5 − x) → x² + 0.04·x − 0.02 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.02))/2 = 0.1228 mol/L

Equilibrium concentration

[PCl₅] = 0.5 − 0.1228 = 0.3772 mol/L

Answer

[PCl5] = 0.3772 mol/L
11.pH & buffers

Answer: pH = 2.228

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/(0.2 − x) → x = 0.005911 mol/L

pH

pH = −log(0.005911) = 2.228

Answer

pH = 2.228
12.pH & buffers

Answer: pH = 3.344

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) = 0.2 − 0.02 = 0.18 mol
n(A⁻) = 0.25 + 0.02 = 0.27 mol

Henderson–Hasselbalch

pH = 3.167 + log(0.27/0.18) = 3.344

Answer

pH = 3.344
13.Reaction kinetics

Answer: t = 46.05 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)/0.05 = 46.05 min

Answer

t = 46.05 min
14.Reaction kinetics

Answer: Ea = 60 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 = 293.1 K, T2 = 323.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.285/(-3.167e-4) = 60000 J/mol
Ea = 60 kJ/mol

Answer

Ea = 60 kJ/mol

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