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(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

20 g of Na and 8 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
2.StoichiometryHard
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 30 g of Mg reacts completely (the other reactants are in excess) and 34.81 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 34.81 g
Y = %
3.Molarity & dilutionMedium
6 mol/LV1​ = ?0.3 mol/L200 mL

You need 200 mL of 0.3 mol/L solution and have a 6 mol/L stock solution. What volume V1 of stock solution do you need?

c1 = 6 mol/Lc2 = 0.3 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
0.2 mol/L200 mL1.2 mol/L500 mLc = ?700 mL

200 mL of a 0.2 mol/L solution is mixed with 500 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).

c1 = 0.2 mol/LV1 = 200 mLc2 = 1.2 mol/LV2 = 500 mL
c = mol/L
5.Ideal gas lawMedium
220 g of N₂P = 500 kPaT = 95 °CV = ?

What volume V does 220 g of N₂ occupy at 500 kPa and 95 °C? Treat it as an ideal gas.

m = 220 gM = 28.01 g/molP = 500 kPaT = 95 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
CO₂, M = 44.01 g/molP = 650 kPaT = 160 °Cρ = ?

Find the density ρ of CO₂ at 650 kPa and 160 °C, treating it as an ideal gas.

M = 44.01 g/molP = 650 kPaT = 160 °CR = 8.314 J/(mol·K)
ρ = kg/m³
7.Calorimetry & enthalpyMedium
Tf​ = ?Aluminiumm = 75 gT = 110 °Cc = 0.897 J/(g·°C)Waterm = 400 gT = 15 °Cc = 4.18 J/(g·°C)

A 75 g block of aluminium (c = 0.897 J/(g·°C)) at 110 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 15 °C in an insulated cup. Find the final temperature Tf.

mmetal = 75 gcmetal = 0.897 J/(g·°C)Tmetal = 110 °Cmwater = 400 gTwater = 15 °Ccwater = 4.18 J/(g·°C)
Tf = °C
8.Calorimetry & enthalpyHard
H25 g C4​H10​2C4​H10​ + 13O₂reactants8CO₂ + 10H₂O(l)productsq = ?

Use the standard enthalpies of formation to find the heat q released when 25 g of C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l). Give q as a positive number.

ΔH°f(C4H10) = −125.6 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​HII0.40.40C−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 = 49. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 49
[H2] = mol/L
10.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.200C−x+x+xE0.2 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].

[PCl₅]₀ = 0.2 mol/LKc = 0.1
[PCl5] = mol/L
11.pH & buffersMedium
02468101214acidicneutralbasicpH = ?

Find the pH pH of a 0.05 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).

c = 0.05 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
02468101214acidicneutralbasicpH = ?

A buffer contains 0.25 mol/L hypochlorous acid and 0.05 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.

[HA] = 0.25[A⁻] = 0.05Ka = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
t (min)C (mol/L)2 minC = ?

A reaction with rate law −rA = k·CA² has k = 0.2 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 2 min.

k = 0.2 L/(mol·min)CA0 = 0.5 mol/Lt = 2 min
CA = mol/L
14.Reaction kineticsHard
CA0​ (mol/L)r (mol/(L·s))run 10.20.002run 20.40.00283n = ?

Two experiments measure the initial rate of A → products. With CA0 = 0.2 mol/L the rate is 0.002 mol/(L·s); with CA0 = 0.4 mol/L it is 0.00283 mol/(L·s). Find the reaction order n with respect to A.

CA0,1 = 0.2 mol/Lr1 = 0.002 mol/(L·s)CA0,2 = 0.4 mol/Lr2 = 0.00283 mol/(L·s)
n =

3Solutions

1.Stoichiometry

Answer: m = 13.19 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(Na) = 20/22.99 = 0.8699 mol
n(Cl2) = 8/70.9 = 0.1128 mol

Limiting reactant

0.8699/2 = 0.435, 0.1128/1 = 0.1128 → Cl2

Mass of product

n(NaCl) = 0.1128·2 = 0.2257 mol
m = 0.2257·58.44 = 13.19 g

Answer

m = 13.19 g
2.Stoichiometry

Answer: Y = 69.99 %

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(Mg) = 30/24.31 = 1.234 mol
m = 1.234·2/2·40.3 = 49.73 g

Percent yield

Y = 34.81/49.73·100 = 69.99%

Answer

Y = 69.99%
3.Molarity & dilution

Answer: V1 = 10 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 = 0.3·200/6 = 10 mL

Answer

V1 = 10 mL
4.Molarity & dilution

Answer: c = 0.9143 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 = 0.2·200 + 1.2·500 = 640 mmol

Concentration of the mixture

c = 640/700 = 0.9143 mol/L

Answer

c = 0.9143 mol/L
5.Ideal gas law

Answer: V = 48.08 L

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

Moles

n = 220/28.01 = 7.854 mol
T = 95 + 273.15 = 368.1 K

Volume

V = nRT/P = 7.854·8.314·368.1/500 = 48.08 L

Answer

V = 48.08 L
6.Ideal gas law

Answer: ρ = 7.944 kg/m³

Ideal gas density: ρ = PM/(RT).

Rearrange the gas law

PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 160 + 273.15 = 433.1 K

Density

ρ = 650·44.01/(8.314·433.1) = 7.944 kg/m³

Answer

ρ = 7.944 kg/m³
7.Calorimetry & enthalpy

Answer: Tf = 18.67 °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

75·0.897·(110 − Tf) = 400·4.18·(Tf − 15)

Solve for the final temperature

Tf = (67.28·110 + 1672·15)/(67.28 + 1672) = 18.67 °C

Answer

Tf = 18.67 °C
8.Calorimetry & enthalpy

Answer: q = 1238 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° = [8·(−393.5) + 10·(−285.8)] − [2·(−125.6) + 13·0] = −5755 kJ

Heat for the given mass

n = 25/58.12 = 0.4301 mol
q = n·|ΔH°/2| = 0.4301·2877 = 1238 kJ

Answer

q = 1238 kJ
9.Chemical equilibrium

Answer: [H2] = 0.08889 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

49 = (2x)²/(0.4 − x)²

Take the square root

7 = 2x/(0.4 − x) → x = 0.4·7/(2 + 7) = 0.3111 mol/L

Equilibrium concentration

[H₂] = 0.4 − x = 0.08889 mol/L

Answer

[H2] = 0.08889 mol/L
10.Chemical equilibrium

Answer: [PCl5] = 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

0.1 = x²/(0.2 − x) → x² + 0.1·x − 0.02 = 0

Solve the quadratic

x = (−0.1 + √(0.1² + 4·0.02))/2 = 0.1 mol/L

Equilibrium concentration

[PCl₅] = 0.2 − 0.1 = 0.1 mol/L

Answer

[PCl5] = 0.1 mol/L
11.pH & buffers

Answer: pH = 2.759

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/(0.05 − x) → x = 0.001744 mol/L

pH

pH = −log(0.001744) = 2.759

Answer

pH = 2.759
12.pH & buffers

Answer: pH = 6.824

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 = 7.523

Henderson–Hasselbalch

pH = 7.523 + log(0.05/0.25) = 6.824

Answer

pH = 6.824
13.Reaction kinetics

Answer: CA = 0.4167 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 + 0.2·2 = 2.4 L/mol
CA = 1/2.4 = 0.4167 mol/L

Answer

CA = 0.4167 mol/L
14.Reaction kinetics

Answer: n = 0.5008

Method of initial rates: for r = k·Cⁿ, comparing two runs that differ only in the starting concentration removes k: r₂/r₁ = (C₂/C₁)ⁿ. Doubling C doubles the rate for first order and quadruples it for second order.

Ratios

r2/r1 = 0.00283/0.002 = 1.415
C2/C1 = 0.4/0.2 = 2

Order

n = ln(1.415)/ln(2) = 0.5008

Answer

n = 0.5008

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