What is stoichiometry and how do you solve stoichiometry problems?

Stoichiometry uses balanced chemical equations and mole ratios to calculate amounts of reactants and products. Steps: (1) Balance the equation, (2) Convert given quantity to moles, (3) Use mole ratio, (4) Convert to desired units.

How do you calculate pH from hydrogen ion concentration?

pH = -log[H+]. For strong acids, [H+] equals the acid concentration. For weak acids, use an ICE table with Ka to find [H+], then calculate pH.

What is the ideal gas law and when do you use it?

PV = nRT relates pressure, volume, moles, and temperature of a gas. R = 0.0821 L·atm/(mol·K). Use it when dealing with gas quantities at known conditions. Always use Kelvin for temperature.

How do you determine the limiting reagent?

Convert all reactants to moles, then divide each by its coefficient in the balanced equation. The reactant with the smallest ratio is the limiting reagent - it determines the maximum product yield.

Chemistry Calculators - CalcREX

Stoichiometry - The Mole, Molar Mass & Reaction Calculations

The quantitative heart of chemistry - connecting atoms, molecules, and grams through balanced equations.

1 The Mole Concept

The mole is chemistry's counting unit. One mole contains exactly 6.022 × 10²³ particles (Avogadro's number). It bridges the atomic world (atoms, molecules) to the macroscopic world (grams, liters).

1 mol = 6.022 × 10²³ particles

1 mol of C atoms = 12.011 g1 mol of H₂O molecules = 18.015 g1 mol of NaCl formula units = 58.44 g

2 Molar Mass - The Bridge Between Grams and Moles

Molar mass (M) is the mass of one mole of a substance in g/mol. For elements, it equals the atomic mass from the periodic table. For compounds, add up all atoms.

n = m / M

n = moles (mol)m = mass (g)M = molar mass (g/mol)

Figure 1: The Mole Map - converting between mass, moles, and number of particles. Molar mass (M) connects grams ↔ moles. Avogadro's number connects moles ↔ particles.

3 Balancing Chemical Equations

A balanced equation has equal numbers of each atom on both sides. Coefficients tell you the mole ratio - the key to all stoichiometric calculations.

2 H₂ + O₂ → 2 H₂O

This tells us: 2 mol H₂ reacts with 1 mol O₂ to produce 2 mol H₂O. The mole ratio is 2:1:2.

Steps to Solve Stoichiometry Problems

1Write and balance the chemical equation.

2Convert given quantity to moles (using M or Avogadro's number).

3Use the mole ratio from the balanced equation to find moles of desired substance.

4Convert moles to requested units (grams, liters, particles).

Worked Example 1

Mass-to-Mass Stoichiometry

Problem: How many grams of water are produced when 10.0 g of hydrogen gas reacts with excess oxygen?
Reaction: 2 H₂ + O₂ → 2 H₂O

Step 1: Convert grams H₂ to moles

M(H₂) = 2 × 1.008 = 2.016 g/mol

n(H₂) = 10.0 / 2.016 = 4.960 mol

Step 2: Use mole ratio (2 mol H₂ → 2 mol H₂O)

n(H₂O) = 4.960 × (2/2) = 4.960 mol

Step 3: Convert moles H₂O to grams

M(H₂O) = 2(1.008) + 16.00 = 18.015 g/mol

m = 4.960 × 18.015

m(H₂O) = 89.36 g

Answer: 89.36 g of water is produced. Notice: the mass of products (89.36 g H₂O + O₂ consumed) equals the mass of reactants - conservation of mass.

4 Limiting Reagent & Percent Yield

The limiting reagent is the reactant that runs out first, determining the maximum product. The other reactant is in "excess." Percent yield measures how successful a reaction was.

% Yield = (Actual Yield / Theoretical Yield) × 100

Worked Example 2

Limiting Reagent Problem

Problem: 5.0 g of Fe reacts with 5.0 g of O₂. Which is the limiting reagent? How much Fe₂O₃ forms?
Reaction: 4 Fe + 3 O₂ → 2 Fe₂O₃

Step 1: Convert both to moles

n(Fe) = 5.0 / 55.845 = 0.0895 mol

n(O₂) = 5.0 / 32.00 = 0.1563 mol

Step 2: Compare to required ratio (4 Fe : 3 O₂)

Fe needs: 0.0895 × (3/4) = 0.0671 mol O₂ ← have 0.1563 ✓

O₂ needs: 0.1563 × (4/3) = 0.2084 mol Fe ← have only 0.0895 ✗

Fe is the limiting reagent

Step 3: Calculate product from limiting reagent

n(Fe₂O₃) = 0.0895 × (2/4) = 0.04475 mol

m(Fe₂O₃) = 0.04475 × 159.69 = 7.15 g

7.15 g of Fe₂O₃ is produced

Answer: Fe is limiting. Maximum Fe₂O₃ = 7.15 g. Excess O₂ remaining: 0.1563 − 0.0671 = 0.0892 mol = 2.85 g unused.

Worked Example 3

Percent Yield

Problem: A student runs the Fe₂O₃ reaction above (theoretical yield = 7.15 g) but only obtains 5.60 g. What is the percent yield?

Step 1: Apply the formula

% Yield = (Actual / Theoretical) × 100

% Yield = (5.60 / 7.15) × 100

% Yield = 78.3%

Answer: 78.3% yield. The remaining 21.7% was lost to side reactions, incomplete reaction, or transfer losses - common in real labs.

Solutions & Concentration - Molarity, Dilution & Solubility

How to express, calculate, and manipulate the concentration of solutions.

1 Types of Concentration

Molarity (M) = n / V

mol/L - most common

Molality (m) = n / kg

mol/kg solvent

Mass % = (m_solute/m_total)×100

g/g percentage

ppm = (mg solute / L solution)

parts per million

Key difference: Molarity uses volume of solution (changes with temperature). Molality uses mass of solvent (doesn't change with temperature). Molality is preferred for colligative property calculations.

2 Molarity - The Most Used Unit

M = n / V = (mass / molar mass) / V

M = molarity (mol/L)n = moles of soluteV = volume of solution in liters

Worked Example 1

Preparing a Solution of Known Molarity

Problem: How many grams of NaOH (M = 40.00 g/mol) are needed to prepare 500 mL of a 0.25 M solution?

Step 1: Find moles needed

n = M × V = 0.25 × 0.500 = 0.125 mol

Step 2: Convert to grams

m = n × molar mass = 0.125 × 40.00

m = 5.00 g NaOH

Answer: Dissolve 5.00 g NaOH in water and dilute to a total volume of 500 mL. Note: always add solute to water, not the reverse (especially for acids).

3 Dilution - M₁V₁ = M₂V₂

When you add water to a solution, the number of moles stays the same - only the volume increases. Since n = MV:

M₁V₁ = M₂V₂

M₁, V₁ = initial (concentrated) molarity and volumeM₂, V₂ = final (diluted) molarity and volume

This equation works for any concentration unit (%, ppm, etc.) as long as you use the same unit on both sides. It only works for dilution - not for mixing two different solutions.

Worked Example 2

Dilution Calculation

Problem: How would you prepare 250 mL of 0.10 M HCl from a 12.0 M stock solution?

Step 1: Apply M₁V₁ = M₂V₂

12.0 × V₁ = 0.10 × 250

V₁ = 25 / 12.0

V₁ = 2.08 mL of stock HCl

Step 2: Procedure

Measure 2.08 mL of 12.0 M HCl and carefully add it to approximately 200 mL of distilled water. Then add water to reach exactly 250 mL total volume.

Answer: Use 2.08 mL of 12.0 M HCl diluted to 250 mL. Safety: always add acid to water ("do as you oughta - add acid to water"), never the reverse.

4 Solubility

Solubility is the maximum amount of solute that dissolves in a given amount of solvent at a specific temperature. Solutions can be unsaturated (can dissolve more), saturated (at limit), or supersaturated (beyond limit - unstable).

Temperature Effect - Solids

Most solid solutes become more soluble as temperature increases (e.g., sugar dissolves faster in hot water).

Temperature Effect - Gases

Gases become less soluble as temperature increases (e.g., warm soda goes flat faster).

Worked Example 3

Concentration from Mass and Volume

Problem: 11.7 g of NaCl (M = 58.44 g/mol) is dissolved to make 400 mL of solution. Find (a) the molarity and (b) the concentration in ppm.

Step 1: Moles of NaCl

n = 11.7 / 58.44 = 0.2002 mol

Step 2: Molarity

M = n / V = 0.2002 / 0.400

M = 0.500 M

Step 3: Parts per million (assuming density ≈ 1 g/mL)

ppm = (11.7 g / 400 mL) × 1000 mg/g × (1000 mL/L) / (1000 mg/L per ppm)

ppm = (11,700 mg) / (0.400 L)

ppm = 29,250 ppm

Answer: (a) 0.500 M. (b) 29,250 ppm. For dilute aqueous solutions, 1 ppm ≈ 1 mg/L.

Acids, Bases & pH - Calculations, Titrations & Buffers

Understanding the proton-transfer chemistry that governs biology, industry, and the environment.

1 pH and pOH

The pH scale measures how acidic or basic a solution is. It runs from 0 (very acidic) to 14 (very basic), with 7 being neutral.

pH = −log[H⁺] | pOH = −log[OH⁻]

pH + pOH = 14.00 (at 25°C)[H⁺] × [OH⁻] = 1.0 × 10⁻¹⁴ (Kw)

0 - Strong Acid 7 - Neutral 14 - Strong Base

Each pH unit = 10× change in [H⁺]. So pH 3 is 10 times more acidic than pH 4, and 100 times more acidic than pH 5.

2 Strong vs. Weak Acids & Bases

Strong (100% dissociation)

Acids: HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄

Bases: NaOH, KOH, Ca(OH)₂, Ba(OH)₂

HCl → H⁺ + Cl⁻ (complete)

Weak (partial dissociation)

Acids: CH₃COOH, HF, H₂CO₃, H₃PO₄

Bases: NH₃, CH₃NH₂, pyridine

CH₃COOH ⇌ H⁺ + CH₃COO⁻

Weak Acid: Ka = [H⁺][A⁻] / [HA]

Ka = acid dissociation constantLarger Ka → stronger weak acid

Worked Example 1

pH of a Strong Acid

Problem: Find the pH and pOH of a 0.0050 M HCl solution.

Step 1: HCl is a strong acid (100% dissociation)

[H⁺] = 0.0050 M = 5.0 × 10⁻³ M

Step 2: Calculate pH

pH = −log(5.0 × 10⁻³) = −(−2.30)

pH = 2.30

Step 3: Calculate pOH

pOH = 14.00 − 2.30

pOH = 11.70

Answer: pH = 2.30, pOH = 11.70. The solution is acidic (pH < 7).

Worked Example 2

pH of a Weak Acid (ICE Table)

Problem: Find the pH of 0.10 M acetic acid (CH₃COOH). Ka = 1.8 × 10⁻⁵.

Step 1: Set up ICE table

CH₃COOH ⇌ H⁺ + CH₃COO⁻

I: 0.10 0 0

C: −x +x +x

E: 0.10−x x x

Step 2: Write Ka expression

Ka = x² / (0.10 − x) = 1.8 × 10⁻⁵

Since Ka is small, assume x ≪ 0.10:

x² ≈ 1.8 × 10⁻⁵ × 0.10 = 1.8 × 10⁻⁶

x = [H⁺] = 1.34 × 10⁻³ M

Step 3: pH

pH = −log(1.34 × 10⁻³)

pH = 2.87

Step 4: Verify assumption

x / 0.10 = 1.34% < 5% ✓ (assumption valid)

Answer: pH = 2.87. Only 1.34% of acetic acid dissociates - that's why vinegar is a weak acid. Compare to HCl at the same concentration which would give pH = 1.00.

3 Titration & Equivalence Point

A titration determines unknown concentration by reacting with a known solution. At the equivalence point, moles of acid = moles of base.

At equivalence: n(acid) = n(base)

M_aV_a = M_bV_b (for monoprotic acids/bases)

Worked Example 3

Acid-Base Titration

Problem: 25.0 mL of unknown HCl is titrated with 0.150 M NaOH. The equivalence point is reached at 18.5 mL NaOH. Find the HCl concentration.

Step 1: Moles of NaOH at equivalence

n(NaOH) = 0.150 × 0.0185 = 0.002775 mol

Step 2: HCl + NaOH → NaCl + H₂O (1:1 ratio)

n(HCl) = n(NaOH) = 0.002775 mol

Step 3: Find concentration

M(HCl) = 0.002775 / 0.0250

M(HCl) = 0.111 M

Answer: The HCl concentration is 0.111 M. For polyprotic acids (like H₂SO₄), you'd need to account for the number of protons donated.

Gas Laws - Ideal Gas Law, Boyle's, Charles's & Dalton's

Predicting the behavior of gases using simple relationships between pressure, volume, temperature, and amount.

1 The Individual Gas Laws

Boyle's: P₁V₁ = P₂V₂

Constant T, n

Charles's: V₁/T₁ = V₂/T₂

Constant P, n

Gay-Lussac's: P₁/T₁ = P₂/T₂

Constant V, n

Avogadro's: V₁/n₁ = V₂/n₂

Constant T, P

Combined: P₁V₁/T₁ = P₂V₂/T₂

Constant n

Always use Kelvin! T(K) = T(°C) + 273.15. Gas laws break down with Celsius because zero Celsius is not zero energy.

2 The Ideal Gas Law

The ideal gas law combines all the individual laws into one equation. It works well for most gases at moderate temperatures and pressures.

PV = nRT

P = pressure (atm), V = volume (L)n = moles, R = 0.0821 L⋅atm/(mol⋅K)T = temperature (K)

Figure 2: All individual gas laws derive from PV = nRT by holding different variables constant.

Worked Example 1

Ideal Gas Law - Finding Volume

Problem: What volume does 0.500 mol of N₂ occupy at 25°C and 1.20 atm?

Step 1: Convert temperature

T = 25 + 273.15 = 298.15 K

Step 2: Solve PV = nRT for V

V = nRT / P = (0.500 × 0.0821 × 298.15) / 1.20

V = 10.2 L

Answer: 10.2 L. At STP (0°C, 1 atm), 1 mol of any ideal gas occupies 22.4 L.

3 Dalton's Law of Partial Pressures

In a mixture of gases, each gas exerts its own pressure independently. The total pressure is the sum of all partial pressures.

P_total = P₁ + P₂ + P₃ + ...

Partial pressure: P_i = χ_i × P_totalχ_i = mole fraction = n_i / n_total

Worked Example 2

Boyle's Law - Pressure Change

Problem: A gas occupies 3.5 L at 1.0 atm. What pressure is needed to compress it to 1.2 L at constant temperature?

Step 1: Apply Boyle's Law

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂ = (1.0 × 3.5) / 1.2

P₂ = 2.92 atm

Answer: 2.92 atm. Reducing volume by ≈ 3× requires ≈ 3× the pressure - this is the inverse relationship of Boyle's Law.

Worked Example 3

Gas Stoichiometry - Volume at STP

Problem: How many liters of CO₂ at STP are produced by decomposing 25.0 g of CaCO₃?
CaCO₃ → CaO + CO₂

Step 1: Moles of CaCO₃

M(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

n = 25.0 / 100.09 = 0.2498 mol

Step 2: Mole ratio (1:1)

n(CO₂) = 0.2498 mol

Step 3: Volume at STP (1 mol = 22.4 L)

V = 0.2498 × 22.4

V = 5.60 L of CO₂

Answer: 5.60 L of CO₂ at STP. Gas stoichiometry uses the 22.4 L/mol shortcut at STP, or PV = nRT at other conditions.

Thermochemistry - Enthalpy, Hess's Law & Calorimetry

The study of heat changes in chemical reactions - why some reactions feel hot and others feel cold.

1 Enthalpy (ΔH) - Heat of Reaction

Enthalpy change (ΔH) is the heat absorbed or released during a reaction at constant pressure. It's the most practical measure of reaction energy.

Exothermic (ΔH < 0)

Releases heat to surroundings. Products have lower energy than reactants. Feels warm.

CH₄ + 2O₂ → CO₂ + 2H₂O

ΔH = −890 kJ/mol

Endothermic (ΔH > 0)

Absorbs heat from surroundings. Products have higher energy than reactants. Feels cold.

NH₄NO₃ → NH₄⁺ + NO₃⁻

ΔH = +25.7 kJ/mol

Sign convention: Negative ΔH = exothermic (system loses heat). Positive ΔH = endothermic (system gains heat). Think: the sign tells you what happens to the system, not the surroundings.

2 Calorimetry - Measuring Heat

A calorimeter measures heat changes. The heat absorbed by water equals the heat released by the reaction (and vice versa).

q = mcΔT

q = heat (J), m = mass (g)c = specific heat capacity (J/g⋅°C)ΔT = temperature change (°C or K)c(water) = 4.184 J/g⋅°C

q_reaction = −q_calorimeter

Heat lost by reaction = heat gained by water (and vice versa)

Worked Example 1

Coffee Cup Calorimetry

Problem: When 50.0 mL of 1.0 M HCl and 50.0 mL of 1.0 M NaOH (both at 25.0°C) are mixed in a calorimeter, the temperature rises to 31.9°C. Find ΔH for the neutralization reaction (assume d = 1.00 g/mL, c = 4.184 J/g⋅°C).

Step 1: Heat absorbed by solution

m = 50.0 + 50.0 = 100.0 g (total solution)

ΔT = 31.9 − 25.0 = 6.9°C

q = mcΔT = 100.0 × 4.184 × 6.9

q = 2887 J = 2.887 kJ

Step 2: Moles of reaction

n = M × V = 1.0 × 0.050 = 0.050 mol

Step 3: ΔH per mole

ΔH = −q / n = −2.887 / 0.050

ΔH = −57.7 kJ/mol

Answer: ΔH = −57.7 kJ/mol (exothermic). The accepted value for strong acid-strong base neutralization is −57.1 kJ/mol. Negative sign because temperature rose (heat was released).

3 Hess's Law

The total enthalpy change for a reaction is the same regardless of the pathway taken. This means you can add, subtract, or reverse known reactions to find unknown ΔH values.

ΔH_rxn = ΣΔH_f°(products) − ΣΔH_f°(reactants)

ΔH_f° = standard enthalpy of formationΔH_f° of elements in standard state = 0

Rules for Manipulating Equations

1If you reverse a reaction, change the sign of ΔH.

2If you multiply a reaction by a factor, multiply ΔH by the same factor.

3Add manipulated equations so intermediates cancel. Sum the ΔH values.

Worked Example 2

Hess's Law - Formation Enthalpies

Problem: Find ΔH° for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).
Given: ΔH_f°[CO₂(g)] = −393.5 kJ/mol, ΔH_f°[H₂O(l)] = −285.8 kJ/mol, ΔH_f°[CH₄(g)] = −74.8 kJ/mol.

Step 1: Apply Hess's Law formula

ΔH° = ΣΔH_f°(products) − ΣΔH_f°(reactants)

Step 2: Sum products

Products = [1(−393.5) + 2(−285.8)]

= −393.5 + (−571.6) = −965.1 kJ

Step 3: Sum reactants

Reactants = [1(−74.8) + 2(0)]

= −74.8 kJ (O₂ is element in std state → ΔH_f° = 0)

Step 4: Calculate

ΔH° = −965.1 − (−74.8)

ΔH° = −890.3 kJ/mol

Answer: ΔH° = −890.3 kJ/mol. Burning 1 mole of methane (16 g, the main component of natural gas) releases 890.3 kJ - enough to heat about 2 liters of water from 25°C to boiling.

Worked Example 3

Bond Energy Calculation

Problem: Estimate ΔH for H₂ + Cl₂ → 2HCl using bond energies:
H−H = 436 kJ/mol, Cl−Cl = 242 kJ/mol, H−Cl = 431 kJ/mol.

Step 1: Energy to break bonds (endothermic, +)

Break: 1(H−H) + 1(Cl−Cl) = 436 + 242 = +678 kJ

Step 2: Energy to form bonds (exothermic, −)

Form: 2(H−Cl) = 2 × 431 = −862 kJ

Step 3: ΔH = bonds broken − bonds formed

ΔH = +678 + (−862)

ΔH = −184 kJ

Answer: ΔH ≈ −184 kJ. Exothermic because more energy is released forming H−Cl bonds than is needed to break H−H and Cl−Cl bonds. Bond energy estimates are approximate.

Atomic Structure & Periodicity - Electron Configuration & Trends

How electrons are arranged in atoms - and why this arrangement determines all chemical behavior.

1 Quantum Numbers & Orbitals

Each electron is described by four quantum numbers that specify its energy level, shape, orientation, and spin.

n (principal)

Energy level: 1, 2, 3...

ℓ (angular momentum)

Shape: 0=s, 1=p, 2=d, 3=f

m_ℓ (magnetic)

Orientation: −ℓ to +ℓ

m_s (spin)

+½ or −½

Pauli Exclusion Principle: No two electrons can share all four quantum numbers. Each orbital holds max 2 electrons with opposite spins.

2 Electron Configuration

Electrons fill orbitals in order of increasing energy following the Aufbau principle:

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d ...

s: max 2e⁻ | p: max 6e⁻ | d: max 10e⁻ | f: max 14e⁻

Hund's Rule: Electrons fill orbitals of the same energy singly first (all with same spin) before pairing up. This minimizes repulsion.

Worked Example 1

Writing Electron Configurations

Problem: Write the full and noble gas shorthand electron configuration for Iron (Fe, Z = 26) and its Fe²⁺ ion.

Step 1: Fe (26 electrons) - fill in order

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

Check: 2+2+6+2+6+2+6 = 26 ✓

Noble gas shorthand: [Ar] 4s² 3d⁶

Step 2: Fe²⁺ (remove 2 electrons)

Important: electrons are removed from the highest n first (4s before 3d)!

Fe²⁺: [Ar] 3d⁶ (remove the two 4s electrons)

Answer: Fe: [Ar] 4s² 3d⁶. Fe²⁺: [Ar] 3d⁶. Common mistake: students remove 3d electrons instead of 4s. Always remove from highest principal quantum number first when forming cations.

3 Periodic Trends

The periodic table is organized by atomic number, and properties follow predictable trends based on electron configuration.

Atomic Radius

↓ increases down, ← increases left

Ionization Energy

↑ increases up, → increases right

Electronegativity

↑ increases up, → increases right

Electron Affinity

Generally increases toward upper right

Why trends go across →

More protons = stronger nuclear pull on electrons = smaller radius, harder to remove electrons, higher electronegativity.

Why trends go down ↓

More electron shells = greater distance from nucleus = larger radius, easier to remove electrons, lower electronegativity.

Worked Example 2

Ranking Periodic Trends

Problem: Rank the following in order of increasing atomic radius: Na, Mg, K, Ca.

Step 1: Identify positions

Na (Period 3, Group 1), Mg (Period 3, Group 2), K (Period 4, Group 1), Ca (Period 4, Group 2)

Step 2: Apply trends

Across a period (left → right): radius decreases. So Na > Mg and K > Ca.

Down a group (top → bottom): radius increases. So K > Na and Ca > Mg.

Smallest → Largest: Mg < Na < Ca < K

Answer: Mg < Na < Ca < K. Potassium is largest because it's in Period 4 and Group 1 (fewest protons in that period). Magnesium is smallest because it's in Period 3 with more nuclear charge than Na.

Worked Example 3

Ionization Energy Comparison

Problem: Which has the higher first ionization energy: (a) Li or Na? (b) N or O? Explain.

(a) Li vs Na

Li and Na are in the same group (1). Going down the group, ionization energy decreases because the outer electron is farther from the nucleus.

Li has higher IE than Na

(b) N vs O

N has a half-filled 2p³ subshell (extra stability). O (2p⁴) has one paired electron, making it slightly easier to remove. This is an exception to the left-right trend.

N has slightly higher IE than O

IE: N = 1402 kJ/mol, O = 1314 kJ/mol

Answer: (a) Li > Na (same group, IE decreases going down). (b) N > O (half-filled subshell stability exception). These exceptions to the general trend are common exam questions.

Comprehensive Chemistry Calculators - Solve Any Chemistry Problem

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How it works

Common Elements

Important Constants

Chemistry Topics - In-Depth Guides with Worked Examples

Stoichiometry - The Mole, Molar Mass & Reaction Calculations

1 The Mole Concept

2 Molar Mass - The Bridge Between Grams and Moles

3 Balancing Chemical Equations

Mass-to-Mass Stoichiometry

4 Limiting Reagent & Percent Yield

Limiting Reagent Problem

Percent Yield

Solutions & Concentration - Molarity, Dilution & Solubility

1 Types of Concentration

2 Molarity - The Most Used Unit

Preparing a Solution of Known Molarity

3 Dilution - M₁V₁ = M₂V₂

Dilution Calculation

4 Solubility

Concentration from Mass and Volume

Acids, Bases & pH - Calculations, Titrations & Buffers

1 pH and pOH

2 Strong vs. Weak Acids & Bases

pH of a Strong Acid

pH of a Weak Acid (ICE Table)

3 Titration & Equivalence Point

Acid-Base Titration

Gas Laws - Ideal Gas Law, Boyle's, Charles's & Dalton's

1 The Individual Gas Laws

2 The Ideal Gas Law

Ideal Gas Law - Finding Volume

3 Dalton's Law of Partial Pressures

Boyle's Law - Pressure Change

Gas Stoichiometry - Volume at STP

Thermochemistry - Enthalpy, Hess's Law & Calorimetry

1 Enthalpy (ΔH) - Heat of Reaction

2 Calorimetry - Measuring Heat

Coffee Cup Calorimetry

3 Hess's Law

Hess's Law - Formation Enthalpies

Bond Energy Calculation

Atomic Structure & Periodicity - Electron Configuration & Trends

1 Quantum Numbers & Orbitals

2 Electron Configuration

Writing Electron Configurations

3 Periodic Trends

Ranking Periodic Trends

Ionization Energy Comparison