Introduction
This interactive chemistry portfolio turns your classroom notes into a live learning tool. Each section breaks down key chemistry concepts and includes interactive calculators to help you check your practice work in real-time.
Nuclear Structure & Isotopes
Core Insights
The atom's nucleus is composed of positive protons (p+) and neutral neutrons (n0). Protons define the atomic number and give an element its unique identity.
Protons naturally repel each other because they have the same positive charge (+/+). They are held tightly together in the nucleus by a very strong, short-range force called the Strong Nuclear Force.
Isotopes are versions of the same element that have different numbers of neutrons. This changes the atom's overall weight (atomic mass) but does not change what element it is.
Interactive Isotope Builder
Problem Set & Notebook Verification
Example Set (Standard Solutions)
Example 1.1: Neutrons of Lithium-7
Calculate the exact number of neutrons in a Lithium-7 atom.
Lithium's atomic number (Z) is 3. The mass number (A) is 7.
Neutrons = Mass Number - Protons = 7 - 3 = 4 neutrons
Example 1.2: Adding a Neutron
If you add 1 neutron to Carbon-12 without changing protons, what changes?
Since the protons do not change, it is still Carbon. However, its mass increases to 13, creating the isotope: Carbon-13.
Example 1.3: Weighted Average Mass
Calculate Chlorine's atomic mass from Cl-35 (34.97 amu, 75.78%) and Cl-37 (36.97 amu, 24.22%).
Average = (0.7578 × 34.97) + (0.2422 × 36.97) = 35.45 amu
Example 1.4: Electrostatic vs Strong Force
Why doesn't the positive charge push the nucleus apart?
Even though positive protons repel each other, the much stronger, short-range Strong Nuclear Force overcomes this and holds the nucleus together.
Example 1.5: Isotopic Notation
Write the complete isotopic notation for Uranium-235.
Protons (Z) = 92, Mass (A) = 235. The chemical symbol is: 23592U.
Practice Set (Student Solutions)
Practice 1.1: Phosphorus-32 Components
Determine the proton and neutron count inside Phosphorus-32.
Phosphorus has an atomic number of 15. Neutrons = 32 - 15 = 17 neutrons.
Practice 1.2: Removing a Proton
If you remove 1 proton from a Beryllium (Z=4) atom, what element do you get?
Removing 1 proton leaves 3 protons, changing the element into Lithium (Li).
Practice 1.3: Neon's Weighted Average
Calculate Neon's average mass using: Neon-20 (19.99 amu, 90.48%) and Neon-22 (21.99 amu, 9.52%).
(0.9048 × 19.99) + (0.0952 × 21.99) = 20.18 amu.
Practice 1.4: Helium-4 Nucleus Charge
Find the overall charge of a Helium-4 nucleus.
It contains 2 protons (+2) and 2 neutrons (0). This results in a net charge of +2.
Practice 1.5: Decimals on Periodic Table
Explain why atomic masses on the periodic table are decimals instead of whole numbers.
Because they are weighted averages calculated from how common each natural isotope is in the world!
Need a quick video refresh? Tyler DeWitt's Introduction to Isotopes is highly recommended.
Open Explainer VideoElectronic Structure & Bohr Models
Core Insights
Electrons occupy specific energy regions outside the nucleus called energy levels. Opposite charges attract (+/-), which keeps the electrons from flying away, while like charges repel (-/-), keeping them spread out.
The maximum number of electrons that can fit in each energy level is strictly limited: 2 e- in Level 1, 8 e- in Level 2, and 8 e- in Level 3.
Electrons in the outermost energy level are called Valence Electrons. These electrons determine how an atom bonds and reacts with other atoms.
Dynamic Bohr Orbit Painter
Problem Set & Notebook Verification
Example Set (Standard Solutions)
Example 2.1: Beryllium Shells
Determine the electron layout and valence electron count of Beryllium (Z=4).
Distribution: Level 1 has 2 e-, Level 2 has 2 e-. Total = 4, Valence = 2 e-
Example 2.2: Level 1 Limits
Why can't Beryllium place all 4 of its electrons in the first energy level?
The first energy level is too small and is physically limited to holding a maximum of 2 electrons.
Example 2.3: Carbon Valence
Find the number of valence electrons in a Carbon atom (Z=6).
Carbon has Z=6, so it places 2 electrons in its inner shell, leaving 4 in the outer shell. Its valence electron count is 4!
Example 2.4: Chlorine Configuration
Evaluate the electron shell setup for Chlorine (Z=17).
Shells: 2, 8, 7. It has 7 valence e- in its outer level.
Example 2.5: Octet Rule Stability
Why are Noble Gases like Neon non-reactive?
Neon has a complete set of 8 valence electrons in its outer shell. This full shell makes the atom incredibly stable and non-reactive.
Practice Set (Student Solutions)
Practice 2.1: Oxygen Configuration
List the energy level layout for Oxygen (Z=8).
Level 1 has 2 e-, Level 2 has 6 e-.
Practice 2.2: Sodium Ion Configuration
Find the total number of electrons in a Sodium ion (Na+).
A neutral Sodium atom (11 e-) loses 1 valence electron to become a positive ion: 10 electrons left.
Practice 2.3: Argon Complete Shells
Determine how many full electron shells Argon (Z=18) has.
Argon's configuration is 2, 8, 8, which means it has 3 fully filled shells.
Practice 2.4: Shared Family Properties
Explain why elements in Group 17 (Halogens) behave so similarly.
They all have the same number of valence electrons (7), which means they react in the exact same way to get a full outer shell.
Practice 2.5: Visualizing Helium-4
Describe the Bohr model layout of a Helium-4 atom.
It has a central nucleus with 2 protons and 2 neutrons, orbited by exactly 2 electrons within a single, fully filled outer energy level!
Ready to build Lewis dots and shell structures? Watch the Bohr Model Masterclass on YouTube.
Open Explainer VideoIonic Bonding & Percent Composition
Core Insights
An Ionic Bond forms when electrons are completely transferred from a metal atom to a non-metal atom.
Metals lose electrons and become positive ions called Cations. Non-metals gain those electrons and become negative ions called Anions.
To write the formula for an ionic compound, we use the criss-cross method. This balances the positive and negative charges so the overall compound has a net charge of zero.
Criss-Cross Formula Constructor
Barium Chloride
1 Ba²⁺ : 2 Cl⁻
Problem Set & Notebook Verification
Example Set (Standard Solutions)
Example 3.1: Barium & Chloride Synthesis
Criss-cross Barium (Ba²⁺) and Chloride (Cl⁻) to write a balanced formula.
Criss-crossing charges gives 1 Barium and 2 Chlorides to balance the charges to zero: BaCl2.
Example 3.2: Mass Percentage of Water (Accurate)
Calculate the exact mass percent of Hydrogen in H₂O.
Total H mass = 2.016g. Water total mass = 18.016g.
% H = (2.016 / 18.016) × 100% = 11.19% H.
Example 3.3: Aluminum Oxide Criss-Cross
Pair Aluminum (Al³⁺) and Oxide (O²⁻) to find the formula.
Criss-crossing the charge numbers as subscripts yields: Al2O3.
Example 3.4: Barium Phosphate Synthesis
Create the formula for Barium Phosphate using polyatomic groups.
Ba²⁺ and PO₄³⁻ criss-cross into Ba₃(PO₄)₂. Remember, parentheses are required around polyatomic ions when you have more than one of them!
Example 3.5: Mass Composition of NaCl
Determine the mass percentage of Sodium in Sodium Chloride (NaCl).
Na = 22.99g. Total NaCl mass = 58.44g.
% Na = (22.99 / 58.44) × 100% = 39.34% Na.
Practice Set (Student Solutions)
Practice 3.1: Magnesium Nitrate Formulas
Find the formula when pairing Magnesium (Mg²⁺) and Nitrate (NO₃⁻).
To balance Mg²⁺, we need two single negative nitrate ions: Mg(NO3)2.
Practice 3.2: Chlorine Fraction in Barium Chloride
Find the mass percent of Chlorine in BaCl₂.
Mass of two Cl = 70.90g. BaCl₂ total mass = 208.23g.
% Cl = (70.90 / 208.23) × 100% = 34.05% Cl.
Practice 3.3: Iron(III) Oxide Synthesis
Combine Iron(III) (Fe³⁺) and Oxide (O²⁻) into a compound.
Criss-crossing their charges gives us: Fe2O3.
Practice 3.4: Charge Neutrality Rationale
Explain why ionic compounds must have an overall charge of zero.
Because atoms naturally seek the most stable, lowest-energy state. Unbalanced positive or negative charges would pull the crystal lattice apart!
Practice 3.5: Calcium Carbonate Fraction
Calculate the mass percent of Carbon in chalk (CaCO₃).
C = 12.01g. CaCO₃ total mass = 100.09g.
% C = (12.01 / 100.09) × 100% = 12.00% C.
Need to review Ionic formulas? Watch Tyler DeWitt's guide on Writing Ionic Formulas with Transition Metals.
Open Explainer VideoCovalent Bonding & the Mole
Core Insights
Covalent Bonds form when non-metal atoms share electron pairs to help each other fill their outer energy levels. We draw Lewis structures to show shared bonds as lines and unshared electrons as lone pairs of dots.
In chemistry, we use a counting unit called the mole to measure atoms by weighing them. 1 mole = 6.02 × 1023 particles (Avogadro's Number).
The "Mole Highway" is a method that allows us to convert back and forth between a measurable mass in grams and the actual number of atoms.
Modern Mole Highway Engine
1.00 Mol
6.02 × 10²³
Problem Set & Notebook Verification
Example Set (Standard Solutions)
Example 4.1: Diatomic Oxygen
Draw the Lewis structure for Oxygen Gas (O₂).
Two oxygen atoms share 4 electrons (a double bond) to complete their octets, leaving 2 lone pairs on each atom: :O = O:.
Example 4.2: Sulfate Ion Lewis Setup
Draw the Lewis structure for the Sulfate ion (SO₄²⁻).
Sulfur sits at the center, single-bonded to 4 Oxygen atoms, surrounded by brackets to show its overall negative charge: [SO4]2-.
Example 4.3: Empirical Formula from % Composition
Find the empirical formula of a compound that is 27.29% Carbon and 72.71% Oxygen.
Carbon: 2.27 moles, Oxygen: 4.54 moles. Dividing both by the smallest gives a 1:2 ratio: CO2.
Example 4.4: Atoms to Mass Conversion
Calculate how many carbon atoms are in 24g of pure Carbon-12.
24g / 12g/mol = 2 moles. 2 × 6.02 × 1023 = 1.20 × 1024 atoms.
Example 4.5: Molar Mass of Water
Determine the molar mass of water (H₂O).
2 × 1.01g (Hydrogen) + 16.00g (Oxygen) = 18.02 g/mol. This is the weight of exactly 1 mole of water!
Practice Set (Student Solutions)
Practice 4.1: Carbon Dioxide Lewis Setup
State the bonding types inside Carbon Dioxide (CO₂).
Carbon shares 4 electrons with each Oxygen atom, forming two double bonds: :O = C = O:.
Practice 4.2: Empirical to Molecular Formulas
Find the molecular formula of CH₂ with a molar mass of 42.08 g/mol.
CH₂ weight = 14.03 g/mol. Ratio = 42.08 / 14.03 = 3. Multiply subscripts by 3: C3H6.
Practice 4.3: Gas Mole Assessment
Calculate the total number of moles in 44g of Carbon Dioxide (CO₂).
CO₂ molar mass = 44.01 g/mol. 44g / 44.01g/mol = 1.00 mole.
Practice 4.4: Atoms to Mass Operations
Find the mass in grams of 3.01 × 1023 Carbon atoms.
Since 3.01 × 10²³ is exactly half of a mole, its mass is half of the molar mass: 0.5 × 12.01g = 6.00 grams of Carbon!
Practice 4.5: Ammonia Lewis Setup
Describe the Lewis structure of Ammonia (NH₃).
Nitrogen is the central atom with one lone pair. It forms three single covalent bonds with three hydrogen atoms.
Need a quick review on drawing Lewis structures? Khan Academy has a fantastic tutorial.
Open Explainer VideoReactions & the Activity Series
Core Insights
Metals react differently depending on their active potential. The Activity Series is a list that ranks metals from most reactive to least reactive.
In a Single Replacement Reaction, an active, uncombined metal can "kick out" and replace a weaker metal that is locked inside an ionic compound.
If the incoming free metal is weaker (less reactive) than the metal already in the compound, no chemical reaction occurs, and we write NR (No Reaction).
Single Replacement Simulator
Checking charge statuses...
Comparing positions on the Activity Series...
Li + CuSO₄ → Li₂SO₄ + Cu
Problem Set & Notebook Verification
Example Set (Standard Solutions)
Example 5.1: Zinc & Copper Sulfate
Solve and balance: Zn(s) + CuSO₄(aq) →
Zinc is higher than Copper on the Activity Series. It replaces Copper to form Zinc Sulfate and solid Copper metal: Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s).
Example 5.2: Copper & Hydrochloric Acid
Solve and balance: Cu(s) + HCl(aq) →
Copper is lower than Hydrogen on the Activity Series, so it cannot displace the hydrogen ions: No Reaction (NR).
Example 5.3: Lithium & Sodium Nitrate
Predict the products: Li(s) + NaNO₃(aq) →
Lithium is extremely active and easily displaces Sodium: Li(s) + NaNO3(aq) → LiNO3(aq) + Na(s).
Example 5.4: State Symbols Rationale
What do (s) and (aq) state symbols mean in single replacement reactions?
They describe what physical state the chemicals are in: solid metals (s) versus dissolved ions floating around in water (aq)!
Example 5.5: Aluminum in Zinc Nitrate
Predict products and balance: Al(s) + Zn(NO₃)₂(aq) →
Aluminum is more active than Zinc. Criss-crossing Al³⁺ and NO₃⁻ gives: 2Al(s) + 3Zn(NO3)2(aq) → 2Al(NO3)3(aq) + 3Zn(s).
Practice Set (Student Solutions)
Practice 5.1: Silver in Sodium Chloride
Determine the outcome: Ag(s) + NaCl(aq) →
Silver is near the bottom of the series and cannot displace Sodium: No Reaction (NR).
Practice 5.2: Magnesium in Hydrochloric Acid
Solve and balance: Mg(s) + HCl(aq) →
Magnesium is more active than Hydrogen, forming Hydrogen gas bubbles: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g).
Practice 5.3: Gold in Lead Nitrate
Predict: Au(s) + Pb(NO₃)₂(aq) →
Gold is highly non-reactive and cannot displace Lead: No Reaction (NR).
Practice 5.4: Displacement Observation
What do you physically see when Zinc is placed in a blue Copper Sulfate solution?
The bright blue solution slowly fades, the shiny zinc dissolves, and a dark, reddish-brown solid copper coats the metal!
Practice 5.5: Iron in Copper Sulfate
Predict products and balance: Fe(s) + CuSO₄(aq) →
Iron is more active than Copper, yielding Copper metal: Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s).
Want to visualize how the activity series works? Watch this Single Replacement Reactions guide.
Open Explainer VideoStoichiometry: Chemical Recipes
Core Insights
A balanced chemical equation is like a cooking recipe. It uses front coefficients to show the exact mole ratio of ingredients we need to make a product.
We must remember the diatomic elements: BRINClHOF (Br₂, I₂, N₂, Cl₂, H₂, O₂, F₂). These elements are so reactive that they always exist in pairs when they are pure.
To convert mass from one chemical to another, we follow the Mole Highway: Grams A → Moles A → Mole Ratio B/A → Grams B.
Aluminum Synthesis Calculator
Equation: 4Al + 3O₂ → 2Al₂O₃. Enter a mass of Aluminum to calculate the theoretical yield of Aluminum Oxide.
Problem Set & Notebook Verification
Example Set (Standard Solutions)
Example 6.1: Mass of Al₂O₃ from 10g Al
Calculate the mass of Al₂O₃ produced from 10g of Aluminum: 4Al + 3O₂ → 2Al₂O₃.
Moles Al = 10g / 26.98g/mol = 0.37 mol.
Theoretical Yield = 0.37 mol × (2/4 ratio) × 101.96g/mol = 18.89g Al₂O₃.
Example 6.2: Synthesis vs Decomposition
Explain the difference between synthesis and decomposition reactions.
Synthesis combines simple reactants into one larger product: A + B → AB.
Decomposition breaks a larger compound down: AB → A + B.
Example 6.3: Diatomic Elements
Why is pure oxygen gas always written as O₂ instead of O?
Oxygen is a diatomic element (BRINClHOF). It is highly reactive alone, so it always shares electrons with another oxygen atom to stay stable!
Example 6.4: Limiting Reactant Calculation
If 5 moles of H₂ react with 3 moles of O₂ to make water, find the limiting reactant.
Equation: 2H₂ + O₂ → 2H₂O. 5 moles of H₂ require 2.5 moles of O₂. Since we have 3 moles of O₂ (an excess), Hydrogen (H₂) is the limiting reactant.
Example 6.5: Percent Yield
Calculate the percent yield if your actual lab yield is 15g and the theoretical yield is 18.89g.
% Yield = (Actual / Theoretical) × 100% = (15.00g / 18.89g) × 100% = 79.41% Yield.
Practice Set (Student Solutions)
Practice 6.1: Moles of Hydrogen Needed
How many moles of H₂ are needed to make 10 moles of water?
Equation: 2H₂ + O₂ → 2H₂O. Since the ratio is 1:1, you need exactly 10 moles of H₂.
Practice 6.2: Balancing Decomposition
Balance: KClO₃ → KCl + O₂
The balanced chemical equation is: 2KClO₃ → 2KCl + 3O₂.
Practice 6.3: Balancing Ammonia Synthesis
Balance the synthesis of ammonia from its pure diatomic gases.
Both starting elements are diatomic: N₂ + 3H₂ → 2NH₃!
Practice 6.4: Conservation of Mass
Explain why the total mass of reactants must equal the total mass of products.
Because atoms are never created or destroyed in a reaction; they are simply rearranged into new bonds.
Practice 6.5: Finding the Limiting Reactant
Find the limiting reactant for: 1 mol Fe + 1 mol O₂ → Fe₂O₃.
Balanced: 4Fe + 3O₂ → 2Fe₂O₃. 1 mol Fe requires only 0.75 mol O₂. Iron is used up first, so Fe is Limiting.
Want to master stoichiometry calculations? Watch the Stoichiometry Proportions Guide on YouTube.
Open Explainer VideoResources & Chemistry Reference Material
These external tools provide helpful chemistry charts, periodic databases, and reference guides to support your studies.