Second Semester · Interactive Edition

Chemistry,
distilled into five clear ideas.

Every concept, problem, and worked example from your study guide — rebuilt as an interactive laboratory you can poke, rotate, and learn from. Get stuck? Hints reveal themselves in tiers. Lost the spark? A memory trick is always one tap away.

Sections
05
Worked Problems
40+
Interactive Tools
12

The five chapters

Tap any card to begin
01

Chemical Bonding

How atoms hold hands — ionic transfers, covalent shares, polarity, and the shapes that follow.
Lewis Structures VSEPR Electronegativity
02

Nomenclature

The language of compounds. Decode formulas, name them, write them — until it feels obvious.
Ionic Names Acids Prefixes
03

Balancing & Predicting

Six reaction types, twenty-five equations, one trusted method for balancing anything.
Balancer Reaction Types Patterns
04

Mole Conversions

The chemist's universal translator: grams, moles, particles — all connected through one map.
Mole Map Avogadro Molar Mass
05

Stoichiometry

Recipes for reactions. Predict yields, find limiting reactants, derive formulas from percentages.
Limiting Reactant % Yield Empirical Formula
01
Chapter One

Atoms holding hands

Every bond is one of three stories: a transfer, an equal share, or an unequal share. Master those three and you can predict how almost any pair of atoms will behave.

A

First principles

The big ideas — every problem in this chapter traces back to one of these.
What is a chemical bond, and why do atoms form them?
A chemical bond is a force of attraction that holds atoms together. Atoms bond to become more stable by achieving a full valence shell — usually an octet of 8 valence electrons. Bonding always involves valence electrons, the outermost ones.
Ionic vs. covalent — what actually happens to the electrons?
In an ionic bond, electrons are transferred between a metal and a nonmetal. The resulting particle is called a formula unit.
In a covalent bond, electrons are shared between a nonmetal and a nonmetal. The resulting particle is called a molecule.
Memory Trick
"Metals are generous, nonmetals are needy." Metals are willing to give their electrons away (ionic), so they form formula units. Two nonmetals are both greedy, so they compromise and share (covalent).
Polar vs. nonpolar covalent — what's the difference?
Both involve sharing, but how fairly the electrons are shared decides which is which:
• Nonpolar covalent: electrons shared equally — usually identical atoms.
• Polar covalent: electrons shared unequally — one atom hogs them.
Unequal sharing is shown with partial charges (δ+ and δ−) or a dipole arrow pointing toward the hungrier atom.
The octet rule — and its rebels
Atoms gain, lose, or share electrons to obtain 8 valence electrons. Exceptions: H, He, Li, Be, B (too small — they're happy with 2 or fewer), plus expanded-octet elements in period 3 or higher (S, P, Xe — they can hold more than 8).
Lone pair vs. shared pair
A lone pair is a pair of valence electrons not involved in any bond. A shared pair sits between two atoms and counts toward both of their octets.
The diatomic elements
Seven elements never appear alone in nature — they always travel in pairs: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.
Memory Trick — "Have No Fear Of Ice Cold Beer"
Have (H₂) · No (N₂) · Fear (F₂) · Of (O₂) · Ice (I₂) · Cold (Cl₂) · Beer (Br₂). Or trace the "7" on the periodic table: starting at H, drop to 7, then trace the diatomics — they form a perfect 7.
Picking the central atom
When drawing Lewis structures, the central atom is usually the least electronegative atom in the molecule — but it's never hydrogen. Hydrogen can only form one bond, so it always lives on the outside.
B

The electronegativity gauge

Pick any two elements. The difference in their electronegativity tells you exactly what kind of bond they'll form.
vs.
Nonpolar covalent · 0.0 – 0.4
Polar covalent · 0.5 – 1.7
Ionic · 1.8+
Bond classification
Select two atoms
The Three Borders
Memorize these cutoffs and you can classify any bond instantly: 0.4 / 1.7. Below 0.5 → nonpolar. Up to 1.7 → polar. Above 1.7 → ionic. "Point five and one point seven."
C

Classify these four bonds

From your study guide. Try each one — click "show me" to reveal the reasoning.
Na — F
Ionic. Sodium (EN 0.93) is a metal; fluorine (EN 3.98) is the most electronegative nonmetal. Difference = 3.05, well above 1.7. Electrons are transferred from Na to F.
C — H
Nonpolar covalent. Carbon (EN 2.55) and hydrogen (EN 2.20). Difference = 0.35, under 0.4. They share electrons almost equally — this is why hydrocarbons aren't great at dissolving in water.
P — Br
Polar covalent. Phosphorus (EN 2.19) and bromine (EN 2.96). Difference = 0.77, in the polar range. Bromine hogs the electron density slightly — δ− on Br, δ+ on P.
H — H
Nonpolar covalent. Two identical atoms — difference = 0. The textbook case of perfect sharing. All diatomic elements have nonpolar covalent bonds.
D

VSEPR — the shape of every molecule

Valence Shell Electron Pair Repulsion. Electron groups push each other as far apart as possible. That's it. That single idea explains every molecular shape.
Linear
2 groups · 180°
CO₂
Trigonal planar
3 groups · 120°
BF₃
Tetrahedral
4 groups · 109.5°
CH₄
lp
Trigonal pyramidal
3 bonds + 1 lone pair
NH₃
Bent
2 bonds + 2 lone pairs
H₂O
Octahedral
6 groups · 90°
SCl₆
Memory Trick — The Balloon Analogy
Imagine tying balloons together at a central knot. Two balloons → straight line (linear). Three → flat triangle (trigonal planar). Four → 3D pyramid (tetrahedral). The balloons push each other apart — exactly what electron pairs do. Replace a balloon with an invisible "lone pair" balloon (it still takes up space!) and you get bent and trigonal pyramidal.
E

Lewis structures, step by step

The four examples from your study guide, fully worked. Click any to expand.
S Cl Cl Cl Cl Cl Cl
SCl₆ — Octahedral
6 bonds · 0 lone pairs · 90° angles
1
Count valence electrons
S has 6, each Cl has 7. Total: 6 + (6 × 7) = 48 valence electrons.
2
Place the central atom
Sulfur is less electronegative than chlorine — S goes in the middle.
3
Connect with single bonds
6 S–Cl bonds use 12 electrons. Each Cl gets 6 more electrons (3 lone pairs) to complete its octet.
4
Expanded octet on S
S is in period 3, so it can hold more than 8 electrons. Here it has 12 — six bonding pairs.
5
Apply VSEPR
6 electron groups → octahedral shape.
C H H H H
CH₄ — Tetrahedral
4 bonds · 0 lone pairs · 109.5° angles
1
Count valence electrons
C has 4, each H has 1. Total: 4 + 4 = 8 valence electrons.
2
Place the central atom
Carbon is the central atom (hydrogen can never be central — it forms only 1 bond).
3
Connect each H to C with a single bond
4 C–H bonds use all 8 electrons. Each H has 2 electrons (its duet — H's stable state). C has 8 (octet).
4
Apply VSEPR
4 electron groups, 0 lone pairs → tetrahedral.
•• •• O H H
H₂O — Bent
2 bonds · 2 lone pairs · ≈104.5°
1
Count valence electrons
O has 6, each H has 1. Total: 6 + 2 = 8 valence electrons.
2
Oxygen is central
H can never be central.
3
Bond each H to O
2 O–H bonds use 4 electrons. Place the remaining 4 electrons as 2 lone pairs on O.
4
Apply VSEPR
4 electron groups total (2 bonds + 2 lone pairs). The lone pairs push the H atoms down → bent. This bend is why water is polar — and why life exists.
C O O •• •• •• ••
CO₂ — Linear
2 double bonds · 180°
1
Count valence electrons
C has 4, each O has 6. Total: 4 + 12 = 16 valence electrons.
2
Carbon is central
C is less electronegative than O.
3
Single bonds aren't enough
Two single bonds leave C with only 4 electrons. Promote each to a double bond (C=O). Now C has 8 — full octet.
4
Lone pairs on oxygens
Each O gets 2 lone pairs (4 electrons) to complete its octet.
5
Apply VSEPR
C has 2 electron groups (both double bonds count as one group each) → linear, 180°.
F

Resonance — when one structure isn't enough

Sometimes a molecule has multiple valid Lewis structures that differ only in where the double bond sits. The truth is a blend.
O O O ↔
O₃ (ozone) — two equivalent resonance structures

Ozone (O₃) has one O=O double bond and one O–O single bond. But which side has the double bond? Both. Neither. Half.

The real molecule is an average of all valid Lewis structures — each oxygen-oxygen bond has 1.5 bonds' worth of strength. The same logic applies to NO₃⁻, where the double bond is "spread" across all three oxygens.

Memory Trick
Think of resonance like a photo taken at long exposure. You can't say where the double bond is — it's everywhere at once. The double-headed arrow ↔ means "the truth is a blend of these," not "switches back and forth."
G

Test yourself

Five rapid questions on bonding. Wrong answer? You'll see exactly why.
02
Chapter Two

The language of compounds

Nomenclature is just a system of decisions. Once you can ask the right questions about a compound, the name writes itself.

A

The naming decision tree

Every compound name comes from answering four questions in order. Ask them in this sequence and you'll never be stuck.
Start with a formula Starts with H + anion? YES It's an acid. -ide → hydro__ic acid -ate → __ic acid · -ite → __ous acid NO Has a metal? YES Ionic compound. Name metal + name anion Transition metal? Add Roman numeral for charge NO Molecular (two nonmetals). Use Greek prefixes for both atoms mono- di- tri- tetra- penta-... (skip "mono-" on first atom) HCl → hydrochloric acid FeCl₃ → iron(III) chloride NaBr → sodium bromide SO₃ → sulfur trioxide
B

Greek prefixes & Roman numerals

Two tools that handle every molecular and transition-metal name.
Greek prefixes (for molecular compounds)
1
mono-
2
di-
3
tri-
4
tetra-
5
penta-
6
hexa-
7
hepta-
8
octa-
9
nona-
10
deca-
Memory Trick
Skip "mono-" on the first atom. CO is "carbon monoxide" — not "monocarbon monoxide." But always include it on the second atom. And drop a vowel for sound: pentaoxide → pentoxide, monooxide → monoxide.
Roman numerals (for transition-metal charges)
I
+1 charge
II
+2 charge
III
+3 charge
IV
+4 charge
V
+5 charge
VI
+6 charge
VII
+7 charge
When to use Roman numerals
Only for metals with variable charges — transition metals (Fe, Cu, Co, Mn, Hg, Sn, Pb), plus a few others. To find the charge: work backward from the anion. FeCl₃ → 3 Cl⁻ totals −3, so Fe must be +3 → iron(III) chloride.
Memory Trick — Polyatomic Ion Patterns
The "-ate" / "-ite" pattern: -ate has more oxygens, -ite has fewer. SO₄²⁻ sulfate · SO₃²⁻ sulfite.
Extend with prefixes: per-...-ate = one more O · hypo-...-ite = one fewer O.
Chlorine series: ClO₄⁻ perchlorate → ClO₃⁻ chlorate → ClO₂⁻ chlorite → ClO⁻ hypochlorite.
C

Practice every compound from the study guide

All 50 compounds. Two modes: formula → name and name → formula. Type your answer — instant feedback.
0 / 25 correct
0 / 25 correct
D

All 50 compounds — full answer key

Click any row to copy. Sort by category using the tabs.
03
Chapter Three

Balancing & predicting

Atoms can't be created or destroyed — only rearranged. Every reaction is just an accounting problem with a pattern hiding inside.

A

Six patterns of change

Every reaction in your study guide fits into one of these six categories. Spot the pattern, predict the products.
Synthesis
A + B → AB
Two or more things combine into one. Always one product on the right.
2K + F₂ → 2KF
Decomposition
AB → A + B
The reverse of synthesis. One compound breaks apart into pieces.
2NaCl → 2Na + Cl₂
Single Replacement
A + BC → AC + B
A lone element kicks out another from its compound. One swap.
Mg + Zn(NO₃)₂ → Mg(NO₃)₂ + Zn
Double Replacement
AB + CD → AD + CB
Two compounds swap partners. Like a chemistry square dance.
NaBr + AgNO₃ → AgBr + NaNO₃
Combustion
CₓHᵧ + O₂ → CO₂ + H₂O
A hydrocarbon burns in oxygen. Products are always CO₂ and H₂O.
CH₄ + 2O₂ → CO₂ + 2H₂O
Neutralization
acid + base → salt + H₂O
A type of double replacement. Always produces water.
Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O
Memory Trick — Quick ID
Look at the left side first. One thing? → decomposition. Hydrocarbon + O₂? → combustion. Element + compound? → single replacement. Two compounds? → double replacement (and if one is an acid, also neutralization). Multiple simple things → synthesis.
B

Interactive equation balancer

Use the arrows to add coefficients. Watch the atom tally — when both sides match, you're balanced.
Strategy: Balance atoms that appear in only one compound on each side first. Save hydrogen and oxygen for last — they're usually in many places. If you get fractional coefficients, multiply everything by the denominator to clear them.
C

All 25 reactions from the study guide

All 25 are balanced. Try to classify each one yourself first, then tap the row to reveal the type and the reasoning. The walkthroughs explain why each pattern is what it is — not just the label.
D

Spot the reaction type

Six questions, each picked to test a different decision point — not the ones from the list above. Before you tap an option, walk through this checklist:
1 · Count the reactants
Just one? It must be decomposition. Done.
2 · Water in the products?
Yes, plus an acid + base on the left → neutralization.
3 · CO₂ AND H₂O on the right?
From a fuel + O₂ → combustion.
4 · Lone element on the left?
Combined with a compound → single replacement. Two elements only → synthesis.
5 · Otherwise — two compounds in, two compounds out?
Partners are swapping → double replacement. (Or, if many in and one out, it's still synthesis — synthesis isn't limited to bare elements.)
04
Chapter Four

The mole, the universal translator

A mole is a chemist's dozen — just bigger. Once you can move fluently between grams, moles, and particles, every quantitative problem opens up.

A

The Mole Map

Three quantities, three conversion factors. Memorize this map and every mole problem becomes a 1, 2, or 3-step journey.
MOLES the universal hub always convert through here Grams mass (g) Particles atoms · molecules formula units ÷ molar mass g → mol × molar mass mol → g × 6.022 × 10²³ mol → particles ÷ 6.022 × 10²³ particles → mol To go from grams to particles (or back), you MUST pass through moles.
Memory Trick — "The Mole is the Middle"
Moles are the chemist's hub airport. You can't fly from grams to particles directly — you always connect through Mole International. Two conversions, two factors: molar mass for the gram side, Avogadro's number for the particle side.
B

Avogadro's number — what does 6.022 × 10²³ actually mean?

It's just a count. A really, really big count.

A "mole" is to chemistry what a "dozen" is to eggs — just a fixed count. A dozen = 12. A mole = 6.022 × 10²³.

Why such a weird number? Because that's how many atoms it takes for the mass in grams to match the atomic mass on the periodic table. One mole of carbon-12 weighs exactly 12 grams. One mole of any element weighs its atomic mass in grams.

How big is it? If you had a mole of marshmallows, they would cover the entire surface of the United States to a depth of 600 miles. A mole of seconds is older than the universe.

C

The six worked problems

Every problem from the study guide, fully worked. Try first — peek when stuck.
Problem 1 · Moles to atoms
How many atoms are in 6.50 mol Ar?
1
Identify what you have and what you need
Have: moles. Need: atoms. Direction: moles → particles. So multiply by Avogadro's number.
Avogadro's number (6.022 × 10²³) is defined as the count of particles in one mole. Going FROM moles, you multiply. Going TO moles, you divide. The direction of your conversion tells you whether to multiply or divide — every time.
2
Set up dimensional analysis
Put moles on top, moles on the bottom of Avogadro's ratio — units cancel.
6.50 mol Ar × (6.022 × 10²³ atoms / 1 mol)
Always write the conversion factor as a fraction with the unit you want to cancel on the bottom. If units don't cancel cleanly, you've set it up wrong — flip the fraction.
3
Multiply
= 3.91 × 10²⁴ atoms
Problem 2 · Moles to formula units
How many formula units are in 6.25 mol AgNO₃?
1
Same conversion, different name
Because AgNO₃ is ionic, we count formula units instead of molecules. The math is identical.
2
Multiply by Avogadro
6.25 mol × (6.022 × 10²³ / 1 mol) = 3.76 × 10²⁴ formula units
Problem 3 · Atoms back to moles
How many moles are in 5.36 × 10²⁴ atoms Al?
1
Going backwards: particles → moles
Divide by Avogadro's number. Flip the conversion factor so atoms cancel.
2
Set up and solve
5.36 × 10²⁴ atoms × (1 mol / 6.022 × 10²³ atoms) = 8.90 mol Al
Problem 4 · Moles to grams
What is the mass of 6.45 mol Co?
1
Look up Co's molar mass
From the periodic table: Co = 58.93 g/mol.
Molar mass tells you how many grams one mole of that element weighs. It's the bridge between mass (grams) and amount (moles). Every element has its own bridge — you find it on the periodic table.
2
Multiply moles by molar mass
6.45 mol × (58.93 g / 1 mol) = 380 g Co
Going moles → grams means multiplying by molar mass (g/mol). The moles cancel, leaving grams. If you set it up the other way (dividing) you'd get an answer in mol²/g — gibberish units are a signal you flipped the ratio.
Problem 5 · Two-step: grams to atoms
How many atoms are in 6.548 g Pb?
1
You can't skip moles
Grams → moles → atoms. Two conversions in one chain.
2
Convert grams to moles
6.548 g × (1 mol / 207.2 g) = 0.0316 mol Pb
3
Convert moles to atoms
0.0316 mol × (6.022 × 10²³ / 1 mol) = 1.90 × 10²² atoms
Problem 6 · Formula units to moles
How many moles are in 1.25 × 10²⁷ formula units NaNO₃?
1
Divide by Avogadro
1.25 × 10²⁷ ÷ (6.022 × 10²³) = 2076 mol NaNO₃
2
Sanity check
2076 moles is a huge number — but so was the particle count (10²⁷ is 10,000× Avogadro's number). The answer is reasonable.
D

Your own mole calculator

Plug in a value and a direction — see the full conversion chain.
05
Chapter Five

Stoichiometry, the recipe book

Chemistry's recipes. Predict how much you'll make, figure out what runs out first, and reverse-engineer formulas from percentages alone.

A

The big idea — coefficients are mole ratios

Every balanced equation is a recipe. The coefficients tell you the ratios — and ratios are the bridge between what you have and what you want.

Look at this balanced equation: 2H₂ + O₂ → 2H₂O. Read it like a recipe: "Two moles of hydrogen react with one mole of oxygen to make two moles of water."

Those coefficients (2, 1, 2) are the only reason stoichiometry works. They give you mole ratios, and a mole ratio lets you convert from anything to anything else in the reaction.

The universal recipe: grams → moles → (mole ratio) → moles → grams. Memorize that chain. Every stoichiometry problem walks that path.

Memory Trick — "The See-Saw"
Moles are the fulcrum of the see-saw. Grams of one substance sit on one end, grams of another on the other. You cannot move grams of A to grams of B directly — you have to drop down to moles, slide across via the coefficient ratio, then climb back up.
B

Percent composition

What fraction of a compound's mass comes from each element? Three steps, every time.
1
Find the total molar mass of the compound
Add up every atom's mass. Don't forget subscripts — they multiply.
2
For each element: (mass of that element in the formula) ÷ (total molar mass)
The mass of an element in the formula = atomic mass × subscript.
3
Multiply by 100 to get a percentage
Sanity check: all percentages should add to ~100%. If they don't, you've made an arithmetic error.
Worked Example · Iron(III) oxide, Fe₂O₃
1
Molar mass of Fe₂O₃
2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70 g/mol
2
% Fe
(111.70 / 159.70) × 100 = 69.9% Fe
3
% O
(48.00 / 159.70) × 100 = 30.1% O
Check: 69.9 + 30.1 = 100.0 ✓
Worked Example · Aluminum bromide, AlBr₃
1
Molar mass
26.98 + 3(79.90) = 26.98 + 239.70 = 266.68 g/mol
2
% Al and % Br
(26.98 / 266.68) × 100 = 10.1% Al
(239.70 / 266.68) × 100 = 89.9% Br
Worked Example · Strontium acetate, Sr(C₂H₃O₂)₂
The parentheses make this trickier. Everything inside them gets doubled.
1
Count atoms after distributing the parentheses
1 Sr, 4 C (2×2), 6 H (3×2), 4 O (2×2).
2
Molar mass
87.62 + 4(12.01) + 6(1.008) + 4(16.00) = 205.71 g/mol
3
Percentages
Sr: 87.62/205.71 × 100 = 40.7%
C: 48.04/205.71 × 100 = 23.4% (key shows 22.3% — small rounding difference)
H: 6.05/205.71 × 100 = 2.9%
O: 64.00/205.71 × 100 = 31.1%
Try your own
C

The three-step universal process

Every mass-to-mass stoichiometry problem walks the same path. Learn the path, and you've solved them all.
1.
Convert given to moles
Grams of known ÷ molar mass of known = moles of known.
2.
Apply the mole ratio
Moles known × (coeff. of unknown / coeff. of known) from the balanced equation.
3.
Convert moles to whatever you want
× molar mass for grams, × 6.022×10²³ for particles.
Memory Trick — "Down, Across, Up"
Down from grams to moles. Across via the coefficient ratio. Up from moles to whatever the question asks. If you ever feel lost mid-problem, ask yourself: which leg of the journey am I on?
D

The mass-mass problems from your study guide

All four worked, step by step. Try first — peek when stuck.
Sodium and Water
2Na + 2H₂O → 2NaOH + H₂
How many moles of Na are needed to produce 4.0 mol H₂?
1
Identify the mole ratio
From the balanced equation: 2 mol Na produces 1 mol H₂. So the ratio is 2 Na : 1 H₂.
Coefficients in a balanced equation ARE the mole ratios. They're the only thing connecting different substances in the reaction. Whenever you cross from one substance to another, you must use these coefficients — there is no other path.
2
Apply the ratio
4.0 mol H₂ × (2 mol Na / 1 mol H₂) = 8.0 mol Na
Put the unit you want to cancel (mol H₂) on the bottom of the ratio. The unit you want to keep (mol Na) goes on top. Units cancel cleanly — that's how you know the setup is right.
Phosphorus Combustion
4P + 5O₂ → P₄O₁₀
Three questions about this reaction:
a) Mass of P needed for 3.25 mol P₄O₁₀?
1
Mole ratio
3.25 mol P₄O₁₀ × (4 mol P / 1 mol P₄O₁₀) = 13.0 mol P
2
Convert moles to grams
13.0 mol × (30.97 g/mol) = 403 g P
b) Mass of O₂ used if 0.489 mol P burns?
1
Mole ratio P → O₂
0.489 mol P × (5 mol O₂ / 4 mol P) = 0.611 mol O₂
2
Mass
0.611 mol × (32.00 g/mol) = 19.6 g O₂
c) Mass of P₄O₁₀ produced from 0.489 mol P?
1
Mole ratio P → P₄O₁₀
0.489 mol P × (1 mol P₄O₁₀ / 4 mol P) = 0.1223 mol P₄O₁₀
2
Mass
0.1223 mol × (283.9 g/mol) = 34.7 g P₄O₁₀
Sodium Carbonate and Nitric Acid
Na₂CO₃ + 2HNO₃ → 2NaNO₃ + CO₂ + H₂O
How many moles of Na₂CO₃ are needed to make 100.0 g NaNO₃?
1
Grams of NaNO₃ → moles
100.0 g ÷ (85.0 g/mol) = 1.176 mol NaNO₃
2
Apply mole ratio
1.176 mol NaNO₃ × (1 mol Na₂CO₃ / 2 mol NaNO₃) = 0.588 mol Na₂CO₃
Bonus: How much CO₂ is produced from 7.50 g Na₂CO₃?
1
g → mol Na₂CO₃
7.50 g ÷ (106 g/mol) = 0.0708 mol
2
Mole ratio Na₂CO₃ → CO₂ is 1:1
= 0.0708 mol CO₂
Silver Nitrate and Magnesium Bromide
2AgNO₃ + MgBr₂ → 2AgBr + Mg(NO₃)₂
How much AgBr is produced from 22.54 g AgNO₃?
1
Convert grams AgNO₃ to moles
22.54 g ÷ (169.9 g/mol) = 0.1327 mol AgNO₃
This is the canonical 3-step recipe — every mass-to-mass problem follows it: Down from grams to moles (using molar mass), Across from one substance to another (using the mole ratio), Up from moles to grams (using molar mass again). You can never go grams-to-grams directly — moles are required as the middleman.
2
Apply mole ratio (2:2 = 1:1)
0.1327 mol AgNO₃ × (2 mol AgBr / 2 mol AgNO₃) = 0.1327 mol AgBr
The 2:2 ratio simplifies to 1:1, meaning every mole of AgNO₃ makes exactly one mole of AgBr. Don't skip writing the full ratio though — it's a habit that prevents mistakes when the ratio is NOT 1:1.
3
Convert moles AgBr to grams
0.1327 mol × (187.8 g/mol) = 24.9 g AgBr
E

Limiting reactant — what runs out first

When both reactants have a starting amount, one of them runs out before the other. That one limits how much product you can make.

Imagine you're making s'mores. You have 10 graham crackers but only 3 marshmallows. You can make 3 s'mores — then you're stuck with 4 leftover graham crackers and no marshmallows. The marshmallows are the limiting reactant.

To find the limiting reactant in a real problem: calculate how much product each reactant could make if it were the only constraint. The one that makes less product is limiting. The other is in excess.

Visualize it · 2ZnS + 3O₂ → 2ZnO + 2SO₂
ZnS available: 1.72 mol
O₂ available: 3.04 mol
Worked Example · 2ZnS + 3O₂ → 2ZnO + 2SO₂
Given 1.72 mol ZnS and 3.04 mol O₂, find (a) the limiting reactant, (b) excess remaining, (c) masses of products.
1
Find how much O₂ ZnS demands
1.72 mol ZnS × (3 mol O₂ / 2 mol ZnS) = 2.58 mol O₂ needed
To find the limiting reactant, pick one reactant and ask: "If all of it reacted, how much of the other reactant would I need?" Then compare to what you actually have.
2
Compare to what's available
We have 3.04 mol O₂ but only need 2.58. So O₂ is in excess — meaning ZnS is the limiting reactant.
ZnS demanded 2.58 mol O₂ and got 3.04 — its demand was met, so it used up everything it had (all 1.72 mol). ZnS stopped the reaction because it ran out, even though more O₂ was still available. Whoever runs out first is limiting.
3
Excess O₂ remaining
3.04 − 2.58 = 0.46 mol O₂ leftover
4
Mass of ZnO produced (use limiting reactant ZnS)
1.72 mol ZnS × (2 mol ZnO / 2 mol ZnS) × 81.38 g/mol = 140 g ZnO
Always use the LIMITING reactant for product calculations. The excess reactant has leftovers, so it doesn't determine the maximum product. Only the one that runs out sets the ceiling.
5
Mass of SO₂ produced
1.72 mol ZnS × (2 mol SO₂ / 2 mol ZnS) × 64.07 g/mol = 110 g SO₂
Memory Trick — "Whoever Makes Less, Wins (the Title of Limiting)"
Calculate how much product each reactant could make. The one that makes less is limiting — it ran out first, so it caps the reaction. Then use that limiting reactant for ALL further calculations of products.
F

Percent yield — theory vs. reality

In a perfect world, every atom would end up where you wanted. The real world is messier.

Theoretical yield is the maximum amount of product the math says you can make. It's calculated from stoichiometry.

Actual yield is what you actually measured in the lab. It's almost always less — some product gets stuck on glassware, a side reaction steals reactants, the reaction doesn't quite finish.

The formula: % yield = (actual / theoretical) × 100

Worked Example · Silver Chromate Yield
K₂CrO₄ + 2AgNO₃ → Ag₂CrO₄ + 2KNO₃
Starting with 0.500 g AgNO₃, the lab produced 0.455 g Ag₂CrO₄. Find (a) theoretical yield, (b) percent yield.
1
Convert AgNO₃ to moles
0.500 g ÷ (169.9 g/mol) = 0.00294 mol AgNO₃
Percent yield is about comparing the theoretical maximum to what you actually got. To find the theoretical max, do a regular stoichiometry calculation — start by converting your starting reactant to moles.
2
Apply mole ratio (2:1)
0.00294 mol × (1 mol Ag₂CrO₄ / 2 mol AgNO₃) = 0.00147 mol Ag₂CrO₄
From the balanced equation, 2 AgNO₃ produces 1 Ag₂CrO₄. So you get half as many moles of product as you used of AgNO₃.
3
Theoretical yield in grams
0.00147 mol × (331.7 g/mol) = 0.488 g Ag₂CrO₄
This 0.488 g is the THEORETICAL yield — the maximum the math says is possible. Reality almost always falls short.
4
Percent yield
(0.455 / 0.488) × 100 = 93.2%
93% is a strong yield. Anything above 90% is excellent; below 50% usually means something went wrong.
Percent yield = (actual / theoretical) × 100. The actual yield is what you measured in the lab; the theoretical is what stoichiometry predicts. Their ratio tells you how efficiently the reaction worked.
Percent yield calculator
G

Empirical & molecular formula

Reverse-engineering a compound from percent composition. Sounds backwards, but the steps are mechanical.

Empirical formula: the simplest whole-number ratio of atoms (e.g. CH for benzene).

Molecular formula: the actual number of atoms in one molecule (e.g. C₆H₆ for benzene).

Molecular = (empirical) × n, where n = molecular mass / empirical mass.

1
Assume 100 g of compound
This turns every percentage into grams directly. 42% becomes 42 g — no algebra required.
2
Convert grams of each element to moles
Divide each element's grams by its atomic mass.
3
Divide every mole count by the smallest one
This gives you the ratio. If you don't get whole numbers, multiply all by 2 or 3 until you do.
4
For molecular formula: divide given molar mass by empirical mass
That number, n, is the multiplier. Multiply every subscript by n.
Worked Example · Empirical CH, molar mass 78
1
Empirical mass
12.01 + 1.008 ≈ 13 g/mol
2
n = molecular / empirical
78 ÷ 13 = 6
3
Multiply subscripts
(CH) × 6 = C₆H₆ (benzene)
Worked Example · The big four-element problem
Given: C = 42.87%, H = 3.5985%, O = 28.55%, N = 25.00%. Molar mass = 168 g/mol. Find empirical and molecular formulas.
1
Assume 100 g — percentages become grams
C: 42.87 g · H: 3.5985 g · O: 28.55 g · N: 25.00 g.
Pretending you have 100 g is a trick that saves arithmetic — 42.87% of 100 g is literally 42.87 g. The empirical formula is a RATIO, so it doesn't matter how much you start with; assume the easiest amount.
2
Convert grams to moles
C: 42.87 ÷ 12.01 = 3.57 mol
H: 3.5985 ÷ 1.008 = 3.57 mol
O: 28.55 ÷ 16.00 = 1.78 mol
N: 25.00 ÷ 14.01 = 1.78 mol
Formulas are about counts of atoms, not masses. Moles are proportional to atom count, but grams are not (because different atoms weigh different amounts). To find a formula, you must work in moles.
3
Divide each by the smallest (1.78)
C: 3.57/1.78 ≈ 2 · H: 2 · O: 1 · N: 1
Empirical formula: C₂H₂NO
Dividing by the smallest mole count gives a ratio normalized to "1" for the least-abundant element. The result is the simplest whole-number ratio — exactly what "empirical" means. If you got 1.5, multiply everything by 2; if 1.33, multiply by 3.
4
Empirical mass
2(12.01) + 2(1.008) + 14.01 + 16.00 ≈ 56 g/mol
The empirical mass is the molar mass of just ONE empirical unit. The actual molecule is some integer multiple of that unit — your job is to find the multiplier.
5
n = 168 ÷ 56 = 3
Multiply every subscript by 3 → C₆H₆N₃O₃
n is how many empirical units fit into one real molecule. If the molecular mass is 3 times the empirical mass, then 3 empirical units make up one real molecule — so multiply every subscript by 3.
Memory Trick — "Percent to Mass to Mole to Pure"
Always start by pretending you have 100 g. That converts percentages directly to grams with zero arithmetic. Then it's just grams → moles → divide by smallest → done. If your final answers aren't clean whole numbers (within ~0.1), multiply everything by 2, then 3, until they are.