1. The mole
A mole is just a counting unit, like a dozen — but much bigger. One mole contains 6.02 × 10²³ particles. This number is called Avogadro's number. We need a unit this large because individual atoms are unimaginably tiny: 12 grams of carbon (about a tablespoon) contains roughly 6 × 10²³ atoms.
2. Molar mass
The molar mass of any element equals its atomic mass on the Periodic Table, read in grams per mole (g/mol). For a compound, add up the molar masses of each atom in the formula.
Calculating molar mass: H₂O
2 H × 1.01 g/mol = 2.02 g/mol
1 O × 16.00 g/mol = 16.00 g/mol
Total = 18.02 g/mol
So one mole of water weighs about 18 grams.
Calculating molar mass: Ca(NO₃)₂
The subscript 2 applies to everything inside the parentheses.
1 Ca × 40.08 = 40.08
2 N × 14.01 = 28.02
6 O × 16.00 = 96.00 (2 nitrate ions × 3 oxygens each)
Total = 164.10 g/mol
3. The mole map (the most useful diagram on the exam)
Every quantitative chemistry problem boils down to three conversions, all going through moles. Once you know the conversion factors, you can solve any problem on Parts B and C.
| Going from… | To… | Multiply by |
|---|---|---|
| Grams | Moles | 1 / molar mass |
| Moles | Grams | molar mass |
| Particles | Moles | 1 / (6.02 × 10²³) |
| Moles | Particles | 6.02 × 10²³ |
| Liters (gas at STP) | Moles | 1 / 22.4 |
| Moles | Liters (gas at STP) | 22.4 |
One-step example: grams → moles
How many moles are in 36.04 g of water?
36.04 g × (1 mol / 18.02 g) = 2.00 mol
Two-step example: grams → particles
How many water molecules are in 9.01 g of H₂O?
9.01 g × (1 mol / 18.02 g) × (6.02 × 10²³ molecules / 1 mol)
= 0.500 mol × 6.02 × 10²³ = 3.01 × 10²³ molecules
4. Percent composition by mass
How much of a compound is made of each element, by mass. The formula (also on Reference Table T) is:
Worked example: % water in CuSO₄·5H₂O
Molar mass of CuSO₄·5H₂O = 63.55 + 32.07 + (4 × 16.00) + 5(18.02) = 249.72 g/mol
Mass of water (5 H₂O) = 5 × 18.02 = 90.10 g/mol
% H₂O = (90.10 / 249.72) × 100 = 36.08%
5. Empirical and molecular formulas
| Empirical formula | The simplest whole-number ratio of atoms in a compound. Example: glucose (C₆H₁₂O₆) has the empirical formula CH₂O. |
| Molecular formula | The actual number of each atom in one molecule. Glucose's molecular formula is C₆H₁₂O₆. The molecular formula is always a whole-number multiple of the empirical formula. |
Going from empirical to molecular
multiplier n = molar mass of molecular formula / molar mass of empirical formula
Example: empirical formula CH₂O has molar mass 30.03 g/mol. If the actual molecule has molar
mass 180.18 g/mol, then n = 180.18 / 30.03 = 6. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
6. Writing chemical formulas
Ionic compounds (metal + nonmetal)
Charges must cancel. The simplest way: crisscross the absolute value of each charge to become the other ion's subscript, then simplify if possible.
Example: aluminum oxide
Al has charge +3, O has charge −2.
Crisscross: Al gets subscript 2, O gets subscript 3.
Formula: Al₂O₃
Example: calcium nitrate
Ca²⁺ and NO₃⁻ (from Reference Table E).
Crisscross: Ca gets subscript 1 (omitted), NO₃ gets subscript 2 (use parentheses since it's a polyatomic ion).
Formula: Ca(NO₃)₂
Covalent compounds (two nonmetals)
Use Greek prefixes from Reference Table C: mono, di, tri, tetra, penta, hexa, etc. The prefix tells you the subscript directly.
Examples
- carbon dioxide → CO₂
- dinitrogen tetraoxide → N₂O₄
- sulfur hexafluoride → SF₆
Note: the prefix "mono" is dropped from the first element. Carbon dioxide, not "monocarbon dioxide."
7. Balancing chemical equations
The Law of Conservation of Mass (and atoms, and charge) requires that the number of atoms of each element be equal on both sides of the equation. You can change coefficients (the numbers in front), never subscripts inside formulas.
Worked example: combustion of methane
Unbalanced: CH₄ + O₂ → CO₂ + H₂O
- Count atoms: Left: 1 C, 4 H, 2 O. Right: 1 C, 2 H, 3 O. Hydrogens and oxygens unbalanced.
- Balance H by putting a 2 in front of H₂O: CH₄ + O₂ → CO₂ + 2 H₂O. Now 4 H on each side. Right side has 4 O total.
- Balance O by putting a 2 in front of O₂: CH₄ + 2 O₂ → CO₂ + 2 H₂O. 4 O on each side. ✓
Balanced: CH₄ + 2 O₂ → CO₂ + 2 H₂O
8. The five reaction types
The Regents will ask you to identify the type and predict products for each.
| Type | General pattern | Example |
|---|---|---|
| Synthesis (combination) | A + B → AB | 2 Mg + O₂ → 2 MgO |
| Decomposition | AB → A + B | 2 H₂O → 2 H₂ + O₂ |
| Single replacement | A + BC → AC + B | Zn + 2 HCl → ZnCl₂ + H₂ |
| Double replacement | AB + CD → AD + CB | AgNO₃ + NaCl → AgCl + NaNO₃ |
| Combustion | hydrocarbon + O₂ → CO₂ + H₂O | C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O |
Using Table J for single replacement
A single replacement reaction only happens if the element by itself is higher on the
activity series (Table J) than the element it's trying to replace.
Example: Zn + 2 HCl → ZnCl₂ + H₂ ✓ (Zn is above H on Table J).
Example: Cu + ZnCl₂ → no reaction (Cu is below Zn on Table J).
9. Stoichiometry: the mole ratio
Once an equation is balanced, the coefficients give you the mole ratio between any two substances in the reaction. This is the heart of every Part C stoichiometry problem.
Worked example: mole-to-mole
Given N₂ + 3 H₂ → 2 NH₃, how many moles of NH₃ form when 6.0 mol of H₂ react completely?
The ratio is 3 mol H₂ : 2 mol NH₃.
6.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 4.0 mol NH₃
Worked example: grams-to-grams (full mole map)
Given 2 H₂ + O₂ → 2 H₂O, how many grams of water form from 8.0 g of H₂?
- Grams H₂ → moles H₂: 8.0 g × (1 mol / 2.02 g) = 3.96 mol H₂
- Moles H₂ → moles H₂O using 2:2 ratio: 3.96 mol H₂O
- Moles H₂O → grams H₂O: 3.96 mol × 18.02 g/mol = 71.4 g H₂O
Key terms to know cold
| Mole | 6.02 × 10²³ particles (Avogadro's number) |
| Molar mass | Mass of one mole of a substance, in g/mol; equals atomic mass for an element |
| Molar volume of a gas at STP | 22.4 L/mol |
| Empirical formula | Simplest whole-number ratio of atoms |
| Molecular formula | Actual number of each atom per molecule |
| Percent composition | % of total mass contributed by each element |
| Coefficient | Number in front of a formula in a balanced equation |
| Subscript | Small number after an element symbol; never changes when balancing |
| Law of Conservation of Mass | Mass cannot be created or destroyed; both sides of a balanced equation have the same atoms |
| Stoichiometry | Calculating amounts of reactants/products using the mole ratio from a balanced equation |
| Synthesis | A + B → AB |
| Decomposition | AB → A + B |
| Single replacement | A + BC → AC + B (check Table J) |
| Double replacement | AB + CD → AD + CB |
| Combustion | Hydrocarbon + O₂ → CO₂ + H₂O |
Practice questions
Q1 · Multiple Choice
What is the gram-formula mass of Ca(OH)₂?
- 57 g/mol
- 74 g/mol
- 96 g/mol
- 112 g/mol
Show answer
(2) 74 g/mol. Ca = 40, O × 2 = 32, H × 2 = 2. Total = 40 + 32 + 2 = 74 g/mol.
Q2 · Multiple Choice
What is the total number of atoms in 0.50 mole of Au?
- 3.01 × 10²³
- 6.02 × 10²³
- 9.85 × 10¹
- 1.97 × 10²
Show answer
(1) 3.01 × 10²³. 0.50 mol × 6.02 × 10²³ atoms/mol = 3.01 × 10²³ atoms.
Q3 · Multiple Choice
When the equation _C₂H₆ + _O₂ → _CO₂ + _H₂O is balanced using the smallest whole-number coefficients, the coefficient of O₂ is
- 2
- 3
- 5
- 7
Show answer
(4) 7. Balanced: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O. Check: 4 C, 12 H, 14 O on each side. ✓
Q4 · Part B-2 / Short response
Determine the percent composition by mass of nitrogen in the fertilizer NH₄NO₃ (ammonium nitrate).
Show answer
Molar mass of NH₄NO₃: 2(14.01) + 4(1.01) + 3(16.00) = 28.02 + 4.04 + 48.00 = 80.06 g/mol.
Mass of N in formula: 2 × 14.01 = 28.02 g/mol.
% N = (28.02 / 80.06) × 100 = 35.0%
Q5 · Part C / Extended response
Given the balanced equation: 2 Al + 3 Cl₂ → 2 AlCl₃
Calculate the mass of AlCl₃ that can be produced from the complete reaction of 5.40 g of Al.
Show your work, including a complete unit-cancellation setup.
Show answer
Step 1: Convert grams Al to moles Al.
5.40 g Al × (1 mol Al / 26.98 g Al) = 0.200 mol Al
Step 2: Use mole ratio (2 Al : 2 AlCl₃, which is 1:1).
0.200 mol Al × (2 mol AlCl₃ / 2 mol Al) = 0.200 mol AlCl₃
Step 3: Convert moles AlCl₃ to grams.
Molar mass of AlCl₃ = 26.98 + 3(35.45) = 133.33 g/mol
0.200 mol × 133.33 g/mol = 26.7 g AlCl₃