PRE MI UM ACT I VE- LE A RN IN G WO RK BOO K
Chem Exam Whiteboard Review
Formulas, Isotopes, Molarity, Redox, Gas Laws, Calorimetry, Lewis Structures, VSEPR & Periodic
Trends
General Chemistry • Active-Learning Edition
How to use this workbook
1. Read the concept explanation. 2. Study the worked example. 3. Try the practice problems yourself.
4. Check your work against the Answer Key. 5. Revisit anything you missed.
, 1. Matter & Naming Compounds
Matter → Mixture (homogeneous/uniform vs heterogeneous/non-uniform) or Element/Compound. Compounds
split into Ionic (metal + nonmetal, named metal + nonmetal-ide, transition metals get a Roman numeral) and
Covalent/Molecular (two+ nonmetals, named with prefixes + name + -ide).
Term Definition
A / Z / X notation A = mass number, Z = atomic number, X = chemical symbol.
Molecular Formula Actual atom counts, found using same steps as empirical formula plus one more.
Empirical Formula Simplest whole-number ratio of atoms.
Prefixes
1-Mono, 2-Di, 3-Tri, 4-Tetra, 5-Penta, 6-Hexa, 7-Hepta, 8-Octa, 9-Nona, 10-Deca
2. Empirical & Molecular Formula (Full Method)
WORKED EXAMPLE — Finding an Empirical Formula (68.4% Cr, 31.6% O)
Step 1 — % to grams: 68.4% → 68.4 g Cr; 31.6% → 31.6 g O.
Step 2 — grams to moles: 68.4 g Cr ÷ 52.00 g/mol = 1.315 mol; 31.6 g O ÷ 16.00 g/mol = 1.975 mol.
Step 3 — divide by smallest mol value: Cr → 1.315/1.315 = 1; O → 1.975/1.315 = 1.5.
Step 4 — multiply to get whole numbers (×2): Cr2O3.
MULTIPLIER RULES (when dividing gives a decimal)
x.5 → multiply everything by 2
x.33 or x.66 → multiply everything by 3
x.25 or x.75 → multiply everything by 4
Anything else → treat as already whole (round)
WORKED EXAMPLE — Finding a Molecular Formula from an Empirical Formula
Given empirical formula C5H7N, and known molecular molar mass = 162 g/mol.
Step 1 — Find the empirical formula's molar mass: C: 5(12.01) + H: 7(1.01) + N: 1(14.01) = 81.13 g/mol.
Step 2 — Divide molecular mass by empirical mass: 162 ÷ 81.13 ≈ 2.
Step 3 — Multiply every subscript in the empirical formula by that whole number: C10H14N2.
✏ TRY IT YOURSELF
1. A compound has an empirical formula of CH2O and a molar mass of 180 g/mol. Find the molecular formula.
2. A compound is 40.0% C, 6.7% H, 53.3% O by mass. Find the empirical formula.
Chem Exam Whiteboard Review
Formulas, Isotopes, Molarity, Redox, Gas Laws, Calorimetry, Lewis Structures, VSEPR & Periodic
Trends
General Chemistry • Active-Learning Edition
How to use this workbook
1. Read the concept explanation. 2. Study the worked example. 3. Try the practice problems yourself.
4. Check your work against the Answer Key. 5. Revisit anything you missed.
, 1. Matter & Naming Compounds
Matter → Mixture (homogeneous/uniform vs heterogeneous/non-uniform) or Element/Compound. Compounds
split into Ionic (metal + nonmetal, named metal + nonmetal-ide, transition metals get a Roman numeral) and
Covalent/Molecular (two+ nonmetals, named with prefixes + name + -ide).
Term Definition
A / Z / X notation A = mass number, Z = atomic number, X = chemical symbol.
Molecular Formula Actual atom counts, found using same steps as empirical formula plus one more.
Empirical Formula Simplest whole-number ratio of atoms.
Prefixes
1-Mono, 2-Di, 3-Tri, 4-Tetra, 5-Penta, 6-Hexa, 7-Hepta, 8-Octa, 9-Nona, 10-Deca
2. Empirical & Molecular Formula (Full Method)
WORKED EXAMPLE — Finding an Empirical Formula (68.4% Cr, 31.6% O)
Step 1 — % to grams: 68.4% → 68.4 g Cr; 31.6% → 31.6 g O.
Step 2 — grams to moles: 68.4 g Cr ÷ 52.00 g/mol = 1.315 mol; 31.6 g O ÷ 16.00 g/mol = 1.975 mol.
Step 3 — divide by smallest mol value: Cr → 1.315/1.315 = 1; O → 1.975/1.315 = 1.5.
Step 4 — multiply to get whole numbers (×2): Cr2O3.
MULTIPLIER RULES (when dividing gives a decimal)
x.5 → multiply everything by 2
x.33 or x.66 → multiply everything by 3
x.25 or x.75 → multiply everything by 4
Anything else → treat as already whole (round)
WORKED EXAMPLE — Finding a Molecular Formula from an Empirical Formula
Given empirical formula C5H7N, and known molecular molar mass = 162 g/mol.
Step 1 — Find the empirical formula's molar mass: C: 5(12.01) + H: 7(1.01) + N: 1(14.01) = 81.13 g/mol.
Step 2 — Divide molecular mass by empirical mass: 162 ÷ 81.13 ≈ 2.
Step 3 — Multiply every subscript in the empirical formula by that whole number: C10H14N2.
✏ TRY IT YOURSELF
1. A compound has an empirical formula of CH2O and a molar mass of 180 g/mol. Find the molecular formula.
2. A compound is 40.0% C, 6.7% H, 53.3% O by mass. Find the empirical formula.
