AP Chemistry: The Mole, Molar Mass, and Avogadro’s Number

AP Chemistry – The Mole, Molar Mass, and Avogadro’s Number

AP Chemistry Study Guide: The Mole, Molar Mass, and Avogadro’s Number

What This Is

The mole is the chemist’s "counting unit" for atoms, molecules, or ions—just like a dozen eggs is 12 eggs, a mole is 6.022 × 10²³ particles (Avogadro’s number). It’s the bridge between the microscopic world (atoms, molecules) and the macroscopic world (grams, liters). Without the mole, we couldn’t measure reactants or predict products in chemical reactions. Real-world example: A single drop of water contains about 1.5 × 10²¹ molecules—that’s trillions of molecules, but only 0.05 moles of water. The mole lets us scale up tiny particles to measurable quantities.

Key Terms & Concepts

• Mole (mol): The SI unit for amount of substance. 1 mole = 6.022 × 10²³ particles (Avogadro’s number).
• Avogadro’s Number (N?): 6.022 × 10²³ particles/mol. Named after Amedeo Avogadro, who proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
• Molar Mass (g/mol): The mass of 1 mole of a substance. Numerically equal to the atomic mass (from the periodic table) but in grams. Example: Carbon’s atomic mass is 12.01 amu, so its molar mass is 12.01 g/mol.
• Atomic Mass Unit (amu): A unit of mass for single atoms/molecules. 1 amu = 1.66 × 10?²? g (mass of 1 proton or neutron).
• Formula Mass: The sum of atomic masses in a compound’s formula (e.g., H?O = 2(1.01) + 16.00 = 18.02 amu).
• Molar Mass of a Compound: The sum of molar masses of all atoms in the formula (e.g., H?O = 18.02 g/mol).
• Percent Composition: The percent by mass of each element in a compound.
• Formula: % element = (mass of element in 1 mol / molar mass of compound) × 100%
• Empirical Formula: The simplest whole-number ratio of atoms in a compound (e.g., CH?O for glucose, C?HO?).
• Molecular Formula: The actual number of atoms in a molecule (e.g., C?HO? for glucose).
• Stoichiometry: Using mole ratios from balanced equations to calculate reactants/products.
• Density (g/mL or g/L): Mass per unit volume. Often used to convert between mass and volume for gases/liquids.
• Standard Temperature and Pressure (STP): 0°C (273 K) and 1 atm. At STP, 1 mole of any gas occupies 22.4 L (molar volume).

Step-by-Step / Process Flow

1. Converting Between Grams and Moles

Problem: How many moles are in 50.0 g of CO
1. Find molar mass of CO?: C (12.01 g/mol) + 2(O) (2 × 16.00 g/mol) = 44.01 g/mol.
2. Use the formula: [ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} ]
3. Plug in numbers: [ \text{moles CO?} = \frac{50.0 \text{ g}}{44.01 \text{ g/mol}} = 1.14 \text{ mol} ]

2. Converting Between Moles and Particles (Atoms/Molecules)

Problem: How many molecules are in 2.5 moles of H?O?
1. Use Avogadro’s number: [ \text{particles} = \text{moles} \times N_A ]
2. Plug in numbers: [ \text{molecules H?O} = 2.5 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.5 \times 10^{24} \text{ molecules} ]

3. Finding Percent Composition

Problem: What is the percent composition of oxygen in CaCO
1. Find molar mass of CaCO?: Ca (40.08) + C (12.01) + 3(O) (3 × 16.00) = 100.09 g/mol.
2. Mass of oxygen in 1 mol CaCO?: 3 × 16.00 = 48.00 g.
3. Calculate % O: [ \% \text{O} = \left( \frac{48.00 \text{ g}}{100.09 \text{ g}} \right) \times 100\% = 47.96\% ]

4. Determining Empirical Formula from Percent Composition

Problem: A compound is 40.0% C, 6.7% H, and 53.3% O by mass. What is its empirical formula?
1. Assume 100 g of compound-40.0 g C, 6.7 g H, 53.3 g O.
2. Convert grams to moles: - C: ( \frac{40.0 \text{ g}}{12.01 \text{ g/mol}} = 3.33 \text{ mol} ) - H: ( \frac{6.7 \text{ g}}{1.01 \text{ g/mol}} = 6.63 \text{ mol} ) - O: ( \frac{53.3 \text{ g}}{16.00 \text{ g/mol}} = 3.33 \text{ mol} )
3. Divide by smallest mole value (3.33): - C: ( \frac{3.33}{3.33} = 1 ) - H: ( \frac{6.63}{3.33} \approx 2 ) - O: ( \frac{3.33}{3.33} = 1 )
4. Empirical formula = CH?O.

5. Finding Molecular Formula from Empirical Formula

Problem: The empirical formula of a compound is CH?O, and its molar mass is 180 g/mol. What is its molecular formula?
1. Find molar mass of empirical formula (CH?O): 12.01 + 2(1.01) + 16.00 = 30.03 g/mol.
2. Divide molecular molar mass by empirical molar mass: ( \frac{180 \text{ g/mol}}{30.03 \text{ g/mol}} \approx 6 ).
3. Multiply subscripts by 6: Molecular formula = C?HO?.

Common Mistakes

Mistake 1: Confusing atomic mass (amu) with molar mass (g/mol).

• Correction: Atomic mass is for one atom (e.g., 12.01 amu for C), while molar mass is for 1 mole of atoms (12.01 g/mol for C). They’re numerically the same but have different units.

Mistake 2: Forgetting to multiply by subscripts in molar mass calculations.

• Correction: For H?O, molar mass = 2(H) + 1(O) = 2(1.01) + 16.00 = 18.02 g/mol, not 1.01 + 16.00.

Mistake 3: Using grams instead of moles in stoichiometry.

• Correction: Balanced equations use mole ratios, not gram ratios. Always convert grams-moles before using coefficients.

Mistake 4: Rounding too early in empirical formula calculations.

• Correction: Keep 3+ decimal places until the final step to avoid rounding errors (e.g., 1.333-4/3, not 1.3).

Mistake 5: Misapplying Avogadro’s number (e.g., using it for grams-moles).

• Correction: Avogadro’s number converts moles-particles, not grams-moles. Use molar mass for grams-moles.

AP Exam Insights

1. Multiple-Choice Traps

• Tricky Distinction: The exam loves to test molar mass vs. atomic mass. Remember:
• Atomic mass (amu) = mass of one atom.
• Molar mass (g/mol) = mass of 1 mole of atoms.
• STP Gas Volume: At STP, 1 mole of any gas = 22.4 L. The exam may give a volume at non-STP conditions to trick you.

2. Free-Response Questions (FRQs)

• Common FRQ Type: Calculate percent composition, empirical formula, or molecular formula from lab data (e.g., combustion analysis).
• Stoichiometry Connection: FRQs often combine moles with limiting reactants or percent yield. Always start with a balanced equation.
• Experimental Design: You might be asked to design a lab to determine the empirical formula of a compound (e.g., using mass loss in a reaction).

3. Calculator Pitfalls

• Sig Figs: The AP exam deducts points for incorrect significant figures. Round only at the end.
• Scientific Notation: Avogadro’s number is 6.022 × 10²³, not 6.02 × 10²³ (unless specified).

4. Real-World Applications

• Medicine: Drug dosages are often calculated in moles (e.g., 500 mg of aspirin = 0.0028 mol).
• Environmental Science: CO? emissions are measured in moles or metric tons (1 ton CO?-22,700 moles).

Quick Check Questions

1. Multiple Choice

What is the molar mass of aluminum sulfate, Al?(SO?) (A) 123.0 g/mol (B) 278.0 g/mol (C) 342.1 g/mol (D) 450.0 g/mol

Answer: (C) 342.1 g/mol Explanation: Al? = 2(26.98), S? = 3(32.07), O = 12(16.00)-53.96 + 96.21 + 192.00 = 342.17 g/mol (rounded to 342.1).

2. Short FRQ

A 2.50 g sample of an unknown hydrocarbon (contains only C and H) is combusted, producing 7.66 g of CO? and 3.15 g of H?O. (a) Calculate the moles of CO? and H?O produced. (b) Determine the empirical formula of the hydrocarbon.

Answer: (a) - Moles CO? = ( \frac{7.66 \text{ g}}{44.01 \text{ g/mol}} = 0.174 \text{ mol} ) - Moles H?O = ( \frac{3.15 \text{ g}}{18.02 \text{ g/mol}} = 0.175 \text{ mol} )

(b) - Moles C = 0.174 mol (from CO?) - Moles H = 2 × 0.175 = 0.350 mol (from H?O) - Ratio C:H = 0.174 : 0.350-1 : 2 - Empirical formula = CH?

3. Multiple Choice

How many atoms are in 0.250 moles of nitrogen gas (N?)? (A) 1.51 × 10²³ atoms (B) 3.01 × 10²³ atoms (C) 6.02 × 10²³ atoms (D) 1.20 × 10²? atoms

Answer: (B) 3.01 × 10²³ atoms Explanation: 0.250 mol N? × 6.022 × 10²³ molecules/mol × 2 atoms/molecule = 3.01 × 10²³ atoms.

Last-Minute Cram Sheet

1. 1 mole = 6.022 × 10²³ particles (Avogadro’s number).
2. Molar mass (g/mol) = atomic mass (amu) but in grams (e.g., O = 16.00 g/mol).
3. Percent composition = (mass of element / molar mass) × 100%.
4. Empirical formula = simplest whole-number ratio (e.g., CH?O for glucose).
5. Molecular formula = empirical formula × n (where n = molar mass / empirical mass).
6. At STP, 1 mole of gas = 22.4 L.
7. Always convert grams-moles before using balanced equations.
8. Don’t round until the final answer (sig figs matter!).
9. Hydrocarbons: C?H?-CO? (for C) and H?O (for H) in combustion analysis.
10. Avogadro’s number is for particles, not grams! (Use molar mass for grams-moles.)