Mole - Molar Mass - Percent Composition - Empirical and Molecular Formulas

The Mole Concept

• A mole is a unit measuring the amount of a substance.
• 1 mole of any substance contains 6.022 × 10^{23} particles (Avogadro’s number).
  • These particles can be atoms, molecules, ions, etc.

Key Formulas:

• \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}
• \text{Particles} = \text{Moles} × \text{Avogadro’s Number}

Example:

• How many moles are in 22.4 g of CO_2?
• CO_2 molar mass ≈ 44 g/mol
• \text{Moles} = \frac{22.4 \text{ g}}{44 \text{ g/mol}} = 0.51 \text{ moles}

Molar Mass

• The mass of 1 mole of a compound, in grams per mole (g/mol).
• Calculated by adding up the atomic masses of each atom in a compound.

Example: Molar mass of Al2(SO4)_3:

• Al: 2 × 26.98
• S: 3 × 32.07
• O: 12 × 16.00
• Total = 342.17 g/mol

Mole-Volume Relationship (Gases)

• Applies to gases at standard temperature and pressure (STP).

At STP:

• Temperature = 0°C (273.15 K)

• Pressure = 1 atm

• Under these conditions: 1 mole of any gas = 22.4 liters

Key Formula:

• Only use if the gas is at STP!

Examples:

• Example 1: How many liters are in 3 moles of O_2 at STP?
  • 3 \text{ mol} × \frac{22.4 \text{ L}}{1 \text{ mol}} = 67.2 \text{ L}
• Example 2: What is the number of moles in 44.8 L of CO_2 gas at STP?
  • 44.8 \text{ L} × \frac{1 \text{ mol}}{22.4 \text{ L}} = 2.0 \text{ mol}

Percent Composition

• Tells you what percent of a compound's mass comes from each element.

Formula:

• \text{Percent Composition} = \frac{\text{Mass of element}}{\text{Total mass of compound}} × 100

Example:

• H_2O: H = 2.016 g, O = 16.00 g
• Molar Mass = 18.016 g
  • % H = \frac{2.016}{18.016} × 100 ≈ 11.2\%
  • % O = \frac{16.00}{18.016} × 100 ≈ 88.8\%

Empirical Formula

• The simplest whole-number ratio of atoms in a compound.

How to Find It:

1. Convert % composition to grams (assume 100g total).
2. Convert grams to moles for each element.
3. Divide all mole values by the smallest mole number.
4. Round or multiply to get whole numbers.

Example:

• 40% C, 6.7% H, 53.3% O
  • 40g C ÷ 12.01 = 3.33 mol
  • 6.7g H ÷ 1.008 = 6.65 mol
  • 53.3g O ÷ 16.00 = 3.33 mol
  • Empirical formula: CH_2O

Molecular Formula

• The actual number of atoms in a molecule (can be a multiple of the empirical formula).

Steps:

1. Find the molar mass of the empirical formula.
2. Divide the given molar mass by the empirical formula mass.
3. Multiply all subscripts in the empirical formula by that number.

Example:

• Empirical: CH_2O (30 g/mol), Molar Mass = 180 g/mol
  • 180 ÷ 30 = 6
  • Molecular formula = C6H{12}O_6

Practice Problems

Mole Concept Problems:

1. How many moles are in 22.4 grams of CO_2?
2. How many molecules are in 3.5 moles of H_2O?
3. How many atoms of hydrogen are in 5 moles of C2H6?

Molar Mass Problems:

4. Calculate the molar mass of:
   • a) NaCl
   • b) C6H{12}O_6
   • c) Al2(SO4)_3
5. How many grams are in 0.25 moles of Ca(NO3)2?

Mole–Volume Relationship Practice Problems (at STP)

• Remember: 1 mole of gas = 22.4 L at STP

Basic Problems

1. How many liters of oxygen gas are in 4.5 moles of O_2 at STP?
2. How many moles of nitrogen gas are in 67.2 L at STP?
3. A balloon contains 11.2 L of helium gas at STP. How many moles of He does it contain?
4. What volume will 0.75 moles of SO_2 gas occupy at STP?
5. How many liters are occupied by 2.25 moles of argon gas at STP?

Intermediate Problems

6. A container holds 134.4 L of CO2 gas at STP. How many moles of CO2 are present?
7. How many liters will 0.125 moles of NH_3 gas occupy at STP?
8. If 3.75 moles of hydrogen gas are released in a reaction, what volume does it occupy at STP?
9. A gas sample contains 6.7 moles of NO_2. What is the volume at STP?
10. How many moles are in 179.2 L of oxygen gas at STP?

Percent Composition Problems:

11. Calculate the percent composition of each element in H2SO4.
12. What is the percent by mass of carbon in C2H6?

Empirical Formula Problems:

13. A compound is found to contain 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find the empirical formula.
14. A compound has the following percent composition: 70% Fe and 30% O by mass. Determine the empirical formula.

Molecular Formula Problems:

15. A compound has an empirical formula of CH_2O and a molar mass of 180 g/mol. What is its molecular formula?
16. A compound has an empirical formula of C2H5 and a molar mass of 58 g/mol. Find the molecular formula.

Answer Keys

Mole Concept

1. How many moles are in 22.4 grams of CO_2?
   • Molar mass of CO_2 = 44.01 g/mol
   • 22.4 g ÷ 44.01 g/mol = 0.509 moles
2. How many molecules are in 3.5 moles of H_2O?
   • 1 mole = 6.022×10^{23} molecules
   • 3.5 mol × 6.022×10^{23} = 2.11 × 10^{24} molecules
3. How many atoms of hydrogen are in 5 moles of C2H6?
   • Each molecule has 6 H atoms
   • 5 mol × 6.022×10^{23} × 6 = 1.81 × 10^{25} H atoms

Molar Mass

4. Calculate the molar mass of:
   • a) NaCl → 22.99 + 35.45 = 58.44 g/mol
   • b) C6H{12}O_6 → (6×12.01) + (12×1.008) + (6×16.00) = 180.16 g/mol
   • c) Al2(SO4)_3 → 2×26.98 + 3×(32.07 + 4×16.00) = 342.17 g/mol
5. How many grams are in 0.25 moles of Ca(NO3)2?
   • Molar mass = 40.08 + 2×(14.01 + 3×16.00) = 164.10 g/mol
   • 0.25 mol × 164.10 g/mol = 41.03 g

Mole–Volume Relationship Practice Problems (at STP)

Basic Problems

1. How many liters of oxygen gas are in 4.5 moles of O_2 at STP?
   • 4. 5×22.4=100.8 L
2. How many moles of nitrogen gas are in 67.2 L at STP?
   • 67. 2÷22.4=3.0 mol
3. 11.2 L of helium gas → How many moles?
   • 11. 2÷22.4=0.5 mol
4. Volume of 0.75 mol SO_2 at STP
   • 0. 75×22.4=16.8 L
5. 2.25 mol of argon gas → Volume?
   • 2. 25×22.4=50.4 L

Intermediate Problems

6. 134.4 L of CO_2 → Moles?
   • 134. 4÷22.4=6.0 mol
7. 0.125 mol NH_3 → Volume?
   • 0. 125×22.4=2.8 L
8. 3.75 mol H_2 → Volume?
   • 3. 75×22.4=84.0 L
9. 7 mol NO_2 → Volume?
   • 6. 7×22.4=150.08 L
10. 2 L O_2 → Moles?
    • 179. 2÷22.4=8.0 mol

Percent Composition

11. H2SO4
    • H: (2×1.008)/98.08 × 100 = 2.06%
    • S: 32.07/98.08 × 100 = 32.69%
    • O: (4×16.00)/98.08 × 100 = 65.25%
12. C2H6
    • C: (2×12.01)/30.08 × 100 = 79.86%

Empirical Formula

13. 40% C, 6.7% H, 53.3% O
    • Assume 100g → C = 40g, H = 6.7g, O = 53.3g → Convert to moles:
      • C = 3.33 mol
      • H = 6.65 mol
      • O = 3.33 mol
    • Simplest ratio: C1H2O1 → CH2O
14. 70% Fe, 30% O
    • Assume 100g → Fe = 70g, O = 30g
      • Fe = 1.25 mol
      • O = 1.875 mol
    • Divide by smallest: Fe = 1, O = 1.5 → Multiply by 2 → Fe2O3

Molecular Formula

15. Empirical: CH_2O, Molar mass: 180 g/mol
    • Molar mass of CH_2O = 30 g/mol
    • 180 ÷ 30 = 6
    • C6H{12}O_6
16. Empirical: C2H5, Molar mass: 58 g/mol
    • Molar mass of C2H5 = 29 g/mol
    • 58 ÷ 29 = 2
    • C4H{10}