1. What is the percentage composition of carbon in methane (CH₄)?
Atomic masses: C = 12.0 g/mol, H = 1.0 g/mol
Solution: Molar mass of CH₄ = 12.0 + (4 × 1.0) = 16.0 g/mol. Percentage of C = (12.0 / 16.0) × 100 = 75.0%

2. Step-by-Step Problem: Calculate the percentage composition of water (H₂O).
Guide: Step 1 - Find molar mass, Step 2 - Find mass of each element, Step 3 - Calculate percentages
Solution: Molar mass of H₂O = (2 × 1.0) + 16.0 = 18.0 g/mol. Percentage of H = (2.0 / 18.0) × 100 = 11.1%, Percentage of O = (16.0 / 18.0) × 100 = 88.9%

3. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Which elements are present? (Check all that apply)
Solution: The elements present are Carbon, Hydrogen, and Oxygen.

4. Guided Problem: Find the empirical formula of a compound with 85.7% C and 14.3% H.
Step 1: Convert percentages to grams: 85.7 g C, 14.3 g H
Step 2: Convert to moles: 85.7 g C / 12.0 g/mol = 7.14 mol C, 14.3 g H / 1.0 g/mol = 14.3 mol H
Step 3: Find simplest ratio: C:H = 7.14:14.3 = 1:2
Empirical formula: CH₂

5. If the empirical formula from question 4 has a molar mass of 28.0 g/mol, what is the molecular formula?
Solution: Molar mass of empirical formula (CH₂) = 12.0 + (2 × 1.0) = 14.0 g/mol. Molecular formula = (28.0 g/mol / 14.0 g/mol) × CH₂ = C₂H₄.

6. Extension Challenge: A compound contains 52.2% C, 13.0% H, and 34.8% O. The molar mass is 46.0 g/mol. Determine both empirical and molecular formulas.
Solution: Convert percentages to grams: 52.2 g C, 13.0 g H, 34.8 g O. Moles: C = 52.2/12.0 = 4.35, H = 13.0/1.0 = 13.0, O = 34.8/16.0 = 2.18. Simplest ratio: C:H:O = 4.35:13.0:2.18 = 2:6:1. Empirical formula = C₂H₆O. Molar mass of empirical formula = 2(12.0) + 6(1.0) + 16.0 = 46.0 g/mol. Molecular formula = C₂H₆O.

7. Format:
• A vitamin C tablet contains 0.18 g of ascorbic acid (C₆H₈O₆)
• Molar mass of ascorbic acid = 176.0 g/mol
• Calculate: How many moles of vitamin C are in one tablet?
Solution: Moles = mass / molar mass = 0.18 g / 176.0 g/mol = 0.00102 mol.

8. Advanced Extension: A hydrocarbon produces fractional mole ratios of C:H = 1.00:2.25.
Show how to convert this to a whole number ratio and determine the empirical formula.
Solution: To convert to whole numbers, multiply by 4: C = 4, H = 9. Empirical formula = C₄H₉.

9. Real-World Application: Explain why knowing the percentage composition of compounds is important in:
a) Medicine: ________________
b) Environmental science: ________________

1. What is the percentage composition of carbon in methane (CH₄)?
Atomic masses: C = 12.0 g/mol, H = 1.0 g/mol
Solution: Molar mass of CH₄ = 12.0 + (4 × 1.0) = 16.0 g/mol. Percentage of C = (12.0 / 16.0) × 100 = 75.0%

2. Step-by-Step Problem: Calculate the percentage composition of water (H₂O).
Guide: Step 1 - Find molar mass, Step 2 - Find mass of each element, Step 3 - Calculate percentages
Solution: Molar mass of H₂O = (2 × 1.0) + 16.0 = 18.0 g/mol. Percentage of H = (2.0 / 18.0) × 100 = 11.1%, Percentage of O = (16.0 / 18.0) × 100 = 88.9%

3. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
Which elements are present? (Check all that apply)
Solution: The elements present are Carbon, Hydrogen, and Oxygen.

4. Guided Problem: Find the empirical formula of a compound with 85.7% C and 14.3% H.
Step 1: Convert percentages to grams: 85.7 g C, 14.3 g H
Step 2: Convert to moles: 85.7 g C / 12.0 g/mol = 7.14 mol C, 14.3 g H / 1.0 g/mol = 14.3 mol H
Step 3: Find simplest ratio: C:H = 7.14:14.3 = 1:2
Empirical formula: CH₂

5. If the empirical formula from question 4 has a molar mass of 28.0 g/mol, what is the molecular formula?
Solution: Molar mass of empirical formula (CH₂) = 12.0 + (2 × 1.0) = 14.0 g/mol. Molecular formula = (28.0 g/mol / 14.0 g/mol) × CH₂ = C₂H₄.

6. Extension Challenge: A compound contains 52.2% C, 13.0% H, and 34.8% O. The molar mass is 46.0 g/mol. Determine both empirical and molecular formulas.
Solution: Convert percentages to grams: 52.2 g C, 13.0 g H, 34.8 g O. Moles: C = 52.2/12.0 = 4.35, H = 13.0/1.0 = 13.0, O = 34.8/16.0 = 2.18. Simplest ratio: C:H:O = 4.35:13.0:2.18 = 2:6:1. Empirical formula = C₂H₆O. Molar mass of empirical formula = 2(12.0) + 6(1.0) + 16.0 = 46.0 g/mol. Molecular formula = C₂H₆O.

7. Format:
• A vitamin C tablet contains 0.18 g of ascorbic acid (C₆H₈O₆)
• Molar mass of ascorbic acid = 176.0 g/mol
• Calculate: How many moles of vitamin C are in one tablet?
Solution: Moles = mass / molar mass = 0.18 g / 176.0 g/mol = 0.00102 mol.

8. Advanced Extension: A hydrocarbon produces fractional mole ratios of C:H = 1.00:2.25.
Show how to convert this to a whole number ratio and determine the empirical formula.
Solution: To convert to whole numbers, multiply by 4: C = 4, H = 9. Empirical formula = C₄H₉.

9. Real-World Application: Explain why knowing the percentage composition of compounds is important in:
a) Medicine: ________________
b) Environmental science: ________________