Chemistry 1/9 - Daniel Davis - Free Printable

Let’s solve each problem step by step. We’ll find the molecular formula using the empirical formula and molecular mass, or from percent composition if needed.

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Problem 1:

Empirical formula = NO₂
Molecular mass = 92 g/mol

Step 1: Find the mass of the empirical formula.

- N = 14 g/mol
- O = 16 g/mol → O₂ = 32 g/mol
- Empirical mass = 14 + 32 = 46 g/mol

Step 2: Divide molecular mass by empirical mass to get the multiplier.

→ 92 ÷ 46 = 2

Step 3: Multiply the subscripts in the empirical formula by 2.

→ N₁×2 O₂×2 = N₂O₄

✔ Final Answer for #1: N₂O₄

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Problem 2:

Empirical formula = CH₂
Molecular mass = 70 g/mol

Step 1: Empirical mass

- C = 12
- H₂ = 2 × 1 = 2
- Total = 12 + 2 = 14 g/mol

Step 2: Multiplier = 70 ÷ 14 = 5

Step 3: Multiply subscripts → C₁×5 H₂×5 = C₅H₁₀

✔ Final Answer for #2: C₅H₁₀

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Problem 3:

Percent composition:
Carbon = 40.0%
Hydrogen = 6.7%
Oxygen = 53.5%
Molecular mass = 60 g/mol

Step 1: Assume 100 g sample → so we have:

- C: 40.0 g
- H: 6.7 g
- O: 53.5 g

Step 2: Convert to moles.

- Moles of C = 40.0 ÷ 12 ≈ 3.33 mol
- Moles of H = 6.7 ÷ 1 = 6.7 mol
- Moles of O = 53.5 ÷ 16 ≈ 3.34 mol

Step 3: Divide all by smallest number (≈3.33)

- C: 3.33 ÷ 3.33 = 1
- H: 6.7 ÷ 3.33 ≈ 2.01 → ~2
- O: 3.34 ÷ 3.33 ≈ 1.003 → ~1

So empirical formula = CH₂O

Step 4: Empirical mass = 12 + 2(1) + 16 = 30 g/mol

Step 5: Multiplier = 60 ÷ 30 = 2

Step 6: Molecular formula = C₁×2 H₂×2 O₁×2 = C₂H₄O₂

✔ Final Answer for #3: C₂H₄O₂

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Problem 4:

Percent composition:
Carbon = 64.9%
Hydrogen = 13.5%
Oxygen = 21.6%
Molecular mass = 74 g/mol

Step 1: Assume 100 g → masses:

- C: 64.9 g
- H: 13.5 g
- O: 21.6 g

Step 2: Moles:

- C: 64.9 ÷ 12 ≈ 5.408 mol
- H: 13.5 ÷ 1 = 13.5 mol
- O: 21.6 ÷ 16 = 1.35 mol

Step 3: Divide by smallest (1.35):

- C: 5.408 ÷ 1.35 ≈ 4.006 → ~4
- H: 13.5 ÷ 1.35 = 10
- O: 1.35 ÷ 1.35 = 1

Empirical formula = C₄H₁₀O

Step 4: Empirical mass = 4×12 + 10×1 + 16 = 48 + 10 + 16 = 74 g/mol

Step 5: Molecular mass is also 74 → multiplier = 1

So molecular formula = same as empirical → C₄H₁₀O

✔ Final Answer for #4: C₄H₁₀O

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Problem 5:

Percent composition:
Carbon = 54.5%
Hydrogen = 9.1%
Oxygen = 36.4%
Molecular mass = 88 g/mol

Step 1: Assume 100 g → masses:

- C: 54.5 g
- H: 9.1 g
- O: 36.4 g

Step 2: Moles:

- C: 54.5 ÷ 12 ≈ 4.542 mol
- H: 9.1 ÷ 1 = 9.1 mol
- O: 36.4 ÷ 16 = 2.275 mol

Step 3: Divide by smallest (2.275):

- C: 4.542 ÷ 2.275 ≈ 2.00 → 2
- H: 9.1 ÷ 2.275 ≈ 4.00 → 4
- O: 2.275 ÷ 2.275 = 1

Empirical formula = C₂H₄O

Step 4: Empirical mass = 2×12 + 4×1 + 16 = 24 + 4 + 16 = 44 g/mol

Step 5: Multiplier = 88 ÷ 44 = 2

Step 6: Molecular formula = C₂×2 H₄×2 O₁×2 = C₄H₈O₂

✔ Final Answer for #5: C₄H₈O₂

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Final Answer:
1. N₂O₄
2. C₅H₁₀
3. C₂H₄O₂
4. C₄H₁₀O
5. C₄H₈O₂