Solved Empirical and Molecular Formula Worksheet Show ALL | Chegg.com - Free Printable

1. a. Ribose, C₅H₁₀O₅, a sugar molecule in RNA.
- Molecular formula: C₅H₁₀O₅
- Empirical formula: CH₂O (divide by 5)
- Answer: Molecular formula and empirical formula (both)

b. Ethyl butanoate, C₆H₁₂O₂, a compound with the odor of pineapple.
- Molecular formula: C₆H₁₂O₂
- Empirical formula: C₃H₆O (divide by 2)
- Answer: Molecular formula and empirical formula (both)

c. Chlorophyll, C₅₅H₇₂MgN₄O₅, part of photosynthesis.
- Molecular formula: C₅₅H₇₂MgN₄O₅
- Empirical formula: cannot be simplified further (no common divisor)
- Answer: Molecular formula only

d. DEET, C₁₂H₁₇ON, an insect repellent.
- Molecular formula: C₁₂H₁₇ON
- Empirical formula: cannot be simplified further
- Answer: Molecular formula only

e. Oxalic acid H₂C₂O₄, found in spinach and tea.
- Molecular formula: H₂C₂O₄
- Empirical formula: HCO₂ (divide by 2)
- Answer: Molecular formula and empirical formula (both)

2. a. 94.1% O, 5.9% H
Assume 100 g sample:
- O: 94.1 g → 94.1 / 16.00 = 5.881 mol
- H: 5.9 g → 5.9 / 1.008 = 5.853 mol
Divide by smallest:
- O: 5.881 / 5.853 ≈ 1.005 ≈ 1
- H: 5.853 / 5.853 = 1
Empirical formula: H₂O? Wait, ratio is 1:1, but H:O = 1:1?
Wait, H: 5.853 mol, O: 5.881 mol → ratio ≈ 1:1
So empirical formula: HO
But HO is not a stable compound. Wait, perhaps I made a mistake.
Wait, 5.9% H, 94.1% O → moles:
H: 5.9 / 1.008 = 5.853
O: 94.1 / 16.00 = 5.881
Ratio H:O = 5.853 : 5.881 ≈ 1:1
So empirical formula: HO
But HO is hydroxyl radical. But in practice, this might be H₂O₂?
Wait, H₂O₂ is 5.9% H, 94.1% O? Let's check:
H₂O₂: 2×1.008 = 2.016, 2×16.00 = 32.00, total = 34.016
%H = 2.016 / 34.016 ≈ 5.92%, %O = 94.08% → matches
So empirical formula: HO, molecular formula: H₂O₂
But the question asks for empirical formula. So empirical formula is HO.
Answer: HO

b. 79.9% C, 20.1% H
Assume 100 g:
- C: 79.9 g → 79.9 / 12.01 = 6.653 mol
- H: 20.1 g → 20.1 / 1.008 = 19.93 mol
Divide by smallest (6.653):
- C: 6.653 / 6.653 = 1
- H: 19.93 / 6.653 ≈ 3.00
Empirical formula: CH₃
Answer: CH₃

3. Methyl butanoate: 58.8% C, 9.8% H, 31.4% O
Gram molecular mass: 102 g/mol
Assume 100 g:
- C: 58.8 g → 58.8 / 12.01 = 4.896 mol
- H: 9.8 g → 9.8 / 1.008 = 9.719 mol
- O: 31.4 g → 31.4 / 16.00 = 1.9625 mol
Divide by smallest (1.9625):
- C: 4.896 / 1.9625 ≈ 2.495 ≈ 2.5
- H: 9.719 / 1.9625 ≈ 4.95 ≈ 5
- O: 1.9625 / 1.9625 = 1
Multiply by 2 to eliminate decimal:
- C: 5, H: 10, O: 2
Empirical formula: C₅H₁₀O₂
Empirical mass: 5×12.01 + 10×1.008 + 2×16.00 = 60.05 + 10.08 + 32.00 = 102.13 g/mol
Gram molecular mass is 102 g/mol → matches empirical mass
So molecular formula = empirical formula = C₅H₁₀O₂
Answer: C₅H₁₀O₂

4. Virus particle diameter: 5 × 10⁻⁶ cm
Avogadro's number: 6.022 × 10²³ particles
Length of line: 6.022 × 10²³ × 5 × 10⁻⁶ cm = 3.011 × 10¹⁸ cm
Convert to km:
1 km = 10⁵ cm → 3.011 × 10¹⁸ cm / 10⁵ = 3.011 × 10¹³ km
Answer: 3.011 × 10¹³ km

5. a. 92.25% C, 7.75% H
Assume 100 g:
- C: 92.25 g → 92.25 / 12.01 = 7.681 mol
- H: 7.75 g → 7.75 / 1.008 = 7.688 mol
Divide by smallest:
- C: 7.681 / 7.681 = 1
- H: 7.688 / 7.681 ≈ 1.001 ≈ 1
Empirical formula: CH
Answer: CH

b. Gram molecular mass: 52.03 g/mol
Empirical formula: CH → empirical mass = 12.01 + 1.008 = 13.018 g/mol
n = 52.03 / 13.018 ≈ 4.00
Molecular formula: (CH)₄ = C₄H₄
Answer: C₄H₄