Worksheet with eight empirical and molecular formula calculation problems.

Let’s solve each problem step by step. We’ll use the standard method for finding empirical and molecular formulas:

Step 1: Assume 100 g of compound → masses become grams.
Step 2: Convert grams to moles using atomic masses (C=12, H=1, O=16, N=14, F=19, P=31, I=127).
Step 3: Divide all mole values by the smallest one to get ratios.
Step 4: If needed, multiply to get whole numbers → that’s the empirical formula.
Step 5: For molecular formula, divide molar mass by empirical formula mass → multiply subscripts.

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Problem 1:

Composition: 48.0% C, 4.0% H, 48.0% O

→ In 100g:
C: 48.0g → 48/12 = 4 mol
H: 4.0g → 4/1 = 4 mol
O: 48.0g → 48/16 = 3 mol

Ratios: C₄H₄O₃ → already whole numbers.

✔ Empirical formula: C₄H₄O₃

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Problem 2:

36.0% C, 4.0% H, 28.0% N, 32.0% O

→ In 100g:
C: 36/12 = 3 mol
H: 4/1 = 4 mol
N: 28/14 = 2 mol
O: 32/16 = 2 mol

Ratios: C₃H₄N₂O₂ → already whole.

✔ Empirical formula: C₃H₄N₂O₂

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Problem 3:

24.0% C, 7.0% H, 38.0% F, 31.0% P

→ In 100g:
C: 24/12 = 2 mol
H: 7/1 = 7 mol
F: 38/19 = 2 mol
P: 31/31 = 1 mol

Ratios: C₂H₇F₂P → already whole.

✔ Empirical formula: C₂H₇F₂P

Wait — answer key says [C₂H₇F₂P] but in image it's written as [C₂H₇F₂P]? Actually, let me check again.

Hold on — 7.0% H → 7g → 7 mol? That seems high compared to others. But math checks out.

But wait — maybe typo in my reading? Let me recheck:

Original: “24.0 % C, 7.0 % H, 38.0 % F and 31.0 % P”

Yes → 24/12=2, 7/1=7, 38/19=2, 31/31=1 → C₂H₇F₂P

But answer key says [C₂H₇F₂P] — yes, matches.

Actually, looking back at user input: it says “[C₂H₇F₂P]” — so correct.

✔ Final: C₂H₇F₂P

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Problem 4:

48.0% C, 8.0% H, 28.0% N, 16.0% O → molar mass = 200 g/mol

→ In 100g:
C: 48/12 = 4
H: 8/1 = 8
N: 28/14 = 2
O: 16/16 = 1

Empirical: C₄H₈N₂O → mass = 4×12 + 8×1 + 2×14 + 16 = 48+8+28+16 = 100 g/mol

Molecular mass = 200 → ratio = 200/100 = 2

Multiply subscripts: C₈H₁₆N₄O₂

✔ Empirical: C₄H₈N₂O, Molecular: C₈H₁₆N₄O₂

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Problem 5:

100 mg sample → burned → 176 mg CO₂, 36 mg H₂O → find empirical formula (contains C,H,O)

First, find mass of C from CO₂:

CO₂ is 44 g/mol → 12g C per 44g CO₂

So, 176 mg CO₂ → (12/44)*176 = (3/11)*176 = 3*16 = 48 mg C

Mass of H from H₂O:

H₂O is 18 g/mol → 2g H per 18g H₂O

36 mg H₂O → (2/18)*36 = (1/9)*36 = 4 mg H

Total mass accounted for: C + H = 48 + 4 = 52 mg

Sample was 100 mg → so O = 100 - 52 = 48 mg

Now convert to moles (in mmol since we’re in mg):

C: 48 mg / 12 mg/mmol = 4 mmol
H: 4 mg / 1 mg/mmol = 4 mmol
O: 48 mg / 16 mg/mmol = 3 mmol

Ratio: C₄H₄O₃

✔ Empirical formula: C₄H₄O₃

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Problem 6:

60.0% C, 12.0% H, 28.0% N → molar mass = 300 g/mol

→ In 100g:
C: 60/12 = 5
H: 12/1 = 12
N: 28/14 = 2

Empirical: C₅H₁₂N₂ → mass = 5×12 + 12×1 + 2×14 = 60+12+28 = 100 g/mol

Molecular mass = 300 → ratio = 3

Multiply: C₁₅H₃₆N₆

✔ Empirical: C₅H₁₂N₂, Molecular: C₁₅H₃₆N₆

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Problem 7:

18.0% C, 2.5% H, 63.5% I, 16.0% O → molar mass = 400 g/mol

→ In 100g:
C: 18/12 = 1.5
H: 2.5/1 = 2.5
I: 63.5/127 = 0.5
O: 16/16 = 1

Divide by smallest (0.5):
C: 1.5/0.5 = 3
H: 2.5/0.5 = 5
I: 0.5/0.5 = 1
O: 1/0.5 = 2

Empirical: C₃H₅IO₂ → mass = 3×12 + 5×1 + 127 + 2×16 = 36+5+127+32 = 200 g/mol

Molecular mass = 400 → ratio = 2

Multiply: C₆H₁₀I₂O₄

✔ Empirical: C₃H₅IO₂, Molecular: C₆H₁₀I₂O₄

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Problem 8:

60.0% C, 8.0% H, 32.0% O → empirical first

→ In 100g:
C: 60/12 = 5
H: 8/1 = 8
O: 32/16 = 2

Empirical: C₅H₈O₂ → mass = 5×12 + 8×1 + 2×16 = 60+8+32 = 100 g/mol

Now, titration part:

100 mg compound neutralized with 10.0 mL of 0.0500 M NaOH, 1:1 stoichiometry.

Moles of NaOH used = volume(L) × concentration = 0.010 L × 0.0500 mol/L = 0.0005 mol

Since 1:1 reaction → moles of compound = 0.0005 mol

Mass of compound = 100 mg = 0.100 g

So, molar mass = mass / moles = 0.100 g / 0.0005 mol = 200 g/mol

Empirical mass = 100 → ratio = 200/100 = 2

Molecular formula: C₁₀H₁₆O

✔ Empirical: C₅H₈O₂, Molecular: C₁₀H₁₆O₄

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Final Answer:
1. C₄H₄O₃
2. C₃H₄N₂O₂
3. C₂H₇F₂P
4. Empirical: C₄H₈N₂O; Molecular: C₈H₁N₄O₂
5. C₄H₄O₃
6. Empirical: C₅H₁₂N₂; Molecular: C₁₅H₃₆N₆
7. Empirical: C₃H₅IO₂; Molecular: C₆H₁₀I₂O₄
8. Empirical: C₅H₈O₂; Molecular: C₁₀H₁₆O₄