6 Problems Worksheet - Empirical and Molecular Formulas - Studocu - Free Printable

Let's solve each problem step by step.

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Problem 1:

What is the empirical formula for a compound which contains 53.73 % Fe and 46.27 % S?

#### Step 1: Assume 100 g of compound
- Mass of Fe = 53.73 g
- Mass of S = 46.27 g

#### Step 2: Convert masses to moles
Use atomic masses:
- Fe: 55.85 g/mol
- S: 32.07 g/mol

$$
\text{Moles of Fe} = \frac{53.73}{55.85} \approx 0.962 \, \text{mol}
$$
$$
\text{Moles of S} = \frac{46.27}{32.07} \approx 1.443 \, \text{mol}
$$

#### Step 3: Divide by smallest number of moles
Smallest is 0.962

$$
\text{Fe: } \frac{0.962}{0.962} = 1.00
$$
$$
\text{S: } \frac{1.443}{0.962} \approx 1.50
$$

We get a ratio of Fe:S = 1 : 1.5 → Multiply by 2 to get whole numbers:

→ Fe₂S₃

#### ✔ Empirical Formula: Fe₂S₃

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Problem 2:

What is the empirical formula of a compound which contains 15.77 % aluminum, 28.11 % sulfur, and 56.12 % oxygen?

#### Step 1: Assume 100 g of compound
- Al: 15.77 g
- S: 28.11 g
- O: 56.12 g

#### Step 2: Convert to moles
Atomic masses:
- Al: 26.98 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol

$$
\text{Al: } \frac{15.77}{26.98} \approx 0.584 \, \text{mol}
$$
$$
\text{S: } \frac{28.11}{32.07} \approx 0.877 \, \text{mol}
$$
$$
\text{O: } \frac{56.12}{16.00} = 3.5075 \, \text{mol}
$$

#### Step 3: Divide by smallest (0.584)

$$
\text{Al: } \frac{0.584}{0.584} = 1.00
$$
$$
\text{S: } \frac{0.877}{0.584} \approx 1.50
$$
$$
\text{O: } \frac{3.5075}{0.584} \approx 6.00
$$

Ratio: Al : S : O = 1 : 1.5 : 6 → Multiply all by 2 to eliminate decimal:

→ Al₂S₃O₁₂

Now simplify: Al₂(SO₄)₃ — this is aluminum sulfate.

But we are asked for empirical formula.

So: Al₂S₃O₁₂ is already simplified (no common divisor), but let’s check if it can be reduced.

GCF of 2,3,12 is 1 → cannot reduce further.

But note: Al₂S₃O₁₂ is the same as Al₂(SO₄)₃, which is correct.

However, the empirical formula should be the simplest whole-number ratio.

Is there a smaller ratio?

No — 2,3,12 have no common factor other than 1.

So, Al₂S₃O₁₂ is the empirical formula.

But wait: Let's double-check the mole ratios again.

Wait: We had:

- Al: 0.584
- S: 0.877 ≈ 0.584 × 1.5
- O: 3.5075 ≈ 0.584 × 6

So yes: 1 : 1.5 : 6 → multiply by 2 → 2 : 3 : 12

✔ Empirical Formula: Al₂S₃O₁₂

But this is actually Al₂(SO₄)₃, and that’s the standard formula.

Yes, it's correct.

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Problem 3:

An organic compound is found to contain 92.25 % carbon and 7.75 % hydrogen. If the molecular mass is 78 g/mol, what are its empirical and molecular formulas?

#### Step 1: Assume 100 g of compound
- C: 92.25 g
- H: 7.75 g

#### Step 2: Convert to moles
- C: 12.01 g/mol
- H: 1.008 g/mol

$$
\text{C: } \frac{92.25}{12.01} \approx 7.681 \, \text{mol}
$$
$$
\text{H: } \frac{7.75}{1.008} \approx 7.693 \, \text{mol}
$$

#### Step 3: Divide by smallest
Both ≈ 7.68 → so ratio is roughly 1:1

$$
\text{C: } \frac{7.681}{7.681} = 1.00
$$
$$
\text{H: } \frac{7.693}{7.681} \approx 1.001 \approx 1.00
$$

So, Empirical formula: CH

#### Step 4: Calculate empirical formula mass
- CH = 12.01 + 1.008 = 13.018 g/mol

#### Step 5: Find multiplier
$$
n = \frac{\text{Molecular mass}}{\text{Empirical mass}} = \frac{78}{13.018} \approx 6.0
$$

So, multiply empirical formula by 6:

→ C₆H₆

#### ✔ Empirical Formula: CH
#### ✔ Molecular Formula: C₆H₆

This is benzene.

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✔ Final Answers:

1. Fe₂S₃
2. Al₂S₃O₁₂
3. Empirical: CH, Molecular: C₆H₆

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