10 Empirical And Molecular Formula Practice Problems With Answers

In the world of chemistry, mastering the concepts of empirical and molecular formulas is essential. Whether you're a student preparing for exams or someone keen on understanding the basics of chemistry, practicing with problems is one of the best ways to solidify your knowledge. This article will present you with 10 practice problems, along with their answers, to help you grasp these concepts better. Let’s dive in!

Understanding Empirical and Molecular Formulas

Before we jump into the practice problems, it’s important to clarify what empirical and molecular formulas are:

• Empirical Formula: This is the simplest whole-number ratio of atoms of each element in a compound. For example, the empirical formula for glucose (C₆H₁₂O₆) is CH₂O, which represents the lowest ratio of carbon, hydrogen, and oxygen.

• Molecular Formula: This formula shows the actual number of atoms of each element in a molecule of the compound. For glucose, the molecular formula is C₆H₁₂O₆, indicating that one molecule contains six carbons, twelve hydrogens, and six oxygens.

Understanding the difference is crucial as it will guide your approach when tackling the practice problems below.

Practice Problems

Here are 10 practice problems focusing on empirical and molecular formulas:

Problem 1

A compound contains 40% sulfur and 60% oxygen. What is the empirical formula?

Answer:

1. Convert percentages to grams (assuming 100g sample):
   • 40g S
   • 60g O
2. Convert grams to moles:
   • Moles of S = 40g / 32.07g/mol = 1.25
   • Moles of O = 60g / 16.00g/mol = 3.75
3. Divide by the smallest number of moles (1.25):
   • S: 1.25/1.25 = 1
   • O: 3.75/1.25 = 3
4. Empirical formula = SO₃

Problem 2

Determine the empirical formula for a compound that contains 75% carbon and 25% hydrogen.

Answer:

1. 75g C and 25g H
2. Moles of C = 75g / 12.01g/mol = 6.24 Moles of H = 25g / 1.008g/mol = 24.80
3. Divide by the smallest number of moles (6.24):
   • C: 6.24/6.24 = 1
   • H: 24.80/6.24 = 3.97 (approx 4)
4. Empirical formula = C₁H₄ or CH₄

Problem 3

A hydrocarbon is found to contain 82.7% carbon and 17.3% hydrogen. What are the empirical and molecular formulas if its molar mass is 78 g/mol?

Answer:

1. Moles of C = 82.7g / 12.01g/mol = 6.89 Moles of H = 17.3g / 1.008g/mol = 17.14
2. Divide by the smallest number of moles (6.89):
   • C: 6.89/6.89 = 1
   • H: 17.14/6.89 = 2.48 (approx 2.5, multiply by 2)
3. Empirical formula = C₂H₅
4. Molecular formula (C₂H₅) × 6 = C₆H₁₄ (because molecular mass is 78g/mol)

Problem 4

What is the empirical formula for a compound containing 60.00% carbon, 10.00% hydrogen, and 30.00% oxygen?

Answer:

1. Moles of C = 60.00g / 12.01g/mol = 4.99 Moles of H = 10.00g / 1.008g/mol = 9.92 Moles of O = 30.00g / 16.00g/mol = 1.88
2. Divide by the smallest number of moles (1.88):
   • C: 4.99/1.88 = 2.65 (approx 2.6, multiply by 2)
   • H: 9.92/1.88 = 5.28 (approx 5.3, multiply by 2)
   • O: 1.88/1.88 = 1
3. Empirical formula = C₆H₁₁O₂

Problem 5

A compound has a molar mass of 60 g/mol and has an empirical formula of CH₂O. What is its molecular formula?

Answer:

1. Find the molar mass of CH₂O = 12.01 + (2 × 1.008) + 16.00 = 30.03 g/mol
2. Divide the molecular mass by empirical mass: 60 g/mol ÷ 30.03 g/mol ≈ 2
3. Molecular formula = (CH₂O)₂ = C₂H₄O₂

Problem 6

A sample of a compound contains 10.0 g of nitrogen and 40.0 g of oxygen. What is the empirical formula?

Answer:

1. Moles of N = 10.0g / 14.01g/mol = 0.714 Moles of O = 40.0g / 16.00g/mol = 2.5
2. Divide by the smallest number of moles (0.714):
   • N: 0.714/0.714 = 1
   • O: 2.5/0.714 = 3.50 (approx 3.5, multiply by 2)
3. Empirical formula = N₂O₇

Problem 7

A compound made from 24% potassium, 18% sulfur, and 58% oxygen has a molar mass of 174 g/mol. What is the empirical formula?

Answer:

1. Assume 100g of compound:
   • 24g K, 18g S, 58g O
2. Moles of K = 24g / 39.10g/mol = 0.613 Moles of S = 18g / 32.07g/mol = 0.562 Moles of O = 58g / 16.00g/mol = 3.625
3. Divide by the smallest number of moles (0.562):
   • K: 0.613/0.562 = 1.09 (approx 1.1, multiply by 10)
   • S: 0.562/0.562 = 1
   • O: 3.625/0.562 = 6.45 (approx 6.5, multiply by 2)
4. Empirical formula = K₁S₁O₇

Problem 8

What is the empirical formula of a compound containing 15 g of iron and 10 g of oxygen?

Answer:

1. Moles of Fe = 15g / 55.85g/mol = 0.268 Moles of O = 10g / 16.00g/mol = 0.625
2. Divide by the smallest number of moles (0.268):
   • Fe: 0.268/0.268 = 1
   • O: 0.625/0.268 = 2.33 (approx 2.3, multiply by 3)
3. Empirical formula = Fe₃O₇

Problem 9

A substance is found to contain 54.5% carbon and 45.5% hydrogen. If the molar mass of the substance is 30 g/mol, what is the molecular formula?

Answer:

1. Moles of C = 54.5g / 12.01g/mol = 4.54 Moles of H = 45.5g / 1.008g/mol = 45.12
2. Divide by the smallest number of moles (4.54):
   • C: 4.54/4.54 = 1
   • H: 45.12/4.54 = 9.93 (approx 10)
3. Empirical formula = CH₁₀
4. Molar mass of empirical formula = 12.01 + (10 × 1.008) = 22.09 g/mol
5. Molecular formula = (CH₁₀) × 30g/mol/22.09g/mol ≈ 1.36 (approx 1.5, multiply by 2)
6. Molecular formula = C₂H₂₀

Problem 10

A compound has a formula of C₃H₈ and a molar mass of 44 g/mol. What is the empirical formula?

Answer:

1. Molar mass of C₃H₈ = (3 × 12.01) + (8 × 1.008) = 44.09 g/mol
2. The empirical formula of propane is C₃H₈; therefore, the empirical formula is C₃H₈ since it's already the lowest ratio.

Common Mistakes to Avoid

• Forgetting to Convert to Moles: Always ensure you convert grams to moles; this is a crucial step in determining formulas.
• Incorrect Ratio Reduction: Sometimes it's tempting to stop once you've found the moles; ensure you find the simplest ratio.
• Assuming Percentages: When dealing with percentages, assume a 100g sample for simplicity.

Troubleshooting Issues

• If you're having trouble determining the empirical formula, double-check your calculations for moles.
• If the ratio seems incorrect, ensure you've divided by the smallest number of moles.

<div class="faq-section"> <div class="faq-container"> <h2>Frequently Asked Questions</h2> <div class="faq-item"> <div class="faq-question"> <h3>What is the difference between empirical and molecular formulas?</h3> <span class="faq-toggle">+</span> </div> <div class="faq-answer"> <p>The empirical formula shows the simplest whole-number ratio of elements, while the molecular formula shows the actual number of atoms in a molecule.</p> </div> </div> <div class="faq-item"> <div class="faq-question"> <h3>How do I determine the empirical formula from percentages?</h3> <span class="faq-toggle">+</span> </div> <div class="faq-answer"> <p>Convert percentages to grams, then to moles, and find the simplest ratio by dividing by the smallest number of moles.</p> </div> </div> <div class="faq-item"> <div class="faq-question"> <h3>Can the empirical formula be the same as the molecular formula?</h3> <span class="faq-toggle">+</span> </div> <div class="faq-answer"> <p>Yes, if the compound exists in its simplest form, the empirical formula and molecular formula can be identical.</p> </div> </div> <div class="faq-item"> <div class="faq-question"> <h3>How do I find the molar mass from the empirical formula?</h3> <span class="faq-toggle">+</span> </div> <div class="faq-answer"> <p>Add up the atomic masses of all the atoms in the empirical formula to find its molar mass.</p> </div> </div> </div> </div>

<p class="pro-note">🌟Pro Tip: Regular practice with these problems will boost your confidence and proficiency in chemistry!</p>