Empirical/molecular Formula Practice Worksheet Answer Key

Empirical and Molecular Formula Practice Worksheet: A Comprehensive Guide with Answer Key

Determining the empirical and molecular formulas of chemical compounds is a fundamental skill in chemistry. This comprehensive guide provides a step-by-step approach to solving problems related to empirical and molecular formulas, along with a detailed answer key for a practice worksheet. We'll cover the concepts, methods, and common pitfalls, ensuring you master this crucial area of chemistry.

Understanding Empirical and Molecular Formulas

Before diving into the practice problems, let's clarify the difference between empirical and molecular formulas.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It doesn't necessarily reflect the actual number of atoms present in a molecule. For example, the empirical formula for glucose is CH₂O, while its molecular formula is C₆H₁₂O₆. The empirical formula shows a 1:2:1 ratio of carbon, hydrogen, and oxygen atoms.

Molecular Formula

The molecular formula represents the actual number of atoms of each element in a molecule of a compound. It's a multiple of the empirical formula. In the case of glucose, the molecular formula (C₆H₁₂O₆) is six times the empirical formula (CH₂O).

Methods for Determining Empirical and Molecular Formulas

The process typically involves two main steps:

1. Determining the Empirical Formula

This step usually starts with the percent composition of the elements in the compound (often given as percentages by mass) or the mass of each element present in a given sample.

Steps:

1. Assume a 100g sample: This simplifies calculations when dealing with percentages.
2. Convert percentages to grams: If given percentages, assume you have 100g of the compound, so the percentages directly translate to grams.
3. Convert grams to moles: Use the molar mass of each element to convert the mass (in grams) to moles. Remember, the molar mass is the atomic weight of the element (found on the periodic table) in grams per mole (g/mol).
4. Determine the mole ratio: Divide the number of moles of each element by the smallest number of moles calculated. This gives you the simplest whole-number ratio.
5. Write the empirical formula: Use the whole-number ratios as subscripts for each element in the formula.

2. Determining the Molecular Formula

To find the molecular formula, you'll need the empirical formula and the molar mass of the compound.

Steps:

1. Calculate the empirical formula mass: Add the molar masses of the atoms in the empirical formula.
2. Determine the whole-number multiple: Divide the molar mass of the compound (given) by the empirical formula mass. This will give you a whole number (or very close to one).
3. Multiply the subscripts: Multiply the subscripts in the empirical formula by the whole-number multiple obtained in step 2. This gives you the molecular formula.

Practice Worksheet and Answer Key

Let's put this into practice with a worksheet and detailed answer key.

Worksheet:

Problem 1: A compound is found to contain 85.6% carbon and 14.4% hydrogen by mass. Its molar mass is approximately 42 g/mol. Determine its empirical and molecular formulas.

Problem 2: A 0.500g sample of a compound contains 0.299g of potassium, 0.134g of sulfur, and 0.067g of oxygen. Determine its empirical formula.

Problem 3: A compound has an empirical formula of CH₂O and a molar mass of 180 g/mol. What is its molecular formula?

Problem 4: Analysis of a compound reveals it contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Answer Key:

Problem 1:

1. Empirical Formula:

   • Assume 100g sample: 85.6g C and 14.4g H
   • Moles of C: 85.6g / 12.01 g/mol = 7.13 mol
   • Moles of H: 14.4g / 1.01 g/mol = 14.3 mol
   • Divide by smallest: C: 7.13/7.13 = 1; H: 14.3/7.13 ≈ 2
   • Empirical formula: CH₂

2. Molecular Formula:

   • Empirical formula mass: 12.01 + (2 * 1.01) = 14.03 g/mol
   • Whole number multiple: 42 g/mol / 14.03 g/mol ≈ 3
   • Molecular formula: C₃H₆

Problem 2:

1. Moles of each element:
   • K: 0.299g / 39.10 g/mol = 0.00765 mol
   • S: 0.134g / 32.07 g/mol = 0.00418 mol
   • O: 0.067g / 16.00 g/mol = 0.00419 mol
2. Mole Ratio:
   • K: 0.00765 / 0.00418 ≈ 1.83
   • S: 0.00418 / 0.00418 = 1
   • O: 0.00419 / 0.00418 ≈ 1
   • The ratios are approximately 2:1:1 (multiplying by 2 to get whole numbers)
3. Empirical Formula: K₂SO

Problem 3:

1. Empirical formula mass: 12.01 + (2 * 1.01) + 16.00 = 30.03 g/mol
2. Whole number multiple: 180 g/mol / 30.03 g/mol ≈ 6
3. Molecular formula: C₆H₁₂O₆

Problem 4:

1. Assume 100g sample: 40.0g C, 6.7g H, 53.3g O
2. Moles of each element:
   • C: 40.0g / 12.01 g/mol = 3.33 mol
   • H: 6.7g / 1.01 g/mol = 6.63 mol
   • O: 53.3g / 16.00 g/mol = 3.33 mol
3. Mole Ratio:
   • C: 3.33 / 3.33 = 1
   • H: 6.63 / 3.33 ≈ 2
   • O: 3.33 / 3.33 = 1
4. Empirical Formula: CH₂O

Common Mistakes to Avoid

• Significant Figures: Pay close attention to significant figures throughout your calculations to avoid errors in your final answer.
• Unit Conversions: Ensure you consistently use the correct units (grams, moles, etc.) and convert units as needed.
• Rounding Errors: Avoid excessive rounding during intermediate steps; round only at the very end to minimize error accumulation.
• Incorrect Molar Masses: Double-check that you're using the correct molar masses from the periodic table.

Conclusion

Mastering the determination of empirical and molecular formulas is crucial for success in chemistry. By following the step-by-step methods outlined in this guide and practicing with the provided worksheet and answer key, you'll build confidence and proficiency in solving these types of problems. Remember to focus on understanding the underlying concepts and avoid common pitfalls to ensure accuracy. Practice regularly and you'll become proficient in this essential chemical skill.