Empirical And Molecular Formulas Worksheet Answer Key

Empirical and Molecular Formulas Worksheet: A Comprehensive Guide with Answers

Understanding empirical and molecular formulas is fundamental in chemistry. This comprehensive guide provides a detailed explanation of both concepts, walks you through solving various problems, and offers an answer key for a practice worksheet. Mastering these concepts is crucial for success in stoichiometry and other advanced chemistry topics.

What are Empirical and Molecular Formulas?

Before diving into problem-solving, let's solidify our understanding of the core concepts:

Empirical Formula

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

Key Characteristics of Empirical Formulas:

• Simplest Ratio: Always expresses the ratio of elements in the lowest whole numbers.
• Not Unique: Several compounds can share the same empirical formula.
• Derived from Experimental Data: Typically determined through elemental analysis, which reveals the mass percentage of each element in a compound.

Molecular Formula

The molecular formula represents the actual number of atoms of each element present in a single molecule of a compound. It provides a complete picture of the molecule's composition. For glucose, the molecular formula is C₆H₁₂O₆, reflecting six carbon atoms, twelve hydrogen atoms, and six oxygen atoms per molecule.

Key Characteristics of Molecular Formulas:

• Actual Composition: Shows the exact number of atoms of each element in a molecule.
• Unique to a Compound: Each compound has a unique molecular formula.
• Related to Empirical Formula: The molecular formula is a whole-number multiple of the empirical formula.

Calculating Empirical Formulas: A Step-by-Step Guide

The calculation of empirical formulas relies on converting the mass percentages or masses of elements into moles and then finding the simplest whole-number ratio. Here's a detailed breakdown of the process:

Step 1: Determine the mass of each element. This information might be given directly as percentages or masses, or you might need to calculate them from given data (like the mass of a compound and the mass of one of its constituent elements).

Step 2: Convert the mass of each element to moles. Use the element's molar mass (atomic weight from the periodic table) to perform this conversion. Remember that moles = mass (g) / molar mass (g/mol).

Step 3: Divide the number of moles of each element by the smallest number of moles. This step normalizes the mole ratios, helping to find the simplest whole number ratio.

Step 4: If the resulting ratios aren't whole numbers, multiply all ratios by a small whole number (e.g., 2, 3) to obtain whole numbers. This is necessary to express the empirical formula using whole-number subscripts.

Calculating Molecular Formulas: Building on the Empirical Formula

Once you've determined the empirical formula, calculating the molecular formula requires additional information: the molar mass of the compound.

Step 1: Calculate the molar mass of the empirical formula. This is done by summing the molar masses of all atoms in the empirical formula.

Step 2: Divide the actual molar mass of the compound by the molar mass of the empirical formula. The result should be a whole number or a number very close to a whole number.

Step 3: Multiply the subscripts in the empirical formula by the whole number obtained in Step 2. This gives you the molecular formula.

Worksheet Problems and Solutions

Let's put our knowledge into practice with a series of problems. This worksheet covers different scenarios, testing your understanding of both empirical and molecular formula calculations.

Problem 1: A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.

Solution:

1. Assume a 100g sample: This simplifies the percentages to grams. We have 40.0g C, 6.7g H, and 53.3g O.

2. Convert to moles:

   • Moles of C = 40.0g / 12.01g/mol ≈ 3.33 mol
   • Moles of H = 6.7g / 1.01g/mol ≈ 6.63 mol
   • Moles of O = 53.3g / 16.00g/mol ≈ 3.33 mol

3. Divide by the smallest number of moles (3.33):

   • C: 3.33 mol / 3.33 mol = 1
   • H: 6.63 mol / 3.33 mol ≈ 2
   • O: 3.33 mol / 3.33 mol = 1

4. Empirical formula: CH₂O

Problem 2: The empirical formula of a compound is CH₂O, and its molar mass is 180.2 g/mol. What is its molecular formula?

Solution:

1. Molar mass of empirical formula (CH₂O): 12.01g/mol + 2(1.01g/mol) + 16.00g/mol = 30.03 g/mol

2. Divide the molar mass of the compound by the molar mass of the empirical formula: 180.2 g/mol / 30.03 g/mol ≈ 6

3. Multiply the subscripts in the empirical formula by 6: C₆H₁₂O₆

Problem 3: A 0.500g sample of a compound contains 0.200g of carbon, 0.067g of hydrogen, and the rest is oxygen. Determine the empirical formula.

Solution:

1. Determine the mass of oxygen: 0.500g (total) - 0.200g (C) - 0.067g (H) = 0.233g O

2. Convert to moles:

   • Moles of C = 0.200g / 12.01g/mol ≈ 0.0167 mol
   • Moles of H = 0.067g / 1.01g/mol ≈ 0.0663 mol
   • Moles of O = 0.233g / 16.00g/mol ≈ 0.0146 mol

3. Divide by the smallest number of moles (0.0146):

   • C: 0.0167 mol / 0.0146 mol ≈ 1.14
   • H: 0.0663 mol / 0.0146 mol ≈ 4.54
   • O: 0.0146 mol / 0.0146 mol = 1

4. Multiply by 2 to get whole numbers:

   • C: 1.14 * 2 ≈ 2
   • H: 4.54 * 2 ≈ 9
   • O: 1 * 2 = 2

5. Empirical formula: C₂H₉O₂

Problem 4: A compound has an empirical formula of NO₂ and a molar mass of 92.0 g/mol. What is its molecular formula?

Solution:

1. Molar mass of empirical formula (NO₂): 14.01g/mol + 2(16.00g/mol) = 46.01 g/mol

2. Divide the molar mass of the compound by the molar mass of the empirical formula: 92.0 g/mol / 46.01 g/mol ≈ 2

3. Multiply the subscripts in the empirical formula by 2: N₂O₄

This worksheet provides a solid foundation for understanding and calculating both empirical and molecular formulas. Remember to practice regularly, and don't hesitate to revisit this guide for clarification. Consistent practice is key to mastering these vital chemical concepts. Further practice problems can be found in your textbook or online chemistry resources. Remember to always double-check your calculations and units for accuracy. Good luck!