Empirical Formula Practice Worksheet With Answers

Empirical Formula Practice Worksheet with Answers: Mastering the Fundamentals of Chemistry

Determining the empirical formula of a compound is a fundamental skill in chemistry. It forms the cornerstone of understanding chemical composition and stoichiometry. This comprehensive guide provides a detailed explanation of the process, followed by a practice worksheet with answers to help you solidify your understanding. We'll explore various examples, tackling different complexities to ensure you're well-prepared for any challenge.

Understanding Empirical Formulas

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 (that's the molecular formula), but rather the ratio between them. For example, the molecular formula of glucose is C₆H₁₂O₆, but its empirical formula is CH₂O. Both formulas tell us that glucose contains carbon, hydrogen, and oxygen, but the empirical formula simplifies this ratio to its smallest whole numbers.

Key Differences: Empirical vs. Molecular Formula

• Empirical Formula: Shows the simplest whole-number ratio of atoms in a compound.
• Molecular Formula: Shows the actual number of atoms of each element in a molecule.

To determine the empirical formula, we typically start with the mass percentages or the mass of each element present in a sample of the compound.

Step-by-Step Process for Determining Empirical Formula

The process involves several key steps:

1. Convert percentages to grams: If given percentages, assume a 100g sample. This converts percentages directly into grams. For example, 40% carbon becomes 40g carbon.

2. Convert grams to moles: Use the molar mass of each element to convert the grams of each element into moles. Remember, molar mass is the atomic weight of an element expressed in grams per mole (g/mol). You can find molar masses on the periodic table.

3. Find the mole ratio: Divide the number of moles of each element by the smallest number of moles calculated in step 2. This will give you the ratio of atoms in the simplest whole number form.

4. Express as a whole-number ratio: If the mole ratios are not whole numbers (e.g., 1.5, 2.33), multiply all the ratios by the smallest integer that converts them into whole numbers (e.g., multiplying by 2 to convert 1.5 to 3).

5. Write the empirical formula: Use the whole-number ratios as subscripts for each element in the formula.

Practice Worksheet: Empirical Formula Determination

Let's put these steps into practice with the following problems. Work through each problem step-by-step, and then check your answers against the solutions provided later.

Problem 1: A compound contains 75% carbon and 25% hydrogen. Determine its empirical formula.

Problem 2: A 2.50g sample of a compound contains 0.900g of calcium, 0.480g of carbon, and 1.12g of oxygen. Find the empirical formula.

Problem 3: A compound is analyzed and found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula?

Problem 4: Analysis of a compound shows it contains 26.6% potassium, 35.4% chromium, and 38.1% oxygen. What is the empirical formula?

Problem 5 (Challenge): A 1.000g sample of a compound is found to contain 0.267g of Na, 0.351g of Cr, and 0.382g of O. Determine the empirical formula.

Answers and Detailed Solutions

Now let's go through the solutions step-by-step for each problem. Remember to follow the process outlined above.

Problem 1 Solution:

1. Grams: 75g C, 25g H
2. Moles: 75g C / 12.01g/mol = 6.25 mol C; 25g H / 1.01g/mol = 24.75 mol H
3. Mole Ratio: 6.25 mol C / 6.25 mol = 1; 24.75 mol H / 6.25 mol = 3.96 ≈ 4
4. Whole Number Ratio: 1:4
5. Empirical Formula: CH₄

Problem 2 Solution:

1. Moles: 0.900g Ca / 40.08g/mol = 0.0225 mol Ca; 0.480g C / 12.01g/mol = 0.0400 mol C; 1.12g O / 16.00g/mol = 0.0700 mol O
2. Mole Ratio: 0.0225 mol Ca / 0.0225 mol = 1; 0.0400 mol C / 0.0225 mol = 1.78 ≈ 2; 0.0700 mol O / 0.0225 mol = 3.11 ≈ 3
3. Whole Number Ratio: 1:2:3
4. Empirical Formula: CaC₂O₃

Problem 3 Solution:

1. Grams: 40.0g C, 6.7g H, 53.3g O
2. Moles: 40.0g C / 12.01g/mol = 3.33 mol C; 6.7g H / 1.01g/mol = 6.63 mol H; 53.3g O / 16.00g/mol = 3.33 mol O
3. Mole Ratio: 3.33 mol C / 3.33 mol = 1; 6.63 mol H / 3.33 mol = 1.99 ≈ 2; 3.33 mol O / 3.33 mol = 1
4. Whole Number Ratio: 1:2:1
5. Empirical Formula: CH₂O

Problem 4 Solution:

1. Grams: 26.6g K, 35.4g Cr, 38.1g O
2. Moles: 26.6g K / 39.10g/mol = 0.680 mol K; 35.4g Cr / 52.00g/mol = 0.681 mol Cr; 38.1g O / 16.00g/mol = 2.38 mol O
3. Mole Ratio: 0.680 mol K / 0.680 mol = 1; 0.681 mol Cr / 0.680 mol = 1; 2.38 mol O / 0.680 mol = 3.50 ≈ 7/2
4. Whole Number Ratio: Multiply by 2: 2:2:7
5. Empirical Formula: K₂Cr₂O₇

Problem 5 Solution (Challenge):

1. Moles: 0.267g Na / 22.99g/mol = 0.0116 mol Na; 0.351g Cr / 52.00g/mol = 0.00675 mol Cr; 0.382g O / 16.00g/mol = 0.0239 mol O
2. Mole Ratio: 0.0116 mol Na / 0.00675 mol = 1.72 ≈ 17/10; 0.00675 mol Cr / 0.00675 mol = 1; 0.0239 mol O / 0.00675 mol = 3.54 ≈ 7/2
3. Whole Number Ratio: Multiply by 10: 17:10:35
4. Empirical Formula: Na₁₇Cr₁₀O₃₅

This worksheet and its solutions provide a solid foundation for understanding empirical formula calculations. Remember to practice consistently to master this essential chemistry skill. By working through numerous examples and focusing on the step-by-step process, you can confidently tackle more complex problems. Further practice with diverse examples will further enhance your proficiency.