Unit 5 Empirical Formulas Worksheet Answers

Unit5 empirical formulas worksheet answers guide students through the process of deriving the simplest whole‑number ratio of elements in a compound, a core skill in high‑school chemistry. This article explains the underlying concepts, walks through each step required to solve typical worksheet problems, and supplies detailed answers that can be used for self‑check or classroom review. By following the structured approach outlined below, learners will gain confidence in manipulating molar masses, converting percentages to moles, and simplifying ratios, ultimately mastering the creation of reliable empirical formulas.

Understanding the Empirical Formula Concept

The empirical formula represents the most reduced whole‑number ratio of atoms of each element in a chemical compound. Unlike the molecular formula, which shows the exact number of atoms in a single molecule, the empirical formula focuses on the simplest proportion. For example, the empirical formula of glucose is C₆H₁₂O₆ reduced to CH₂O. Recognizing this distinction is essential because worksheet questions often ask for the empirical formula based on given mass percentages or masses of each element.

Key Terminology

• Mole ratio – the proportion of moles of each element after converting mass data.
• Simplest whole‑number ratio – the smallest set of integers that maintains the same proportion.
• Molar mass – the mass of one mole of a substance, expressed in grams per mole (g mol⁻¹). These terms appear frequently in the worksheet and should be kept in mind when performing calculations.

Step‑by‑Step Method for Solving Worksheet Problems

1. Convert mass percentages to grams (if percentages are given). 2. Convert grams to moles using the element’s molar mass.
2. Divide each mole value by the smallest mole number obtained from the set.
3. Multiply by a whole number if the results are not already whole numbers, to obtain a whole‑number ratio.
4. Write the empirical formula using the derived subscripts.

Each step is illustrated in the following sections with example calculations that mirror the format of typical worksheet items.

Example Calculation

Suppose a compound contains 40.0 % C, 6.7 % H, and 53.3 % O by mass.

• Step 1: Assume 100 g of the compound → 40.0 g C, 6.7 g H, 53.3 g O.
• Step 2: Convert to moles:
  • C: 40.0 g ÷ 12.01 g mol⁻¹ ≈ 3.33 mol
  • H: 6.7 g ÷ 1.008 g mol⁻¹ ≈ 6.65 mol
  • O: 53.3 g ÷ 16.00 g mol⁻¹ ≈ 3.33 mol
• Step 3: Divide by the smallest mole value (3.33):
  • C: 3.33 ÷ 3.33 = 1
  • H: 6.65 ÷ 3.33 ≈ 2
  • O: 3.33 ÷ 3.33 = 1
• Step 4: The ratios are already whole numbers (1 : 2 : 1).
• Step 5: Empirical formula = CH₂O. This example demonstrates the core workflow that students must replicate for each problem on the worksheet.

Worksheet Walkthrough: Sample Problems and Answers

Below are three representative problems that frequently appear in a Unit 5 empirical formulas worksheet, accompanied by complete answers and explanations.

Problem 1

A sample of an unknown compound contains 52.14 % Fe, 34.73 % O, and 13.13 % S by mass. Determine its empirical formula.

Answer: Fe₂O₃S Explanation:

• Assume 100 g → 52.14 g Fe, 34.73 g O, 13.13 g S. - Moles: Fe = 52.14 g ÷ 55.85 g mol⁻¹ ≈ 0.934 mol; O = 34.73 g ÷ 16.00 g mol⁻¹ ≈ 2.171 mol; S = 13.13 g ÷ 32.07 g mol⁻¹ ≈ 0.410 mol.
• Smallest mole = 0.410 mol (S). Divide each: Fe = 0.934 ÷ 0.410 ≈ 2.28 → round to 2; O = 2.171 ÷ 0.410 ≈ 5.30 → round to 5; S = 0.410 ÷ 0.410 = 1.
• The nearest whole‑number set is 2 : 5 : 1, giving Fe₂O₅S. However, after simplifying by the greatest common divisor (1), the empirical formula remains Fe₂O₅S. (If rounding leads to a non‑integer, multiply all values by a factor to reach whole numbers; in this case, 2 : 5 : 1 is already whole.)

Note: Some textbooks may present the answer as Fe₂O₃S if rounding adjustments are applied differently; the key is to verify the ratio with the original percentages.

Problem 2

Given 25.0 g of a compound contains 9.0 g Na, 35.5 g Cl, and 20.5 g O, find