Pogil Relative Mass And The Mole Answer Key

The POGIL relative mass and themole activity is a widely used classroom tool that guides students through the connection between atomic mass units, the mole, and Avogadro’s number. By working in small groups, learners interpret data, construct models, and derive the fundamental relationship that one mole of any substance contains the same number of entities as there are atoms in exactly 12 grams of carbon‑12. The answer key for this activity provides clear, step‑by‑step solutions that reinforce conceptual understanding and help instructors assess student progress.

Introduction to the POGIL Relative Mass and the Mole Worksheet The worksheet begins with a simple comparison: the mass of a single hydrogen atom versus the mass of a dozen hydrogen atoms. Students quickly see that measuring individual atoms is impractical, prompting the need for a counting unit—the mole. Through a series of guided questions, the activity leads them to:

1. Determine the relative mass of elements using the periodic table. 2. Convert between grams, moles, and number of particles.
2. Apply Avogadro’s number (6.022 × 10²³ mol⁻¹) as the bridge between the macroscopic and microscopic worlds.

The answer key mirrors each question, showing the expected reasoning and numerical results. It also highlights common misconceptions, such as confusing atomic mass units with grams per mole, and offers brief explanations to correct them.

Step‑by‑Step Walkthrough of the Activity

Part A: Exploring Relative Mass

Question 1: What is the atomic mass of carbon as shown on the periodic table?
Answer Key: 12.01 amu (atomic mass units).
Explanation: The value listed under each element symbol represents the weighted average mass of its isotopes, expressed in amu.

Question 2: If you have 12.01 grams of carbon, how many moles of carbon atoms do you possess?
Answer Key: 1.00 mol.
Explanation: By definition, the molar mass of an element (grams per mole) is numerically equal to its atomic mass in amu. Thus, 12.01 g ÷ 12.01 g mol⁻¹ = 1.00 mol.

Question 3: How many carbon atoms are present in that 12.01‑gram sample?
Answer Key: 6.022 × 10²³ atoms.
Explanation: Multiply the number of moles by Avogadro’s number: 1.00 mol × 6.022 × 10²³ mol⁻¹ = 6.022 × 10²³ atoms.

Part B: Applying the Concept to Other Elements Question 4: Calculate the number of moles in 35.5 grams of chlorine (Cl).

Answer Key: 0.500 mol. Explanation: Chlorine’s atomic mass is 35.45 amu → molar mass ≈ 35.45 g mol⁻¹. 35.5 g ÷ 35.45 g mol⁻¹ ≈ 1.00 mol × (35.5/35.45) ≈ 0.500 mol (rounded to three significant figures).

Question 5: How many chlorine molecules (Cl₂) are in that sample?
Answer Key: 1.51 × 10²³ molecules.
Explanation: First find moles of Cl₂: 0.500 mol Cl atoms ÷ 2 = 0.250 mol Cl₂. Then multiply by Avogadro’s number: 0.250 mol × 6.022 × 10²³ mol⁻¹ = 1.506 × 10²³ ≈ 1.51 × 10²³ molecules.

Question 6: If you have 2.00 grams of sulfur (S), what mass of oxygen (O) would combine with it to form SO₂ according to the law of definite proportions? Answer Key: 3.20 grams of O.
Explanation: The formula SO₂ indicates a 1 : 2 mole ratio of S to O. Moles of S = 2.00 g ÷ 32.07 g mol⁻¹ = 0.0624 mol. Required moles of O = 2 × 0.0624 = 0.1248 mol. Mass of O = 0.1248 mol × 16.00 g mol⁻¹ = 1.997 g ≈ 2.00 g. However, because each S atom bonds with two O atoms, the total oxygen mass needed is double that: ≈ 4.00 g. The answer key clarifies that the question asks for the mass of O that combines with the given S to produce one mole of SO₂, yielding 3.20 g (0.0624 mol S × 2 × 16.00 g mol⁻¹).

Part C: Conceptual Synthesis

Question 7: Explain why the mole is considered a “chemical dozen”.
Answer Key: The mole groups an enormous number of particles (6.022 × 10²³) into a single, manageable unit, just as a dozen groups 12 items. It allows chemists to relate measurable masses to numbers of atoms or molecules without counting each individually.

Question 8: Describe how isotopic abundance affects the atomic mass listed on the periodic table.
Answer Key: The atomic mass is a weighted average of all naturally occurring isotopes, weighted by their relative abundances. For example, chlorine’s atomic mass of 35.45 amu reflects that about 75 % of chlorine atoms are ^35Cl (34.97 amu) and 25 % are ^37Cl (36.97 amu).

Scientific Explanation Behind the Answers

The core idea linking relative mass and the mole is the definition of the mole itself: one mole of any substance contains exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions, etc