Open with a real-world provocation: hold up (or reference on the slide) a shiny US penny and a small vial of copper(II) sulfate pentahydrate. Tell students: 'This penny is about 2.5 g of Zn coated in Cu. This vial holds 2.5 g of CuSO₄·5H₂O. Which sample contains more atoms — and by how many?' Take 30 seconds of silent think, then a quick show of hands: (a) penny more, (b) vial more, (c) equal. Do NOT resolve it yet — write the vote tally where students can see it and promise, 'By minute 55 today, every one of you will not only pick the right answer, you'll defend it on an AP free-response rubric.' This targets SP 6 Argumentation from the first minute: they will need a claim + evidence + reasoning by the end of class.

1. 7m

   The mole is a count, not a mass

   Content

   A mole is simply a counting number, exactly like 'dozen' or 'ream' — one mole of anything is 6.022 × 10²³ representative particles of that thing. We use it because atoms are absurdly small: a single carbon atom has a mass of 1.99 × 10⁻²³ g, which no balance can read, but 6.022 × 10²³ carbon atoms have a mass of 12.01 g, which any balance can read. That is the whole point of Avogadro's number — it is the scaling factor that turns atomic-scale masses (amu) into laboratory-scale masses (grams). Numerically, 1 amu per particle × 6.022 × 10²³ particles/mol = 1 g/mol. This is why the molar mass of an element in g/mol is the SAME NUMBER as the atomic mass of one atom in amu. The representative particle depends on what the substance is: atoms for elements (Fe, Ne), molecules for molecular compounds (H₂O, CO₂), and formula units for ionic compounds (NaCl, Al₂(SO₄)₃).

   Delivery

   Anchor the deck's mole-map/scale visual by pointing out that the number 6.022 × 10²³ is not magical — it is the exact bridge between amu and grams. Have students turn to a neighbor and finish this sentence: 'One mole of ___ contains 6.022 × 10²³ ___.' Cold-call two pairs, one with an element and one with an ionic compound, so 'atoms' vs 'formula units' surfaces. Pre-empt the biggest misconception up front: molar mass is NOT a count — it is a mass. Say it out loud: 'Molar mass has units of g/mol. Avogadro's number has units of particles/mol. They are not interchangeable.' Targets SP 1 Models and Representations.

2. 7m

   Molar mass: g/mol vs amu/particle

   Content

   Molar mass is the mass in grams of exactly one mole of a substance and always carries units of g/mol. For an element it equals the atomic mass on the periodic table (C = 12.01 g/mol). For a compound it is the sum of the molar masses of every atom in the formula, with subscripts multiplying each contribution. Worked example — Al₂(SO₄)₃: 2(26.98) + 3(32.06) + 12(16.00) = 53.96 + 96.18 + 192.00 = 342.14 g/mol. Notice we counted TWELVE oxygens, not four — the subscript 4 inside the parentheses is multiplied by the subscript 3 outside. A single formula unit of Al₂(SO₄)₃ has a mass of 342.14 amu; one mole of formula units has a mass of 342.14 g. Same number, different units, different scale. When you set up a unit-cancelation calculation, you MUST use g/mol, not amu, or your dimensional analysis is meaningless.

   Delivery

   Walk the Al₂(SO₄)₃ calculation slowly, writing 2, 3, 12 as the mole ratios OF EACH ELEMENT PER MOLE OF COMPOUND — this seeds the fourth misconception (miscounting O). Ask, 'How many moles of O are in one mole of Al₂(SO₄)₃?' Expected: 12. Then 'How many moles of O in 0.50 mol of Al₂(SO₄)₃?' Expected: 6.0 mol. Explicitly write both units side by side: 342.14 amu per formula unit, 342.14 g per mole. Tell students bluntly: 'If I see amu in the denominator of a mole conversion on the exam, that's an automatic point off.' Targets SP 5 Mathematical Routines.

3. 6m

   The mole map: grams ⇌ moles ⇌ particles

   Content

   Every mole calculation on the AP exam is a walk on the same three-node map. The center node is MOLES. To move from grams to moles, divide by molar mass (g/mol). To move from moles to particles, multiply by Avogadro's number (6.022 × 10²³ /mol). Reverse directions invert the operation. There is no shortcut from grams directly to particles — you must pass through moles. Worked example: How many oxygen atoms are in 5.00 g of Al₂(SO₄)₃? Step 1 (g → mol compound): 5.00 g × (1 mol / 342.14 g) = 0.01461 mol Al₂(SO₄)₃. Step 2 (mol compound → mol O atoms): 0.01461 mol × (12 mol O / 1 mol Al₂(SO₄)₃) = 0.1754 mol O. Step 3 (mol O → atoms O): 0.1754 mol × (6.022 × 10²³ /mol) = 1.06 × 10²³ oxygen atoms. Three sig figs, because the mass was given to three.

   Delivery

   Point at the mole-map visual and have students narrate each arrow with the conversion factor. Do the 5.00 g Al₂(SO₄)₃ → O atoms problem live, pausing at each arrow. The critical checkpoint is Step 2 — the stoichiometric factor 12 mol O / 1 mol Al₂(SO₄)₃ — because this is where students who ignore subscripts lose the problem. Ask, 'Where did the 12 come from?' before you write it. Close by returning to the hook: preview that the penny vs CuSO₄·5H₂O question is exactly this map, run twice. Targets SP 4 Model Analysis and SP 5.

1. 30m

   Bean-Mole Analogy + AP-Style Mole Map Problem SetLab

   Split the block into Part A (hands-on, 10 min) and Part B (problem set, 20 min). Circulate constantly during Part B — the two most common errors you will see are (1) students converting grams directly to particles without stopping at moles, and (2) students forgetting to multiply by the subscript when asked for atoms of a specific element. When you spot either, stop the pair and have them narrate the mole map arrows before recalculating. Part A targets SP 1 (Models) and SP 3 (Representing Data); Part B targets SP 4, SP 5, and SP 6. Debrief the last 3 minutes by resolving the hook penny-vs-CuSO₄·5H₂O question as a whole class using a student volunteer's setup. Student handout — distribute at the start of the activity: Part A — The Bean Mole (10 minutes, with a partner) You will build a physical analogy for molar mass. In this analogy, one 'bean-mole' = 50 beans (instead of 6.022 × 10²³ particles). 1. Zero the balance with an empty weigh boat. 2. Count out exactly 50 pinto beans into the weigh boat. 3. Record the total mass of the 50 beans: ______ g 4. Divide by 50 to find the average mass of one bean: ______ g 5. This total mass IS the 'bean-molar-mass' — the mass of one bean-mole. Record it: ______ g/bean-mole Answer with your partner: - Q1. If another pair reports a bean-molar-mass that differs from yours by 0.5 g, does that mean one of you counted wrong? Explain in one sentence. - Q2. A bag on the counter is labeled '84.0 g of pinto beans.' Using YOUR bean-molar-mass, how many bean-moles are in the bag? How many individual beans? - Q3. In this analogy, what plays the role of Avogadro's number? What plays the role of molar mass? The bean-molar-mass depends on the average bean — the real molar mass depends on the average atom (isotopes weighted by abundance). Part B — AP-Style Mole Map Problems (20 minutes) Show full dimensional-analysis setups. Report answers to the correct number of significant figures. Molar masses: H = 1.008, C = 12.01, N = 14.01, O = 16.00, Na = 22.99, S = 32.06, Cl = 35.45, Cu = 63.55, Zn = 65.38, Al = 26.98. 1. Determine the molar mass of aluminum sulfate, Al₂(SO₄)₃. Show each element's contribution. - Molar mass = ______ g/mol 2. A student masses 4.28 g of Al₂(SO₄)₃. - a) How many moles of Al₂(SO₄)₃ are in the sample? ______ - b) How many moles of sulfate ions (SO₄²⁻) are in the sample? ______ - c) How many oxygen ATOMS are in the sample? ______ 3. Which of the following samples contains the greatest number of atoms? Justify your choice with a calculation for each — do NOT guess from mass alone. - A) 10.0 g of He - B) 10.0 g of H₂O - C) 10.0 g of CO₂ - D) 10.0 g of NaCl 4. AP-style claim-evidence-reasoning: A classmate says, 'One mole of CO₂ and one mole of H₂O must have the same mass because they both have one mole of molecules.' In 2–3 sentences, state whether the classmate is correct and use specific numerical evidence to justify your reasoning. 5. Return to the hook. A US penny has a mass of 2.500 g and is essentially pure Zn. A vial contains 2.500 g of CuSO₄·5H₂O (molar mass 249.68 g/mol). - a) How many TOTAL atoms are in the penny? - b) How many TOTAL atoms are in the vial? (Remember: each formula unit of CuSO₄·5H₂O contains 1 + 1 + 4 + 5(3) = 21 atoms — 1 Cu, 1 S, 4 O in sulfate, and 5 H₂O each with 3 atoms.) - c) Which sample has more atoms, and by roughly what factor?

   Materials

   • Electronic balance (0.01 g precision), 1 per pair
   • Weigh boat or small paper cup
   • ~200 dried pinto beans per pair (pre-counted bags of 50 work best)
   • Calculator
   • Periodic table
   • Printed handout (content below)

   Example outputs

   • Part A Q3: Avogadro's number is played by '50 beans per bean-mole'; molar mass is played by the measured mass of those 50 beans (e.g., 22.5 g/bean-mole).
   • Part B Q2c: 4.28 g ÷ 342.14 g/mol = 0.01251 mol Al₂(SO₄)₃; × 12 mol O / 1 mol compound = 0.1501 mol O; × 6.022 × 10²³ = 9.04 × 10²² O atoms.
   • Part B Q3: A) 10.0/4.003 × 6.022e23 = 1.50 × 10²⁴ atoms; B) 10.0/18.02 × 3 × 6.022e23 = 1.00 × 10²⁴ atoms; C) 10.0/44.01 × 3 × 6.022e23 = 4.11 × 10²³ atoms; D) 10.0/58.44 × 2 × 6.022e23 = 2.06 × 10²³ atoms. Answer: A (He) has the most atoms.
   • Part B Q5: Penny — 2.500/65.38 × 6.022e23 = 2.303 × 10²² Zn atoms. Vial — 2.500/249.68 × 21 × 6.022e23 = 1.266 × 10²³ atoms. The vial has ~5.5× more atoms even though the masses are equal.

1. Which of the following 5.00 g samples contains the greatest number of atoms of oxygen? (A) H₂O (B) CO₂ (C) O₂ (D) Al₂(SO₄)₃

   multiple choiceD. Compute mol O in each 5.00 g sample: H₂O → 5.00/18.02 × 1 = 0.277 mol O; CO₂ → 5.00/44.01 × 2 = 0.227 mol O; O₂ → 5.00/32.00 × 2 = 0.313 mol O; Al₂(SO₄)₃ → 5.00/342.14 × 12 = 0.175 mol O. Wait — O₂ gives the most (0.313 mol). Correct answer: C (O₂). This item targets SP 5 and pre-empts the 'more O in the formula = more O in the sample' error; a student who picks D fell into that misconception.

2. A student writes the following setup to find the number of molecules in 9.00 g of H₂O: '9.00 g × (1 mol / 18.02 amu) × (6.022 × 10²³ / 1 mol) = ...' Identify the specific error in the setup and write the correct expression. (Targets SP 5, SP 6.)

   short answerThe unit in the denominator of the first conversion factor is wrong — it should be 18.02 g, not 18.02 amu. Molar mass has units of g/mol, not amu/mol. Correct setup: 9.00 g × (1 mol / 18.02 g) × (6.022 × 10²³ molecules / 1 mol) = 3.01 × 10²³ molecules. The amu-per-particle value cannot cancel grams.

3. Calculate the number of moles of oxygen ATOMS in 12.5 g of Ca(NO₃)₂. Molar mass of Ca(NO₃)₂ = 164.10 g/mol. Show your dimensional analysis. (Targets SP 5.)

   calculation12.5 g × (1 mol Ca(NO₃)₂ / 164.10 g) × (6 mol O / 1 mol Ca(NO₃)₂) = 0.457 mol O. Key step: each formula unit contains 2 nitrate groups × 3 O = 6 O, so the stoichiometric factor is 6, not 3.

4. Claim–Evidence–Reasoning: A student argues, 'Equal moles of any two substances must occupy equal volumes because they contain the same number of particles.' Is the claim correct for solids and liquids at room temperature? Justify with a specific numerical comparison. (Targets SP 6.)

   short answerIncorrect for solids and liquids. Equal moles contain equal numbers of particles, but the volume depends on how tightly the particles pack and their individual sizes/masses. Example: 1 mol H₂O(l) has a mass of 18.02 g and occupies ~18.0 mL; 1 mol Hg(l) has a mass of 200.6 g and occupies ~14.8 mL; 1 mol NaCl(s) is 58.44 g and occupies ~27 mL. Same particle count, very different volumes and masses. (The claim is approximately true only for ideal gases at the same T and P — Avogadro's law — which is a separate result.)

• ☐Electronic balances (0.01 g), 1 per pair
• ☐Weigh boats or small cups
• ☐Pre-counted bags of 50 dried pinto beans per pair
• ☐Calculators
• ☐Periodic tables
• ☐Printed student handouts (Parts A and B)
• ☐One US penny and a small labeled vial of CuSO₄·5H₂O for the hook display