Open by holding up (or referring to) two things students recognize: a smoke detector and a bag of chips with a 'best by' date. Tell them both depend on isotopes. Americium-241 in smoke detectors and carbon-14 dating both work because certain isotopes of an element behave differently from the ordinary form. Then pose the puzzle: uranium-235 fuels nuclear reactors, but uranium-238 does not — yet both are uranium. Ask: 'If they're the same element, what's actually different between them?' Take 2-3 quick student guesses (do not correct yet), then tell them by the end of class they'll be able to answer that precisely and calculate the number on the periodic table under any element's symbol.

1. 6m

   The three subatomic particles: location, charge, mass

   Content

   Every atom is built from three particles. Protons carry a +1 charge and sit inside the tiny, dense nucleus at the center of the atom. Neutrons sit alongside protons in the nucleus and carry no charge (0). Both the proton and the neutron have a mass of about 1 atomic mass unit (amu), so essentially all of an atom's mass is packed into the nucleus. Electrons carry a −1 charge and move around the nucleus in a diffuse electron cloud — they are about 1/1836 the mass of a proton, so we usually treat their mass as negligible. In a neutral atom, the number of protons equals the number of electrons, so the charges cancel. The image on the slide shows a dense nucleus (red protons, gray neutrons) with a fuzzy electron cloud around it — notice how much bigger the cloud is than the nucleus, but how little it contributes to mass.

   Delivery

   Emphasize the two big takeaways: (1) charge comes from protons and electrons, (2) mass comes from protons and neutrons. Ask a cold-call check: 'Which particle determines what element this is?' (Proton.) 'Which particle has essentially no mass?' (Electron.) Pre-empt the misconception that electrons weigh 'about the same' as protons — write the 1/1836 ratio so they see how extreme it is. This slide is the reference students will keep coming back to for the rest of the lesson.

2. 5m

   Atomic number, mass number, and isotope notation

   Content

   Two numbers describe any specific atom. The atomic number (Z) is the number of protons — it defines which element you have. Change Z and you change the element. The mass number (A) is the total count of nucleons: protons + neutrons. So neutrons = A − Z. We write this in isotope notation as ᴬZX, for example ²³⁵U and ²³⁸U. Both have Z = 92 (both are uranium), but ²³⁵U has 235 − 92 = 143 neutrons, and ²³⁸U has 238 − 92 = 146 neutrons. Same element, different neutron count — that is exactly what an isotope is. On the periodic-table cell shown on the slide, the whole number above the symbol is the atomic number Z, and the decimal number below the symbol is the weighted average atomic mass (we'll get to that in a moment) — do not confuse them.

   Delivery

   Work the ²³⁵U vs ²³⁸U example live: write A, subtract Z, get neutrons. Then have students do ¹²C and ¹⁴C on whiteboards — both carbon (Z = 6), 6 vs 8 neutrons. Pre-empt the misconception that 'isotope' means 'radioactive.' Say plainly: ¹²C is a stable isotope; it's the most common carbon on Earth. Every element has isotopes; most aren't radioactive.

3. 5m

   Atoms vs ions vs isotopes — what changed?

   Content

   Changing different particles changes the atom in different ways. Changing the number of protons changes the element itself — sodium (11 p) becoming magnesium (12 p) is a nuclear change, not something that happens in ordinary chemistry. Changing the number of neutrons keeps you in the same element but gives you a different isotope, like ¹²C vs ¹⁴C. Changing the number of electrons keeps you in the same element but gives you an ion. A neutral Na atom has 11 protons and 11 electrons; if it loses one electron it becomes Na⁺, a cation, with 11 p and 10 e. A neutral Cl atom has 17 p and 17 e; if it gains one electron it becomes Cl⁻, an anion, with 17 p and 18 e. Notice: gaining electrons makes the charge more negative because electrons are negative.

   Delivery

   Use the framing question 'What changed?' every time. Walk through the Na, Na⁺, Cl⁻ comparison and have students narrate what changed in each case. Head off the biggest misconception right here: 'If I add an electron to sodium, do I still have sodium?' Yes — element identity is protons only. Losing/gaining electrons only changes charge. Losing/gaining neutrons only changes mass. Gaining/losing protons is the only thing that would change the element.

4. 6m

   Weighted average atomic mass

   Content

   The decimal number under an element's symbol on the periodic table is not the mass of a single atom. It's a weighted average of the masses of that element's naturally occurring isotopes, weighted by how common each one is. The formula is: average atomic mass = Σ (fractional abundance × isotope mass). Fractional abundance means percent divided by 100. Worked example — chlorine has two natural isotopes: ³⁵Cl at 75.77% abundance with mass 34.969 amu, and ³⁷Cl at 24.23% with mass 36.966 amu. Average = (0.7577)(34.969) + (0.2423)(36.966) = 26.50 + 8.957 = 35.45 amu. That is exactly the value shown for Cl on the periodic table. Notice the answer is closer to 35 than to 37 — that's because the lighter isotope is much more abundant. The weighted average always pulls toward the most common isotope.

   Delivery

   Work the chlorine example step by step on the board and have students copy every line. Emphasize converting percent → decimal before multiplying. Pre-empt two misconceptions: (1) 'Just average 35 and 37 to get 36' — no, that's an unweighted average and it ignores abundance; (2) 'The atomic mass is the mass of one atom' — no, no single chlorine atom has mass 35.45; every real Cl atom is either ~35 or ~37 amu. Ask students: 'Which isotope is more abundant, and how do you know from the periodic-table value 35.45?' (³⁵Cl, because the average is close to 35.)

1. 25m

   Beanium Lab: modeling isotope abundance and weighted averageLab

   Setup (teacher, before class): Prepare one plastic cup of 'Beanium' for each lab group of 3 students. Each cup contains a fixed mixture of three bean 'isotopes': pinto, lima, and black. Suggested per cup: about 20 pinto, 12 lima, 8 black (40 total), but any mix works — the point is that groups have different abundances so results vary. Place one electronic balance and one weigh boat per group at each station. Run-time (25 min total): Distribute the handout below, one cup, one balance, one weigh boat per group. Circulate to make sure students (a) tare the weigh boat, (b) mass ALL beans of one isotope at a time, not one bean at a time, and (c) compute fractional abundance as count / total, not as a percent left over 100. After ~18 minutes, stop the class and have two groups put their results on the board so students see that different abundance mixes give different 'atomic masses' for beanium — that's the point. Student handout — Beanium Lab You have a cup of the element Beanium. Beanium has three isotopes: pinto-Bn, lima-Bn, and black-Bn. Your job is to determine the weighted average atomic mass of your sample of Beanium. Part 1 — Count and mass each isotope - Pour your cup out onto the weigh boat's tray area and sort the beans into three piles by type. - For each isotope, do the following and record it in the data table: - Count the number of beans of that type. - Tare the weigh boat on the balance, place ALL beans of that type in the boat, and record the total mass in grams. - Divide total mass by count to get the average mass of ONE bean of that isotope (this is that isotope's 'atomic mass'). Data table - Pinto-Bn: count = __, total mass = g, mass of one = __ g - Lima-Bn: count = __, total mass = g, mass of one = __ g - Black-Bn: count = __, total mass = g, mass of one = __ g - Total number of beans (all three added) = ____ Part 2 — Fractional abundance For each isotope, fractional abundance = (count of that isotope) / (total number of beans). - Pinto-Bn fractional abundance = ____ (should be a decimal between 0 and 1) - Lima-Bn fractional abundance = ____ - Black-Bn fractional abundance = ____ - Check: your three fractional abundances must add to 1.00 (±0.01). If they don't, recount. Part 3 — Weighted average atomic mass of Beanium Use the formula: average atomic mass = Σ (fractional abundance × isotope mass) Show your work: = (__ × ) + ( × ) + ( × __) = __ + + __ = ____ g ← this is the average atomic mass of one Beanium bean Part 4 — Analysis questions (answer in complete sentences) 1. Which isotope of Beanium is most abundant in YOUR sample? Is your weighted average closer to that isotope's mass, or to the other isotopes' masses? Why does that make sense? 2. Another group had a different mixture of pinto, lima, and black beans. Would they get the same weighted average atomic mass as you? Explain using the words 'abundance' and 'isotope.' 3. In real chemistry, why doesn't the periodic table list a different atomic mass for every sample of chlorine, no matter where on Earth it comes from? 4. If you added 10 more black beans (the heaviest isotope) to your cup, would the weighted average atomic mass go up, go down, or stay the same? Justify. Clean-up: Return ALL beans to the cup — do NOT throw any away or eat them. Return the cup, weigh boat, and balance to the front table.

   Materials

   • electronic balances
   • mixed dried beans of 2-3 varieties (pinto, lima, black)
   • weigh boats
   • plastic cups

   Example outputs

   • Group A sample: 20 pinto (0.35 g each), 12 lima (0.85 g each), 8 black (0.55 g each), total 40 beans. Fractional abundances: 0.50 pinto, 0.30 lima, 0.20 black (sum = 1.00). Weighted average = (0.50)(0.35) + (0.30)(0.85) + (0.20)(0.55) = 0.175 + 0.255 + 0.110 = 0.540 g per bean.
   • Group B sample (different mix): 10 pinto (0.35 g), 20 lima (0.85 g), 10 black (0.55 g), total 40. Fractional abundances: 0.25, 0.50, 0.25. Weighted average = (0.25)(0.35) + (0.50)(0.85) + (0.25)(0.55) = 0.0875 + 0.425 + 0.1375 = 0.650 g. Correct student analysis: 'Group B's average is higher because they had more lima beans, the heaviest isotope. Different abundance mixes give different weighted averages — just like different samples would give if isotope abundance varied.'

   No-equipment fallback

   If balances are unavailable, provide each group with a pre-made data card listing counts and per-bean masses for a fictional Beanium sample (e.g., 20 pinto at 0.35 g each, 12 lima at 0.85 g each, 8 black at 0.55 g each) and have them complete Parts 2-4 as a paper calculation.

1. An atom of ⁵⁶Fe (iron, Z = 26) is neutral. How many protons, neutrons, and electrons does it have?

   short answerProtons = 26 (equal to Z). Neutrons = 56 − 26 = 30. Electrons = 26 (neutral atom, so equal to protons).

2. An ion of sulfur is written S²⁻. Sulfur has atomic number 16 and this isotope has mass number 32. How many protons, neutrons, and electrons does S²⁻ have? Is it a cation or an anion?

   short answerProtons = 16 (Z, unchanged — S is still sulfur). Neutrons = 32 − 16 = 16. Electrons = 16 + 2 = 18 (gained 2 electrons to give a 2− charge). It is an anion (negative ion).

3. Are ³⁵Cl and ³⁷Cl the same element, isotopes of each other, ions of each other, or two different elements? Justify in one sentence.

   multiple choiceIsotopes of each other. Both have Z = 17 (both are chlorine), but different mass numbers because they have different numbers of neutrons (18 vs 20).

4. Copper has two natural isotopes: ⁶³Cu with mass 62.93 amu and 69.17% abundance, and ⁶⁵Cu with mass 64.93 amu and 30.83% abundance. Calculate the weighted average atomic mass of copper.

   calculationConvert percents to decimals: 0.6917 and 0.3083. Average = (0.6917)(62.93) + (0.3083)(64.93) = 43.53 + 20.02 = 63.55 amu This matches the value on the periodic table for Cu.

• ☐electronic balances (1 per group)
• ☐mixed dried beans — pinto, lima, and black (~40 beans per group)
• ☐weigh boats (1 per group)
• ☐plastic cups (1 per group, pre-filled with the bean mixture)
• ☐student handouts (Beanium Lab, 1 per student)
• ☐calculators
• ☐whiteboards / student notebooks for practice problems