04/11: The Mole

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CHEMISTRY 10
21: The Mole and the Avogadro constant 
Week 13
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ChemistrySecondary Education

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CHEMISTRY 10
21: The Mole and the Avogadro constant 
Week 13

Slide 1 - Diapositive

W E L C O M E !       Today is the 4th of November, 2024.

  • Where will you put your learning materials: notebook, coursebook, pen and pencil, and fully-charged laptop?
  • Have you kept your phone in your locker?
  • Have you gone to the bathroom before the class starts?
  • Have you checked any announcements or missing work on our Google Classroom?
  • Are you seated in your assigned chair?
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2:00

Slide 2 - Diapositive

  • define what a mole is.
  • calculate the (a) amount of substance, (b) mass, (c) molar mass, (d) relative atomic mass or relative molecular / formula mass, and (e) number of particles using the value of the Avogadro constant.

Slide 3 - Diapositive

  • Which is easier to count: A or B ?

Slide 4 - Diapositive

Which is easier to count: A or B ?

Slide 5 - Sondage

  • Which is easier to count: A or B ?

Slide 6 - Diapositive

  • Which is easier to count: A or B ?

Slide 7 - Diapositive

Slide 8 - Diapositive

Mole (symbol: mol)
  • amount of substance that contains as many particles (atoms, molecules, ions formula units, or electrons) as precisely 12 g of carbon-12.

Slide 9 - Diapositive

How many atoms are there in each mass of element?

Slide 10 - Diapositive

Molar Mass (symbol: Mr)
  • the mass, in grams, of 1 mole of a substance

Slide 11 - Diapositive

  • How would you calculate the number of particles in a given mole?

Slide 12 - Diapositive

  • How would you calculate the number of particles in a given mole?

Slide 13 - Diapositive

Sample
  • 2Mg (s) + O2 (g) → 2MgO (s)
Strategy:
  • 2Mg = 2 x (6.02 x 10²³ atoms) =  1.20x10²⁴ atoms
  • 1O2 = 1 x (6.02 x 10²³ molecules) = 6.02 x 10²³ molecules
  • 2MgO = 2 x (6.02 x 10²³ formula units) = 1.20x10²⁴ formula units
  • 2Mg²⁺ = 2 x (6.02 x 10²³ Mg²⁺ cations) = 1.20x10²⁴ Mg²⁺ cations
  • 2O²⁻ = 2 x (6.02 x 10²³ O²⁻ anions) = 1.20x10²⁴ O²⁻ anions

Slide 14 - Diapositive

Guided Practice
  • Continue on page 10 of your learning module.
  • Study the sample question.
  • Do (a) to (e).
  • If you have any questions, please raise your hand.
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Slide 15 - Diapositive

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CHEMISTRY 10
21: The Mole and the Avogadro constant 
Week 13

Slide 16 - Diapositive

Think about these:
  • 1 mole of CO₂ = 6.02 x 10²³ molecules
  • 1 mole of CO₂ = 44.0 g
  • 44.0 g of CO₂ has 6.02 x 10²³ molecules
How many carbon dioxide molecules and moles are in 22.0 g of carbon dioxide?

Slide 17 - Diapositive

Slide 18 - Diapositive

Sample
  • A flask contains 0.750 mol of carbon dioxide gas, CO₂. What mass of carbon dioxide gas is in this sample?
Strategy:
  • CO₂ n = 0.750 mol
  • CO₂ Mᵣ = ?    C = 12.0 g ; O₂ = 2 x 16.0                   Total Mᵣ = 12.0 + 32.0 = 44.0 g / mol
  • CO₂ m = ?
  • CO₂ m = n x Mᵣ
  • CO₂ m = 0.750 mol x 44.0 g / mol
  • CO₂ m = 33.0 g
  •                                                Answer: The mass of 0.75 mol of carbon dioxide is 33.0 g.

Slide 19 - Diapositive

Guided Practice
  • Continue on pages 13-14 of your learning module.
  • Study the sample question.
  • Do (Activity 4) to (Activity 5).
  • If you have any questions, please raise your hand.
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10:00

Slide 20 - Diapositive

Share your thought process.

Slide 21 - Diapositive

Slide 22 - Diapositive


I can define what a mole is.
Yes, I can confidently do it.
Yes, I can but need more practice.
No, I lack confidence so I need more practice or ask help from my teacher.

Slide 23 - Sondage


I can calculate the (a) amount of substance, (b) mass, (c) molar mass, (d) relative atomic mass or relative molecular / formula mass, and (e) number of particles using the value of the Avogadro constant.
Yes, I can confidently do it.
Yes, I can but need more practice.
No, I lack confidence so I need more practice or ask help from my teacher.

Slide 24 - Sondage

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Slide 25 - Diapositive

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STEM Project
Building a Futuristic City Scale Model
Week 11

Slide 26 - Diapositive