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Unit 2 – Section C

Unit 2 – Section C. Conserving Matter. HW 1. Read & take notes on Section C.1. C.1 – Keeping Track of Atoms. The law of conservation of matter – in a chemical reaction matter is neither created nor destroyed.

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Unit 2 – Section C

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  1. Unit 2 – Section C Conserving Matter

  2. HW 1 Read & take notes on Section C.1

  3. C.1 – Keeping Track of Atoms The law of conservation of matter – in a chemical reaction matter is neither created nor destroyed. +  C O2 CO2 1 Carbon 1 oxygen 1 carbon dioxide atom (C) molecule (O2) molecule (CO2) Molecules can be converted and decomposed by chemical processes: but atoms are forever.

  4. C.1 – Keeping Track of Atoms(continued) Reactants are placed on the left of the arrow; Products are placed on the right. +  C O2 CO2 1 Carbon 1 oxygen 1 carbon dioxide atom (C) molecule (O2) molecule (CO2) In a balanced chemical equation, the number of atoms for left side equals the number for the right side.

  5. C.1 – Keeping Track of Atoms(continued) Coefficients indicate the relative number of each unit involved. +  Cu (s) O2 (g) CuO (s) 2 Copper oxygen 2 copper oxide atoms (Cu) molecule (O2) molecules (CuO)

  6. C.1 – Keeping Track of Atoms(continued) Chemists use the term formula unit when referring to the smallest unit of an ionic compound. +  Cu (s) O2 (g) CuO (s) 2 Copper oxygen 2 copper oxide atoms (Cu) molecule (O2) molecules (CuO)

  7. Classwork Answer questions 1-5 in Section C.2 pg 155

  8. C.2 – Accounting for Atoms • Methane burning with oxygen CH4 + 2 O2 CO2 + 2 H2O Reactants Products C C H H O O

  9. C.2 – Accounting for Atoms(continued) • Hydrobromic acid reacting with magnesium HBr + Mg  H2 + MgBr2 Reactants Products H H Br Br Mg Mg

  10. C.2 – Accounting for Atoms(continued) • Hydrogen sulfide and metallic silver react 4 Ag + 4 H2S + O2 2 Ag2S + 4 H2O Reactants Products Ag Ag H H S S O O

  11. C.2 – Accounting for Atoms(continued) • Cellulose burns to form carbon dioxide and water vapor. C6H10O5 + 6 O2 6 CO2 + 5 H2O Reactants Products C C H H O O

  12. C.2 – Accounting for Atoms(continued) • Nitroglycerin decomposes explosively to form nitrogen, oxygen, carbon dioxide and water vapor. 2 C3H5(NO3) 3 3 N2 + O2 + 6 CO2 + 5 H2O Reactants Products C C H H N N O O

  13. HW 2 Read & take notes on Section C.3 Address & answer questions 1-6 in section C.4

  14. C.3 – Nature’s Conservation: Balancing Chemical Equations • If polyatomic ions (examples NO3-, CO32-) appear as both reactants and product treat them as units. • If water is involved, balance the hydrogen and oxygen atoms last. • Recountall atoms after you think an equation is balanced.

  15. C.4 – Writing Chemical Equations Writing to balance the chemical equations… Methane Chlorine Chloroform Hydrogen chloride __ CH4 + __ Cl2 __ CHCl3 + __ HCl Products Reactants C H Cl C H Cl

  16. C.4 – Writing Chemical Equations(continued) Products Reactants __ C + __ O2 __ CO 1a. C O C O

  17. C.4 – Writing Chemical Equations(continued) Products Reactants __ Fe2O3 + __ CO  __ Fe + __ CO2 1b. C Fe O C Fe O

  18. C.4 – Writing Chemical Equations(continued) Products Reactants __ CuO + __ C  __ Cu + __ CO2 2. C Cu O C Cu O

  19. C.4 – Writing Chemical Equations(continued) Products Reactants __ O3 __ O2 3. O O

  20. C.4 – Writing Chemical Equations(continued) Products Reactants __ NH3 + __ O2 __ NO2 + __ H2O 4. NHO NHO

  21. C.4 – Writing Chemical Equations(continued) Products Reactants __ Cu + __ AgNO3 __ Cu(NO3)2 + __ Ag 5. CuAgNO CuAgNO

  22. C.4 – Writing Chemical Equations(continued) Products Reactants __ C8H18 + __ O2 __ CO2 + __ H2O 6. C H O C H O

  23. HW 3 Read & take notes on Section C.5

  24. C.5 – Introducing the Mole Concept Chemist have created a counting unit for elements called the mole (symbolized mol). 602 000 000 000 000 000 000 000 particles 6.02 x 1023 One mole of ANY element or molecule contains

  25. C.5 – Introducing the Mole Concept(continued) Furthermore, the atomic weight of elements can be used to find the molar mass of a substance. One mole of boron atoms (6.02 x 1023) would have a molar mass of 10.81 g

  26. C.5 – Introducing the Mole Concept(continued) More examples… One mole of carbon atoms (6.02 x 1023) would have a molar mass of _______ g

  27. C.5 – Introducing the Mole Concept(continued) More examples… One mole of copper atoms (6.02 x 1023) would have a molar mass of ______ g One mole of silver atoms (6.02 x 1023) would have a molar mass of ______ g One mole of gold atoms (6.02 x 1023) would have a molar mass of ______ g

  28. C.5 – Introducing the Mole Concept(continued) (Curve ball) How about the molar mass of oxygen gas (O2)? One mole of oxygen gas (O2) molecules (6.02 x 1023) would have a molar mass of _______ g

  29. C.5 – Introducing the Mole Concept(continued) (Curve ball 2) How about the molar mass of water (H2O)? One mole of water (H2O) molecules (6.02 x 1023) would have a molar mass of _______ g

  30. C.6 – Molar Masses HW Questions 1-4,6,8 pg 163

  31. C.6 – Molar Masses 1. One mole of nitrogen (N) atoms (6.02 x 1023) would have a molar mass of _______ g 2. One mole of nitrogen (N2) molecules (6.02 x 1023) would have a molar mass of _______ g

  32. C.6 – Molar Masses(continued) 3. One mole of table salt (NaCl) molecules (6.02 x 1023) would have a molar mass of _______ g

  33. C.6 – Molar Masses(continued) 4. One mole of table sugar (C12H22O11) molecules (6.02 x 1023) would have a molar mass of _______ g

  34. C.6 – Molar Masses(continued) 6. One mole of magnesium phosphate Mg3(PO4)2 molecules (6.02 x 1023) would have a molar mass of ______ g

  35. C.6 – Molar Masses(continued) 8. One mole of calcium hydroxyapatite Ca10(PO4)6(OH)2 molecules (6.02 x 1023) would have a molar mass of ______ g

  36. HW 5 Read & take notes on Section C.7

  37. C.7 – Equations and Molar Relationships Let’s revisit copper-refining… 2 CuO(s) + C(s)  2 Cu(s) + CO2(g) Alternatively stated… 2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2 In this example, for every two moles of CuO that react, one mole of CO2 is produced.

  38. C.7 – Equations and Molar Relationships(continued) X mol CuO 1 mol CuO = 955.0 g CuO 79.55 g CuO Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction? Using . . . 2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2 We can reason that . . .

  39. C.7 – Equations and Molar Relationships(continued) X mol CuO 1 mol CuO = 955.0 g CuO X 1 mol CuO = X mol CuO 955.0 g CuO 79.55 g CuO 79.55 g CuO Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction? (continued) Solving for X . . . 955.0 g CuO X 1 mol CuO = 79.55 g CuO X X mol CuO NOTICE . . .

  40. C.7 – Equations and Molar Relationships(continued) 1 mol CuO 955.0 g CuO X 1 mol CuO = X mol CuO 79.55 g CuO 79.55 g CuO Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction? (continued) Solving for X . . . 12.01 mol CuO = X This all started with . . . A proportion we created called a conversion factor, both referring to the same number of particles.

  41. C.7 – Equations and Molar Relationships(continued) 1 mol C 2 mol CuO Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction? (continued) So . . . The refiner knows there are 12.01 mol CuO in 955.0 g. Remembering the equation we started with . . . 2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO2 12.01 mol CuO X = 6.005 mol C

  42. C.7 – Equations and Molar Relationships(continued) Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction? (continued) So . . . Once we know The refiner knows we need 6.005 mol C to refine 955 g of CuO we can calculate the actual mass of C needed . . . Appropriate conversion factors . . . 12.01 g C 6.005 mol C X = 72.12 g C 1 mol C

  43. C.8 – Molar Relationships HW 6 Questions 1-4 pg 166

  44. C.8 – Molar Relationships 1. CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O(l) Molar mass of each…

  45. C.8 – Molar Relationships(continued) 2. Mass (in grams) of: a. 1.0 mol HCl b. 5.0 mol HCl c. 0.50 mol CuO

  46. C.8 – Molar Relationships (continued) 3. # of moles represented by a. 941.5 g CuCl2 b. 201.6 g CuCl2 c. 73.0 g HCl

  47. C.8 – Molar Relationships 4. CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O(l) a. How many moles of CuO are needed to react with 4 mol HCl? a. How many moles of HCl are needed to react with 4 mol CuO?

  48. Mole Quiz (explained) Start with

  49. Mole Quiz (continued) 955.0 g B2O3 X1 mol B2O3 ------------------- = 13.72 mol B2O3 69.619 g B2O3 6 mol K 39.098 g K -------------X13.72 mol B2O3 = 82.32 mol K X --------------- = 1 mol B2O3 1 mol K 3219 g K

  50. Mole Quiz (continued) 1000. g Al2S3 X1 mol Al2S3 ------------------- = 6.660 mol Al2S3 150.16 g Al2S3 3 mol S8 256.53 g S8 -------------X6.660 mol Al2S3 = 2.498 mol S8 X ---------------- = 8 mol Al2S3 1 mol S8 640.8 g S8

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