Presentation is loading. Please wait.

Presentation is loading. Please wait.

Unit 2 – Section C Conserving Matter. HW 1 Read & take notes on Section C.1.

Similar presentations


Presentation on theme: "Unit 2 – Section C Conserving Matter. HW 1 Read & take notes on Section C.1."— Presentation transcript:

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 O 2 CO 2 1 Carbon 1 oxygen 1 carbon dioxide atom (C) molecule (O 2 ) molecule (CO 2 ) 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 O 2 CO 2 1 Carbon 1 oxygen 1 carbon dioxide atom (C) molecule (O 2 ) molecule (CO 2 ) In a balanced chemical equation, the number of atoms for left side equals the number for the right side.

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

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

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

8 C.2 – Accounting for Atoms 1)Methane burning with oxygen CH 4 + 2 O 2  CO 2 + 2 H 2 O ReactantsProductsCHO

9 C.2 – Accounting for Atoms (continued) 2)Hydrobromic acid reacting with magnesium HBr + Mg  H 2 + MgBr 2 ReactantsProductsHBrMg

10 C.2 – Accounting for Atoms (continued) 3.Hydrogen sulfide and metallic silver react 4 Ag + 4 H 2 S + O 2  2 Ag 2 S + 4 H 2 O ReactantsProductsAgHSO

11 C.2 – Accounting for Atoms (continued) 4.Cellulose burns to form carbon dioxide and water vapor. C 6 H 10 O 5 + 6 O 2  6 CO 2 + 5 H 2 O Reactants ProductsCHO

12 C.2 – Accounting for Atoms (continued) 5. Nitroglycerin decomposes explosively to form nitrogen, oxygen, carbon dioxide and water vapor. 2 C 3 H 5 (NO 3 ) 3  3 N 2 + O 2 + 6 CO 2 + 5 H 2 O Reactants ProductsCHNO

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 NO 3 -, CO 3 2- ) appear as both reactants and product treat them as units.  If water is involved, balance the hydrogen and oxygen atoms last.  Recount all atoms after you think an equation is balanced.

15 C.4 – Writing Chemical Equations Writing to balance the chemical equations… MethaneChlorineChloroformHydrogen chloride __ CH 4 + __ Cl 2  __ CHCl 3 + __ HCl Reactants Products C H Cl C H Cl

16 C.4 – Writing Chemical Equations (continued) 1a. __ C + __ O 2  __ CO Reactants Products CO CO CO CO

17 C.4 – Writing Chemical Equations (continued) 1b. __ Fe 2 O 3 + __ CO  __ Fe + __ CO 2 Reactants Products C Fe O C Fe O

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

19 C.4 – Writing Chemical Equations (continued) 3. __ O 3  __ O 2 Reactants Products O O

20 C.4 – Writing Chemical Equations (continued) 4. __ NH 3 + __ O 2  __ NO 2 + __ H 2 O Reactants Products NHO NHO NHO NHO

21 C.4 – Writing Chemical Equations (continued) 5. __ Cu + __ AgNO 3  __ Cu(NO 3 ) 2 + __ Ag Reactants Products Cu Ag N O Cu Ag N O

22 C.4 – Writing Chemical Equations (continued) 6. __ C 8 H 18 + __ O 2  __ CO 2 + __ H 2 O Reactants Products CHO CHO CHO CHO

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). One mole of ANY element or molecule contains 602 000 000 000 000 000 000 000 particles 6.02 x 10 23

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 10 23 ) 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 10 23 ) 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 10 23 ) would have a molar mass of ______ g One mole of silver atoms (6.02 x 10 23 ) would have a molar mass of ______ g One mole of gold atoms (6.02 x 10 23 ) 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 (O 2 )? One mole of oxygen gas (O 2 ) molecules (6.02 x 10 23 ) 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 (H 2 O)? One mole of water (H 2 O) molecules (6.02 x 10 23 ) 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 10 23 ) would have a molar mass of _______ g 2. One mole of nitrogen (N 2 ) molecules (6.02 x 10 23 ) would have a molar mass of _______ g

32 C.6 – Molar Masses (continued) 3. One mole of table salt (NaCl) molecules (6.02 x 10 23 ) would have a molar mass of _______ g

33 C.6 – Molar Masses (continued) 4. One mole of table sugar (C 12 H 22 O 11 ) molecules (6.02 x 10 23 ) would have a molar mass of _______ g

34 C.6 – Molar Masses (continued) 6. One mole of magnesium phosphate Mg 3 (PO 4 ) 2 molecules (6.02 x 10 23 ) would have a molar mass of ______ g

35 C.6 – Molar Masses (continued) 8. One mole of calcium hydroxyapatite Ca 10 (PO 4 ) 6 (OH) 2 molecules (6.02 x 10 23 ) 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) + CO 2 (g) Alternatively stated… 2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO 2 In this example, for every t wo moles of CuO that react, o ne mole of CO 2 is produced.

38 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? Using... 2 mol CuO + 1 mol C  2 mol Cu + 1 mol CO 2 We can reason that... 1 mol CuO 79.55 g CuO X mol CuO 955.0 g CuO =

39 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) Solving for X... 1 mol CuO 79.55 g CuO X mol CuO 955.0 g CuO = 955.0 g CuO X 1 mol CuO = 79.55 g CuO X X mol CuO X mol CuO 955.0 g CuO X1 mol CuO 79.55 g CuO = NOTICE...

40 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) Solving for X... 1 mol CuO 79.55 g CuO X mol CuO 955.0 g CuO X1 mol CuO 79.55 g CuO = 12.01 mol CuO = X This all started with... A proportion we created called a c onversion factor, both referring to the same number of particles.

41 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... 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 CO 2 12.01 mol CuO X 1 mol C 2 mol CuO = 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... 6.005 mol C X 12.01 g C 1 mol C = 72.12 g C

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

44 C.8 – Molar Relationships 1.CuO(s) + 2 HCl(aq)  CuCl 2 (aq) + H 2 O (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 CuCl 2 b. 201.6 g CuCl 2 c. 73.0 g HCl

47 C.8 – Molar Relationships 4.CuO(s) + 2 HCl(aq)  CuCl 2 (aq) + H 2 O (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 6 K + B 2 O 3  3 K 2 O + 2 B Please write out the conversion factor for the reactants and products of the above chemical equation. 1) 1 mol K ------------------- 39.098 g K 2) 1 mol B 2 O 3 ------------------- 69.619 g B 2 O 3 3) 1 mol K 2 O ------------------- 94.194 g K 2 O 4) 1 mol B ------------------- 10.811 g B

49 6 K + B 2 O 3  3 K 2 O + 2 B A processor needs to convert 955.0 g B2O3 to pure B. What mass of K is needed for this reaction? (show all your work) Mole Quiz (continued) 955.0 g B 2 O 3 X 1 mol B 2 O 3 ------------------- = 13.72 mol B 2 O 3 69.619 g B 2 O 3 6 mol K 39.098 g K ------------- X 13.72 mol B 2 O 3 = 82.32 mol K X --------------- = 1 mol B 2 O 3 1 mol K 3219 g K

50 16 Al + 3 S 8  8 Al 2 S 3 A processor plans to create 1000. g of Al2S3. What mass of pure S is needed for this reaction? (show all your work) Mole Quiz (continued) 1000. g Al 2 S 3 X 1 mol Al 2 S 3 ------------------- = 6.660 mol Al 2 S 3 150.16 g Al 2 S 3 3 mol S 8 256.53 g S 8 ------------- X 6.660 mol Al 2 S 3 = 2.498 mol S 8 X ---------------- = 8 mol Al 2 S 3 1 mol S 8 640.8 g S 8

51 HW 7 Read & take notes on Section C.9

52 C.9 – Compositions of Materials The percent mass of each material found in an item is called the percent composition. Hint: remember solution concentration Example: post-1982 penny has a mass of 2.500 g has 2.4375 g zinc & 0.0625 g copper. What is the percent composition? 2.4375 g zinc 2.500 g total X 100% =97.50 % zinc

53 C.9 – Compositions of Materials (continued) Example: post-1982 penny has a mass of 2.500 g has 2.4375 g zinc & 0.0625 g copper. What is the percent composition? 0.0625 g copper 2.500 g total X 100% =2.50 % copper

54 C.9 – Compositions of Materials (continued) Why are the ideas of molar mass & percentage composition so important? Mass of copper Mass of Cu 2 S X 100% =% copper Some Copper-containing Minerals Common Name Formula Chalcocite Cu 2 S Chalcopyrite CuFeS 2 Malachite Cu 2 CO 3 (OH) 2 It helps us determine which is more profitable to mine.

55 C.10 – Percent Composition HW 8 – Questions 1-2 on pg 168

56 C.10 – Percent Composition 1.An atom inventory for Cu 3 (CO 3 ) 2 (OH) 2 There are : 3 Cu atoms 2 C atoms 8 O atoms 2 H atoms

57 C.10 – Percent Composition (continued) 2. Percent copper in ? a. Chalcopyrite CuFeS 2 63.55 g + 55.85 g + (2 x 32.07g) Start with... = 183.54 g 63.55 g 183.54 g x 100 % = 34.62 % Cu

58 (2 x 63.55 g) + 12.01 g + (5 x 16.00 g) + (2 x 1.008) C.10 – Percent Composition (continued) 2. Percent copper in ? b. malachite Cu 2 CO 3 (OH) 2 c. Chalcocite, at 79.85% copper, would be the most profitable to mine. = 221.13 g (2 x 63.55 g) 221.13 g x 100 % = 57.48 % Cu

59 C.11 – Retrieving Copper HW 9 – please pre-read the lab.

60 HW 10 Read & take notes on Section C.12

61 C.12 – Conservation in the Community Resources… Renewable Fresh water, Air, Fertile soil, Plants, and Animals EVENTUALLY replenished by natural processes. Nonrenewable Metals, Natural gas, coal and petroleum CANNOT be readily replenished.

62 C.13 – Rethinking, Reusing, Replacing & Recycling HW 11 Questions 1 & 2 on pg. 176

63 C.14 – The Life Cycle of Material Not covered C.15 – Copper Life-Cycle Analysis


Download ppt "Unit 2 – Section C Conserving Matter. HW 1 Read & take notes on Section C.1."

Similar presentations


Ads by Google