1. Chapter 3.1 to 3.5 Student Notes Stoichiometry

2. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighin3.1 Counting by Weighing 3.2 Atomic Masse3.2 Atomic Masses 3.3 The Mol3.3 The Mole 3.4 Molar Mas3.4 Molar Mass 3.5Percent Composition of Compound3.5Percent Composition of Compounds 3.6Determining the Formula of a Compoun3.6Determining the Formula of a Compound 3.7Chemical Equation3.7Chemical Equations 3.8Balancing Chemical Equation3.8Balancing Chemical Equations 3.9Stoichiometric Calculations: Amounts of Reactants and Product3.9Stoichiometric Calculations: Amounts of Reactants and Products 3.10The Concept of Limiting Reagen3.10The Concept of Limiting Reagent

3. 3 Chapter 3 Table of Contents Pick up calculator, notes packet, and pre-lab packet. CW: Notes 3.1 to 3.5 CW: Finish Test part 2 ch.1-2 if not done yet CW/HW: Read lab and complete Pre-lab questions. Note that the reading helps with the pre-lab questions due Thursday for lab. Try to be here by 7:45 for setup to give us time to discuss pre-lab and complete lab. HW: Section 3.1-3.5 pg. 117 #21, 26, 27, 29, 30, 31, 33, 35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due Monday 9/23/13 to go over. TUESDAY - B DAY - SEPT. 17, 2013

4. Chapter 3 Copyright © Cengage Learning. All rights reserved 3 Chemical Stoichiometry Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Some types of stoichiometry types: –moles to moles, moles to mass, moles to #particles –mass to mass, mass to particles, or mass to moles –particles (Avogadro’s #) –also for gases at STP, volume conversions can be added as 22.4 L = 1 mole of any gas at STP –Video link only if need some review –http://www.youtube.com/watch?v=poRTFPE1jUkhttp://www.youtube.com/watch?v=poRTFPE1jUk

5. Section 3.1 Counting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 5 Exercise A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile?

6. Section 3.1 Counting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 4 Need average mass of the object. Objects behave as though they were all identical. Carbon-12 standard “An atomic weight (relative atomic mass) of an element from a specified source is the ratio of the average mass per atom of the element to 1/12 of the mass of 12 C" in its nuclear and electronic ground state.

7. Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 6 Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12 C 1.11% 13 C < 0.01% 14 C (which is too – small to affect the overall average)

8. Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 7 98.89% of 12 amu + 1.11% of 13.0034 amu = Average Atomic Mass for Carbon CALCULATING THE AVERAGE ATOMIC MASS (0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.01 amu exact number

9. Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 8 Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon. Average Atomic Mass for Carbon

10. Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 9 Schematic Diagram of a Mass Spectrometer Figure 3.1 (left) A Scientist Injecting a Sample into a Mass Spectrometer. (right) Schematic Diagram of a Mass Spectrometer *Mass spectrometer - Most accurate method for comparing the masses of atoms and determining the isotopic composition of a natural element.

11. 11 Section 3.2 Atomic MassesCounting by Weighing Return to TOC Mass Spectrometer - Evaluating the data For this element, there are 3 isotopes with a mass of 20, 21, and 22. From the 2nd graph, we determine that 91% is 20 amu, 0.3% 21 amu, and 9% 22 amu. What element do you think we have? Did you say Neon? You would be correct!

12. 12 Section 3.2 Atomic MassesCounting by Weighing Return to TOC CALCULATING AVERAGE ATOMIC MASS EXAMPLE 1 The natural abundance for boron isotopes is: 19.9% 10 B (10.013 amu) and 80.1% 11 B (11.009amu). Calculate the atomic weight of boron. Atomic Mass = [(mass of isotope) (%abundance) ] + [(mass of isotope) (%abundance)] + [….] % abundance is expressed as a decimal and the sum is continued for any other existing isotopes. (0.199) (10.013) + (0.801) (11.009) = 10.810796 = 10.8 g/mol See periodic table to see if this matches up. It should!

13. Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 10 Exercise - Complete and show work in notes. An element consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element. Show work in notes before checking answer.

14. Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 10 Exercise - Complete and show work in notes. ANSWER BELOW IN GREEN. You must show your work. An element consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element. Show work in notes. 186.2 amu Rhenium (Re)

15. 15 Section 3.2 Atomic MassesCounting by Weighing Return to TOC AP CHEM STYLE QUESTION AP EXAMPLE 1 Copper exists as two isotopes: 63 Cu (62.9298 amu) and 65 Cu (64.9278 amu). What are the percent abundances of the isotopes? Try to do in your notes with pencil before checking answers.

16. 16 Section 3.2 Atomic MassesCounting by Weighing Return to TOC AP Example 1 continued with answer Since the overall atomic weight for copper is not given in the problem, you must look it up in the periodic table to work this solution. Atomic mass for Cu = 63.546 Assign x to one of the isotopes: 65 Cu % = x then 63 Cu % = 1-x The other isotope is 1-x (1 whole representing 100% but using 1 is easier.) 63.546 = [(1-x)(62.9298)] + [(x)(64.9278)] 63.546 = 62.9298 - 62.9298x + 64.9278x.6162 = 1.998x x =.3084 convert to % by x 100 = 30.84% which was assigned to Cu-65 Cu-63 then is 1-.3084 (or 100-30.84%) = 69.16%

17. 17 Section 3.2 Atomic MassesCounting by Weighing Return to TOC AP CHEM Example type 2 The atomic mass of lithium is 6.94, the naturally occurring isotopes are 6 Li = 6.015121 amu, and 7 Li = 7.016003 amu. Determine the percent abundance of each isotope. 6.94 =[(% 6 Li)(6.015121)] + [(% 7 Li)(7.016003)] Assigning variables for unknown %: Let % 6 Li = x then % 7 Li = 1-x (where 1 is 1 whole or 100% you could do as 100-x also if you like) 6.94 = [(x)(6.015121)] +[(1-x)(7.016003)] 6.94 = 6.015121x + 7.016003 - 7.016003x (combine like terms) -0.076003 = -1.000882 x x = 0.075936 = 7.59% Li-6 and 1-x or 0.924064 = 92.4% Li-7

18. Moles and Formula Mass

19. 19 Mole (mol): amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon-12 isotope. Atoms are small, so this is a BIG number … Avogadro’s number ( N A ) = 6.022 × 10 23 units/mol 1 mol = 6.022 × 10 23 “units” (atoms, molecules, ions, formula units, oranges, etc.) –A mole of oranges would weigh about as much as the earth! Mole is NOT abbreviated as either M or m The Mole & Avogadro’s Number

20. The Mole 1 dozen = 1 gross = 1 ream = 1 mole = 12 144 500 6.022 x 10 23 There are exactly 12 grams of carbon-12 in one mole of carbon-12.

21. Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 11 The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number). 1 mole C = 6.022 x 10 23 C atoms = 12.01 g C

22. Avogadro’s Number 6.022 x 10 23 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855). Amadeo Avogadro I didn’t discover it. Its just named after me!

23. Copyright © Houghton Mifflin Company. All rights reserved.3– Table 3.1 Comparison of 1 Mole Samples of Various Elements

24. 24

25. 25 Conversions with moles, atoms, molar mass, and volume of a gas at STP Strategies 25

26. 26 The Mole Bridge Rules for Mole Conversions using the Mole Bridge 1. You can only go from left to right. 2. The great divide side, you will divide. 3. On Rabbitville side, you will multiply. (rabbits multiply)

27. Calculations with Moles: Converting moles to grams with Dim. Analysis 1.How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li = g Li 1 mol Li 6.94 g Li 24.3

28. Calculations with Moles: Converting grams to moles 2. How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62

29. Calculations with Moles: Using Avogadro’s Number 3. How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol = atoms 1 mol 6.02 x 10 23 atoms 2.11 x 10 24

30. Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 12 Concept Check 4. Calculate the number of iron atoms in a 4.48 mole sample of iron. Show your work in your notes before checking answer.

31. Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 12 Concept Check ANSWER IN GREEN 4. Calculate the number of iron atoms in a 4.48 mole sample of iron. Show your work in your notes before checking answer. 2.70×10 24 Fe atoms

32. 32 Section 3.3 The MoleCounting by Weighing Return to TOC Definitions of Similar Terms Molar Mass is the mass of one mole of a substance (6.02 x 10 23 formula units). The unit for molar mass (note it is the mass of a mole) is grams/mole. Atomic Weight is the molar mass of an element. Formula Weight is the molar mass of an IONIC compound. Molecular Weight is the molar mass of a COVALENT compound. Molecular Mass is the mass of a given molecule (NOT MOLES of molecules). The units are atomic mass units (amu). 1amu =1.660 538 782×10 –27 kg 1 amu is 1/12 the mass of a carbon-12 atom which has a mass of 12.0000.

33. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 13 Molar Mass Mass in grams of one mole of the substance: Atomic Weight of Si = 28.09 g/mol Molecular mass of N 2 = 28.0 g/mol Molar Mass of H 2 O = 18.02 g/mol (2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO 3 ) 2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g)

34. 34 Molecular mass : sum of the masses of the atoms represented in a molecular formula. Simply put: the mass of a molecule. Molecular mass is specifically for molecules. Ionic compounds don’t exist as molecules; for them we use … Formula mass : sum of the masses of the atoms or ions present in a formula unit. Molecular Masses and Formula Masses

35. 35 Determining the Formula Mass of Ammonium Sulfate

36. 36 One Mole of Four Elements One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury?

37. 37 Molar mass is the mass of one mole of a substance. Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However … … the units of molar mass are grams (g/mol). Examples: 1 atom Na = 22.99 u 1 mol Na = 22.99 g 1 formula unit KCl = 74.56 u The Mole and Molar Mass 1 mol CO 2 = 44.01 g 1 molecule CO 2 = 44.01 u 1 mol KCl = 74.56 g

38. 38 CALCULATE MOLAR MASS Practice calculating the molar mass of the following compounds. You may round to nearest hundredths when using the mass numbers from the periodic table. You can stop at Cu(NO 3 ) 2.

39. 39 We can use these equalities to construct conversion factors, such as: Note: preliminary and follow-up calculations may be needed. 1 mol Na ––––––––– 22.99 g Na Conversions involving Mass, Moles, and Number of Atoms/Molecules 1 mol Na = 6.022 × 10 23 Na atoms = 22.99 g Na 22.99 g Na ––––––––– 1 mol Na –––––––––––––––––– 6.022 × 10 23 Na atoms

40. 40 We can read formulas in terms of moles of atoms or ions.

41. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 15 Concept Check Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×10 23 Cu atoms

42. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 14 Concept Check Which of the following is closest to the average mass of one atom of copper? a) 63.55 g b) 52.00 g c) 58.93 g d) 65.38 g e) 1.055 x 10 -22 g 63.546 g/mol so 1 mol/6.022 x 10 23 atoms which is g/atom

43. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 16 Concept Check Which of the following 100.0 g samples contains the greatest number of atoms? a) Magnesium b) Zinc c) Silver

44. 44 Section 3.6 Determining the Formula of a Compound Return to TOC Assignments Monday 9/23/13 Pick up handouts for use later & have out ch. 3 notes and HW ch. 3 CW: Notes 3.5-3.6 finished together CW: Empirical and Molecular Formula Race Game CW/HW: Notes 3.7-3.8 with computers. HW: Formal Lab report due on Wednesday. Be sure you completed homework 3.1-3.5 previously assigned and turn in by Wednesday. HW: Empirical Formula and Balancing Equations w/sheets due on Friday. HW: Be reading chapter 3. The next test is over ch.3-4 and is tentatively scheduled for October 16th. Ch. 4 is mostly new material not covered in Gen. Chemistry. 44

45. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 16 Concept Check Which of the following 100.0 g samples contains the greatest number of atoms? a) Magnesium greatest atoms would be greatest moles, so convert to moles first 100 g/24.3 g/mol (about 4 mol) b) Zinc 100g/65 g/mol = between 1 and 2 mol c) Silver 100 g/107 g/mol = less than 1 mol

46. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 17 Exercise Rank the following according to number of atoms (greatest to least): a) 107.9 g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) c)

47. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 17 Exercise Rank the following according to number of atoms (greatest to least): a) 107.9 g of silver change to moles 107.9/107.9=1mol b) 70.0 g of zinc 70/65.4 = > 1 mol; 1.07 mol c) 21.0 g of magnesium 21/24.3 = less than 1 mol b) a) c)

48. 48 Section 3.4 Molar Mass Return to TOC Stopped here with presentation on Tuesday stopped to let others finish test and then work on homework problems or pre-lab... (could have probably gone longer because most didn’t need much time to finish.

49. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 18 Exercise Consider separate 100.0 gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2  Rank them from greatest to least number of oxygen atoms. H 2 O, CO 2, C 3 H 6 O 2, N 2 O

50. Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 18 Exercise Consider separate 100.0 gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 18g/mol 44 g/mol 74 g/mol 44 g/mol 5.5+ 2.3 1.35 2.3 5.5+ 2.3 x2 so 2.7 x2 so 4.6  Rank them from greatest to least number of oxygen atoms. H 2 O, CO 2, C 3 H 6 O 2, N 2 O

51. 51 The mass percent composition of a compound refers to the proportion of the constituent elements, expressed as the number of grams of each element per 100 grams of the compound. In other words … Mass Percent Composition from Chemical Formulas X g element X % element = –––––––––––––– OR … 100 g compound g element % element = ––––––––––– × 100 g compound Notes 3.5

52. Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 20 Mass percent of an element: For iron in iron(III) oxide, (Fe 2 O 3 ):

53. Calculating Molar Mass Calculate the molar mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mol

54. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00 24.31 84.32 28.83% Mg 12.01 84.32 14.24% C 48.00 84.32 56.93% O x 100 Total % should be close to

55. Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 21 Exercise Consider separate 100.0 gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2  Rank them from highest to lowest percent oxygen by mass. H 2 O, CO 2, C 3 H 6 O 2, N 2 O

56. Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 21 Exercise Consider separate 100.0 gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 16/18 16/44 32/74 32/44 89% 36% 43% 73%  Rank them from highest to lowest percent oxygen by mass. H 2 O, CO 2, C 3 H 6 O 2, N 2 O

57. 57 Section 3.5 Percent Composition of Compounds Return to TOC ASSIGNMENTS Ch. 3 Homework - Tuesday - 9/17/13 HW: Section 3.1-3.5 pg. 117 #21, 26, 27, 29, 30, 31, 33, 35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due Monday 9/23/13 to go over. You must show your work for most of these with calculations and units. No work for math problems will receive no points. CW: Finish test part 2 if you haven’t. HW: Read and complete Pre-lab Determination of the Molecular Weight of an Acid for lab on Thursday.