1. 3-1 CHEM 100, Fall 2013 LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@coes.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m Tu,Th,F 8:00 - 10:00 a.m. Or by appointment Test Dates : Chemistry 100(02) Fall 2013 September 30, 2013 (Test 1): Chapter 1 & 2 October 21, 2013 (Test 2): Chapter 3 & 4 November 13, 2013 (Test 3) Chapter 5 & 6 November 14, 2013 (Make-up test) comprehensive: Chapters 1-6 9:30-10:45:15 AM, CTH 328 CHEM 100, Fall 2012 LA TECH

2. 3-2 CHEM 100, Fall 2013 LA TECH REQUIRED : Textbook: Principles of Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase the Mastering Chemistry Group Homework, Slides and Exam review guides and sample exam questions are available online: http://moodle.latech.edu/ and follow the course information links. http://moodle.latech.edu/ OPTIONAL : Study Guide: Chemistry: A Molecular Approach, 2nd Edition- Nivaldo J. Tro 2nd Edition Student Solutions Manual: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro 2nd Text Book & Resources

3. 3-3 CHEM 100, Fall 2013 LA TECH 3.1 Hydrogen, Oxygen, and Water…………………………….78 3.2 Chemical Bonds……………………………………………80 3.3 Representing Compounds: Chemical Formulas and Molecular Models..82 3.4 An Atomic-Level View of Elements and Compounds……………..84 3.5 Ionic Compounds: Formulas and Names…………………… 87 3.6 Molecular Compounds: Formulas and Names………………………93 3.7 Formula Mass and the Mole Concept for Compounds………… 97 3.8 Composition of Compounds…………………………….. 100 3.9 Determining a Chemical Formula from Experimental Data……… 105 3.10 Writing and Balancing Chemical Equations…………………… 110 3.11 Organic Compounds………………………. 114 Chapter 3. Molecules, Compounds, and Chemical Equations

4. 3-4 CHEM 100, Fall 2013 LA TECH  Classifying Substances as Atomic Elements, Molecular Elements, Molecular Compounds, or Ionic Compounds (3.4)  Writing Formulas for Ionic Compounds (3.5)  Naming Simple Ionic Compounds (3.5)  Naming Ionic Compounds Containing Polyatomic Ions (3.5)  Writing Molecular and Empirical Formulas (3.3)  Naming Molecular Compounds (3.6)  Naming Acids (3.6)  Calculating Formula Mass (3.7)  Using Formula Mass to Count Molecules by Weighing (3.7)  Calculating Mass Percent Composition (3.8)  Using Mass Percent Composition as a Conversion Factor (3.8)  Using Chemical Formulas as Conversion Factors (3.8)  Obtaining an Empirical Formula from Experimental Data (3.9)  Calculating a Molecular Formula from an Empirical Formula and Molar Mass (3.9)  Obtaining an Empirical Formula from Combustion Analysis (3.9)  Balancing Chemical Equations (3.10) Chapter 3. KEY CONCEPTS

5. 3-5 CHEM 100, Fall 2013 LA TECH 1) Glucose has a molecular formula of C 6 H 12 O 6 (M.W. 180.16 g/mol). a. How many grams of C are available in 1 mole of glucose? b. How many grams of H are available in 1 mole of glucose? c. How many grams of O are available in 1 mole of glucose?

6. 3-6 CHEM 100, Fall 2013 LA TECH Percentage Composition description of a compound based on the percent relative amounts of each element in the compound

7. 3-7 CHEM 100, Fall 2013 LA TECH n x Gram Atomic weight % mass = --------------------------------------- x 100 formula weight (GMW, GFW) n = subscript of the element in the formula % Element Composition in Compounds from Formula

8. 3-8 CHEM 100, Fall 2013 LA TECH 2) What are the mass % of carbon and hydrogen in carbon dioxide CO 2 ? (CO 2 ; M.W. = 44.01 g/mole) % C: % O: CHEM 100, Fall 2012 LA TECH 2-8

9. 3-9 CHEM 100, Fall 2013 LA TECH Example: What is the percent composition of carbon in chloroform, CHCl 3, a substance once used as an anesthetic? = 1(gaw) C + 1(gaw) H + 3(gaw) Cl MM = (12.01 + 1.00797 + 3 x 35.453) amu = 119.377amu 1 x (12.011) %C = x 100 = 10.061% C 119.377 1(1.00797) 1(1.00797) = x 100 = 0.844359% H 119.377 119.377 3(35.453) 3(35.453) = x 100 = 89.095% Cl 119.377 119.377 %H %C

10. 3-10 CHEM 100, Fall 2013 LA TECH 3) Bicarbonate of soda (sodium hydrogen carbonate) has ionic formula: NaHCO 3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. Molar mass: %Na: % C: % O: %H

11. 3-11 CHEM 100, Fall 2013 LA TECH Example: What is the percent composition of chloroform, CHCl 3, a substance once used as an anesthetic? %C = 10.061% C %H = 0.844% H %Cl = 89.095% Cl 100.00

12. 3-12 CHEM 100, Fall 2013 LA TECH Mass percent of element in C 6 H 12 O 6 Molar Mass = 180.16 g/mol 6 x 12 %C = x 100 = 40.00% C 180.16 12 x 1.01 12 x 1.01 % H = x 100 = 6.73% H 180.16 6 x 16.00 6 x 16.00 %O = x 100 = 53.29% O 180.16 180.16 Total =100.02%

13. 3-13 CHEM 100, Fall 2013 LA TECH 4) NH 3 (M.W. 17.03 g/mole) and NH 4 NO 3 (F.W. 80.05 g/mole) used as nitrogen fertilizer. Which one will provide more nitrogen for the same weight? NH 3 NH 4 NO 3 NH 3 NH 4 NO 3

14. 3-14 CHEM 100, Fall 2013 LA TECH Simple whole number ratio of each atom expressed in the subscript of the formula. Molecular Formula = C 6 H 12 O 6 of glucose Empirical Formula = CH 2 O Empirical formula is calculated from % composition What is Empirical Formula?

15. 3-15 CHEM 100, Fall 2013 LA TECH How do you get Empirical Formula from % composition and vice versa?

16. 3-16 CHEM 100, Fall 2013 LA TECH Example: The burning of fossil fuels in air produces a brown-colored gas, a major air pollutant, that contains 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? %N = x 100 = 30.5% N %O = x 100 = 69.5% O

17. 3-17 CHEM 100, Fall 2013 LA TECH %N = 30.5% N %O = 69.5% O To get Relative # Atoms N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 1.00 1 O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99 1.99 2 Empirical Formula NO 2 Empirical Formula Weight = 46.0 (%/gaw), Divide by Smaller and by Multiply Integer Empirical Formula from % composition

18. 3-18 CHEM 100, Fall 2013 LA TECH Molecular Formula = n x empirical Formula Molecular from Empirical Formula Molecular weight 180

19. 3-19 CHEM 100, Fall 2013 LA TECH Molecular Formula Weight 180 Molecular Formula Weight 180 n = ------------------------- = ---- = 6 n = ------------------------- = ---- = 6 Empirical Formula Weight 30 Empirical Formula Weight 30 Molecular Formula = C 6 H 12 O 6 of glucose Empirical Formula Weight 30 Molecular Formula Weight = 180 Molecular Formula = (CH 2 O) n = (CH 2 O) 6 Molecular Formula from Empirical

20. 3-20 CHEM 100, Fall 2013 LA TECH Example: A colorless liquid used in rocket engines, whose empirical formula is NO 2, has a molar mass (MW) of 92.0. What is the molecular formula? FW = 1(gaw) N + 2(gaw) O = 46.0

21. 3-21 CHEM 100, Fall 2013 LA TECH 5) A carbohydrate contains 38.71% weight of C, 9.71% weight of H and 51.58% weight of O. What is the empirical formula? Moles of C: Moles of H: Moles of O: Mole ration of C:= H:= O: = Simple mole ratio of C:= H:= O: = Empirical formula =

22. 3-22 CHEM 100, Fall 2013 LA TECH 6) Combustion Analysis gives the following mass %: 26.7% C; 2.2% H; 71.1% O If a separate analysis determined the molecular mass of the compound to be 90 g/mole. Determine the Empirical Formula and the Molecular Formula of the compound.  Moles of C:  Moles of H:  Moles of O:  Mole ration of C:= H:= O: =  Simple mole ratio of C:= H:= O: =  Empirical formula =

23. 3-23 CHEM 100, Fall 2013 LA TECH 6) Combustion Analysis gives the following mass %: 26.7% C; 2.2% H; 71.1% O If a separate analysis determined the molecular mass of the compound to be 90 g/mole. Determine the Empirical Formula and the Molecular Formula of the compound.  Empirical formula =  Empirical formula mass  Molar mass/empirical formula mass =  Molecular formula =

24. 3-24 CHEM 100, Fall 2013 LA TECH Combustion Analysis

25. 3-25 CHEM 100, Fall 2013 LA TECH 7) When 5.00 g of a compound containing only carbon and hydrogen is burned completely, 15.7 g of CO 2 and 6.45 g of H 2 O is produced. What is the empirical formula?  Moles of C:  Moles of H:  Mole ratio of C:= H:=  Simple mole ratio of C:= H:=  Empirical formula of the compound =

26. 3-26 CHEM 100, Fall 2013 LA TECH Example Benzoic acid is known to sample contain only C, H, and O. A 6.49-mg of benzoic acid was burned completely in a C H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? (16.4-mg of CO 2 )(12.01-mg C) #mg C = = 4.48-mg C (44.01-mg CO 2 ) (44.01-mg CO 2 ) 4.48-mg C %C = x 100 = 68.9% C 6.49-mg sample 6.49-mg sample

27. 3-27 CHEM 100, Fall 2013 LA TECH (2.85-mg of H 2 O )(2.02-mg H 2 ) #mg H = = 0.319-mg H (18.02-mg H 2 O) (18.02-mg H 2 O) 0.319 mg H 0.319 mg H %C = x 100 = 4.92% H 6.49-mg sample Example Benzoic acid is known to sample contain only C, H, and O. A 6.49-mg of benzoic acid was burned completely in a C H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid?

28. 3-28 CHEM 100, Fall 2013 LA TECH Example Benzoic acid is known to sample contain only C, H, and O. A 6.49-mg of benzoic acid was burned completely in a C H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? % O = (100 - (68.9% C + 4.92% H) = 26.2% O 68.9% C + 4.92% H + ? % O = 100%

29. 3-29 CHEM 100, Fall 2013 LA TECH Chemical Equations P 4 O 10 (s) + 6H 2 O (l) = 4 H 3 PO 4 (l) Reactants enter into a reaction. Products are formed by the reaction. Parenthesis represent physical state Stoichiometric Coefficients are numbers in front of chemical formula gives the amounts (moles) of each substance used and each substance produced. Equation Must be balanced!

30. 3-30 CHEM 100, Fall 2013 LA TECH ----->, or ? Chemical Reactions Could be described in words Chemical Reactants?Products? Reaction conditions? equation: Stoichiometric coefficients? Number in front of substances representing moles, atoms, molecules

31. 3-31 CHEM 100, Fall 2013 LA TECH Balanced Chemical Equation Representation of a chemical uses coefficients (prefix numbers known as stoichiometric coefficients) to represent the relative amounts of reactants and reaction which products

32. 3-32 CHEM 100, Fall 2013 LA TECH Writing and Balancing Chemical Equations. Write a word Convert word Balance the formula equation by the use of prefixes (coefficients) to balance the number of each type equation. of atom on the reactant and equation. equation into formula equation product sides of the

33. 3-33 CHEM 100, Fall 2013 LA TECH Types of Chemical Reactions

34. 3-34 CHEM 100, Fall 2013 LA TECH Reaction of H 2 and I 2

35. 3-35 CHEM 100, Fall 2013 LA TECH Synthesis or Combination Reactions Formation of a compound from simpler compounds or elements. Decomposition reactions Compound breaks up to from simp er compounds or el ements. Displacement reactions acement: In a compound Single displ is replaced by another element. Exchange reactions n a compound group or ion is replaced by another group or ion in another compound. Double displacement: an el ement

36. 3-36 CHEM 100, Fall 2013 LA TECH Formation Reactions Formation of a compound from elements at standardstate. Combustion Reactions Compound reacts with oxygen to produce oxid es: water and carbon dioxide for organic compounds Acid/Base (Neutralization)Reactions An acid and a base react to form water and salt ( most ionic compounds are sa lts) Precipitation Reactions when two aqueous salt sol utions are mixed an insolu ble salt is formed when two aqueous salt solutions are mixed.

37. 3-37 CHEM 100, Fall 2013 LA TECH Balancing Chemical Equations 1. Check for Diatomic Molecules - H 2 - N 2 - O 2 - F 2 - Cl 2 - Br 2 - I 2 if these elements appear By Themselves in an equation, they Must be written with a subscript of 2 2. Balance Metals 3. Balance Nonmetals 4. Balance Oxygen 5. Balance Hydrogen 6.Recount All Atoms 7. If EVERY coefficient will reduce, rewrite in the simplest whole-number ratio

38. 3-38 CHEM 100, Fall 2013 LA TECH Balance following reactions: a) MgO(s) + Si(s)  Mg(s) + SiO 2 (s) First balance Metals (Mg and Si): MgO(s) + Si(s)  Mg(s) + SiO 2 (s) Then balance Nonmetals(O): MgO(s) + Si(s)  Mg(s) + SiO 2 (s) Recount All Atoms: Reactant side: Mg, O, Si. Product side : Mg, O, Si. b) P 4 O 10 (s) + H 2 O(l)  H 3 PO 4 (l ) First balance P and H: P4O10(s) + H 2 O(l) H 3 PO 4 (l) Then balance O: P 4 O 10 (s) + H 2 O(l) H 3 PO 4 (l)

39. 3-39 CHEM 100, Fall 2013 LA TECH 8) Balance following reactions: Combustion reactions of organic compounds f) C 3 H 8 (g) + O 2 (g) = CO 2 (g) + H 2 O(l) g) C 4 H 10 (g) + O 2 (g) = CO 2 (g) + H 2 O(l)