1. Stoichiometry1 Chem 120 Sec. 03 Fall 2003 me: C. Chieh Ph. D. (Dr. Jay, Prof. J, or Peter if U ps c120) Cyberspace Chemistry (CaCt) http://www. science.uwaterloo.ca/~cchieh/cact/ Office: C2-263 Phone: 888-4567 ext. 5816 e-mail: cchieh@uwaterloo.ca Please provide your ID number in your e-mail. Try to know me and want me to know you, then we probably will be friends.

2. Stoichiometry2 Students Responsibility Plan to fit the class – survive in society. Solve suggested problems - do them regularly. Ask for help before it’s too late – we are here to help. Read all instructions – distinguish assumptions and fact. Learn basics and techniques of science, and apply them to solve problems and achieve your goals. Know what to learn, how to organize, what to ask, where to find answers, know if you know your stuff, and how to learn in general.

3. Stoichiometry3 University Students You are a proud student of the university – very different from your past, very difficult at first, and very independent eventually. You are responsible for everything you do, and no one else but you are responsible for you. You have to do all things on your own – no one tells you what to learn or what to do. You have to know the right from wrong – think independently, do not completely trust anyone, have sound judgments, and make wise decisions. Get not just the look, but what under the hood.

4. Stoichiometry4 Announcements Labs starts this week for odd-numbered lab sections (1, 3, 5, … 27) Labs starts this week for odd-numbered lab sections (2, 4, 6, … 28) Purchase safety goggles and lab manual before first lab from ChemStores, ESC 109, across from the first-year lab. Must have a valid WHMIS sticker to work in the lab. If not, attend a WHMIS training session before your second lab period. Tutorials will start next week, no tutorials this week. Sign up for a Study Skill (4 weekly sessions: time management, note taking, reading, preparation for exams) workshop at http://www. adm.uwaterloo.ca/infocs/

5. Stoichiometry5 Announcements cont. Science 101 – Student Handbook (useful information and survival tips) is available for pick up in the Science Undergrad Office, ESC 253. Unlike years in the past, students who will fail Chem120 will not be allowed to take Chem123. They can repeat Chem120 in the Winter term by DE courses or the next fall. (serious)

6. Stoichiometry6 Properties of Matter Matter, energy & man-made things Physical properties: b.p., m.p., density, color, hardness, dielectric constant, heat of fusion, heat of vaporization, heat of sublimation malleability, brittle … Chemical properties: composition, bonding formula, structural, 3-D structure oxidation, reduction, combustion, chemical changes, decomposition, acidity, stability Biological properties: toxicity, chemical changes in biological systems

7. Stoichiometry7 Classification of Matter Mixture (substances) homogeneous: gasoline, perfume, wine, drinks hetrogeneous: milk, orange juice Compounds (pure) water H 2 O, methane CH 4, carbon dioxide CO 2, … Elements hydrogen, helium, lithium, nitrogen, oxygen, … uranium Make a diagram to classify matter

8. Stoichiometry8 SI Units and Measurements 7 Base Quantities Length: m mass:kg time: s temperature: K amount of substance: mol electric current: A luminous intensity: cd Derived Units 1000 mm = 1 m 1 L = 1 dm 3 Concentration: mol L – 1 Density: g cm – 3 speed: km/hr Always use units to specify quantities!

9. Stoichiometry9 Unit Conversions 1 m 1 km --------------- 1000000 mm 1000 mm ------------ 1 m = 0.001 km What’s the mass of 250 mL ethanol, whose density = 0.789 g / mL? 0.789 g ------------ 1 mL = ____ g ethanol250 mL ethanol Calculate volume of 250 g ethanol

10. Stoichiometry10 Unit Conversions – cont. Convert 50 miles per hour to meters per second (1 mile = 1.609 km) 50 mile --------- 1 hr 1.609 km ------------ 1 mile 1000 m ------------ 1 km 1 hr ------------ 60 min 1 min ------------ 60 s = 22 m s – 1 Pay some attention to significant figures, precision, and accuracy. 1 hr -------------- 3600 s 1609 km -------------- 1 mile

11. Stoichiometry11 Atoms and Atomic Theory (1808) 1.Each element is made up of tiny individual particles called atoms. 2.Atoms are indivisible; they cannot be created or destroyed. 3.All atoms of each element are identical in every respect. 4.Atoms of one element are different from atoms of any other element. 5.Atoms of one element may combine with atoms of another element, usually in the ratio of small whole numbers, to form chemical compounds. Picture atoms in your mind!

12. Stoichiometry12 Rutherford Atoms Following the discovery of electrons and radioactivity Rutherford concluded that atoms consists of heavy dense nuclei whose diameters are 10 – 5 times that of the atom from his alpha scattering experiment. Following the discovery of protons and neutrons, we know that atomic nuclei consists of neutrons and protons. Explain the following: you need to understand them to solve problems chemical elements atomic structure isotopes atomic mass, atomic weight, atomic radius molar mass, molecular weight

13. Stoichiometry13 Natural Potassium Isotope 39 K 1940 K 1941 K 19 atomic mass38.9637069 39.9639987 40.9618260 abundance93.2581%0.0117% 6.7302% stable Half-life: ( radioactive ) stable 1.277e9 years Beta Decay Energy 1311.093 keV Evaluate the atomic weight from the given data

14. Stoichiometry14 The Mole (mol) How many 40 K (0.012% of natural K) atoms are in 500 mg of KCl? Avogadro’s proposed the concept of molecules N A = 6.022e23 (x10 23 ) 0.5 g KCl 0.00012 mol 40 K -------------------- 1 mol K 1 mol K --------------- (39.1+35.5) g 6.022e23 atoms -------------- 1 mol = 4.8e17 40 K atoms

15. Stoichiometry15 Chemical Formulas Molecular formula: H 2 O H 2 O 2 C 6 H 12 Empirical formula: -HOCH 2 Structural formula:H-O-HH-O-O-H Structure Explain these terms yourself

16. Stoichiometry16 Determine Chemical Formulas The experiment to determine chemical composition (formula) C x H 2y =(burned in O 2 )=> x CO 2 + y H 2 O After the combustion, CO 2 and H 2 O are absorbed by some chemicals and their mass determined. From their masses, relative compositions of H and C are determined. Other method is used to determine N and O. When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of CO 2 and 1.29 g of H 2 O is produced. What is the empirical formula?

17. Stoichiometry17 When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of CO 2 and 1.29 g of H 2 O is produced. What is the empirical formula? Solution: 12 g C 1 mol C 3.14 g CO 2 ------------- ------------- = 0.0714 mol C 44 g CO 2 12 g C 2 g H 1 mol H 1.29 g H 2 O ------------- ------------- = 0.143 mol H 18 g H 2 O 1 g H Thus, mole ratio of C : H is 0.0714 : 0.143 = 1 : 2. Therefore, the empirical formula is CH 2 What quantities do you need in order to find the molecular formula? What is the weight % of C and H in this compound?

18. Stoichiometry18 Oxidation States 1.The oxidation state of any element such as Fe, H 2, O 2, P 4, S 8 is zero (0). 2.The oxidation state of oxygen in its compounds is -2, except for peroxides like H 2 O 2, and Na 2 O 2, in which the oxidation state for O is -1. 3.The oxidation state of hydrogen is +1 in its compounds, except for metal hydrides, such as NaH, LiH, etc., in which the oxidation state for H is -1. 4.The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion. 5.The following elements usually have the same oxidation states in their compounds: +1 for alkali metals - Li, Na, K, Rb, Cs +2 for alkaline earth metals - Be, Mg, Ca, Sr, Ba -1 for halogens except when they form compounds with oxygen or one another

19. Stoichiometry19 Oxidation states – cont. What are the oxidation states of N in these compounds? NH 3 N 2 H 4 NH 2 OH N 2 N 2 O NO NO 2 – NO 2 NO 3 – Determine oxidation states of all elements of any formula you see anywhere. What are the oxidation states of Cl in these compounds? Cl – Cl 2 ClO – ClO 2 – ClO 2 ClO 2 – ClO 3 – ClO 4 –

20. Stoichiometry20 Chemical Names Chemicals are named in a systematic way according to certain rules. Knowing the rules allows you to name a substance, and lets you understand the substance from the name. Thus, it’s a good idea to review, or learn, the rules for naming organic and inorganic compounds. Its not fun to teach these rules, but you may have fun reading them to find out how much you already know.

21. Stoichiometry21 Chemical & Physical Reactions Chemical reactions: 4 NH 3 + 3 O 2  2 N 2 + 6 H 2 O Balance: (atoms cannot be destroyed nor created) __ Al + __ O 2  __ Al 2 O 3 __ C 6 H 14 O 4 + __ O 2  __ CO 2 + __ H 2 O __ H 3 PO 4 + __ CaO  __ Ca 3 (PO 4 ) 2 + __ H 2 O Physical reactions: (s = solid; l = liquid; g = gas) H 2 O(s)  H 2 O(l) (energy is involved in H 2 O(s)  H 2 O(g) phase transitions) Stoichiometric coefficient Study states of matter & phase transition

22. Stoichiometry22 Reaction Stoichiometry Stoichiometry : quantitative relationship among reactants and products in a balanced reaction equation. Quantities may be in mole, mass, weight, or volume. How much KCl and O 2 are produced by decomposing 123.0 g of KClO 3 ? 2 KClO 3 (s)  2 KCl (s) + 3 O 2 (g) 245.2 g149.2 g96 g 123.0xy use proportionality to get x = 74.8 g KCl; y = 48.2 g O 2

23. Stoichiometry23 Stoichiometry Calculations 123.0 g KClO 3 How much KCl and O 2 are produced by decomposing 123.0 g of KClO 3 ? 2 KClO 3 (s)  2 KCl (s) + 3 O 2 (g) 245.2 g149.2 g96 g 123.0xy use proportionality to get x = 74.8 g KCl; y = 48.2 g O 2 96.0 g O 2 245.2 g KClO 3 1 mol O 2 32.0 g O 2 22.4 L O 2 1 mol O 2 Versatile, getting the amount in various units. Find amount of KCl in g, mol, and volume (specific gravity = 1.984) 123.0 g KClO 3

24. Stoichiometry24 Solutions and Concentrations Concentrations are expressed in many ways, g / 100 mL, g / L, mol / L (= M), mole fraction, weight fraction, percentage etc. Molarity is the expression of concentration in mole per liter. Explain solvent, solute, solution, and mixture For quantities in reactions: Amount = concentration * volume C 1 V 1 = C 2 V 2

25. Stoichiometry25 Excess and Limiting Reagent The reactant that is completely reacted is the limiting reagent or reactant. Reactants left behind due to a limiting reagent are excess reagents or reactants. How much AgNO 3 should be added in order to completely precipitate all the chloride ions in 25.00 mL of 0.1000 M NaCl solution? 25.00 mL 0.1000 mol NaCl 1000 mL 1 mol AgCl 1 mol NaCl 169.9 g AgNO 3 1 mol AgCl = 0.4248 g AgNO 3 Practice this method

26. Stoichiometry26 How much AgNO 3 should be added in order to completely precipitate all the chloride ions in 25.00 mL of 0.1000 M NaCl solution? Excess and Limiting Reagent - cont If less than 0.4248 g AgNO 3 is used, AgNO 3, what is the limiting reagent.? What if more than 0.4248 g AgNO 3 is used? If 0.3000 g AgNO 3 is used, how much AgCl is formed? = 0.4248 g AgNO 3 0.3000 g AgNO 3 143.4 g AgCl 169.9 g AgNO 3 = 0.2532 g AgCl Explain excess and limiting reagent

27. Stoichiometry27 Theoretical, actual, and percent yields = 0.2532 g AgCl0.3000 g AgNO 3 From the previous example, + 25.00 mL 0.1000 M NaCl This amount is the theoretical yield, what you get, usually less, is the actual yield. Percent yield = Actual yield * 100 % Theoretical yield Explain

28. Stoichiometry28 Percent in a Mixture When 2.0 g of NaCl and NaNO 3 mixture is dissolved in water, sufficient amount of AgNO 3 is added. The amount of dry AgCl is1.0 g. What is the percentage of NaCl in the mixture? Chemistry: Na +, Cl –, and NO 3 – ions are present in solution; so do Ag + and NO 3 – Ag + + Cl –  AgCl (s, a white ppt) 1.0 g AgCl 58.5 g NaCl 143.5 g AgCl 100 % NaCl 2.0 g sample = 20.4 % NaCl 1.0 g AgCl Is it possible to get more than 5.0 g AgCl in this experiment? Why or why not?