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CHEM 120: Introduction to Inorganic Chemistry

This course provides an introduction to inorganic chemistry, covering topics such as calculations, chemical equations, compounds, and the mole concept. The course is instructed by Dr. Upali Siriwardane and includes tests and homework assignments. Office hours are available for additional assistance.

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CHEM 120: Introduction to Inorganic Chemistry

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  1. CHEM 120: Introduction to Inorganic Chemistry Instructor: Upali Siriwardane (Ph.D., Ohio State University) CTH 311, Tele: 257-4941, e-mail: upali@chem.latech.edu Office hours: 10:00 to 12:00 Tu & Th ; 8:00-9:00 and 11:00-12:00 M,W,& F

  2. Chapters Covered and Test dates • Tests will be given in regular class periods  from  9:30-10:45 a.m. on the following days: September 22,     2004 (Test 1): Chapters 1 & 2 • October 8,         2004(Test 2):  Chapters  3, & 4 • October 20,         2004 (Test 3): Chapter  5 & 6 • November 3,        2004 (Test 4): Chapter  7 & 8 • November 15,      2004 (Test 5): Chapter  9 & 10 • November 17,      2004 MAKE-UP: Comprehensive test (Covers all chapters • Grading: • [( Test 1 + Test 2 + Test3 + Test4 + Test5)] x.70 + [ Homework + quiz average] x 0.30 = Final Average •                               5

  3. Chapter 5. Calculations and the Chemical Equation 5.1 The Mole Concept and Atoms The Mole and Avogadro's Number Calculating Atoms, Moles, and Mass 5.2 Compounds The Chemical Formula 5.3 The Mole Concept Applied to Compounds 5.4 The Chemical Equation and The Information It Conveys A Recipe for Chemical Change Features of a Chemical Equation The Experimental Basis of a Chemical Equation 5.5 Balancing Chemical Equations 5.6 Calculations Using the Chemical Equation General Principles Use of Conversion Factors Theoretical and Percent Yield Pharmaceutical Chemistry: The Practical Significance of Percent Yield

  4. The mole concept and atoms • In ch 1 we learned that 1 amu = 1.661 x 10-24 g • So if the average mass of a gold atom is196.97 amu x 1.661 x 10-24 g = 3.27 x 10-22 g 1 amu • a very small no.

  5. Where did I get the 196.97 anu for the mass of one Au atom? • From the ____________________!!! • If I write amu after these nos. it implies that I have the mass of ______ atom of that element (in amu). • But 3.27 x 10-22 g is too small an amt to work with in the lab. • What to do? • Scale up to quantities that we can handle by

  6. Avogadro’s number • Defining one mole (mol) as amt of substance that contains as many elementary entities (atoms, molecules, ions , etc) as there are in atoms in exactly 12 g of the carbon-12 isotope. This is determined experimentally and is…...

  7. Useful relationship • # moles X = #g X/molar mass X

  8. Some problems • How many atoms are there in 5.10 moles of sulfur? What’s the mass of 5.10 moles of S? • How many moles of calcium atoms are in 1.16 x 1024 atoms of Ca? How many grams?

  9. Which of the following has more atoms: 1.10g of hydrogen atoms or 14.7 g of chromium atoms? • How many moles are in 0.040 kg Na?

  10. What’s the mass, in grams, of one atom of potassium? • One atom of some element has a mass of 1,45 x 10-22 g. Identify the element.

  11. Compounds • The chemical formula: • MgO (ion pair) • H2O • C12H22O11 • Ca3(PO4)2 • CuSO4.5H2O vs CuSO4

  12. The mole concept applied to compounds • The formula weight of a species is the sum of atomic masses (amu) of the atoms in a species. • Formula weight of NH3 For an ionic compound MgF2 =

  13. Covalent Cpds Molecular weight moles  (Molar Mass) Ionic Cpds Formula weight  Formula units  (Molar Mass) • In general we talk about • moles for covalent compounds • formula units rather than moles of ionic compounds.

  14. Molar mass • Mass of one mole of NH3: • Mass of 6.022 x 1023 molecules of NH3 is • Mass of one molecule of NH3 is.

  15. Mass of one mole of MgF2 is • Mass of one formula unit of MgF2 is • Mass of 6.022 x 1023 formula units of MgF2 is

  16. Calc the molar mass of Ca(NO3)2. • Calc the molar mass of a compound if 0.372 mol of it has a mass of 152g.

  17. 5.38. How many grams of each are required to have 0.100 mol of • A. NaOH • B. H2SO4 • C. C2H5OH • D. Ca3(PO4)2

  18. 5.40. How many moles are in 50.0 g of • A. CS2 • B. Al2(CO3)3 • C. Sr(OH)2 • D. LiNO3

  19. Calc the no. of C, H, and O atoms in 1.50 g of glucose (C6H12O6). • What is the average mass of one C3H8 molecule? • What is the mass of 5.00 x 1024 molecules of NH3?

  20. Law of conservation of mass • Mass is neither created nor destroyed in an ordinary chemical rxn. • Or the sum of the masses of the reactants is equal to the sum of the masses of the products

  21. mercury + oxygen ---> mercury(II)oxide • 10.03g ? 10.83g • Easier to use symbols for chem eqns.

  22. reactants  products • lhs rhs • may indicate physical state by (s), (g), (l), (aq)-aqueous solution • Remember that H2,N2,O2,F2,Cl2,Br2,I2 occur as diatomics in nature and are used as diatomics in chemical eqns

  23. To balance: Have to have same no of each kind of atom on both sides of the eqn. The bonding arrangement changes, but the no of each kind of atom doesn’t change.

  24. (D)

  25. H O 2 F e B r ( a q ) + H ( g ) 2 2 (aq) = aqueous, dissolved in water Chemical equations • Indicate phases of reactants and products Fe(s) + 2 HBr(aq) Conditions etc. (s) = solid (g) = gas (l) = liquid

  26. Balancing chemical eqns • Use correct formulas for the reactants and products (if word eqn at start) • Balance by putting coefficients (nos) in front of the formulas. You may not change the formulas! These coefficients are called the stoichiometric (measure of mass) coefficients. • By convention use the lowest set of whole no. coefficients to balance.

  27. Start by balancing elements that appear only once on each side of the equation • Balance remaining elements • Check your balanced equation! • To predict products--do an experiment

  28. To balance • hydrogen + nitrogen  ammonia • 1. write the symbols for the species in the rxn

  29. Now figure out how to get the same no of atoms of each kind on both sides by using whole no coefficients in front of the species. • As H2 + _N2 NH3, then • H2 + _N2 _ NH3, then • _ H2 + _N2 _ NH3 • Now have _ H’s, _N’s on both sides and the lowest set of whole no coefficients have been used. The equation is balanced.

  30. 3H2 + N2  2NH3 • 3 mol of H2 reacts with 1 mol of N2 to form 2 mol of NH3 • 3 molecules of H2 reacts with 1 molecule of N2 to form 2 molecules of NH3 • 6H + 2N reacts to give 6H and 2N • 6g of H2 reacts with 28 g of N2 to form 34g of NH3 • Note that

  31. Balance • C2H6 + O2 CO2 + H2O • H2O2g H2O + O2 • C2H5OH + O2g CO2 + H2O • KOH + H3PO4g K3PO4 + H2O • N2O5g N2O4 + O2

  32. Balance • NH4NO3g N2O + H2O • NH4NO2g N2 + H2O • Be2C + H2O g Be(OH)2 + CH4 • NH3 + CuO g Cu + N2 + H2O

  33. Balance • S2Cl2(s) + NH3(g) g N4S4(s) + NH4Cl(s) + S8(s)

  34. Calculations using the chemical eqn • Quantitative study of reactants and products in a chemical reaction • How much product will be formed? • How much reactant is needed? • Use coefficients in a balanced equation to convert between moles of different substances in a chemical reaction.

  35. Chemical Reaction • 3H2(g) + N2(g) ------> 2NH3(g) • 3 mol H2 (reactant) = 1 mol N2 (reactant) consumed • 3 mol H2 (reactant) = 2 mol NH3 (products) produced • 1 mol N2 (reactant) = 2 mol NH3 (products) produced • 3 x 2 (6) g H2 (reactant) = 1x 28 (28)mol N2 (reactant) consumed • 3 x 2 H2 (6) (reactant) = 2x 17 (34) NH3 (products) produced • 1 x 28 (28) g N2 (reactant) = 2 x 17 (34) NH3 (products) produced

  36. Chemical reactions • Hydrogen reacts with nitrogen to form NH3. • theoretically it is should be that when 6 g of hydrogen reacts completely with 28 g of nitrogen, 34 g of ammonia is formed. • However in real chemical reactions actual__ g of hydrogen reacting with __ g of nitrogen, __ g of ammonia is produced need be experimentally determined.

  37. 2H2 + O2g 2H2O

  38. 2H2 + O2g 2H2O • How many moles of H2 is needed to completely react with19.8 mol O2? • How many moles of H2O are formed when 25.4 mol of H2 react?

  39. 2H2 + O2g 2H2O • How many moles of H2 react with 38 g of O2? • What mass of H2O is formed when 59.0g of H2 reacts completely with O2? How much O2 reacted in this case?

  40. Mass relationships in chemical equations • Mole-to-mole conversions • use mole ratios as conversion factors • Mass-to-mole and mole-to-mass conversions • use molecular weights as conversion factors • Mass-to-mass conversions • do in multiple steps

  41. General prescription

  42. Problems • How many grams of Al2O3 can be produced from 15.0 g Al? • 4 Al(s) + 3O2(g) g 2Al2O3(s)

  43. C3H8 + O2 CO2 + H2O balance • How many mol of O2 does it take to completely burn 7.0 mol of C3H8? • How many mol each of CO2 and H2O are produced? • How many grams of oxygen does it take to completely burn 25.0 g of C3H8? • How many grams each of CO2 and H2O are produced when 25.0 g of C3H8 is burned?

  44. A 4.00 g sample of Fe3O4 reacts with O2 to produce Fe2O3. • 4Fe3O4(s) + O2(g)g 6Fe2O3(s) • Determine the no. of grams of Fe2O3 produced.

  45. Theoretical and percent yield • How good an experimentalist are you? • What if 100% of reactants are not converted to desired products? • Frequently happens because of “side reactions” (other products), handling, etc. • 100% amount is theoretical yield • Amount obtained is actual yield

  46. Theoretical yield - amount of product that would result if all limiting reagent gave only product • Actual yield - the amount of product actually obtained from a reaction (almost always less than the theoretical yield) • Percent yield - calculated by • % yield = actual yield  100% • theoretical yield

  47. Theoretical yield is what we calculate assuming 100% conversion of reactants to products. • In the combustion of 33.5g of C3H6, 16.1 g of H2O is isolated. What is the percent yield?

  48. If the % yield of Fe2O3 in problem was 90.0% what was the actual yield of Fe2O3?

  49. A 3.5 g sample of water reacts with PCl3 according to : 3H2O + PCl3g H3PO3 + 3HCl. • How many grams of H3PO3 are produced?

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