3. The study of chemical change is the heart of chemistry.

4. L AW OF C ONSERVATION OF M ASS “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789 “Atoms are neither created nor destroyed during any chemical reaction.” during any chemical reaction.”

5. 3.1 C HEMICAL E QUATIONS  Consider a simple chemical equation: 2 H 2 + O 2 → 2 H 2 O + : “reacts with”, → : “produces” reactants and products coefficients balanced

6. 3.1 C HEMICAL E QUATIONS  The difference between a subscript and a coefficient  Subscripts should not be change when balancing equation  B ALANCING E QUATIONS

7. 3.1 C HEMICAL E QUATIONS  Consider a chemical reaction:  B ALANCING E QUATIONS

8. 3.1 C HEMICAL E QUATIONS  Express the chemical reaction  Balance carbon and hydrogen  Complete the chemical equation by balancing oxygen  B ALANCING E QUATIONS

9. 3.1 C HEMICAL E QUATIONS

10.  I NDICATING THE STATES OF REACTANTS AND PRODUCTS

11. 3.2 S OME S IMPLE P ATTERNS OF C HEMICAL R EACTIVITY  C OMBINATION AND DECOMPOSITION REACTIONS

12. 3.2 S OME S IMPLE P ATTERNS OF C HEMICAL R EACTIVITY  C OMBINATION AND DECOMPOSITION REACTIONS  Figure 3.7

13. 3.2 S OME S IMPLE P ATTERNS OF C HEMICAL R EACTIVITY

14.  C OMBUSTION IN AIR  When hydrocarbons are combusted in air, they react with O 2 to form CO 2 and H 2 O.  Combustion of hydrocarbon derivatives CH 3 OH, C 2 H 5 OH Glucose (C 6 H 12 O 6 ) – oxidation reaction

15. 3.3 F ORMULA W EIGHTS  How do we relate the numbers of atoms or molecules to the amounts we measure in the laboratory?  Although we can not directly count atoms or molecules, we can indirectly determine their numbers if we know their masses.

16. 3.3 F ORMULA W EIGHTS  F ORMULA AND M OLECULAR W EIGHTS

17. 3.3 F ORMULA W EIGHTS  P ERCENTAGE COMPOSITION FROM FORMULAS  Needed for identifying unknown samples

18. 3.4 A VOGADRO ’ S N UMBER & THE M OLE  The definition of a mole The amount of matter that contains as many objects (atoms, molecules, etc) as the number of atoms in exactly 12 g of isotopically pure 12 C.  Avogadro’s number The number of objects in 1 mole of matter. 6.0221421 × 10 23

19. 3.4 A VOGADRO ’ S N UMBER & THE M OLE

20.  M OLAR M ASS  The mass of a single atom of an element (in amu) is numerically equal to the mass (in grams) of 1mol of that element.

21. 3.4 A VOGADRO ’ S N UMBER & THE M OLE  M OLAR M ASS  The molar mass of a substance is the mass in grams of one mole of the substance.  Figure 3.10

22. 3.4 A VOGADRO ’ S N UMBER & THE M OLE 0.02989 mol 6.05 mol 180.0 g/mol 84.0 g/mol

23. 3.4 A VOGADRO ’ S N UMBER & THE M OLE 71.1 g (a) 532 g, (b) 0.0029 g 164.1 g/mol 84.0 g/mol 98.1 g/mol

24. 3.4 A VOGADRO ’ S N UMBER & THE M OLE Molecules C 6 H 12 O 6

25. 3.5 E MPIRICAL F ORMULAS FROM A NALYSES  The ratio of the number of moles of each element in a compound gives the subscript in a compound’s empirical formula.  Consider a compound containing Hg & Cl (MW Hg 200.6, MW Cl 35.5) (73.9% Hg, 26.1% Cl by mass) How to get the empirical formula for the compound? HgCl 2

26.  Procedure for calculating an empirical formula from percentage composition. 3.5 E MPIRICAL F ORMULAS FROM A NALYSES 40.92 g C, 4.58 g H, and 54.50 g O. C:H:O = 3(1:1.33:1) = 3:4:3 C 3 H 4 O 3

27. 3.5 E MPIRICAL FORMULAS FROM ANALYSES  The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula.  M OLECULAR FORMULA FROM EMPIRICAL FORMULA The formula weight of the empirical formula C 3 H 4 is 3(12.0 amu) + 4(1.0 amu) = 40.0 amu the molecular formula: C 9 H 12

28. 3.5 E MPIRICAL FORMULAS FROM ANALYSES  A common technique used for the determination of the empirical formula for compounds containing principally carbon and hydrogen.  C OMBUSTION A NALYSIS Fig 3.14 Apparatus for combustion analysis.

29. 3.5 E MPIRICAL FORMULAS FROM ANALYSES  C OMBUSTION A NALYSIS Mass of O = mass of sample - (mass of C + mass of H) = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O C3H8OC3H8O

30. 3.6 Q UANTITATIVE I NFORMATION FROM B ALANCED E QUATIONS FROM B ALANCED E QUATIONS

31. 3.6 Q UANTITATIVE I NFORMATION FROM B ALANCED E QUATIONS FROM B ALANCED E QUATIONS

32. 3.6 Q UANTITATIVE I NFORMATION FROM B ALANCED E QUATIONS FROM B ALANCED E QUATIONS  Procedure for calculating amounts of reactants consumed or products formed in a reaction

33. KClO 3 122.55, KCl 74.5, O 2 32.00 Answer: 1.76 g

34. MW LiOH 23.95 2 LiOH(s) + CO 2 (g) → Li 2 CO 3 (s) + H 2 O(l) Answer: 3.64 g

36. 3.7 L IMITING R EACTANTS Let’s consider a sandwich-making process: If you have 10 slices of bread and 7 slices of cheese,  You will have 5 sandwiches and 2 slices of cheese leftover.  In this case, Limiting reactant (limiting reagent) : Bd Excess reactant (excess reagent) : Ch

37. 3.7 L IMITING R EACTANTS

38. Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting Reactant The most important commercial process for converting N 2 from the air into nitrogen- containing compounds is based on the reaction of N 2 and H 2 to form ammonia (NH 3 ): N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) How many moles of NH 3 can be formed from 3.0 mol of N 2 and 6.0 mol of H 2 ?

39. Answer: (a) AgNO 3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn Zn: 65.39 g/mol AgNO3: 169.87 g/mol Ag: 107.9 g/mol Zn(NO3)2: 189.39 g/mol

40. 3.7 L IMITING R EACTANTS  T HEORETICAL Y IELD  The quantity of product that is calculated to form when all of the limiting reactant reacts (a) The theoretical yield is 146.14 g/mol 84.16 g/mol

41. 3.7 L IMITING R EACTANTS  T HEORETICAL Y IELD  The quantity of product that is calculated to form when all of the limiting reactant reacts 146.14 g/mol 84.16 g/mol

42. E XERCISES

44. 3.94

45. E XERCISES Atomic number 57, Lanthanum 3.96

46. E XERCISES 3.101