1. Chemical Quantities and Chemical Reactions

2. Chemical Quantities

3. Measuring Matter Often measure the amount of something by one of three different methods: –count –mass –volume

4. Measuring Matter By count: - Example: 1 dozen apples = 12 apples By mass: - Example: 1 dozen apples = 2.0 kg apples By volume: - Example: 1 dozen apples = 0.20 bushel apples

5. Measuring Matter Can convert among units: Example: 1 dozen apples 2.0 kg apples 12 apples 1 dozen apples 1 dozen apples 0.20 bushel apples

6. Finding Mass from a Count What is the mass of 90 average-sized apples if 1 dozen of the apples has a mass of 2.0 kg? # apples dozens apples mass apples Conversion factors: 1 dozen apples 2.0 kg apples 12 apples1 dozen apples

7. Finding Mass from a Count 90 apples X 1 dozen apples X 2.0 kg apples 12 apples 1 dozen apples = 15 kg apples

8. What is a Mole? mole - (mol) 6.02  10 23 representative particles of a substance - SI unit for measuring the amount of a substance -The number 6.02  10 23 is called Avogadro’s number (Italian scientist) - helped clarify difference between atoms and molecules

9. What is a Mole? Mole of any substance contains Avogadro’s number of representative particles, or 6.02  10 23 representative particles Representative particle - refers to species present in a substance - usually atoms, molecules, or formula units

10. What is a Mole?

11. Converting Number of Atoms to Moles Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium is 1.25 X 10 23 atoms of magnesium? Conversion factors: 1 mol Mg 6.02 X 10 23 atoms Mg

12. Converting Number of Atoms to Moles 1.25 X 10 23 atoms Mg X 1 mol Mg 6.02 X 10 23 atoms Mg = 2.08 X 10 -1 mol Mg = 0.208 mol Mg

13. What is a Mole? Converting Moles to Number of Particles

14. Converting Moles to Number of Atoms Propane is a gas used for cooking and heating. How many atoms are in 2.12 mol of propane (C 3 H 8 )? Conversion factors: 6.02 X 10 23 molecules C 3 H 8 1 mol C 3 H 8 11 atoms 1 molecule C 3 H 8

15. Converting Moles to Number of Atoms Example Continued: 2.12 mol C 3 H 8 X 6.02 X 10 23 molecules C 3 H 8 1 mol C 3 H 8 X 11 atoms 1 molecule C 3 H 8 = 1.4039 X 10 25 atoms = 1.40 X 10 25 atoms

16. The Mass of a Mole of an Element Atomic mass of an element expressed in grams is the mass of a mole of the element Molar mass- mass of a mole of an element - Examples: molar mass of Carbon = 12.0 g molar mass of Hydrogen = 1.0 g

17. The Mass of a Mole of an Element

18. One molar mass of carbon, sulfur, mercury, and iron are shown

19. The Mass of a Mole of an Element You must know formula of compound to calculate the mass of a mole of an element To calculate the molar mass of a compound: - find the number of grams of each element in one mole of the compound - add the masses of the elements in the compound

20. Finding the Molar Mass of a Compound The decomposition of hydrogen peroxide (H 2 O 2 ) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide? 1 mole H = 1.0 g H 1 mol O = 16.0 g O

21. Finding the Molar Mass of a Compound Example Continued: 2 mol H X 1.0 g H = 2.0 g H 1 mol H 2 mol O X 16.0 g O = 32.0 g O 1 mol O Molar mass of H 2 O 2 = 2.0 g + 32.0 g = 34.0 g

22. The Mole-Mass Relationship It’s very important to use the formula of the substance when calculating molar mass Use molar mass of an element or compound to convert between the mass of a substance and the moles of a substance

23. Converting Moles to Mass The aluminum satellite dishes are resistant to corrosion because aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al 2 O 3 ). This tough, resistant coating prevents any further corrosion. What is the mass of 9.45 mol of aluminum oxide? Moles = 9.45 mol Al 2 O 3 1 mol Al 2 O 3 = 102.0 g Al 2 O 3

24. Converting Moles to Mass Example Continued: mass = 9.45 mol Al 2 O 3 X 102.0 g Al 2 O 3 1 mol Al 2 O 3 = 964 g Al 2 O 3

25. Converting Mass to Moles When iron is exposed to air, it corrodes to form red-brown rust. Rust is iron(III) oxide (Fe 2 O 3 ). How many moles of iron(III) oxide are contained in 92.2 g of pure Fe 2 O 3 ? Mass = 92.2 g Fe 2 O 3 1 mol Fe 2 O 3 = 159.6 g Fe 2 O 3

26. Converting Mass to Moles Example Continued: Moles = 92.2 g Fe 2 O 3 X 1 mol Fe 2 O 3 159.6 g Fe 2 O 3 = 0.578 mol Fe 2 O 3

27. Mole-Volume Relationship Avogadro’s hypothesis – equal volumes of gases at same temperature and pressure contain equal numbers of particles

28. Mole–Volume Relationship Volume of a gas varies with temperature and pressure - Example: balloon filled with helium shrinks when outside on a cold day - The volume of a gas is usually measured at a standard temperature and pressure Standard temperature and pressure (STP) - means a temperature of 0°C and a pressure of 101.3 kPa, or 1 atmosphere (atm)

29. Mole-Volume Relationship At STP, 1 mol or, 6.02  10 23 representative particles, of any gas occupies a volume of 22.4 L Molar volume – quantity of 22.4 L

30. Calculating the Volume of Gas at STP Sulfur dioxide (SO 2 ) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters of 0.60 mol SO 2 gas at STP. Moles = 0.60 mol SO 2 1 mol SO 2 = 22.4 L SO 2 Conversion factor: 22.4 L SO 2 1 mol SO 2

31. Calculating the Volume of a Gas at STP Example Continued: Volume = 0.60 mol SO 2 X 22.4 L SO 2 1 mol SO 2 = 13 L SO 2

32. The Mole–Volume Relationship Calculating Molar Mass from Density:

33. Calculating Molar Mass of a Gas at STP The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass of the compound? Density = 1.964 g/L 1 mol (gas at STP) = 22.4 L Molar mass = 1.964 g X 22.4 L 1 L1 mol = 44.0 g/mol

34. The Mole Road Map

38. The Percent Composition of a Compound Percent composition- percent by mass of each element in the compound Percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%

39. Percent Composition

40. Calculating Percent Composition from Mass Data When a 13.60-g sample of a compound containing only magnesium and oxygen is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Mass = 13.60 g Mass of oxygen = 5.40 g Mass of magnesium = 13.60g – 5.40 g = 8.20 g Mg

41. Calculating Percent Composition from Mass Data Example Continued: % Mg = 8.20 g X 100% = 60.3% 13.60 g % O = 5.40 g X 100% = 39.7% 13.60 g 60.3% + 39.7% = 100%

42. The Percent Composition of a Compound Percent Composition from the Chemical Formula

43. Calculating Percent Composition from a Formula Propane (C 3 H 8 ), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Mass of C in 1 mol C 3 H 8 = 36.0g Mass of H in 1 mole C 3 H 8 = 8.0g Molar mass of C 3 H 8 = 44.0g/mol

44. Calculating Percent Composition from a Formula Example Continued: % C = 36.0 g X 100% = 81.8% 44.0g % H = 8.0g X 100% = 18% 44.0g Add up to 100% when expressed as 2 significant figures

45. The Percent Composition of a Compound Percent Composition as a Conversion Factor –You can use percent composition to calculate the number of grams of any element in a specific mass of a compound.

46. The Percent Composition of a Compound Propane (C 3 H 8 ) is 81.8% carbon and 18% hydrogen. You can calculate the mass of carbon and the mass of hydrogen in an 82.0 g sample of C 3 H 8.

47. Empirical Formulas Empirical formula- gives the lowest whole-number ratio of the atoms of the elements in a compound –The empirical formula of a compound shows the smallest whole-number ratio of the atoms in the compound

48. Determining Empirical Formula A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Percent of nitrogen = 25.9% N Percent of oxygen = 74.1% O

49. Determining Empirical Formula Since percent means parts per 100, you can assume that 100.0 g of the compound contains 25.9 g N and 74.1 g O. 25.9 g N X 1 mol N = 1.85 mol N 14.0 g N 74.1 g O X 1 mol O = 4.63 mol O 16.0 g O

50. Determining Empirical Formula Divide each molar quantity by the smaller number of moles: 1.85 mol N = 1 mol N 1.85 4.63 mol O = 2.50 mol O 1.85

51. Determining Empirical Formula Can’t have fraction : N 1 O 2.5 Obtain lowest whole-number ratio by multiplying each part of the ratio by the smallest whole number that will convert both subscripts into whole numbers 1 mol N X 2 = 2 mol N 2.5 mol O X 2 = 5 mol O Empirical Formula = N 2 O 5

52. Empirical Formulas Ethyne (C 2 H 2 ) is a gas used in welder’s torches. Styrene (C 8 H 8 ) is used in making polystyrene These two compounds of carbon have the same empirical formula (CH) but different molecular formulas

53. Molecular Formulas Molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole- number multiple of its empirical formula - must know compound’s molar mass to determine molecular formulas

54. Molecular Formulas Methanal, ethanoic acid, and glucose all have the same empirical formula— CH 2 O.

55. Molecular Formulas

56. Finding Molecular Formula Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N. Empirical formula = CH 4 N Molar mass = 60.0 g/mol Empirical formula mass = 30.0 g/mol 60.0/30.0 = 2 (multiply empirical by 2) Molecular formula = C 2 H 8 N 2

57. Chemical Reactions

58. Writing Chemical Equations Word Equations –To write a word equation: write the names of the reactants to the left of the arrow separated by plus signs write the names of the products to the right of the arrow, also separated by plus signs Reactant + Reactant  Product + Product

59. Writing Chemical Equations Methane + Oxygen  Carbon dioxide + Water

60. Writing Chemical Equations iron + oxygen  iron(III) oxide

61. Writing Chemical Equations Hydrogen Peroxide  Water and Oxygen

62. Writing Chemical Equations Chemical Equations – Chemical equation- is a representation of a chemical reaction the formulas of the reactants (on the left) are connected by an arrow with the formulas of the products (on the right)

63. Writing Chemical Equations Skeleton equation- a chemical equation that does not indicate the relative amounts of the reactants and products Here is the equation for rusting: Fe + O 2  Fe 2 O 3

64. Writing Chemical Equations You can indicate the physical states of substances by using symbols in a chemical reaction: - use (s) for a solid - use (l) for a liquid - use (g) for a gas - use (aq) for aqueous or a substance dissolved in water Fe(s) + O 2 (g)Fe 2 O 3 (s)

65. Writing Chemical Equations

66. Catalyst- is a substance that speeds up the reaction but is not used up in the reaction Without Catalyst With Catalyst

67. Writing Chemical Equations A catalyst is neither a reactant or product - its formula is written above the arrow in a chemical equation Example: H 2 O 2 (aq)H 2 O(l) + O 2 (g) MnO 2

68. Writing a Skeleton Equation Hydrochloric acid and solid sodium hydrogen carbonate are reacted together. The products formed are aqueous sodium chloride, water, and carbon dioxide gas. Write a skeleton equation for this reaction. Solid sodium hydrogen carbonate: NaHCO 3 (s) Hydrochloric acid: HCl(aq) Water: H 2 O(l) Carbon dioxide gas: CO 2 (g)

69. Writing a Skeleton Equation Example Continued: NaHCO 3 (s) + HCl(aq)NaCl(aq) + H 2 O(l) + CO 2 (g)

70. Balancing Chemical Equations This is a balanced equation for making a bicycle: Coefficients—small whole numbers that are placed in front of the formulas in an equation in order to balance it

71. Balancing Chemical Equations A chemical reaction is also described by a balanced equation Balanced equation- each side of the equation has the same number of atoms of each element and mass is conserved

72. Balancing Chemical Equations To write a balanced chemical equation: - write the skeleton equation - use coefficients to balance the equation so that it obeys the law of conservation of mass

73. Writing a Balanced Chemical Equation Hydrogen and oxygen react to form water. The reaction releases enough energy to launch a rocket. Write a balanced equation for the reaction. Determine the correct formulas for all of the reactants and products –Hydrogen = H 2 –Oxygen = O 2 –Water = H 2 O

74. Writing a Balanced Chemical Equation Write the skeleton equation: H 2 (g) + O 2 (g) H 2 O(l) Determine the number of atoms of each element between the reactants and products: - reactants = 2 hydrogen, 2 oxygen - products = 2 hydrogen, 1 oxygen

75. Writing a Balanced Chemical Equation Balance the elements one at a time If no coefficient is written, it is assumed to be 1 H 2 (g) + O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g) 2H 2 O(l)

76. Writing a Balanced Chemical Equation Practice Problem: Balance the following: Cu(s) + AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + Ag(s) Answer: Cu(s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag(s)

77. Classifying Reactions The five general types of reaction are: -combination -decomposition -single-replacement -double-replacement -combustion

78. Classifying Reactions Combination reaction- is a chemical change in which two or more substances react to form a single new substance Example: 2Mg(s) + O 2 (g) 2MgO(s)

79. Writing Combination Reactions Copper and sulfur are reactants in a combination reaction. Complete the equation for the reaction: Cu(s) + S(s)? Cu(s) + S(s)CuS(s)

80. Classifying Reactions Decomposition reaction- is a chemical change in which a single compound breaks down into two or more simpler products

81. Decomposition Reaction Example: 2HgO(s)2Hg(l) + O 2 (g) Practice: Decomposition of water using electrolysis H 2 O(l)? H 2 O(l)H 2 (g) + O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g) (balanced) electricity

82. Classifying Reactions Single-replacement reaction- is a chemical change in which one element replaces a second element in a compound

83. Single-Replacement Reaction Example: 2K(s) + 2H 2 O(l) 2KOH(aq) + H 2 (g) Practice: Write a balanced equation for the following: Zn(s) + H 2 SO 4 (aq)? Zn(s) + H 2 SO 4 (aq)ZnSO 4 (aq) + H 2 (g)

84. Classifying Reactions Activity series of metals lists metals in order of decreasing reactivity

85. Classifying Reactions Double-replacement reaction- is a chemical change involving an exchange of positive ions between two compounds

86. Double-replacement reaction Example: Na 2 S(aq) + Cd(NO 3 ) 2 (aq) CdS(s) + 2NaNO 3 (aq) Practice: Write the equation for the following double-replacement reaction: CaBr 2 (aq) + AgNO 3 (aq)? CaBr 2 (aq) + AgNO 3 (aq)AgBr(s) + Ca(NO 3 ) 2 (aq) CaBr 2 (aq) + 2AgNO 3 (aq)2AgBr(s) + Ca(NO 3 ) 2 (aq)

87. Classifying Reactions Combustion reaction- is a chemical change in which an element or a compound reacts with oxygen, often producing energy in the form of heat and light

88. Combustion Reactions Example: Combustion of gasoline 2C 8 H 18 (l) + 25O 2 (g) 16CO 2 (g) + 18H 2 O(l) Practice: write the combustion of benzene C 6 H 6 (l) C 6 H 6 (l) + O 2 (g)CO 2 (g) + H 2 O(g) 2C 6 H 6 (l) + 15O 2 (g)12CO 2 (g) + 6H 2 O(g)

89. Predicting the Products of a Chemical Reaction The number of elements and/or compounds reacting is a good indicator of possible reaction type and thus possible products

90. Net Ionic Equations Complete ionic equation is an equation that shows dissolved ionic compounds as dissociated free ions

91. Net Ionic Equations Spectator ion - ion that appears on both sides of an equation and is not directly involved in the reaction Net ionic equation- is an equation for a reaction in solution that shows only those particles that are directly involved in the chemical change - balanced with respect to both mass and charge

92. Net Ionic Equations Sodium ions and nitrate ions are not changed during the chemical reaction of silver nitrate and sodium chloride so the net ionic equation is:

93. Writing Net Ionic Equations Aqueous solutions of iron(III) chloride and potassium hydroxide are mixed. A precipitate of iron(III) hydroxide forms. Identify the spectator ions and write a balanced net ionic equation. Fe 3+ (aq) + 3Cl - (aq) + 3K + (aq) + 3OH - (aq) Fe(OH) 3 (s) + 3K + (aq) + 3Cl - (aq) Spectators = K + and Cl - Fe 3+ (aq) + 3OH - (aq)Fe(OH) 3 (s)

94. Predicting the Formation of a Precipitate You can predict the formation of a precipitate by using the general rules for solubility of ionic compounds