Chemistry

1. Chemistry Mr. Blake First Semester Final Review

2. Significant Figures • Any digit that is not zero is significant • 1.234 kg 4 significant figures • Zeros between nonzero digits are significant • 606 m 3 significant figures • Zeros to the left of the first nonzero digit are not significant • 0.08 L 1 significant figure • If a number is greater than 1, then all zeros to the right of the decimal point are significant • 2.0 mg 2 significant figures • If a number is less than 1, then only the zeros that are at the end and in the middle of the number are significant • 0.00420 g 3 significant figures

3. How many significant figures are in each of the following measurements? 24 mL 2 significant figures 4 significant figures 3001 g 0.0320 m3 3 significant figures 6.4 x 104 molecules 2 significant figures 560 kg 2 significant figures

4. 89.332 + 1.1 one significant figure after decimal point two significant figures after decimal point 90.432 round off to 90.4 round off to 0.79 3.70 -2.9133 0.7867 Significant Figures Addition or Subtraction The answer cannot have more digits to the right of the decimal point than any of the original numbers.

5. 3 sig figs round to 3 sig figs 2 sig figs round to 2 sig figs Significant Figures Multiplication or Division The number of significant figures in the result is set by the original number that has the smallest number of significant figures 4.51 x 3.6666 = 16.536366 = 16.5 6.8 ÷ 112.04 = 0.0606926 = 0.061

6. 6.64 + 6.68 + 6.70 = 6.67333 = 6.67 = 7 3 Significant Figures Exact Numbers Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures The average of three measured lengths; 6.64, 6.68 and 6.70? Because 3 is an exact number

7. move decimal left move decimal right Scientific Notation 568.762 0.00000772 n > 0 n < 0 568.762 = 5.68762 x 102 0.00000772 = 7.72 x 10-6 Addition or Subtraction • Write each quantity with the same exponent n • Combine N1 and N2 • The exponent, n, remains the same 4.31 x 104 + 3.9 x 103 = 4.31 x 104 + 0.39 x 104 = 4.70 x 104

9. Converting within metric system using dimensional analysis 1. Convert to base unit by canceling units (Top unit cancels with bottom unit). 2. Place the multiplier with the base unit. 3. Place a 1 in front of the unit with prefix. 4. To enter multiplier into the calculator, use a 1 before the exponent key. Example: 10 -6 = 1 EE/EXP - 6 5. Metric dimensional analysis examples: a. Convert 3.6 nm to m. b. Convert 0.456 dag to pg. 3.6 nm 10 -9 m = 3.6 x 10 -9 m 1 nm 0.456 dag 10 1 g 1 pg = 4.56 x 10 12 pg 1 dag 10 -12 g

10. 1000 mL 1L L2 1.63 L x = 1630 mL mL 1L 1.63 L x = 0.001630 1000 mL Dimensional Analysis Method of Solving Problems • Determine which unit conversion factor(s) are needed • Carry units through calculation • If all units cancel except for the desired unit(s), then the problem was solved correctly. How many mL are in 1.63 L? 1 L = 1000 mL

11. mass density = volume A piece of platinum metal with a density of 21.5 g/cm3 has a volume of 4.49 cm3. What is its mass? m m d = d = V V Unit 2 Density – SI derived unit for density is kg/m3 1 g/cm3 = 1 g/mL = 1000 kg/m3 = 21.5 g/cm3 x 4.49 cm3 = 96.5 g m = d x V

12. Unit 3: Properties of Matter elements compounds 1. Substances: a. Substances can be either pure _________ or _____________. b. Substances are identified by enumerating their physical and chemical properties. • All specimens of a given substance will have the ____ chemical and physical properties. d. Examples of substances: 2. Mixtures: a. Mixtures: Two or more pure substances. 1). Homogeneous mixtures -- 2). Heterogeneous mixtures -- b. Examples of mixtures: same elements compounds Al, C, S, Au NaCl, H2O uniform nonuniform Homogeneous Heterogeneous salt water, shampoo granite, sand

13. Physical Properties of substances Include those features which definitely distinguish one substance from another. a. Density- b. Specific gravity- c. Hardness- Ability to resist scratching. MOH hardness scale is used as a basis of comparing the hardness of substances. 1). Low numbers = relatively soft substances. 2). High numbers = relatively hard substances. MOH scale: Diamond 10 d. Odor e. Color f. Taste g. Solubility in solvents h. Physical state: M.P. F.P. B.P. i. Properties of metals- 1). Malleability- hammered in a thin sheet 2). Ductility - stretched into a wire 3). Conduction of heat j. Accidental physical properties: mass & volume

14. Properties/Change Chemical Properties- Describe the ability of a substance to form other substances under given conditions. a. Chemical change: A change from one substance into another. b. Chemical properties may be considered to be a listing of all the chemical reactions of a substance. Changes in Matter: Matter undergoes physical and chemicals changes. • Energy may be defined as the ability to ________. • Matter always possesses energy in one form or another. Physical Change: The composition of a substance is ____________ and the substance ________ its own identity. Examples: Ans. Melting, boiling, freezing, crushing, cutting & breaking Chemical Change: The substance loses its ________ and the new substance formed has ______ chemical and physical properties. Evidence of chemical changes: Ans. Examples: Ans. do work not changed retains identity new Color, gas, solid, & heat/temperature change Rusting, photosynthesis, burning, respiration, oxidizing, & spoiling

15. Accuracy – how close a measurement is to the true value Precision – how close a set of measurements are to each other accurate & precise precise but not accurate not accurate & not precise

16. Calculating Error Absolute error (Ea) Ea = O - A Percent (relative) error (Er) Er = O - A A X 100

17. Unit 4: Ionic Nomenclature Ionic Compound • Metal / Non-metal ( Metal always are written first). • One element gains e - and the other loses e -. • Transfer of e- • Charged ions attract one another (opposites attract). Ex. Na + Cl – Ca 2+ Br – Shortcut to determining formula ( Criss - Cross ). 1). Number from charge becomes the subscript. 2). All ionic compounds are neutral (no + or -). 3). Subscripts are written in lowest possible ratio. 4). The number “1” is never written (It is implied). Ex. Lithium & oxide Ex. Aluminum & oxide Li + O 2- Al 3+ O 2- NaCl CaBr2 Li2O Al2O3

18. Helpful Hints on Oxy-Anions 1. -ite : smaller # of oxygen 2. -ate : larger # of oxygen Ex. NO3- NO2- SO42- SO32- 3. Hypo ite ClO- ite ClO2- chlorite ate ClO3- chlorate per ate ClO4- 4. thio = sulfur replacing an oxygen. Ex. OCN- SCN- SO42- S2O32- nitrate nitrite sulfate sulfite hypochlorite perchlorate cyanate thiocyanate sulfate thiosulfate

19. Nomenclature of binary ionic compounds (bi = 2) 1. Cation is named first (name of atom). 2. Anion is named second, ending changed to ide. 3. If the metal (cation) can have multiple charges, the charge is written as a roman numeral (IUPAC). 4. Formula to name: Li2O Fe2O3 5. Name to formula: Beryllimum fluoride Tin (II) oxide Lithium oxide Iron (III) oxide BeF2 SnO

20. Polyatomic Ions Group of atoms with a single charge. Ex. (1) CN- = (2) NH4+ = (3) OH- = 1. Polyatomic ions will always stay together as a group. • If there is more than one polyatomic ion, it must be placed in parentheses. • Examples: Fe 2+ OH - Fe(OH)2 Na + CN - NaCN cyanide ammonium hydroxide Iron (ll) hydroxide Sodium cyanide

21. Ternary Compounds • Fe(NO3)3 • Li2SO4 • Potassium thiocyanate • Aluminum permanganate • Lead (IV) acetate Iron (III) nitrate Lithium sulfate KSCN Al(MnO4)3 Pb(C2H3O2)4

22. Nomenclature of Hydrates 1. Hydrate: Ionic compound with water molecules stuck in the crystal lattice. The water is included in the name and formula. Ex. ZnSO4• 7 H2O CaCO3• 3 H2O Zinc sulfate heptahydrate Calcium carbonate trihydrate

23. Molecular Compounds 1. Molecular (covalent) compounds a. Non-metal to non-metal. Right of staircase including hydrogen b. Sharing of electrons. c. Non-metals can often combine in several different ways. 2. Nomenclature of binary molecular compounds: a. Greek prefixes are used: b. The prefix mono is omitted for the 1st element. Ex. CO = c. For oxides the ending a is omitted. 1). N2O = 2). N2O3 = Carbon monoxide Dinitrogen monoxide Dinitrogen trioxide

24. Nomenclature (Acids) 1. Acids: Compounds that contain hydrogen as the positive ion (H+). 2. Exceptions: water & hydrogen peroxide. 3. Binary Acids: Acids that do not contain oxygen. a. Use prefix hydro b. Add stem or full name of anion. c. Add suffix ic. d. Add the word acid. Ex. HBr = 4. Ternary Acids: Contain 3 or more elements including oxygen. a. Acids formed with anions that contain -ate become ic acids. Ex. HNO3 H2PO4 b. Acids formed with anions that contain -ite become -ous acids. Ex. HNO2 H2SO3 Hydrobromic acid Nitric acid Phosphoric acid Nitrous acid Sulfurous acid

25. Chemical Nomenclature • Ionic Compounds • often a metal + nonmetal • anion (nonmetal), add “ide” to element name Barium chloride BaCl2 Potassium oxide K2O Magnesium hydroxide Mg(OH)2 KNO3 Potassium nitrate

26. Transition metal ionic compounds • indicate charge on metal with Roman numerals Iron(II) chloride FeCl2 2 Cl- -2 so Fe is +2 FeCl3 3 Cl- -3 so Fe is +3 Iron(III) chloride Cr2S3 3 S-2 -6 so Cr is +3 (6/2) Chromium(III) sulfide

27. TOXIC! Laughing Gas Molecular Compounds HI hydrogen iodide NF3 nitrogen trifluoride SO2 sulfur dioxide N2Cl4 dinitrogen tetrachloride NO2 nitrogen dioxide N2O dinitrogen monoxide

28. Dozen = 12 Pair = 2 Unit 5 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.022 x 1023 Avogadro’s number (NA)

29. Do You Understand Molar Mass? 1 mol K 6.022 x 1023 atoms K x x = 1 mol K 39.10 g K How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 0.551 g K 8.49 x 1021 atoms K

30. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

31. Do You Understand Molecular Mass? 8 mol H atoms 6.022 x 1023 H atoms 1 mol C3H8O x x x = 1 mol C3H8O 1 mol H atoms 60 g C3H8O How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 72.5 g C3H8O 5.82 x 1024 atoms H

32. 2 x (12.01 g) 6 x (1.008 g) 1 x (16.00 g) n x molar mass of element %C = %H = %O = x 100% = 34.73% x 100% = 13.13% x 100% = 52.14% x 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0%

33. A + B C S + O2 SO2 C A + B 2KClO3 2KCl + 3O2 Unit 6 Types of Oxidation-Reduction Reactions Combination Reaction Decomposition Reaction

34. A + BC AC + B Sr + 2H2O Sr(OH)2 + H2 TiCl4 + 2Mg Ti + 2MgCl2 Cl2 + 2KBr 2KCl + Br2 Types of Oxidation-Reduction Reactions Displacement Reaction Hydrogen Displacement Metal Displacement Halogen Displacement

35. 3 ways of representing the reaction of H2 with O2 to form H2O reactants products A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction

36. C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 Balancing Chemical Equations • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water • Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

37. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Balancing Chemical Equations • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3

38. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Balancing Chemical Equations • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2

39. 4 C (2 x 2) 4 C Reactants Products 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation.

40. Mass Changes in Chemical Reactions • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units

41. Molar Molar Mass Mass C. Road Map (Memorize) A + B → C + D Mass (g) A Mass (g) D Mol to mol ratio!!! Moles A Moles D Coefficients From Balanced Particles A Molecules A Atoms A 6.022 x 1023 6.022 x 1023 Chemical Equations Particles D Molecules D Atoms D

42. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O 4 mol H2O 18.0 g H2O 1 mol CH3OH x = x x 2 mol CH3OH 32.0 g CH3OH 1 mol H2O Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH Coefficients from balanced chemical equation molar mass H2O 209 g CH3OH 235 g H2O

43. How many grams of ammonium sulfate can be produced if 30.0 mol of sulfuric acid react with excess NH3 according to the equation: 2 NH3 + H2SO4 --> (NH4)2SO4 30.0 mol H2SO4 1 mol (NH4)2SO4 132.16 g (NH4)2SO4 1 mol H2SO4 1 mol (NH4)2SO4 = 3964.8 g (NH4)2SO4 = 3960 g (NH4)2SO4

44. What mass of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation: 2 Na + 2 H2O --> 2 NaOH + H2 20.0 g Na 1 mol Na 2 mol NaOH 40.00 g NaOH 22.99 g Na 2 mol Na 1 mol NaOH = 34.8 g NaOH

45. 6 green used up 6 red left over Limiting Reagents

46. Do You Understand Limiting Reagents? 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent

47. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al

48. % Yield = x 100 Actual Yield Theoretical Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction.

49. Unit 7 Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough.

50. Properties of Waves Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n