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Chapter 7

Chapter 7. Chemical Formulas & Chemical Compounds Ch 7: Problem Set: pg 231: 2, 4 pg 251: 1,3,4,6,16,17,23,24,33,36,37,42, 50. 7.1 Chemical Names and Formulas. Naming monatomic ions End in -ide Look up the charges on the periodic table F vs F -1 F F -1 S vs S -2 S S -2. Fluorine.

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Chapter 7

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  1. Chapter 7 Chemical Formulas & Chemical Compounds Ch 7: Problem Set: pg 231: 2, 4 pg 251: 1,3,4,6,16,17,23,24,33,36,37,42, 50

  2. 7.1 Chemical Names and Formulas • Naming monatomic ions • End in -ide • Look up the charges on the periodic table • F vs F-1 • F • F-1 • S vs S-2 • S • S-2 Fluorine Fluoride Sulfur Sulfide

  3. Writing Formulas for Ionic Compounds • Cations – positive ions (metal) • metal • Anions – negative ions • nonmetal • Charge on compound = 0

  4. Writing Formulas for Ionic Compounds • Binary ionic compounds • Rules • if the charges on the ions are the same, drop ‘em • if the charges are different, criss-cross • Same charges – • Na+1 Cl-1 - • Mg+2 O-2- • Different charges- • Na+1 S-2- • Mg+2 Cl-1- NaCl Sodium Chloride MgO Magnesium Oxide Na2S Sodium Sulfide MgCl2 Magnesium Chloride

  5. Writing Formulas for Ionic Compounds • Ternary Ionic compounds • Metal + (Polyatomic ion) • When naming, do not use the ending –ide • Sodium Nitrate • Sodium Carbonate • Aluminum Nitrate Na+1 NO3-1 Na(NO3) Na+1 CO3-2 Na2(CO3) Al+3 NO3-1 Al(NO3)3

  6. Writing Formulas for Ionic Compounds • Aluminum Phosphate • Aluminum Bihypophosphite • Calcium Phosphite Al+3 PO4-3 Al(PO4) Al2(HPO2)3 Al+3 HPO2-2 Ca3(PO3)2 Ca+2 PO3-3

  7. Writing Names for Ionic Compounds • Front name – positive (cation – metal) • Back name – negative (anion – nonmetal) • Binary ionic compounds – composed of only 2 types of elements ( M + NM) – end in -ide • NaCl • MgCl2 • Al2O3 • NaH Sodium Chloride Magnesium Chloride Aluminum Oxide Sodium Hydride

  8. The BIG Lie • Stock System – use Roman Numerals for naming compounds with metals that have multiple charges (the transitions!) Cory Matthews Loves Topanga Ingrid

  9. The BIG Lie • More exceptions to the Lie! • Ag is always = +1 charge • DON’T write Ag I • Zn always = +2 charge • DON’T write Zn II

  10. Practice! • Sn2N2 • AgOH • PbCO3 • Zn(OH)2 • Fe2(SO4)3 Tin (II) Nitride Silver Hydroxide Lead (II) Carbonate Zinc Hydroxide Iron (III) Sulfate

  11. More Practice!! • Cu(HSO2)2 • CuSO2 • CuHSO2 Copper (II) Bihyposulfite Copper (II) Hyposulfite Copper (I) Bihyposulfite

  12. Practice! • LiClO3 • LiClO2 • CaCO3 • Ca(HCO2)2 • Fe(NO3)3 Lithium Chlorate Lithium Chlorite Calcium Carbonate Calcium Bicarbonite Iron (III) Nitrate

  13. Writing Names for Molecular Compounds • Molecular Compounds – covalent compounds • 2 nonmetals • To name, we use prefixes don’t use the prefix Mono on the first atom

  14. Writing Names for Molecular Compounds • CO • CO2 • PCl3 • CBr4 • N2O5 • SF6 Carbon Monoxide Carbon Dioxide Phosphorous trichloride Carbon tetrabromide Dinitrogen pentoxide Sulfur hexafluoride

  15. Writing Formulas for Molecular Compounds • The prefixes = the subscripts. • Do NOTlook at the charges. • Sulfur Dioxide • Disulfur Trioxide • Dinitrogen pentoxide SO2 S2O3 N2O5

  16. Naming Acids • Acid - when a solution yields H+ ions in solution • 2 types • Binary • H and one other type of atom • ternary (sometimes called oxy) • acids that have H with a polyatomic ion

  17. Naming Binary Acids • Rules • Hydro__(beginning of name)__ic acid • Ex. HCl • Hydrochloric acid • HBr • HF • H2S • H3P Hydrobromic acid Hydrofluoric acid Hydrosulfuric acid Hydrophosphoric acid

  18. Writing Formulas from Names for Acids • Do the criss-cross • Ex. Hydronitric acid • H+1 N-3 H3N • Hydroiodic acid • H+1 I-1 HI • Hydrosulfuric acid • H+1 S-2 H2S

  19. Naming Ternary Acids • H + polyatomic • Rules • Do NOT start with hydro- • If the ending of polyatomic is –ate -ic + acid • If the ending of polyatomic is –ite -ous + acid • Ate/ite ic/ous • Example: • H2SO4 • H+ and SO4-2 – sulfate • sulfuric acid

  20. Naming Ternary Acids • H2SO3 • HClO4 • HClO3 • HClO2 • HClO

  21. Formulas for Ternary Acids • Use the criss-cross method • Nitric acid • Phosphorous acid

  22. 7.2 Oxidation Numbers • Since electrons are shared, there is no definite charge - we assign the more electronegative element the “apparent” negative charge - this is known as the oxidation # • oxidation numbers can also be positive. • oxidation # - a number assigned to an atom to show the distribution of electrons

  23. Oxidation Numbers • Rules • Free elements = 0 • Ex. Mg = O • Ions= charges • F = -1 • S = -2 • Oxygen (O) = -2 • except in peroxides (H2O2) O = -1 • H = +1 • except in metal hydrides (MgH, NaH) H = -1

  24. Oxidation Numbers • ….Rules • More electronegative atom gets a (-) charge • Ox #’s add up to 0 in compounds • Ox #’s = the charge in polyatomic ions

  25. Oxidation # Practice • FeO (Iron II Oxide) O = -2 Fe = ? • Fe2O3 (Iron III oxide) O = -2 Fe = ? -2 + x = 0 x = 2 3(-2) + 2x = 0 x = 3

  26. Oxidation # Practice • H2SO4 (Hydrogen Sulfate or Sulfuric Acid) O = -2 H = +1 S = SO4-2 X + 4(-2) = -2 X = 6

  27. Oxidation # Practice • H2SO3 (Hydrogen sulfite or sulfurous acid) • H = • S = • O = +1 +4 -2 SO3-2 X + 3(-2) = -2 X = 4

  28. Oxidation # Practice • H2Cr2O7 (Hydrogen Dichromate or Dichromic Acid • H = +1 • Cr = • O = -2 • NO3-1 (Nitrate) • N = • O = -2 • MgH2 (Magnesium hydride) • Mg = • H = -1 +6 +4 +2

  29. 7.3 Using Chemical Formulas • Step 1 – be able to calculate molar mass (aka – formula mass, molecular weight, atomic weight, atomic mass, gram formula weight, etc.) • Add atomic weights from the periodic table • round to the nearest 10th place • Examples • CH4 • MgSO4· 7H2O 12.0 + 4(1.0) = 16.0 g/mol 24.3 + 32.1 + 4(16.0) + 14(1.0) + 7(16.0) = 246.4 g/mol

  30. Liters 22.4 L Mole Molar Mass 6.022 x1023 Atoms, molecules, particles Grams 7.3 Using Chemical Formulas • Step 2 – be able to convert between grams, moles, particles, and liters

  31. Using Chemical Formulas • Convert 32.0 g of CH4 to moles, liters, molecules, total atoms, atoms of H • Moles • Liters • Molecules • Atoms • Atoms H

  32. Conversions • Moles CH4 • 32.0 g 1 mole 1 16.0g CH4 • Liters • 32.0g 1 mole 22.4L 1 16.0g 1mole 2.00moles 44.8 Liters

  33. Conversions • Molecules • 32.0g 1 mole 6.022 x 1023molecules 1 16.0g CH4 1mole • Atoms • 32.0g 1 mole 6.022 x 1023atoms 5 1 16.0g 1mole • Atoms H • 32.0g 1 mole 6.022 x 1023atoms 4 1 16.0g 1mole 1.20 x 1024molecules 6.00 x 1024molecules 4.80 x 1024molecules

  34. Percent Composition • Percentage Composition - every compound has a certain percentage of each type of atom (we measure it by mass) • Formula • % composition = mass element mass compound X 100 =

  35. Practice - % Composition • Calculate % composition if a compound contains 24 g of Carbon and 64 g of Oxygen • % composition = mass element mass compound • Total mass compound = 24 + 64 = 88 g • % Composition C = 24 x 100 = 27 % C 88 • % Composition O = 64 x 100 = 73% O 88 X 100 =

  36. Practice - % Composition • What is the % composition of Ba(OH)2?

  37. Molecular Mass of Ba(OH)2 Ba= 1x 137= 137 g O= 2x16= 32 g H=2x1= 2 g =171 g Ba: 137/171 = O: 32/171= H: 2/171 = X 100 = 80.1% X 100 = 18.7% X 100 = 1.2%

  38. Empirical Formula- the lowest whole number ratio of atoms in a compound (simplest formula) Three basic steps to the problems: • 1. Divide (%’s or grams) by the gram atomic mass • 2. Divide the resulting #’s by the smallest of those numbers • 3. Multiply by 2 or 3 only if a whole number ratio isn’t the result of step 2 • Ex1:Calculate the empirical formula if there is 52.17 % C, 13.04% H, and 34.78 % O. • C  52.17/12.0 = 4.3475  4.3475/2.17375 = 2 • H  13.04/1.0 = 13.04  13.04/2.17375 = 5.998849914 6 • O  34.78/16.0 = 2.17375  2.17375 /2.17375 = 1 Thus, the empirical formula = C2H6O

  39. Ex2 Calculate the empirical formula if there is 26.56 % K, 35.41 % Cr, and 38.03 % O. • K  26.56/39.1 = .679283887  .679283887/.679283887 = 1 • Cr  35.41/52.0 = .680961538  .680961538 /.679283887 = 1.002469  1 • O  38.03/16.0 = 2.376875  2.376875 /.679283887 = 3.499089 • The formula is K1Cr1O3.5 (cannot have half of an atom) • So, multiply each element by 2 • The empirical formula = K2Cr2O7

  40. Ex 3: Find the empirical formula if a sample contains 5.6 g N and 12.8 g O. • N  5.6/14.0 = .4  .4/.4 = 1 • O  12.8/16.0 = .8  .8/.4 = 2 • The empirical formula = NO2

  41. Finding Molecular Formulas: • The same steps as empirical formula with one additional step - use the gram formula mass of the empirical formula and its relationship to the gram formula • mass of the molecular formula to find what number to multiply • the empirical • formula by to find the molecular formula (sounds more complicated • than it is).

  42. Ex4 - Find the molecular formula of a compound (mw = 144.0 g) with 66.67 % C, 11.11 % H, and 22.22 % O C  66.67/12.0 = 5.5558333  5.5558333 /1.38875 = 4.000600  4 H  11.11/1.0 = 11.11  11.11/1.38875 = 8 O  22.22/16.0 = 1.38875  1.38875 /1.38875 = 1 The empirical formula = C4H8O Check the empirical formula’s mass: C  4 (12.0) = 48.0 H  8 (1.0) = 8.0 O  1 (16.0) = 16.0 72.0 (see next slide)

  43. EX 4: continued The compound needs to have a mass of 144.0 grams. C4H8O only has a mass of 72.0 grams. By dividing 144.0 by 72.0, it is found that C4H8O is off by a factor of 2. So, multiply each element in C4H8O by 2. The molecular formula now equals C8H16O2

  44. EX 5: 19.80% C, 2.50% H, 66.10% O, 11.6% N and MW=242.0 (C=12.0, O=16.0, H=1.0 and N=14.0)

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