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Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds

Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds. A chemical formula indicates the kind and relative number of atoms in a chemical compound. C 8 H 18 (octane) has 8 carbon and 18 hydrogen atoms. Forming Ionic Compounds.

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Modern Chemistry Chapter 7 Chemical Formulas & Chemical Compounds

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  1. Modern Chemistry Chapter 7Chemical Formulas & Chemical Compounds • A chemical formula indicates the kind and relative number of atoms in a chemical compound. • C8H18 (octane) has 8 carbon and 18 hydrogen atoms.

  2. Forming Ionic Compounds • Compounds that have the elements held together by ionic bonds are called ionic compounds. • For an ionic compound to exist, the algebraic sum of the positive and the negative charges of the ions MUST = 0. • For instance, when a calcium atom becomes an ion, it has an overall 2+ charge which must be neutralized by ion(s) that have a 2- charge. • IF a Ca2+ cation forms an ionic bond with an O2- anion, the resulting compound will be neutral and the formula would be CaO. • However, if the Ca2+ bonds with a F- anion, it would require two F- ions to neutralize the Ca2+  CaF2

  3. Calcium ( Ca2+ ) combines with oxygen ( O2- )  CaO: + ---------- - Ca2+ O2- + ---------- - Calcium (Ca2+ ) combines with fluorine (F1- )  CaF2: + ---------- - F1- Ca2+ + ---------- - F1-

  4. Binary Ionic Compounds monatomic ions- ions formed from a single atom • IF the ion has a positive charge, use the name of the element • IF the ion has a negative charge, replace the ending of the element name with “ide”.

  5. Binary Ionic Compounds • binary compound- a compound composed of two elements • Writing binary ionic compound formulas: • Write the symbols for the ions side by side with the cation being first. • IF the charges of the two ions do not add to zero, cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion so the algebraic sum of the ions equals zero. • Check the subscripts and make sure they are in the smallest whole number ratio possible. e.g. aluminum oxide Al3+O2-  Al2O3

  6. Naming Binary Ionic Compounds • nomenclature- a naming system • Naming ionic compounds: • Write the name of the cation in the formula. • Write the name of the anion in the formula. Al2O3 aluminum oxide Do practice problems #1 & 2 on page 223.

  7. Problems- page 223 • a- potassium (K+) & iodide (I-)  KI b- magnesium (Mg2+) & chloride (Cl-)  MgCl2 c- sodium (Na+) & sulfide (S2-)  Na2S d- aluminum (Al3+) & sulfide (S2-)  Al2S3 e- aluminum (Al3+) & nitride (N3-)  AlN

  8. #2 a) AgCl silver chloride b) ZnO  zinc oxide c) CaBr2  calcium bromide d) SrF2  strontium fluoride e) BaO  barium oxide f) CaCl2  calcium chloride

  9. Stock System of Nomenclature • Some metallic elements that form cations such as chromium, cobalt, copper, iron, lead, manganese, mercury, nickel, and tin can form cations of more than one charge. (See ion chart) • For cations that can have multiple ionic charges, place a Roman numeral in parentheses that is equal to the ionic charge after the name of the metal. Cu1+  copper (I) Fe2+ iron (II) Cu2+  copper (II) Fe3+  iron (III)

  10. Using the Stock System • Write the formula of the ionic compound. • Use the charge of the anion to determine the charge of the cation. • Write the name of the cation with the charge followed by the name of the anion. CuCl copper (I) chloride CuCl2 copper (II) chloride Do practice problems #1 & 2 on page 225.

  11. Practice- page 225 #1 a) Cu2+ & Br-  CuBr2  copper II bromide b) Fe 2+ & O2-  FeO  iron II oxide c) Pb2+ & Cl-  PbCl2  lead II chloride d) Hg 2+ & S2-  HgS  mercury II sulfide e) Sn2+ & F-  SnF2  tin II fluoride f) Fe 3+ & O2-  Fe2O3  iron III oxide

  12. Practice- page 225 #2 a) CuO copper II oxide b) CoF3  cobalt III fluoride c) SnI4  tin IV iodide d) FeS  iron II sulfide

  13. Polyatomic Ions • polyatomic ion- a group of covalently bonded atoms with an ionic charge • oxyanion- a negatively charged polyatomic ion that contains oxygen

  14. Ionic Compounds & Polyatomic Ions • Writing and naming strategies are the same for ionic compounds with polyatomic ions. However, if more than one polyatomic ion is needed in the formula, the formula of the polyatomic ion is placed in parentheses and a subscript is used outside the parenthesis to show how many of the polyatomic ions are needed. e.g. iron (II) nitrate  Fe(NO3)2 • Do practice problems #1 & 2 on page 227.

  15. Practice Problems #1 page 227 • a- sodium iodide NaI b- calcium chloride CaCl2 c- potassium sulfide K2S d- lithium nitrate LiNO3

  16. e- copper (II) sulfate CuSO4 f- sodium carbonate Na2CO3 g- calcium nitrite Ca(NO2)2 h- potassium perchlorate KClO4

  17. 2a- Ag2O silver oxide b- Ca(OH)2 calcium hydroxide c- KClO3 potassium chlorate d- NH4OH ammonium hydroxide e- Fe2(CrO4)3 iron (III) chromate f- KClO potassium hypochlorite

  18. Practice Do the following formulas match the names given? IF they do not match, provide the CORRECT name or formula. CuSO4  copper I sulfate Fe2(SO4)3  iron III sulfate FeSO4  iron II sulfate copper I nitrate  CuNO3 copper II nitrate  Cu2NO3

  19. Practice Do the following formulas match the names given? CuSO4  copper I sulfate NO [copperII] Fe2(SO4)3  iron III sulfate YES FeSO4  iron II sulfate YES copper I nitrate  CuNO3 YES copper II nitrate  Cu2NO3 NO [Cu(NO3)2]

  20. Ionic Compound Nomenclature Name the following compounds: • MgBr2 • CuO • Cu2O • FeSO4 • Fe2(SO4)3 • CaSO4 • Cu2SO4 • CuSO4 • FePO4 • Fe3(PO4)2

  21. Ionic Compound Nomenclature Name the following compounds: MgBr2 magnesium bromide CuO copper II oxide Cu2O copper I oxide FeSO4 iron II sulfate Fe2(SO4)3 iron III sulfate

  22. CaSO4 calcium sulfate Cu2SO4 copper I sulfate CuSO4 copper II sulfate FePO4 iron III phosphate Fe3(PO4)2 iron II phosphate

  23. Ionic Compound Nomenclature Write the formulas of the following ionic compounds: • aluminum nitrate • aluminum nitride • magnesium phosphate • magnesium bromide • copper I sulfate • copper II sulfate • iron II nitrate • iron III fluoride • calcium hydroxide • calcium phosphate

  24. aluminum nitrate Al 3+ NO31- Al(NO3)3 aluminum nitride Al 3+ N 3- AlN magnesium phosphate Mg 2+ PO43- Mg3(PO4)2

  25. magnesium bromide Mg 2+ Br 1- MgBr2 copper I sulfate Cu 1+ SO4 2- Cu2SO4 copper II sulfate Cu 2+ SO42- CuSO4 iron II nitrate Fe 2+ NO31- Fe(NO3)2

  26. iron III fluoride Fe 3+F 1- FeF3 calcium hydroxide Ca 2+ OH 1- Ca(OH)2 calcium phosphate Ca 2+ PO43- Ca3(PO4)2

  27. Binary Molecular Compounds For this course, molecular compounds consist of two non-metals. For our purposes, hydrogen will be considered a non-metal. The ratio of the elements is NOT determined by their individual ionic charges. e.g. CO & CO2 or H2O & H2O2

  28. Naming of Binary Molecular Compounds From Formulas • Write the name of the first element in the formula. • Write the name of the second element using the suffix “ide”. • Use numerical prefixes to show the number of atoms of each element. e.g. P2O5 diphosphoruspentoxide 1 = mono 6 = hexa 2 = di 7 = hepta 3 = tri 8 = octa 4 = tetra 9 = nona 5 = penta 10 = deca

  29. Binary Molecular Compounds P4O10 tetra + phosphorus & dec + oxide tetraphosphorusdecoxide CO  carbon & mon + oxide carbon monoxide CO2  carbon & di + oxide carbon dioxide

  30. Formulas for Molecular Compounds • The element with the smaller group number is usually given first. If both elements are in the same group, the element with the larger period number is given first. This element is given a prefix ONLY if it contributes more than one atom to the molecule of the compound. • The second element is named by combining a prefix for the number of atoms of the element in the compound, the root of the name of the element, and the suffix “ide”. • The “o” or the “a” at the end of a prefix is usually dropped when the word following the prefix begins with another vowel.

  31. Writing Molecular Formulas • Write the formula of the first element in the compound name followed by the numerical subscript that shows how many there are (if there is no numerical prefix, there is one atom of the element). • Write the formula of the second element in the compound name followed by a subscript that shows how many atoms of the element are designated by the numerical prefix in the name. carbon dioxide  CO2 Do practice problems #1 & 2 on page 229.

  32. Practice Problems #1 & 2 page 229 1- a- SO3 sulfur trioxide b- ICl3 iodine trichloride c- PBr5 phosphorus pentabromide 2- a- carbon tetraiodide CI4 b- phosphorus trichloride PCl3 c- dinitrogen trioxide N2O3

  33. Molecular Compound Nomenclature Name the following molecular compounds. • N2O5 • SO2 • P4O10 • CO • CO2 • SiO2 • H2O2 • CF4 • PBr3 • SF2

  34. Name the following molecular compounds. • N2O5dinitrogenpentoxide • SO2 sulfur dioxide • P4O10tetraphosphorusdecoxide • CO carbon monoxide • CO2 carbon dioxide • SiO2 silicon dioxide • H2O2dihydrogen dioxide • CF4 carbon tetrafluoride • PBr3 phosphorus tribromide • SF2 sulfur difluoride

  35. Molecular Compound Nomenclature Write the formula for the following compounds. • carbon tetraiodide • trinitrogen heptoxide • triphosphorus hexasulfide • oxygen dichloride • disilicon triphosphide • tetranitrogen heptoxide • carbon disulfide • dihydrogen monosulfide • trihydrogen monophosphide • silicon disulfide

  36. Molecular Compound Nomenclature Write the formula for the following compounds. • carbon tetraiodide CI4 • trinitrogenheptoxide N3O7 • triphosphorushexasulfide P3S6 • oxygen dichloride OCl2 • disilicontriphosphide Si2P3 • tetranitrogenheptoxide N4O7 • carbon disulfide CS2 • dihydrogenmonosulfide H2S • trihydrogenmonophosphide H3P • silicon disulfide SiS2

  37. Section Review Problem #2 page 231 2- a- aluminum + bromine  AlBr3 b- sodium + oxygen  Na2O c- magnesium + iodine  MgI2 d- lead (II) + oxygen  PbO e- tin (II) + iodine  SnI2 f- iron (III) + sulfur  Fe2S3 g- copper (II) + nitrate  Cu(NO3)2 h- ammonium + sulfate  (NH4)2SO4

  38. Section Review Problem #3 page 231 • a- NaI sodium iodide b- MgS magnesium sulfide c- CaO calcium oxide d- K2S  potassium sulfide e- CuBr copper (I) bromide f- FeCl2 iron (II) chloride

  39. Section Review Problem #4 (a-e) page 231 • a- sodium hydroxide  NaOH b- lead (II) nitrate  Pb(NO3)2 c- iron (II) sulfate  FeSO4 d- diphosphorus trioxide  P2O3 e- carbon diselenide CSe2

  40. Oxidation Numbers • oxidation numbers (oxidation states)- assigned to the atoms composing a molecular compound or polyatomic ion that indicate the general distribution of electrons among the bonded atoms in the compound or ion

  41. Assigning Oxidation Numbers • The atoms in a pure element are assigned an oxidation number of zero. • The more electronegative (second) element in a binary molecular compound is assigned the number equal to the negative charge it would have if it were an anion. • Fluorine always has an oxidation number of -1 because it is the most electronegative element. • Oxygen has an oxidation number of -2 in almost all compounds.

  42. 5) Hydrogen has an oxidation number of +1 in compounds where it is listed first and -1 when it is listed last in the compound formula. 6) The algebraic sum of all oxidation numbers in a neutral compound is equal to zero. 7) The algebraic sum of the oxidation numbers of the atoms in a polyatomic ion equal the ion’s charge. 8) Oxidation numbers can also be assigned to atoms in an ionic compound.

  43. Using Oxidation Numbers • Do practice problem #1 on page 234.

  44. Practice #1 pg 234 a) HCl H = 1+ Cl = 1- b) CF4 C = 4+ F = 1- c) PCl3 P = 3+ Cl = 1- d) SO2 S = 4+ O = 2- e) HNO3 H = 1+ N = 5+ O = 2- f) KH K = 1+ H = 1- g) P4O10 P= 5+ O = 2- h) HClO3 H = 1+ Cl = 5+ O = 2- i) N2O5 N = 5+ O = 2- j) GeCl2 Ge = 2+ Cl = 1-

  45. Oxidation Number problems What would be the oxidation number of each element in the following compounds & polyatomic ions? H2O H = O = H2SO4 H = S = O = N2O5 N = O = SO42- S = O = PO43- P = O =

  46. What would be the oxidation number of each element in the following compounds & polyatomic ions? H2O H = 1+ O = 2- H2SO4 H = 1+ S = 6+ O = 2- N2O5 N = 5+ O = 2- SO42- S = 6+ O = 2- PO43- P = 5+ O = 2-

  47. Oxidation Numbers & the Stock System • We can use oxidation numbers assigned to the less electronegative (first) element to name binary molecular compounds by using the oxidation number as if it were a cation. PCl3 phosphorus trichloride phosphorus (III) chloride Do section review problems #1-2 on page 235.

  48. Problems page 235 #1a- HF H = 1+ F = 1- b- CI4 C = 4+ I = 1- c- H2O H = 1+ O = 2- d- PI3 P = 3+ I = 1- e- CS2 C = 4+ S = 2- f- This is a rare case when O = 1-. g- H2CO3 H = 1+ C = 4+ O = 2- h- NO21- N = 3+ O = 2-

  49. Problems page 235 #2a- CI4 carbon (IV) iodide b- SO3  sulfur (VI) oxide c- As2S3  arsenic (III) sulfide d- NCl3  nitrogen (III) chloride

  50. Oxidation Numbers & the Stock System • Using oxidation numbers & the stock system, what would be the names of the following binary molecular compounds? (fill in the blank with the roman numeral) N2O5 nitrogen __ oxide SiO2 silicon __ oxide CF4 carbon __ fluoride PI3 phosphorus __ iodide SiBr4 silicon __ bromide

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