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1 Ions Ions: A charged particle formed when a neutral atom or group of atoms gain or lose one or more electrons. Example Na  Na + + e - F + e -  F -

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Presentation on theme: "1 Ions Ions: A charged particle formed when a neutral atom or group of atoms gain or lose one or more electrons. Example Na  Na + + e - F + e -  F -"— Presentation transcript:

1 1 Ions Ions: A charged particle formed when a neutral atom or group of atoms gain or lose one or more electrons. Example Na  Na + + e - F + e -  F -

2 2 Cations Mg  Mg 2+ + 2e - Al  Al 3+ + 3e - Cation: A positively-charged ion. One or more electrons are lost from a neutral atom  oxidation

3 3 Anions I + e -  I - O + 2e -  O 2- O + 2e -  O 2- S + 2e -  S 2- S + 2e -  S 2- Anion: A negatively-charged ion. Electrons are gained by a neutral atom  reduction

4 4 Ions Ion charges can be predicted from the Periodic Table Ion charges can be predicted from the Periodic Table Main Group metal (IA-IVA) ion-charges correspond to group number Main Group metal (IA-IVA) ion-charges correspond to group number Sodium (Na) in IA  +1 Sodium (Na) in IA  +1 Exceptions Exceptions Tl, Sn, Pb, Sb, and Bi Tl, Sn, Pb, Sb, and Bi We’ll talk about non-main group metal ion-charges later We’ll talk about non-main group metal ion-charges later Main Group non-metal (IIIA-VIIIA) ion-charges correspond to (group# - 8) Main Group non-metal (IIIA-VIIIA) ion-charges correspond to (group# - 8) Fluorine (F) in VIIA  (7-8) = -1 Fluorine (F) in VIIA  (7-8) = -1 Boron is exception  -3 not -5 Boron is exception  -3 not -5 All this has to do with electron configuration All this has to do with electron configuration More anon More anon See website as well: http://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htm See website as well: http://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htm http://web.clark.edu/aaliabadi/CHEM131ions%20to%20memorize.htm

5 5 Compounds That Contain Ions  Require metal and non-metal  Form ionic bonds  Called an ionic compound Characteristic Properties 1. Very high melting points 2. Conduct an electric current when melted or when dissolved in water

6 6 Ionic compound The number of cations and anions must have a net charge of zero. The number of cations and anions must have a net charge of zero.

7 7 Compounds that Contain Ions Writing Formulas for Ionic Compounds Give the formulas for the compounds that contain the following pairs of ions: (a) K and I (b) Mg and N (c) Al and O

8 8 Types of Ionic Compounds Type I Compounds: The metal present forms only one type of cation. Type I Compounds: The metal present forms only one type of cation. Examples: the main group metals Examples: the main group metals Type II Compounds: The metal present can form two or more cations that have different charges or oxidation states Type II Compounds: The metal present can form two or more cations that have different charges or oxidation states Oxidation state = ionic charge, if any, on species Oxidation state = ionic charge, if any, on species Examples include Cr 2+, Cr 3+, Cu +, Cu 2+, etc. Examples include Cr 2+, Cr 3+, Cu +, Cu 2+, etc.

9 9 Naming Ionic Compounds 1. The cation is always named first and the anion second. 2. The cation takes its name from the name of the element. 3. The anion is named by taking the first part of the element name and adding –ide.

10 10 Naming Ionic Compounds Name the following Type I compounds NaCl NaCl RaBr 2 RaBr 2 Rb 2 O Rb 2 O AlI 3 AlI 3 K 3 N K 3 N Cs 4 Si Cs 4 Si Give the chemical formula for the following Type I compounds Strontium phosphide Strontium phosphide Calcium fluoride Calcium fluoride Beryllium carbide Beryllium carbide Lithium hydride Lithium hydride Barium sulfide Barium sulfide Magnesium telluride Magnesium telluride

11 11 Naming Ionic Compounds Type II compounds need to be identified by a Roman numeral  (I), (IV), etc. Type II compounds need to be identified by a Roman numeral  (I), (IV), etc. Represents oxidation state of cation Represents oxidation state of cation Not how many cations are present in compound! Not how many cations are present in compound! Example: NaCl  sodium (I) chloride is INCORRECT Example: NaCl  sodium (I) chloride is INCORRECT Example: SnCl 4  tin (IV) chloride is correct Example: SnCl 4  tin (IV) chloride is correct

12 12 Naming Ionic Compounds Type II Ionic Compounds FeCl 2 and FeCl 3 PbO and PbO 2 MnS and Mn 2 S 7

13 13 A Mixed Bag PbBr 2 and PbBr 4 PbBr 2 and PbBr 4 Aluminum arsenide Aluminum arsenide FeS and Fe 2 S 3 FeS and Fe 2 S 3 Thallium (III) boride Thallium (III) boride Mercury (II) carbide Mercury (II) carbide Na 2 S Na 2 S CoCl 3 CoCl 3 Cerium (IV) phosphide Cerium (IV) phosphide ScF 3 ScF 3 Gold (I) selenide Gold (I) selenide Vanadium (V) telluride Vanadium (V) telluride

14 14 Naming Compounds that Contain Polyatomic Ions Polyatomic Ion: An ion that contains more than one atom. They are charged entities composed of several atoms bound together. Consult my website for the list that must be memorized: Consult my website for the list that must be memorized: http://web.clark.edu/aaliabadi/CHEM131ions %20to%20memorize.htm http://web.clark.edu/aaliabadi/CHEM131ions %20to%20memorize.htm

15 15 Trends Sulfide, sulfite, sulfate Sulfide, sulfite, sulfate Nitride, nitrite, nitrate Nitride, nitrite, nitrate Phosphide, phosphite, phosphate Phosphide, phosphite, phosphate Chloride, hypochlorite, chlorite, chlorate, perchlorate Chloride, hypochlorite, chlorite, chlorate, perchlorate Parenthesis required if more than one polyatomic ion present Parenthesis required if more than one polyatomic ion present Ca(IO 3 ) 2 is correct Ca(IO 3 ) 2 is correct Ca(I) 2 is INCORRECT Ca(I) 2 is INCORRECT

16 16 Naming Compounds that Contain Polyatomic Ions Name or provide the chemical formula for each of the following compounds: (a) Ca(OH) 2 (e) Co(ClO 4 ) 2 (b) Sodium phosphate (f) platinum (IV) bicarbonate (c) KMnO 4 (g) Cu(NO 2 ) 2 (d) Ammonium carbonate (h) nickel (III) sulfite

17 17 Naming Compounds that Contain Polyatomic Ions Name or provide each of the following compounds: Name or provide each of the following compounds: (a) calcium carbonate (e) MoO (b) BaSO 4 (f) Iridium (VII) acetate (c) CsClO 4 (g) ZnO 2 (d) Zirconium (IV) bisulfite (h) lithium cyanide

18 18 Naming Acids Acids: A substance that yields hydrogen ions (protons, H + ) when dissolved in water. HCl (aq)  H + (aq) + Cl - (aq) HCl (aq)  H + (aq) + Cl - (aq) H 3 PO 4(aq)  3H + (aq) + PO 4 3- (aq) H 3 PO 4(aq)  3H + (aq) + PO 4 3- (aq)

19 19 Rules for naming acids If the formula does not contain oxygen the prefix of the acid is hydro and the suffix –ic is attached to the root name for the element. If the formula does not contain oxygen the prefix of the acid is hydro and the suffix –ic is attached to the root name for the element. Ex: HCl = hydrochloric acid, H 2 S = hydrosulfuric acid Ex: HCl = hydrochloric acid, H 2 S = hydrosulfuric acid When the anion contains oxygen, the acid name is formed from the anion name. The suffix –ic or –ous is added. When the anion contains oxygen, the acid name is formed from the anion name. The suffix –ic or –ous is added. When the anion ends in –ate, the suffix –ic is used. When the anion ends in –ate, the suffix –ic is used. H 2 CO 3 = carbonic acid H 2 CO 3 = carbonic acid When the anion ends in –ite, the suffix –ous is used. When the anion ends in –ite, the suffix –ous is used. H 2 SO 3 = sulfurous acid H 2 SO 3 = sulfurous acid

20 20 More Anion Acid ClO 4 - ______________ HClO 4 _______________ ClO 3 - ______________ HClO 3 _______________ ClO 2 - ______________ HClO 2 _______________ ClO - _______________ HClO ________________

21 21 Naming Compounds that Contain Only Nonmetals: Type III Type III compounds contain only nonmetals. Type III compounds contain only nonmetals. Form covalent bonds Form covalent bonds  share electrons  share electrons Rules for Naming Type III Binary Compounds 1. The first element in the formula is named first, and the full element name is used. 2. The second element is named as though it were an anion. 3. Prefixes are used to denote the numbers of atoms present. 4. The prefix mono- is never used for naming the first element. 5. Drop the “a” when it is followed by an “o”  Tetraoxide should be tetroxide

22 22 Naming Compounds that Contain Only Nonmetals: Type III Prefixes Used to Indicate Numbers in Chemical Names Prefix Number Indicated Prefix Number Indicated mono- 1 mono- 1 di- 2 di- 2 tri- 3 tri- 3 tetra- 4 tetra- 4 penta- 5 penta- 5 hexa- 6 hexa- 6 hepta- 7 hepta- 7 octa- 8 octa- 8 nona- 9 nona- 9 deca- 10 deca- 10

23 23 Practice CCl 4 CCl 4 Silicon dioxide Silicon dioxide NO 2 NO 2 Sulfur trioxide Sulfur trioxide P 2 O 5 P 2 O 5 Iodine pentafluoride Iodine pentafluoride Dinitrogen tetroxide Dinitrogen tetroxide SeI 2 SeI 2 Xenon hexafluoride Xenon hexafluoride

24 24 The Name Game Extra credit opportunity Extra credit opportunity Mix-and-match Mix-and-match

25 25 Remember the mole? Mole = amt that contains as many “things” as there are atoms of 12 g of C-12 Mole = amt that contains as many “things” as there are atoms of 12 g of C-12 1 mole = 6.022 x 10 23 particles 1 mole = 6.022 x 10 23 particles Molar mass (MM) = mass in grams per 1 mole of particle (g/mol) Molar mass (MM) = mass in grams per 1 mole of particle (g/mol)

26 26 Molecular Mass Summation of molar masses from Periodic Table based on molecular formula Summation of molar masses from Periodic Table based on molecular formula NaCl NaCl I 2 I 2 V 2 O 5 V 2 O 5

27 27 Solutions

28 28 Molar Mass Example Calculate the mass of 30.0 moles of polyvinyl chloride (PVC), C 2 H 3 Cl.

29 29 Solution

30 30 Molar Mass Example A sample of Na 2 SO 4. with a mass of 300.0 grams represents what number of moles of Na 2 SO 4 ? Example Calculate the number of grams of caffeine, C 8 H 10 N 4 O 2, in 8.13 x 10 23 molecules.

31 31 Solution

32 32 Percent Composition of Compounds: NaCl

33 33 Percent Composition of Compounds Example Compute the mass percent of each element in sodium sulfide, Na 2 S. Example Compute the mass percent of each element in nitric acid, HNO 3(aq).

34 34 Solution

35 35 Solution

36 36 Empirical Formulas Empirical Formula: or the simplest formula; the smallest whole-number ratio of the atoms present. Molecular Formula: the actual formula of a compound. It gives the composition of the molecules that are present. Empirical Formula Molecular Formula CHC 6 H 6 CHC 6 H 6 CH 2 O C 6 H 12 O 6 CH 2 O C 6 H 12 O 6 H 2 OH 2 O H 2 OH 2 O

37 37 Calculation of Empirical Formulas Steps for Determining the Empirical Formula of a Compound 1. Obtain the mass of each element present (in grams). 2. Determine the number of moles of each type of atom present. 3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers are integers (whole numbers), these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go to step 4. 4. Multiply the numbers derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula.

38 38 Calculation of Empirical Formulas Experiment: Suppose we weigh out 6.50 grams of Cr. We decide to heat this Cr in the air so that the Cr can react with O to form Cr x O y. After the sample cools, we weigh it again and find its mass to be 9.50 grams. How do we find the mass of oxygen? What is the empirical formula of this compound? Let’s work on this together.

39 39 Calculation of Empirical Formulas Example Nylon-6 consists of 63.68% C, 12.38% N, 9.80% H, and 14.14%O. Calculate the empirical formula for Nylon-6.

40 40 Solution

41 41 Calculation of Molecular Formulas We need to know the empirical formula and molar mass of the molecular compound. Molecular Formula = (empirical formula) n where n is a small whole number. Molecular Formula = n x Empirical Formula Molar Mass = n x Formula Weight  n = Molar Mass/Formula Weight

42 42 Calculation of Molecular Formulas Example A compound shows the following percentage compostion: 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96 g/mol. Determine the empirical formula and the molecular formula for this compound.

43 43 Solution

44 44 Calculation of Molecular Formulas Example Vitamin C consists of 40.92% C, 4.58% H, and 54.50% O on a mass basis, and has a molar mass of 176.12 g/mol. Determine the molecular formula of the compound.

45 45 Solution


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