2. Objectives: 1. Explain the significance of a chemical formula. 2. Determine the formula of an ionic compound formed between two given ions. 3. Name an ionic compound given its formula. 4. Using prefixes, name a binary molecular compound from its formula. 5. Write the formula of a binary molecular compound given its name.

3. Chemical names and Formulas  There are millions of natural and synthetic chemical compounds  Calcium carbonate – limestone  Sodium chloride – table salt  Dihydrogen monoxide – water  These are their chemical and common names  Chemical names help to describe the atomic makeup of the compounds

4. Significance of a Chemical Formula  Chemical formula  Indicates the relative number of atoms of each kind in a chemical compound.  Molecular formula  Indicates the relative number of atoms of each kind in a molecule. (Covalently bonded) C 8 H 18 Subscript indicates there are 8 atoms of carbon in a molecule of octane Subscript indicates there are 18 atoms of hydrogen in a molecule of octane

5.  Chemical formula for ionic compound  Ionic compound consists of lattice of positive and negative ions held together by mutual attraction.  Chemical formula represented by one formula unit  Simplest ratio of the compounds positive and negative ions  Aluminum sulfate below consists of aluminum cations and sulfate anions Al 2 (SO 4 ) 3 Subscript 2 refers to 2 Aluminum atoms Subscript 4 refers to 4 oxygen atoms in the sulfate ion Subscript 3 refers to everything inside the parentheses giving 3 sulfate ions, with a total of 3 sulfur atoms and 12 oxygen atoms Note: when you only have one of an atom, no subscript is used Note: parentheses are used to identify polyatomic ion as one unit

6. Monatomic ions  Ions formed from a single atom  Examples  Na +1 lose one electron  Mg 2+  S 2- gain two electrons  N -3  Cl 1-  Not all main-group elements readily form ions  Examples  Carbon & Silicon form covalent bonds  d-block elements form variable charges  examples  Copper, can be Cu +1 or Cu +2  Iron, can be Fe +2 or Fe +3  Lead, can bePb +2, Pb +3, or Pb +4

7. Naming Monatomic ions  Positive ions  Name of element  Ex: K + Potassium Mg +2 Magnesium Al +3 __________ Sr +2 __________  Negative ions  Base of element + -ide ending  Ex: F -1 Fluoride N -3 Nitride O -2 _______ Br -1 _______

8. Binary Ionic Compounds  Compounds composed of two different elements  Total # of positive charges must be equal to total # of negative charges  Writing formulas, Ex: Aluminum oxide 1. Write the symbols for ions (Cation first) Al +3 O -2 2. Cross over the charges as subscripts Al 2 O 3 3. Check to make sure total charges are equal, divide by largest number, to give smallest whole-number ratio Al 2 O 3 +3 -2 2 x (+3) = +6 3 x (-2) = -6

9. Naming binary ionic compounds  Nomenclature  Naming system  Name Al 2 O 3  Name cation first : full name of cation  Aluminum  Name Anion last : base of anion + -ide  oxide Al 2 O 3 aluminum oxide

10. Practice Naming and Writing Formulas  Name 1. AgCl 2. ZnO 3. SrF 2  Write the formulas for 1. Zinc iodide 2. Zinc sulfide 3. Aluminum sulfide silver chloride zinc oxide strontium fluoride ZnI 2 ZnS Al 2 S 3

11. Stock System of Nomenclature  Some cations may have two or more different charges  Use stock system of naming (usually with d-block elements)  Roman numeral represents charge in parentheses  Fe +2 Fe +3 Iron(II) Iron(III)  Some cations that commonly form only one cation  Do not use roman numerals ( main group elements)  No Anions form more than one charge CuCl 2 +2 copper(II)chloride

12. Practice stock system  Write formula and give name for compound formed by ions Cr +3 and F -1 CrF 3 chromium(III) fluoride  Write formulas and give name for the following ionic compounds: CuBr 2 copper(II) bromideCu +2 and Br -1 Fe +2 and O -2 Fe +3 and O -2 FeO iron(II) oxide Fe 2 O 3 iron(III) oxide

13. Compounds containing Polyatomic Ions  All but NH 4 +, ammonium ion, are negatively charged  Most are oxyanions  Examples  NO 3 -1 NO 2 -1 nitratenitrite  Most common anion has –ate ending  Anion with one less oxygen has –ite ending  Anion with two less oxygen has hypo prefix and –ite ending  Anion with one extra oxygen has per prefix and –ate ending ClO 3 -1 ClO 2 -1 ClO -1 ClO 4 -1 chloratechlorite hypochloriteperchlorate

14. Naming Compounds with Polyatomic Ions  Same as naming for ionic compounds except  Name polyatomic ion as one unit  Example:  AgNO 3  Use parentheses if more than one polyatomic ion  Example  Al 2 (SO 4 ) 3 silver nitrate Show 2 Al +3 ions and 3 SO 4 -2 ions aluminum sulfate

15. Writing Formulas and Naming Compounds with Polyatomic Ions  Write the formula for these: 1. tin(IV) sulfate 2. calcium chloride 3. lithium nitrate 4. calcium nitrite 5. potassium perchlorate  Write the names for these: 1. Ag 2 O 2. Ca(OH) 2 3. NH 4 OH 4. FeCrO 4 5. KClO silver oxide calcium hydroxide potassium hypochlorite ammonium hydroxide iron(II) chromate Sn(SO4) 2 CaCl 2 LiNO 3 Ca(NO 2 ) 2 KClO 4

16. Naming Binary Molecular Compounds  May use stock system to name these  New system – must understand oxidation numbers  Prefix system  Old system – must know numerical prefixes 1. mono- 2. di- 3. tri- 4. tetra- 5. penta- 6. hexa- 7. hepta- 8. octa- 9. nona- 10. deca-

17. Rules for prefix system of Nomenclature 1. less-electronegative element is given first 2. Second element is named by combining a. Prefix indicating number of atoms b. Root of name of second element c. -ide ending 3. The o or a at the end of a prefix is usually dropped when the word following the prefix begins with another vowel Ex: P 4 O 10 Ex:tetra First element only gets a prefix if it has more than one phosphorusdecoxide

18. The 6 binary compounds of Nitrogen and Oxygen N 2 O NO NO 2 N 2 O 3 N 2 O 4 N 2 O 5 dinitrogenmonoxide nitrogen dinitrogen dioxidenitrogen monoxide dinitrogen trioxide tetroxide pentoxidedinitrogen Name the following molecular compounds 1.SO 3 2.ICl 3 3.PBr 5 1.carbon tetraiodide 2.phosphorus trichloride 3.oxygen difluoride Write formulas for the following molecular compounds sulfur trioxide iodine trichloride Phosphorus pentabromide CI 4 PCl 3 OF 2

19. Oxidation Numbers

20.  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

21. Assigning Oxidation Numbers 1. The atoms in a pure element are assigned an oxidation number of zero. 2. 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. 3. Fluorine always has an oxidation number of -1 because it is the most electronegative element. 4. Oxygen has an oxidation number of -2 in almost all compounds.

22. 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. 9) A monatomic ion has an oxidation number equal to the charge of the ion

23. Using Oxidation Numbers  Do practice problem #1 on page 234.

24. Practice #1 pg 234 a)HClH = 1+Cl = 1- b)CF 4 C = 4+F = 1- c)PCl 3 P = 3+Cl = 1- d) SO 2 S = 4+O = 2- e)HNO 3 H = 1+N = 5+O = 2- f)KHK = 1+H = 1- g)P 4 O 10 P= 5+O = 2- h)HClO 3 H = 1+Cl = 5+O = 2- i)N 2 O 5 N = 5+O = 2- j)GeCl 2 Ge = 2+Cl = 1-

25. Oxidation Number problems What would be the oxidation number of each element in the following compounds & polyatomic ions? H 2 OH = O = H 2 SO 4 H = S = O = N 2 O 5 N = O = SO 4 2- S =O = PO 4 3- P =O =

26. What would be the oxidation number of each element in the following compounds & polyatomic ions? H2OH2O H = 1+O = 2- H 2 SO 4 H = 1+S = 6+O = 2- N2O5N2O5 N = 5+O = 2- SO 4 2- S = 6+O = 2- PO 4 3- P = 5+O = 2-

27. 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. PCl 3  phosphorus trichloride  phosphorus (III) chloride Do section review problems #1-2 on page 235.

28.  Problems page 235 #1a-HF H = 1+F = 1- b-CI 4 C = 4+I = 1- c-H 2 O H = 1+O = 2- d-PI 3 P = 3+I = 1- e-CS 2 C = 4+S = 2- f-Na 2 O 2 Na= 1+O=1- This is a rare case when O = 1-. g-H 2 CO 3 H = 1+ C = 4+ O = 2- h-NO 2 1- N = 3+O = 2- i- SO S= 6+O=2-

29.  Problems page 235 #2a-CI 4  carbon (IV) iodide b-SO 3  sulfur (VI) oxide c-As 2 S 3  arsenic (III) sulfide d-NCl 3  nitrogen (III) chloride

30. 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) N 2 O 5  nitrogen __ oxide SiO 2  silicon __ oxide CF 4  carbon __ fluoride PI 3  phosphorus __ iodide SiBr 4  silicon __ bromide

31.  Using oxidation numbers & the stock system, what would be the names of the following binary molecular compounds? N 2 O 5  nitrogen V oxide SiO 2  silicon IV oxide CF 4  carbon IV fluoride PI 3  phosphorus III iodide SiBr 4  silicon IV bromide

32. Using Chemical Formulas

33. To find molar mass: add the masses of the elements in the formula of the compound. To find number of grams (mass): multiply # of moles times the molar mass of the compound. To find the number of moles: divide the number of grams by the molar mass of the compound.

34. Using Chemical Formulas  formula mass- the sum of the average atomic masses of all atoms represented in its formula  Do practice #1 on page 238  molar mass- the mass of one mole of an element or a compound (equal to the formula mass expressed in grams)  Do practice problems #1 & 2 on page 239.

35.  Practice #2 page 239 a) Al 2 S 3  2 Al x 27.0 = 54.0 3 S x 32.1 = 96.3 54.0 + 96.3 = 150.3 g/mol b) NaNO 3  1 Na x 23.0 = 23.0 1 N x 14.0 = 14.0 3 O x 16.0 = 48.0 23.0 + 14.0 + 48.0 = 85.0 g/mol c) Ba(OH) 2  1 Ba x 137.3 = 137.3 2 O x 16.0 = 32.0 2 H x 1.0 = 2.0 137.3 + 32.0 + 2.0 = 171.3 g/mol

36.  Problem #1 page 242 a) 6.60 g (NH 4 ) 2 SO 4 N = 2 x 14.0 = 28.0 H = 8 x 1.0 = 8.0 S = 1 x 32.1 = 32.1 O = 4 x 16.0 = 64.0 129.1 6.60/129.1 = 0.051 mol (NH 4 ) 2 SO 4 b) 4.5 kg = 4500 g Ca(OH) 2 Ca = 1 x 40.1 = 40.1 O = 2 x 16.0 = 32.0 H = 2 x 1.0 = 2.0 74.1 4500/74.1 = 60.7 mol Ca(OH) 2

37. To calculate % composition by mass: 1- find the molar mass of a compound 2- divide the mass of each element by the molar mass of the compound 3- multiply by 100 to convert each ratio to a percent

38. Percentage Composition  percentage composition- the percentage of the total mass of each element in a compound mass of element in 1 mole x 100% molar mass of compound eg. CO 2 mass C = 1 x 12.0 = 12.0 mass O = 2 x 16.0 = 32.0 molar mass of CO 2 = 44.0 g/mol %C = 12.0/44.0 (100) = 27.3% %O = 32.0/44.0 (100) = 72.7%

39. eg H 2 O = H = 2 x 1.0 = 2.0 O = 1 x 16.0 = 16.0 18.0 g/mol %H in H 2 O = 2.0 x 100 = 11.1% 18.0 % O in H 2 O = 16.0 x 100 = 88.9% 18.0

40. % composition by mass practice Do Practice problems #1-3 on page 244. Do Section Review problems #1, 3, & 5 on page 244.

41.  Problem #1 page 244 a) PbCl 2 Pb = 1 x 207.2 = 207.2 Cl = 2 x 35.5 = 71.0 278.2 Pb = 207.2 x 100 = 74.5% 278.2 Cl = 71.0 x 100 = 25.5% 278.2

42.  Problem #2 page 244  ZnSO 4 ·7H 2 OZn = 1 x 65.4 = 65.4 S = 1 x 32.1 = 32.1 O = 4 x 16.0 = 64.0 H 2 O = 7 x 18.0 = 126.0 287.5 %H 2 O = 126.0 x 100 = 43.8% 287.5

43.  Section Review #3 mass of 3.25 mol Fe 2 (SO 4 ) 3 ? Fe = 2 x 55.8 = 111.6 S = 3 x 32.1 = 96.3 O = 12 x 16.0 = 192.0 399.9 g/mol 3.25 mol x 399.9 g/mol = 1299.7 g

44.  Section Review #5 % composition of each element of (NH 4 ) 2 CO 3 N = 2 x 14.0 = 28.0 H = 8 x 1.0 = 8.0 C = 1 x 12.0 = 12.0 O = 3 x 16.0 = 48.0 96.0 g/mol %N = 28.0 x 100 = 29.2% 96.0 %H = 8.0 x 100 = 8.3% 96.0 %C = 12.0 x 100 = 12.5% 96.0 %O = 48.0 x 100 = 50.0% 96.0

45. Empirical Formula

46.  Empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole- number mole ratio of the ratio of the different atoms in the compound

47.  The empirical formula for the gas diborane is BH 3 but the molecular formula is B 2 H 6  The molecular formula corresponds to the empirical ratio multiplied by two

48.  Assume you have a 100.0g sample of the compound  The percent composition for diborane 78.1% B and 21.9% H 78.1 g B x 1 mol B = 7.22 mol B 10.81 g B 21.9 g H x 1 mol H = 21.7 mol H 1.01 g H  To find a ratio of smallest whole numbers, divide each number of moles by the smallest number 7.22 mol B: 21.7 mol H= 1 mole B:3.01 mol H 7.22 7.22

49. x( empirical formula) = molecular formula In an experiment, the mass for diborane was 27.67 amu The mass for the emperical formula is 13.84 amu x= 27.67 amu = 2.000 13.84 amu The molecular formula is B 2 H 6. 2(BH 3 ) = B 2 H 6