1. 1 TOPIC 5-CHEMICAL COMPOUNDS CONTENTS Types of Chemical Compounds and Their Formulas The Mole Concept and Chemical Compounds Composition of Chemical Compounds Oxidation States Naming Chemical Compounds

2. 2 CHEMICAL COMPOUNDS They are composed of two or more different elements A molecule of a compound is a group of bonded atoms that actually exists and can be identified as a distinct entity Chemical formula indicates The elements present and The relative numbers of atoms of each element in the compound

3. 3 Molecular Compounds A molecular compound is a compound Comprised of discrete molecules The forces that hold atoms together in molecules are known covalent bonds –A formula unit is the smallest collection of atoms on which a formula can be based. –Water  H 2 O –Carbon dioxide  CO 2 –Sodium chloride  NaCl –Iron (II) sulfide  FeS Subscript numbers indicate the relative numbers of atoms Types of Chemical Compounds and Their Formulas

4. 4 Empirical Formula,Molecular Formula And Structural Formula The empirical formula is the simplest formula of a compound.It tells us the types of atoms present and their relative numbers Exp: The empirical formula of glucose is CH 2 O. The molecular formula is based on an actual molecule of the compound. In some cases the empirical and molecular formulas are identical such as Formaldehyde(CH 2 O). In other cases the molecular formula is a «multiple» of the empirical formula such as acetic acid(C 2 H 4 O 2 ) and glucose(C 6 H 12 O 6 ) A structural formula show the order in which atoms are bonded together in a molecule and by what types of bonds

5. 5 Empirical formula CH 2 O Molecular formula C 2 H 4 O 2 Structural formula Molecular Model (Ball and stick) Molecular Model (Space filling) CH 3 COOH CH 3 CO 2 H Linear formula Oleic acid (Omega-9) C 18 H 36 Acetic acid CH 3 CO 2 H

6. 6 Ionic Compounds  An important feature of the metallic elements isthe tendency of their atoms to lose one or more electrons when they combine with nonmetal atoms. In these combinations the nonmetal atoms display a tendency to gain one or more electrons.  The ionic compounds are the compounds comprised of positive and negative ions joined together by electrostatic forces of attraction.  As a result of the electron transfer, the metal atom becomes a positive ion and is a cation.  And the nonmetal atom becomes a negative ion and is called an anion.

7. 7 In some cases electrons are transferred from one atom to another. Positive and negative ions are formed and attract each other through electrostatic forces called Ionic bonds Na [Ne] 3s 1 +Cl [Ne] 3s 2 3p 5 Na [Ne] + +Cl [Ar] - Cl - Na + “Cation” - Metal atoms lose electrons and build up the “ + ” charged ions. “Anion” – Nonmetal atoms gain electrons and build up the “ - ” charged ions Sodium combines with chlorine.As a result of this combination occurs an ionic compound Sodiumchloride

8. 8 Na:Cl = 1:1  formula unit = NaCl Mg +2 Na + + Cl - Na Cl NaCl is a compound with a neutral electrical charge Cl - Mg +2 + 2Cl - MgCl 2 MgCl 2 is a compound with a neutral electrical charge formula unit Formula unit is the smallest collection of the atoms on which the formula is based Formula Unit

9. 9 The Mole Concept and Chemical Compounds Formula Mass: The mass of a compound in a formula unit as in amu(atomic mass unit). Molecular mass: The mass of a molecule relative to the amu. (Formula mass= Molecular mass) 1 mole of compound is an amount of compound containing 6,02214 x10 23 formula unit or molecules Molar mass: The mass of one mol of the compound- one mol of molecules of a molecular compound and one mol of formula units of an ionic compound

10. 10 Formula mass: is the total mass of the atoms, building up a compound, in amu. Calculate the formula mass of these compounds: NaCl, H 2 O, ve H 3 PO 4 NaCl 1 Na 22,9898 amu= 22,9898 amu 1 Cl 35,4527 amu= 35,4527 amu 1 NaCl= 58,4425 amu 2 H 1,0079 amu= 2,0158 amu 1 O 15,9994 amu= 15,9994 amu 1 H 2 O= 18,0152 amu H 3 PO 4 3 H 1,0079 amu= 3,0237 amu 1 P 30,9738 amu= 30,9738 amu 4 O 15,9994 amu= 63,9976 amu 1 H 3 PO 4 = 97,9951 amu H2OH2O

11. 11 MatterFormula massMolecular mass H 2 Fe H 3 PO 4 SO 3 NaCl 2,0158 amu2,0158 g 55,8470 amu55,8470 g 97,9952 amu97,9952 g 80,0642 amu80,0642 g 58,4425 amu58,4425 g

12. 12 HalotanC 2 HBrClF 3 M(C 2 HBrClF 3 ) = 2M C + M H + M Br + M Cl + 3M F = (2 x 12,01) + 1,01 + 79,90 + 35,45 + (3 x 18,99) = 197,38 g/mol Composition of Chemical Compounds Mole raten C /n halotan Mass rate M C /M halotan

13. 13 2 mol N x 14,0067 g/mol = 28,0134 g N 4 mol H x 1,0079 g/mol = 4,0316 g H 3 mol O x 15,9994 g/mol = 47,9982 g O 80,0432 g/mol %N = x 100% = % 35,00 28,0134 g N 2 80,0432 g %H = x 100% = % 5,04 4,0316 g H 2 80,0432 g %O = x 100% = % 59,96 47,9982 g O 2 80,0432 g % 100,00 The molar mass of NH 4 NO 3 and the percent composition of the atoms

14. 14 Molar mass of sulfuric acid = 2(1,008g) + 1(32,07g) + 4(16,00g) = 98,09 g/mol %H = x 100% = % 2,06 H 2(1,008g H 2 ) 98,09g %S = x 100% = % 32,69 S 1(32,07g S) 98,09g %O = x 100% = % 65,25 O 4(16,00g O) 98,09 g % 100,00 The percent composition of Sulfuric acid

15. 15 1. Assume that you have a sample of 100 g 2. Convert the masses of the elements in the 100 g sample to amounts into moles 3. Write a tentative formula based on the number of moles just determined. 4. Attempt to convert the subscripts in the tentative formula to small whole numbers. This require dividing each of the subscripts by the smallest one 5. If the subscript at this point differ only very slightly from whole numbers, round them off to whole numbers. 6. Multiply all subscripts by a small whole number chosen to make all subscripts integral. Solution in 6 steps: Determination of the empirical and molecular formula

16. 16 Step 1 : Determine the mass of each element in a 100 g sample C: 62,58 g H: 9,63 g O: 27,79 g Dibutyl succinate is an insect repellant used against household ants and roaches. Its composition is 62,58 % carbon, 9,63% hydrogen and 27,9% oxygen. Its experimentally determined molecular mass is 230 amu. What are the empirical and molecular formulas of dibutyl succinate Example

17. 17 Step 2: Convert each of these masses to an amount in moles. Step 3 : Write a tentative formula based on these numbers of moles Step 4: Divide each of the subscripts of the tentative formula by the smallest. C 5,21 H 9,55 O 1,74 C 2,99 H 5,49 O

18. 18 Step 5: Multiply all subscripts by a small number chosen to make all subscripts integral(here by 2). C 5,98 H 10,98 O 2 Empirical formula: C 6 H 11 O 2 Step 6: Determination of the molecular formula. If the empirical formula is 115 amu(12,011 x 6 +11x1,008+2x15,99), Molecular formula is 230 amu just twice as the empirical formula: Molecular formula: C 12 H 22 O 4

19. 19 Assume that the sample is 100 g ! C = 56,8 g C/(12,01 g C/ mol C) = 4,73 mol C H = 6,50 g H/( 1,008 g H / mol H) = 6,45 mol H O = 28,4 g O/(16,00 g O/ mol O) = 1,78 mol O N = 8,28 g N/(14,01 g N/ mol N) = 0,591 mol N Divide by 0,591 = C = 8,00 mol C = 8,0 mol C H = 10,9 mol H = 11,0 mol H O = 3,01 mol O = 3,0 mol OC 8 H 11 O 3 N N = 1,00 mol N = 1,0 mol N It is found that adrenalin an important compound for our body has a mass percent composition as 56,8% C, 6,5% H, 28,4% O an 8,28% N. Calculate and write down the empirical formula of adrenalin Example

20. 20 Combustion Analysis C n H m + (n+ ) O 2 = n CO (g) + H 2 O (g) m 2 m 2 Oxygen flow The sample containing C, H and O Magnesiumperchlorate H 2 O absorber Sodiumhydroxide CO 2 absorber

21. 21 The combustion of a 6,49 mg sample of ascorbic acid containing C, H and O yields 9,74 mg CO 2 and 2,64 mg H 2 O. Determine the empirical formula of ascorbic acid. C: 9,74 x10 -3 g CO 2 x(12,01 g C/44,01 g CO 2 ) = 2,65 x 10 -3 g C H: 2,64 x10 -3 g H 2 O x (2,016 g H 2 /18,02 gH 2 O) = 2,92 x 10 -4 g H Mass of Oxygen = 6,49 mg – 2,65 mg – 0,30 mg = 3,54 mg O C = 2,65 x 10 -3 g C / ( 12,01 g C / mol C ) = = 2,21 x 10 -4 mol C H = 0,295 x 10 -3 g H / ( 1,008 g H / mol H ) = = 2,92 x 10 -4 mol H O = 3,54 x 10 -3 g O / ( 16,00 g O / mol O ) = = 2,21 x 10 -4 mol O Each result is divided by 2,21 x 10 -4 : C = 1,00 multiplied by 3 = 3,00 = 3,0 H = 1,32 = 3,96 = 4,0 O = 1,00 = 3,00 = 3,0 C3H4O3C3H4O3 Example

22. 22 Metals tend to lose electron Na Na + + e - Nonmetals tend to gain electron Cl + e - Cl - Oxidation state designates the number of electrons that an atom loses, gains or otherwise uses in joining with other atoms in compounds. Oxidation States

23. 23 1.The oxidation state(O.S.) of an atom in the free(uncombined) element is 0. 2.The total of the oxidation states of all the atoms in a molecule or formula unit is 0. For an ion this total is equal to the charge of the ion. 3. In their compounds the alkali metals have O.S. +1 and the alkaline earth metals +2. 4. In its compounds, hydrogen has generally O.S. +1(but sometimes -1); fluorine, -1. 5.In its compounds oxygen has O.S. -2. 6.In their binary(two-element) compounds with metals; i.Halogenes (Group 7A elements ) have O.S. = -1, ii. Group 6A elements have O.S. = -2 and iii. Group 5A elements have O.S. = -3. Rules for assigning oxidation state Whenever two rules appear to contradict one another(which they often will), follow the rule that appears higher in the list.

24. 24 Determination of O.S. : What is the oxidation state of the underlined element in each of the following? a) P 4 ; b) Al 2 O 3 ; c) MnO 4 - ; d) NaH a)P 4 is the formula of a molecule of elemental phosphorus. For an atom of a free element the O.S. = 0 b)Al 2 O 3 : The total of the oxidation states of all the atoms in this formula unit is 0(Rule 2) The O.S. for oxygen = -2. The total for three O atoms is -6.The total for two Al atoms is -6. The O.S. of Al = +3 c) MnO 4 - : The total of the oxidation states of all the atoms in the ion is -1, the total for the four O atoms is -8. The O.S. of Mn = +7. d) NaH: Rule 3 states Na should have O.S +1. Rule 4indicates that H should also have O.S. +1. If both atoms had O.S. +1, the total for the formula unit would be +2.This violates Rule 2. Rule 2 and rule 3 take precedence over rule 4. Na has O.S. +1; the total for the formula unit is 0and the O.S. of H is -1. Examples

25. 25 Naming Inorganic Compounds Organic Compounds: The compounds formed by carbon and hydrogen or carbon and hydrogen together with oxygen,nitrogen and a few other elements. Inorganic Compounds: Compounds that do not fit the description above

26. 26 Naming Inorganic Compounds Binary compounds are those formed between two elements (If one of the elements is a metal and the other a nonmetal, the binary compound is usually comprised of ions, that is a binary ionic compound) NaCl=sodium chloride The name of metal is unchanged Ionic compounds must be electrically neutral MgI 2 =Magnesium iodide Al 2 O 3 =Aluminium oxide Na 2 S=Sodium sulfide The name of the nonmetal modified to end in «ide»

27. 27 Some Simple Ions Cations Anions Charge Symbol Name 1+ H + Hydrogen 1- H - Hydride Li + Lithium F - Floride Na + Sodium Cl - Chloride K + Potassium Br - Bromide Cs + Cesium I - Iodide Ag + Silver 2+ Mg 2+ Magnesium 2- O 2 - Oxide Ca 2+ Calcium S 2 - Sulfide Sr 2+ Strontium Ba 2+ Barium Zn 2+ Zink Cd 2+ Cadmium 3+ Al 3+ Aluminium 3- N 3 - Nitride

28. 28 Sodium and Oxygen Zink and Chlorine Calcium and Fluorine Strontium and Nitrogen Hydrogen and Iodine Scandium and Sulfur

29. 29 a) Sodium and Oxygen Na 2 O Sodium oxide b) Zink and Chlorine ZnCl 2 Zink chloride c) Calcium and Fluorine CaF 2 Calcium fluoride d) Strontium and Nitrogen Sr 3 N 2 Strontium nitride e) Hydrogen and Iodine HI Hydrogen iodide f) Scandium and Sulphur Sc 2 S 3 Scandium sulfide

30. 30 Some Metals with more than one O.S. ElementSymbolNomenclature ChromiumCr +2 Chromium (II) Cr +3 Chromium (III) CobaltCo +2 Cobalt (II) Co +3 Cobalt (III) CopperCu +1 Copper (I) Cu +2 Copper (II) IronFe +2 Iron (II) Fe +3 Iron (III) LeadPb +2 Lead (II) Pb +4 Lead (IV) ManganeseMn +2 Manganese (II) Mn +3 Manganese (III) MercuryHg 2 +2 Mercury (I) Hg +2 Mercury (II) TinSn +2 Tin (II) Sn +4 Tin (IV)

31. 31 Binary Compounds of Two Nonmetals Molecular Compound We first write the element with the (+) O.S. HCl Hydrogen chloride mono1penta5 di2hexa6 tri3hepta7 tetra4octa8 Some pairs of nonmetals form more than a single binary molecular compound. We use the prefixes written below:

32. 32 Exercise-2 BCl 3 CCl 4 CO CO 2 NO NO 2 N 2 O

33. 33 Exercise-3 Dinitrogen trioxide Dinitrogen tetroxide Dinitrogen pentoxide Phosphorus trichloride Phosphorus pentachloride Sulphur hexafluoride

34. 34 FormulaName BCl 3 Boron trichloride CCl 4 Carbon tetrachloride COCarbon monoxide CO 2 Carbon dioxide NONirogen monoxide NO 2 Nitrogen dioxide N2ON2ODinitrogen monoxide N2O3N2O3 Dinitrogen trioxide N2O4N2O4 Dinitrogen tetroxide N2O5N2O5 Dinitrogen pentoxide PCl 3 Phosphorus trichloride PCl 5 Phosphorus pentachloride SF 6 Sulphur hexafluoride Nomenclature of Binary Compounds

35. 35 Nitrite Nitrate Oxalate Permanganate Phosfate Hydrogen phosphate Dihydrogen phosphate Sulfite Hydrogen sulfite or bisulfite Sulfate

36. 36 Hydrogen sulfate or bisulfate Thiosulfate Acetate Carbonate Hydrogen carbonate or bicarbonate Hypochlorite Chlorite Chlorate Perchlorate Chromate Dichromate Cyanide Hydroxide Ammonium

37. 37 Some Common Polyatomic Ions NameFormulaTypical Compound Cation AmmoniumNH 4 + NH 4 Cl Anions AcetateC2H3O2-C2H3O2- NaC 2 H 3 O 2 CarbonateCO 3 2- Na 2 CO 3 Hydrogen carbonate or bicarbonate HCO 3 - NaHCO 3 HypochloriteClO - ClO ChloriteClO 2 - NaClO 2 ChlorateClO 3 - NaClO 3 PerchlorateClO 4 - NaClO 4 ChromateCrO 4 2- Na 2 CrO 4 DichromateCr 2 O 7 2- Na 2 Cr 2 O 7 CyanideCN - NaCN HydroxideOH - NaOH

38. 38 Some Common Polyatomic Ions NameFormulaTypical Compound NitriteNO 2 - NaNO 2 NitrateNO 3 - NaNO 3 OxalateC 2 O 4 2- Na 2 C 2 O 4 PermanganateMnO 4 - NaMnO 4 PhosfatePO 4 3- Na 3 PO 4 Hydrogen phosphate HPO 4 2- Na 2 HPO 4 Dihydrogen phosphate H 2 PO 4 - NaH 2 PO 4 SulfiteSO 3 2- Na 2 SO 3 Hydrogen sulfite or bisulfite HSO 3 - NaHSO 3 SulfateSO 4 2- Na 2 SO 4 Hydrogen sulfate or bisulfate HSO 4 - NaHSO 4 ThiosulfateS 2 O 3 2- Na 2 S 2 O 3

39. 39 Nomenclature of some Oxoacids and their salts Oxidation State Formula of acid Name of acid Formula of salt Name of salt Cl: +1HClO Hypochlorous acid NaHClOSodium hypochlorite Cl: +3HClO 2 Chlorous acidNaClO 2 Sodium chlorite Cl: +5HClO 3 Chloric acidNaClO 3 Sodium chlorate Cl: +7HClO 4 Perchloric acidNaClO 4 Sodium perchlorate N: +3HNO 2 Nitrous acidNaNO 2 Sodium nitrite N: +5HNO 3 Nitric acidNaNO 3 Sodium nitrate S: +4H 2 SO 3 Sulfurous acidNa 2 SO 3 Sodium sulfite S: +6H 2 SO 4 Sulfuric acidNa 2 SO 4 Sodium sulfate

40. 40 Binary acids Definition: Certain compounds of hydrogen with other nonmetal atoms Examples: HF= hydrofluoric acid HCl= hydrochloric acid HBr= hydrobromic acid HI = hydroiodic acid H 2 S = hydrosulfuric acid

41. 41 Some Compounds of Greater Complexity Hydrate: Each formula unit of the compound has associated with it a certain number of water molecules. The water molecules are incorporated in the solid structure of the compound Example: CoCl 2. 6 H 2 O (Cobalt(II) chloride hexahydrate) Formula mass= 129.8 u+(6x18.02 u)=237.9 u When the water is totally removed from the hydrates,the resulting compound is said to be anhydrous(without water). Anhydrous compounds can be used as water absorbers,as in the use of anhydrous magnesium perchlorate in combustion analysis Anhydrous CoCl 2 is blue, the hexahydrate is pink