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TOPIC 5-CHEMICAL COMPOUNDS

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. CHEMICAL COMPOUNDS. They are composed of two or more different elements

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TOPIC 5-CHEMICAL COMPOUNDS

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  1. TOPIC 5-CHEMICAL COMPOUNDS CONTENTS • Types of ChemicalCompoundsandTheirFormulas • TheMoleConceptandChemicalCompounds • Composition of ChemicalCompounds • OxidationStates • NamingChemicalCompounds

  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. MolecularCompounds A molecularcompoundis a compound Comprised of discretemolecules Theforcesthatholdatomstogether in moleculesareknowncovalentbonds A formulaunitis thesmallestcollection of atoms on which a formula can be based. Water H2O Carbondioxide CO2 SodiumchlorideNaCl Iron (II) sulfideFeS Types of Chemical Compounds and Their Formulas Subscript numbers indicate the relative numbers of atoms

  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 CH2O. 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(CH2O). In other cases the molecular formula is a «multiple» of the empirical formula such as acetic acid(C2H4O2) and glucose(C6H12O6) A structural formula show the order in which atoms are bonded together in a molecule and by what types of bonds

  5. Empirical formula CH2O Molecular formula C2H4O2 Structural formula Molecular Model (Space filling) Molecular Model (Ball and stick) CH3COOH CH3CO2H Acetic acid CH3CO2H Linear formula Oleic acid (Omega-9) C18H36

  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. 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 Sodium combines with chlorine.As a result of this combination occurs an ionic compound Sodiumchloride Na [Ne] 3s1 + Cl [Ne] 3s2 3p5 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

  8. Formula Unit Formula unit is the smallest collection of the atoms on which the formula is based Na:Cl = 1:1 formula unit = NaCl Na+ +Cl - NaCl NaCl is a compound with a neutral electrical charge Cl- MgCl2 is a compound with a neutral electrical charge Mg+2 Cl- Mg+2 + 2Cl - MgCl2 formula unit

  9. 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 x1023 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 The Mole Concept and Chemical Compounds

  10. Formula mass:is the total mass of the atoms, building up a compound, in amu. Calculate the formula mass of these compounds: NaCl, H2O, ve H3PO4 NaCl H2O 1 Na 22,9898amu= 22,9898amu 1 Cl 35,4527amu= 35,4527amu 1 NaCl= 58,4425amu 2 H 1,0079 amu= 2,0158 amu 1 O 15,9994 amu= 15,9994 amu 1 H2O= 18,0152amu H3PO4 3 H 1,0079 amu= 3,0237amu 1 P 30,9738amu= 30,9738amu 4 O 15,9994 amu= 63,9976 amu 1 H3PO4= 97,9951amu

  11. Matter Formula mass Molecular mass 2,0158 amu 2,0158 g 55,8470amu 55,8470 g 97,9952 amu 97,9952 g 80,0642amu 80,0642 g 58,4425amu 58,4425 g H2 Fe H3PO4 SO3 NaCl

  12. M(C2HBrClF3) = 2MC + MH + MBr + MCl + 3MF = (2 x 12,01) + 1,01 + 79,90 + 35,45 + (3 x 18,99) = 197,38 g/mol Composition of Chemical Compounds Halotan C2HBrClF3 Mole rate nC/nhalotan Mass rate MC/Mhalotan

  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 The molar mass of NH4NO3 and the percent composition of the atoms 28,0134 g N2 80,0432 g %N = x 100% = % 35,00 4,0316 g H2 80,0432 g %H = x 100% = % 5,04 47,9982 g O2 80,0432 g %O = x 100% = % 59,96 % 100,00

  14. The percent composition of Sulfuric acid Molar mass of sulfuric acid = 2(1,008g) + 1(32,07g) + 4(16,00g) = 98,09 g/mol 2(1,008g H2) 98,09g %H = x 100% =% 2,06 H 1(32,07g S) 98,09g %S = x 100% = %32,69 S 4(16,00g O) 98,09 g %O = x 100% = % 65,25 O % 100,00

  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. Determination of the empirical and molecular formula Solution in 6 steps:

  16. Example 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 Step 1 : Determine the mass of each element in a100 g sample C: 62,58 g H: 9,63 g O: 27,79 g

  17. Step 2: Convert each of these masses to an amount in moles. C5,21H9,55O1,74 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. C2,99H5,49O

  18. Step 5: Multiply all subscripts by a small number chosen to make all subscripts integral(here by 2). C5,98H10,98O2 Empirical formula:C6H11O2 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:C12H22O4

  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 OC8H11O3N N = 1,00 mol N = 1,0 mol N Example 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

  20. Combustion Analysis Sodiumhydroxide CO2 absorber The sample containing C, H and O MagnesiumperchlorateH2O absorber Oxygen flow m 2 m 2 CnHm + (n+ ) O2 = n CO(g) + H2O(g)

  21. The combustion of a 6,49 mg sample of ascorbic acid containing C, H and O yields 9,74 mg CO2 and 2,64 mg H2O . Determine the empirical formula of ascorbic acid. Example C: 9,74 x10-3g CO2 x(12,01 g C/44,01 g CO2) = 2,65 x 10-3 g C H: 2,64 x10-3g H2O x (2,016 g H2/18,02 gH2O) = 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 C3H4O3

  22. Oxidation States 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.

  23. Theoxidationstate(O.S.) of an atom in thefree(uncombined) element is 0 . The total of theoxidationstates of alltheatoms in a moleculeorformulaunit is 0. For an ionthis total is equaltothecharge of theion. Intheircompoundsthe alkali metalshave O.S. +1 andthe alkaline earth metals +2. 4. Initscompounds, hydrogen has generally O.S. +1(but sometimes -1); fluorine, -1. Initscompoundsoxygen has O.S. -2. Intheirbinary(two-element) compoundswithmetals; Halogenes (Group 7Aelements ) haveO.S. =-1, Group 6AelementshaveO.S. = -2 and Group 5AelementshaveO.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. Examples Determination of O.S. : What is the oxidation state of the underlined element in each of the following? a) P4; b) Al2O3; c) MnO4-; d) NaH • P4is the formula of a molecule of elemental phosphorus. For an atom of a free element the O.S. = 0 • Al2O3: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) MnO4-: 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.

  25. 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 Naming Inorganic Compounds

  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) The name of the nonmetal modified to end in «ide» NaCl =sodiumchloride Ionic compounds must be electrically neutral The name of metal is unchanged MgI2=Magnesiumiodide Al2O3=Aluminiumoxide Na2S = Sodiumsulfide

  27. Some Simple Ions Cations Anions ChargeSymbol Name Charge SymbolName 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+ Mg2+Magnesium 2- O 2 - Oxide Ca2+Calcium S2 -Sulfide Sr2+ Strontium Ba2+Barium Zn2+Zink Cd2+ Cadmium 3+ Al3+Aluminium 3-N 3 - Nitride

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

  29. a) Sodium and Oxygen Na2O Sodium oxide b) Zink and Chlorine ZnCl2 Zink chloride c) Calcium and Fluorine CaF2 Calcium fluoride d) Strontium and Nitrogen Sr3N2 Strontium nitride e) Hydrogen and Iodine HI Hydrogen iodide f) Scandium and Sulphur Sc2S3 Scandium sulfide

  30. Some Metals with more than one O.S. Element Symbol Nomenclature Chromium Cr+2 Chromium (II) Cr+3 Chromium (III) Cobalt Co+2 Cobalt (II) Co+3 Cobalt (III) Copper Cu+1 Copper (I) Cu+2 Copper (II) Iron Fe+2 Iron (II) Fe+3 Iron (III) Lead Pb+2 Lead (II) Pb+4 Lead (IV) Manganese Mn+2 Manganese (II) Mn+3 Manganese (III) Mercury Hg2+2 Mercury (I) Hg+2 Mercury (II) Tin Sn+2 Tin (II) Sn+4 Tin (IV)

  31. Molecular Compound We first write the element with the (+) O.S. HCl Hydrogenchloride Binary Compounds of Two Nonmetals Some pairs of nonmetals form more than a single binary molecular compound. We use the prefixes written below: mono 1 penta 5 di 2 hexa 6 tri 3 hepta 7 tetra 4 octa 8

  32. Exercise-2 • BCl3 • CCl4 • CO • CO2 • NO • NO2 • N2O

  33. Exercise-3 • Dinitrogentrioxide • Dinitrogentetroxide • Dinitrogenpentoxide • Phosphorustrichloride • Phosphoruspentachloride • Sulphurhexafluoride

  34. Nomenclature of Binary Compounds

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

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

  37. Some Common Polyatomic Ions

  38. Some Common Polyatomic Ions

  39. Nomenclature of some Oxoacids and their salts

  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 H2S = hydrosulfuric acid

  41. Some Compounds of Greater Complexity Hydrate:Eachformulaunit of thecompound has associatedwith it a certainnumber of watermolecules. • Thewatermoleculesareincorporated in thesolidstructure of thecompound • Example: CoCl2. 6 H2O (Cobalt(II) chloridehexahydrate) • Formula mass= 129.8 u+(6x18.02 u)=237.9 u • Whenthewater is totallyremovedfromthehydrates,theresultingcompound is saidto be anhydrous(withoutwater). Anhydrouscompounds can be used as waterabsorbers,as in theuse of anhydrousmagnesiumperchlorate in combustionanalysis • Anhydrous CoCl2 is blue, thehexahydrate is pink

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