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International University for Science & Technology College of Pharmacy

International University for Science & Technology College of Pharmacy. General Chemistry (Students of Dentistry) Prof. Dr. M. H. Al-Samman. Chapter 3. CHEMISTRY. Covalent Bond and Molecular Compounds. The Covalent Bond:

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International University for Science & Technology College of Pharmacy

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  1. International University for Science & TechnologyCollege of Pharmacy General Chemistry (Students of Dentistry) Prof. Dr. M. H. Al-Samman

  2. Chapter 3 CHEMISTRY

  3. Covalent Bond and Molecular Compounds The Covalent Bond: A single covalent bond: it is the bond formed by a shared pair of electrons between two atoms, one electron of each. H – H , Cl – Cl ,F – F A double covalent bond: it is the bond formed by sharing two pairs of electron between two atoms, two electron of each. O=O , A triple covalent bond : it is the bond formed by sharing three pairs of electrons between two atoms, three electrons of each.

  4. Covalent Bond and Molecular Compounds Notice: -Triple bond is shorter and stronger than double bond, double bond is shorter and stronger than single bond. -We call the compounds with covalent bonds Molecular Compounds. -Coordinate Bond: it is the bond formed by a shared pair of electrons between two atoms, one of them donates the lone pair of the electrons while the other atom provides the empty orbital.

  5. Molecules and Formulas A molecular formula gives the number of each kind of atom in a molecule. An empirical formula simply gives the (whole number) ratio of atoms of elements in a compound.

  6. Ball-and-stick model vs. Space-filling model EOS Molecular Compounds

  7. The elemental ratio C:H:O is 1:2:1, so the empirical formula is CH2O EOS Empirical and Molecular Formulas Empirical formula: the simplest whole number ratio of elements in a compound Example: Molecular formula of glucose – C6H12O6 EOS

  8. Structural Formulas Shows how atoms are attached to one another. EOS

  9. Mass Percent Compositionfrom Chemical Formulas The mass percent composition of a compound refers to the proportion of the constituent elements, expressed as the number of grams of each element per 100 grams of the compound. -Each g amount of the compound contains g amount of the element -Each 100 g of compound contains x g of the element .

  10. Percentage Composition of Butane

  11. molecular formula mass = integer (nearly) empirical formula mass Relating Molecular Formulasto Empirical Formulas A molecular formula is a simple integer multiple of the empirical formula. That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc. So: we find the molecular formula by: We then multiply each subscript in the empirical formula by the integer.

  12. Mass Percent Compositionand Chemical Formulas Example : Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula. Answer: Moles of C percent :76.57/12.011 = 6.374 % Moles of H percent : 6.43/1.0079 = 6.379 % Moles of O percent :17.00/15.998=1.062 % - We divide each figure by the smallest figure to obtain the ratio between the elements in the formula: The number of O atoms = 1.062/1.062 = 1 atom The number of C atoms = 6.347/ 1.062 = 6 atom The number of H atoms = 6.379/1.062 = 6 atom The empirical formula of phenol is : C₆H₆O

  13. Mass Percent Compositionand Chemical Formulas Example: Diethylene glycol, used in antifreeze, as a softening agent for textile fibers and some leathers, and as a moistening agent for glues and paper, has the composition 38.70% C, 9.67% H, and 51.61% O by mass. 1-Determine its empirical formula. 2-Determine its molecular formula if the molecular mass of the compound is 62 u Answer: Moles of C % = 38.70/12.011 = 3.225 mol Moles of H % = 9.67/1.0079 = 9.594 mol Moles of O % = 51.61/15.998 = 3.226 mol

  14. Mass Percent Compositionand Chemical Formulas Example: cont. -Number of O atoms = 1 atom -Number of C atoms = 1 atom -Number of H atoms = 2.97=3 atom -The empirical formula is : CH₃O • The empirical formula mass: 12 + ( 3x1) + 16 = 31 u • The integer number : 62/31 =2 • The molecular formula : C₂H₆O₂

  15. Mass Percent Compositionand Chemical Formulas Example: The empirical formula of hydroquinone, a chemical used in photography, is C3H3O, and its molecular mass is 110 u. What is its molecular formula? Answer: The empirical formula mass : (12x3)+ (3x1) + 16 = 55 u The integer number : 110/55 = 2 The molecular formula : C₆H₆O₂

  16. Mass Percent Compositionand Chemical Formulas Example : Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment shows that the molecular mass of the compound is 90 u. Determine (a) the mass percent composition, (b) the empirical formula, and (c) the molecular formula of the compound. Mass of carbon=0.1935x12/44=0.052779 Mass of hydrogen=0.1000x2/18=0.11111 Percent c=(0.052779/0.1000)x100=52.77% Percent H=(0.0555/0.1000)x100=11.11% Percent of O=100 – 63.88 = 36.12% C mol%=52.77/12=4.3975 mol H mol %=5.55/1=5.55 mol O mole %=36.12/16=2.25mol

  17. Mass Percent Compositionand Chemical Formulas Then by dividing each molar percent by the smallest figure which is 2.25 we get: -Number of O atoms=1 atom -Number of C atoms =2 atom -Number of H atom=5 atom -The empirical formula : C₂H₅O -Empirical formula mass : 24 + 5+ 16 = 45 u • Integer number: 90/45 = 2 • The molecular formula : C₄H₁₀O₂

  18. BalancingChemical Equations Example : Balance the equation Fe + O2 Fe2O3 (not balanced) 2Fe + 3/2 O₂ Fe2O3 4Fe + 3 O₂ 2Fe2O3 Example : Balance the equation C2H6 + O2  CO2+ H2O C2H6 + 7/2 O2  2CO2+ 3 H2O 2C2H6 + 7 O2  4 CO2+ 6 H2O Example : Balance the equation H3PO4 + NaCN  HCN + Na3PO4 H3PO4 + 3 NaCN  3 HCN + Na3PO4

  19. BalancingChemical Equations Example: When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is C3H8+ 5 O2 3 CO2+ 4 H2O 1mole ----5mole 0.105------? Number of O moles = 0.105 x 5 /1 = 0.525 mol

  20. BalancingChemical Equations Q- When 11.6 g of butane is burned in an excess of oxygen. • how many grams of oxygen are consumed? • The reaction is C₄H₁₀+ 13/2 O2 4 CO2+ 5 H2O 58 (13/2)x32 11.6g x The mass of consumed oxygen: 41.6 g

  21. Limiting Reactants If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). Limiting Reactant: it is the substance which is completely consumed during the reaction and limits the reaction.

  22. Limiting Reactants

  23. Limiting Reactants The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. Theoretical yield: it is the yield which we predict by mathematical calculations. The actual yield: it is the yield which we actually ( really ) obtain. The percent yield :relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

  24. Determining Limiting Reagents Practice Problem Example: During a process we burned 11.60 g of butane C₄H₁₀ with 80 g of oxygen , we obtained 30 g of carbon dioxide . 1- write the equation of the reaction. 2- balance the equation. 3- assign the limiting reactant in the reaction. 4- calculate the theoretical yield of carbon dioxide. 5- calculate the percentage yield of the reaction.

  25. Determining Limiting Reagents Practice Problem Answer: 1- C₄H₁₀ + O₂ → CO₂ + H₂O 2- C₄H₁₀ + O₂ → 4 CO₂ + 5 H₂O C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O 3- to assign the limiting reagent, we write: C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O 58 g 208 g 11.6 g x g The needed amount of oxygen= (11.6x208)/58 = 41.6g So, Butane is the limiting reactant .

  26. Determining Limiting Reagents Practice Problem Answer: 4- To calculate the theoretical yield, we write: C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O 58 g 4x 44 g 11.6 g y g The theoretical yield= (11.6 x 176)/ 58= 35.2 g 5- to calculate the percentage yield, we write: The percentage yield=(actual yield/ theoretical yield)x100 The percentage yield = (30/35.2)x100 = 85.22%

  27. Determining Limiting Reagents Guided Practice Problem • Example: • Part of the SO2 that is introduced into the atmosphere ends up being converted to sulfuric acid, H2SO4. The net reaction is: • 2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq) • Answer: • How much H2SO4 can be formed from 5.0 mol of SO2, 1.0 mol O2, and an unlimited quantity of H2O? • Assign the limiting reagent of the reaction

  28. Determining Limiting Reagents Guided Practice Problem • 2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq) • 2mol 1 mol 2 mol • 1 mol 2mol • - The amount of H2SO4 in grams= 2x98 = 196 g • - Oxygen is the limiting reagent in the reaction.

  29. Determining Limiting Reagents Guided Practice Problem • Consider the following reaction: • 2Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq) • Suppose that a solution containing 3.50 g of Na3PO4 is mixed with a solution containing 6.40 g of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed? What is the % yield, if experimentally, only 4.70 g were obtained from the reaction? • Solution • At first, it must determine which of reactants is completely • consumed and is therefore the limiting reactant. The • quantity of this reactant, in turn, will determine the quantity of • Barium Phosphate Ba3(PO4)2

  30. It is needed a grams-to-moles conversion factor to convert from the given reactant masses and moles-to-grams factor to convert to the desired product mass. The quantity of excess reactant can be calculated as the difference between the given mass of this reactant and the mass consumed (reacted) in the reaction • The balanced equation is given as following: 2 Na3PO4(aq)+ 3Ba(NO3)2(aq) Ba3(PO4)2(s)+ 6NaNO3(aq) • It can identify the limiting reactant by finding the number of moles Barium Phosphate Ba3(PO4)2produced by assuming first one reactant, and then the other is as the limiting reactant. No. moles Na3PO4 = 0.021 moles (MW of Na3PO4 = 164) No. moles Ba(NO3)2 = 0.025 moles (MW of Ba(NO3)2 = 261) • 0.021 moles Na3PO4 produce 0.011 moles of Ba3(PO4)2 • 0.025 moles Ba(NO3)2 produce 0.008 moles of Ba3(PO4)2 Because the amount of product in the second calculation [0.008 mol Ba3(PO4)2] is smaller than (0.021 mol Ba3(PO4)2), thus the Barium Nitrate is the limiting reactant. So thus when 0.008 mol of Ba3(PO4)2 has been formed, a quantity 0.025 mol of Ba(NO3)2is completely Consumed and the reaction stops, producing a specific mass of Ba3(PO4)2

  31. Determining Limiting Reagents Guided Practice Problem • Having found that the amount of product is 0.0083 mol Ba3(PO4)2, thus the mass of Ba3(PO4)2 is 3.7 grams. Number of moles of reacted Na3PO4 is 0.017 moles. The mass of Na3PO4 is 2.8 grams the unreacted of Na3PO4 is 0.71 gr. • The theoretical yield = 4.99 gr. Ba3(PO4)2 • [MW ofBa3(PO4)2= 601] • The experimental yield = 4.70 gr.Ba3(PO4)2 • The percentage yield = (4.70 / 4.99)x100 = 94.2 %

  32. The End of Chapter 3The test will cover Chapters 1-3, Scheduled Homework: 3.9, 3.11, 3.15, 3.17, 3.19, 3.21, 3.25, 3.27, 3.31, 3.33, 3.43, 3.47

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