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

2. Chapter 3 CHEMISTRY

3. 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. Covalent Bond and Molecular Compounds

4. 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. Covalent Bond and Molecular Compounds

5. 5 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. CompoundMolecular formula Empirical formula Hydrogen peroxideH2O2H2O2 HO OctaneC 8 H 18 ???? Molecules and Formulas

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

7. 7 Empirical and Molecular Formulas EOS Empirical formula: the simplest whole number ratio of elements in a compound Example: Molecular formula of glucose – C 6 H 12 O 6 EOS The elemental ratio C:H:O is 1:2:1, so the empirical formula is CH 2 O

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

9. 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. Mass Percent Composition from Chemical Formulas

10. Percentage Composition of Butane

11. A molecular formula is a simple integer multiple of the empirical formula. That is, an empirical formula of CH 2 means that the molecular formula is CH 2, or C 2 H 4, or C 3 H 6, or C 4 H 8, etc. So: we find the molecular formula by: = integer (nearly) molecular formula mass empirical formula mass We then multiply each subscript in the empirical formula by the integer. Relating Molecular Formulas to Empirical Formulas

12. 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 : C ₆ H ₆ O Mass Percent Composition and Chemical Formulas

13. 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 Mass Percent Composition and Chemical Formulas

14. 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 : CH ₃O -The empirical formula mass: 12 + ( 3x1) + 16 = 31 u -The integer number : 62/31 =2 -The molecular formula : C ₂ H₆O ₂ Mass Percent Composition and Chemical Formulas

15. Example: The empirical formula of hydroquinone, a chemical used in photography, is C 3 H 3 O, 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 : C ₆ H ₆ O ₂ Mass Percent Composition and Chemical Formulas

16. Example : Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO 2 and 0.1000 g H 2 O. 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 Mass Percent Composition and Chemical Formulas

17. 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 : C ₂ H₅O -Empirical formula mass : 24 + 5+ 16 = 45 u -Integer number: 90/45 = 2 -The molecular formula : C ₄ H ₁ ₀O ₂ Mass Percent Composition and Chemical Formulas

18. Example : Balance the equation Fe + O 2  Fe 2 O 3 (not balanced) 2Fe + 3/2 O ₂  Fe 2 O 3 4Fe + 3 O ₂  2Fe 2 O 3 Example : Balance the equation C 2 H 6 + O 2  CO 2 + H 2 O C 2 H 6 + 7/2 O 2  2CO 2 + 3 H 2 O 2C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O Example : Balance the equation H 3 PO 4 + NaCN  HCN + Na 3 PO 4 H 3 PO 4 + 3 NaCN  3 HCN + Na 3 PO 4 Balancing Chemical Equations

19. Example: When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is C 3 H 8 + 5 O 2  3 CO 2 + 4 H 2 O 1mole ----5mole 0.105------? Number of O moles = 0.105 x 5 /1 = 0.525 mol Balancing Chemical Equations

20. Q- When 11.6 g of butane is burned in an excess of oxygen. a)how many grams of oxygen are consumed? b) The reaction is C ₄H₁₀ + 13/2 O 2  4 CO 2 + 5 H 2 O 58 (13/2)x32 11.6g x The mass of consumed oxygen: 41.6 g Balancing Chemical Equations

21. 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. Limiting Reactants

23. 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: Limiting Reactants

24. Determining Limiting Reagents Practice Problem Example: During a process we burned 11.60 g of butane C ₄H ₁ ₀ 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- C ₄H ₁ ₀ + O₂ → CO₂ + H₂O 2- C ₄H ₁ ₀ + O₂ → 4 CO₂ + 5 H₂O C ₄H ₁ ₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O 3- to assign the limiting reagent, we write: C ₄H ₁ ₀ + 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: C ₄H ₁ ₀ + 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 SO 2 that is introduced into the atmosphere ends up being converted to sulfuric acid, H 2 SO 4. The net reaction is: 2SO 2 (g) + O 2 (g) + 2H 2 O(l)  2H 2 SO 4 (aq) Answer: How much H 2 SO 4 can be formed from 5.0 mol of SO 2, 1.0 mol O 2, and an unlimited quantity of H 2 O? Assign the limiting reagent of the reaction

28. Determining Limiting Reagents Guided Practice Problem 2SO 2 (g) + O 2 (g) + 2H 2 O(l)  2H 2 SO 4 (aq) 2mol 1 mol 2 mol 1 mol 2mol - The amount of H 2 SO 4 in grams= 2x98 = 196 g - Oxygen is the limiting reagent in the reaction.

29. Consider the following reaction: 2Na 3 PO 4(aq) + 3Ba(NO 3 ) 2(aq)  Ba 3 (PO 4 ) 2(s) + 6NaNO 3(aq) Suppose that a solution containing 3.50 g of Na 3 PO 4 is mixed with a solution containing 6.40 g of Ba(NO 3 ) 2. How many grams of Ba 3 (PO 4 ) 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 Ba 3 (PO 4 ) 2 Determining Limiting Reagents Guided Practice Problem

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 Na 3 PO 4 (aq) + 3Ba(NO 3 ) 2 (aq)  Ba 3 (PO 4 ) 2 (s) + 6NaNO 3 (aq) It can identify the limiting reactant by finding the number of moles Barium Phosphate Ba 3 (PO 4 ) 2 produced by assuming first one reactant, and then the other is as the limiting reactant. No. moles Na 3 PO 4 = 0.021 moles (MW of Na 3 PO 4 = 164) No. moles Ba(NO 3 ) 2 = 0.025 moles (MW of Ba(NO 3 ) 2 = 261) -0.021 moles Na 3 PO 4 produce 0.011 moles of Ba 3 (PO 4 ) 2 -0.025 moles Ba(NO 3 ) 2 produce 0.008 moles of Ba 3 (PO 4 ) 2 Because the amount of product in the second calculation [0.008 mol Ba 3 (PO 4 ) 2 ] is smaller than (0.021 mol Ba 3 (PO 4 ) 2 ), thus the Barium Nitrate is the limiting reactant. So thus when 0.008 mol of Ba 3 (PO 4 ) 2 has been formed, a quantity 0.025 mol of Ba(NO 3 ) 2 is completely Consumed and the reaction stops, producing a specific mass of Ba 3 (PO 4 ) 2

31. Having found that the amount of product is 0.0083 mol Ba 3 (PO 4 ) 2, thus the mass of Ba 3 (PO 4 ) 2 is 3.7 grams. Number of moles of reacted Na 3 PO 4 is 0.017 moles. The mass of Na 3 PO 4 is 2.8 grams the unreacted of Na 3 PO 4 is 0.71 gr. The theoretical yield = 4.99 gr. Ba 3 (PO 4 ) 2 [MW of Ba 3 (PO 4 ) 2 = 601] The experimental yield = 4.70 gr. Ba 3 (PO 4 ) 2 The percentage yield = (4.70 / 4.99)x100 = 94.2 % Determining Limiting Reagents Guided Practice Problem

32. The End of Chapter 3 The 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