1. General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

2. فصل سوم: استوکیومتری I ترکیبات شیمیائی

3. Contents 3-1Molecular and Ionic Compounds 3-2Molecular Mass 3-3Composition

4. Molecules, Compounds, and Formulas Molecule – smallest identifiable unit in which a pure substance can be divided and still retain the composition and chemical properties of the substance

5. Formulas Molecular Formula – formula that expresses the number of atoms of each type within one molecule of a compound, e.g. H 2 O 2, P 4. Condensed Formula – variation of the molecular formula that shows groups of atoms showing how atoms are grouped e.g. CH 3 CH 2 OH. –C 2 H 6 O could be ethanol or dimethylether. Condensed formula tells us the difference.

6. Formulas Structural Formula – shows how the atoms in a compound are connected –Lines between atoms represent chemical bonds A chemical bond is the interaction between two or more atoms that holds them together

8. Models Since the particulate level is to small to be seen, we use models to represent particles –Models are a representation of something else Two common models –Ball-and-stick model Atoms are balls Electrons are sticks that connect atoms –Space-filling model Shows outer boundries of the particle in 3D space

9. Models (continued) Ball and Stick Model Space-filling model

10. How to look at molecules Different ways to look at compounds –Name –Molecular formula –Condensed formula –Structural formula –Molecular model Example –Ethanol –C 2 H 6 O –CH 3 CH 2 OH

11. Standard color scheme

12. Some molecules H2O2H2O2 CH 3 CH 2 ClP 4 O 10 CH 3 CH(OH)CH 3 HCO 2 H

13. Inorganic molecules S8S8 P4P4

14. An atom that either gains or loses electron(s) is an ion. There is no change in the number of protons or neutrons in the nucleus of the atom. Cation – has a positive charge from loss of electron(s). Anion – has a negative charge from gain of electron(s). Ions and Ionic Compounds

15. Cations If an atom loses an electron (the electron is transferred to an atom of another element in a reaction), the atom has one more proton than electrons. –When this happens, a cation is formed –Cation is the positively charged ion –Li 1 e - + Li +

16. Anions If an atom gains an electron (the electron is transferred to the atom from another element in a reaction), the atom has one more electron than protons. –When this happens, an anion is formed –Anion is the negatively charged ion –O + 2e - O 2-

17. How to determine if atom will gain or lose electrons. Metals generally lose electrons in the course of their reactions to form cations Nonmetals frequently gain one or more electrons to form anions in the course of their chemical reactions.

19. Sodium chloride Simplest formula unit is NaCl Extended array of Na+ and Cl- ions

20. Monoatomic Ions Monoatomic ions – single atoms that have lost or gained electrons Metals form ions having a positive charge of their group number Transition metals form cations, but they can form several different cations Nonmetals form ions having a negative charge = 8 minus the group number Noble gases lose or gain electrons very rarely

22. Polyatomic Ions Polyatomic ions are made up of two or more atoms –Example: CO 3 2-, so it has a carbon and 3 Oxygens

24. Atoms and the Mole Mole (mol) – the amount of substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope –So, one mole has the same amount of particles no matter what substance we are talking about Avogadro’s number - –6.022 x 10 23 atoms/mol

25. Molar Mass Molar mass – mass in grams of one mole of atoms –Sodium (Na) – 22.99 g/mol = 6.022 x 10 23 Na atoms –Lead (Pb) – 207.2 g/mol = 6.022 x 10 23 Pb atoms

26. Molecular mass Molecular formula C 6 H 12 O 6 Empirical formulaCH 2 O Glucose 6 x 12.01 + 12 x 1.01 + 6 x 16.00 Molecular Mass:Use the naturally occurring mixture of isotopes, = 180.18 Exact Mass:Use the most abundant isotopes, 6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915 = 180.06339

27. Example: Determine (a) the number of NH 4 + ions in a 264 g sample of (NH 4 ) 2 SO 4 and (b) the volume of 1,2,3- propanetriol (glycerol, d = 1.261 g/mL) that contains 1.00 mol O atoms. Example: An Estimation Example Which of the following is a reasonable value for the number of atoms in 1.00 g of helium? (a) 4.1 × 10 –23 (c) 1.5 × 10 23 (b) 4.0 (d) 1.5 × 10 24

28. Percent Composition Law of constant composition – any sample of a pure compound always consists of the same elements combined in the same proportion by mass Three ways to express this –Number of atoms of each type per molecule –Mass of each element per mole of compound –Mass percent - mass of each element in the compound relative to the total mass Percent Composition – mass percent of each element

29. 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. In other words … Mass Percent Composition from Chemical Formulas g element X % element X = –––––––––––––– OR … 100 g compound g element % element = ––––––––––– × 100 g compound

30. Percentage Composition of Butane

31. HalothaneC 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 19.00) = 197.38 g/mol Chemical Composition Mole ration C /n halothane Mass ratiom C /m halothane

32. Example Calculating the Mass Percent Composition of a Compound Calculate the molecular mass M(C 2 HBrClF 3 ) = 197.38 g/mol For one mole of compound, formulate the mass ratio and convert to percent:

33. Example

34. Example: Calculate, to four significant figures, the mass percent of each element in ammonium nitrate. Example: How many grams of nitrogen are present in 46.34 g ammonium nitrate? Example: An Estimation Example Without doing detailed calculations, determine which of these compounds contains the greatest mass of sulfur per gram of compound: barium sulfate, lithium sulfate, sodium sulfate, or lead sulfate.

35. Formulas From Percent Composition Just the REVERSE of what we did weight % mass to to mass moles Mole ratio Ratio gives formula %A gA x mol A X mol A Y mol B AxByAxBy %B gB y mol B

36. Formulas From Percent Composition This tells us the ratio not the total number of atoms In a molecular formula, we must know the total number of atoms So, NH 2 is the empirical formula of hydrazine –Hydrazine could be NH 2, N 2 H 4, N 3 H 6 …. –If given, molar mass of hydrazine is 32.0 g/mol, we know the correct molecular formula is N 2 H 4

37. Empirical formula 1.Choose an arbitrary sample size (100g). 2.Convert masses to amounts in moles. 3.Write a formula. 4.Convert formula to small whole numbers. 5.Multiply all subscripts by a small whole number to make the subscripts integral. 5 Step approach:

38. 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

39. Example 3.11 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?

40. Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition. Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Step 1: Determine the mass of each element in a 100g sample. C 62.58 g H 9.63 g O 27.79 g Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate? Example:

41. Step 2: Convert masses to amounts in moles. Step 3: Write a tentative formula. Step 4: Convert to small whole numbers. C 5.21 H 9.55 O 1.74 C 2.99 H 5.49 O Example:

42. Step 5: Convert to a small whole number ratio. Multiply x 2 to get C 5.98 H 10.98 O 2 The empirical formula is C 6 H 11 O 2 Step 6: Determine the molecular formula. Empirical formula mass is 115 u. Molecular formula mass is 230 u. The molecular formula is C 12 H 22 O 4 Example:

43. Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula. 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 45.27% C, 9.50% H, and 45.23% O by mass. Determine its empirical formula.

44. … is one method of determining empirical formulas in the laboratory. This method is used primarily for simple organic compounds (that contain carbon, hydrogen, oxygen). –The organic compound is burned in oxygen. –The products of combustion (usually CO 2 and H 2 O) are weighed. –The amount of each element is determined from the mass of products. Elemental Analysis …

45. Elemental Analysis (cont’d) The sample is burned in a stream of oxygen gas, producing … … H 2 O, which is absorbed by MgClO 4, and … … CO 2, which is absorbed by NaOH.

46. Elemental Analysis (cont’d) If our sample were CH 3 OH, every two molecules of CH 3 OH … … would give two molecules of CO 2 … … and four molecules of H 2 O.

47. 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.

48. Chapter 3 Questions 1, 4, 19, 30, 35, 40, 41, 46, 54, 57, 60, 69