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General Chemistry

General Chemistry. M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology. فصل س و م :. استوکیومتری. Contents. 3-1 Molecular and Ionic Compounds 3-2 Molecular Mass 3-3 Composition. Molecules, Compounds, and Formulas.

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General Chemistry

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

  2. فصل سوم: استوکیومتری

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

  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. H2O2, P4. • Condensed Formula – variation of the molecular formula that shows groups of atoms showing how atoms are grouped e.g. CH3CH2OH. • C2H6O 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

  7. 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

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

  9. Different ways to look at compounds Name Molecular formula Condensed formula Structural formula Molecular model Example Ethanol C2H6O CH3CH2OH How to look at molecules

  10. Standard color scheme

  11. H2O2 CH3CH2Cl P4O10 CH3CH(OH)CH3 HCO2H Some molecules

  12. P4 S8 Inorganic molecules

  13. Ions and Ionic Compounds • 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).

  14. 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+

  15. 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- O2-

  16. 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.

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

  18. 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

  19. Polyatomic Ions • Polyatomic ions are made up of two or more atoms • Example: CO32- , so it has a carbon and 3 Oxygens

  20. 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 1023 atoms/mol

  21. Molar Mass • Molar mass: mass in grams of one mole of atoms • Sodium (Na): 22.99 g/mol = 6.022 x 1023 Na atoms • Lead (Pb): 207.2 g/mol = 6.022 x 1023 Pb atoms

  22. Molecular Mass: Use the naturally occurring mixture of isotopes, 6 x 12.01 + 12 x 1.01 + 6 x 16.00 = 180.18 Molecular mass Glucose Molecular formula C6H12O6Empirical formula CH2O

  23. Example: Determine (a) the number of NH4+ ions in a 264 g sample of (NH4)2SO4 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) 1.5×1023 (b) 4.0 (d) 1.5×1024

  24. 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

  25. 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. In other words … g element X % element X = –––––––––––––– OR … 100 g compound g element % element = ––––––––––– × 100 g compound

  26. Percentage Composition of Butane

  27. M(C2HBrClF3) = 2MC + MH + MBr + MCl + 3MF = (2 x 12.01) + 1.01 + 79.90 + 35.45 + (3 x 19.00) = 197.38 g/mol Chemical Composition Halothane C2HBrClF3 Mole ratio nC/nhalothane Mass ratio mC/mhalothane

  28. Example Calculating the Mass Percent Composition of a Compound Calculate the molecular mass M(C2HBrClF3) = 197.38 g/mol For one mole of compound, formulate the mass ratio and convert to percent:

  29. Example

  30. 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.

  31. Formulas From Percent Composition • Just the REVERSE of what we did

  32. 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, NH2 is the empirical formula of hydrazine • Hydrazine could be NH2, N2H4, N3H6 …. • If given, molar mass of hydrazine is 32.0 g/mol, we know the correct molecular formula is N2H4

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

  34. 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.

  35. Example 3.11 The empirical formula of hydroquinone, a chemical used in photography, is C3H3O, and its molecular mass is 110 u. What is its molecular formula?

  36. Example: 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? 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

  37. Example: Step 2: Convert masses to amounts in moles. Step 3: Write a tentative formula. C5.21H9.55O1.74 Step 4: Convert to small whole numbers. C2.99H5.49O

  38. Multiply x 2 to get C5.98H10.98O2 The empirical formula is C6H11O2 Empirical formula mass is 115 u. Molecular formula mass is 230 u. The molecular formula is C12H22O4 Example: Step 5: Convert to a small whole number ratio. Step 6: Determine the molecular formula.

  39. Example: 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.

  40. Elemental Analysis … • … 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 CO2 and H2O) are weighed. • The amount of each element is determined from the mass of products.

  41. Elemental Analysis (cont’d) … H2O, which is absorbed by MgClO4, and … The sample is burned in a stream of oxygen gas, producing … … CO2, which is absorbed by NaOH.

  42. Elemental Analysis (cont’d) If our sample were CH3OH, every two molecules of CH3OH … … would give two molecules of CO2 … … and four molecules of H2O.

  43. 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.

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

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