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Chemistry-140 Lecture 6

Chemistry-140 Lecture 6 . Chapter Highlights define molecules definitions of molecular formula & structural formula understand formation of cations & anions learn names & formula for polyatomic ions molar mass of compounds percent composition empirical formula.

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Chemistry-140 Lecture 6

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  1. Chemistry-140 Lecture 6 • Chapter Highlights • define molecules • definitions of molecular formula & structural formula • understand formation of cations & anions • learn names & formula for polyatomic ions • molar mass of compounds • percent composition • empirical formula Chapter 3: Molecules & Compounds

  2. Chemistry-140 Lecture 6 • Molecules are tightly bound assemblies of two or more atoms. This "package" behaves as a single unit. • Some elements exist as discrete molecules; • H2, C, O2 (O3), N2, F2, Cl2, Br2, I2, P4, S8 • Remember: Although He, Ne, Ar, Kr, Xe, Rn are gases they are not really molecules since they are monatomic Molecules

  3. Chemistry-140 Lecture 6 • Chemical formula: a collection of elemental symbols with subscripts that indicate the relative number of atoms of each element in the substance • Molecular formula: the chemical formula of a molecular compound Formulas • Example: • the molecular formula for sucrose is C12H22O11

  4. Chemistry-140 Lecture 6 • Compounds are pure substances that can be decomposed into one or more different pure substances Molecular Compounds Example: 1 molecule of sucrose 12 atoms of C + 11 molecules of water side 4: frames 05413-05903

  5. Chemistry-140 Lecture 6 • Structural formula: emphasizes how atoms are connected and shows any chemically active groups (functional group) Formulas • Example: • the molecular formula for ethanol is C2H6O • the structural formula for ethanol is CH3CH2OH • OH (alcohol functional group) is an important chemically active group

  6. Chemistry-140 Lecture 6 Molecular Models • Ball & stick model for ethanol: • gray = carbon • white = hydrogen • red = oxygen Models of methane CH4 ball &stick perspective drawing space filling

  7. Chemistry-140 Lecture 6 • Atoms of almost all elements can gain or lose electrons to form ions (charged species) • Compounds composed of ions are known as ionic compounds • Cations are positively charged ions • Anions are negatively charged ions Ions • Example: • sodium chloride, NaCl is composed of • sodium cations = Na+ and chloride anions = Cl-

  8. Chemistry-140 Lecture 6

  9. Chemistry-140 Lecture 6 Monatomic and Polyatomic Ions • Monatomic ions are comprised of single atoms, while polyatomic ions are comprised of several atoms • Examples: • monatomic: Na+ Ca2+ Fe3+ S2- Cl- N3- • polyatomic: NH4+ SO42- ClO3- PO43-

  10. Chemistry-140 Lecture 6 Predicting the Charges on Ions • The charge of an ion can be predicted from the elements position in the periodic table. • Monatomic ions: atoms either gain or lose electrons until they have the same number of electrons as the nearest noble gas. • metals lose electrons to form cations • nonmetals gain electrons to form anions • For example: • Ar (18 electrons), K (19 electrons) and Cl (17 electrons). • K K+ • Cl Cl- • in order to have 18 electrons

  11. Chemistry-140 Lecture 6 The Monatomic Anions hydride oxide nitride carbide fluoride chloride phosphide bromide sulfide selenide iodide telluride

  12. Chemistry-140 Lecture 6 Ionic Compounds • Ionic compounds are those compounds formed from the combination of ions. • Please Remember!! • total cationic charge = total anionic charge • overall the material is neutral • the TOTAL charge on the compound = ZERO

  13. Chemistry-140 Lecture 6 Question: What ionic compound would you expect from the combination of Mg and N?

  14. Chemistry-140 Lecture 6 Answer: Mg Mg2+ (magnesium ion) N N3- (nitride ion) In order to obtain overall neutrality 3 Mg2+ combine with 2 N3- to yield Mg3N2 (magnesium nitride)

  15. Chemistry-140 Lecture 6 Naming Compounds • Positive ions. • Name plus ion, for example, aluminum ion. • Specify the charge on the ion, for example • cobalt (II) and cobalt (III) ions • Ammonium, carbonium and oxonium ions all refer to different types of positive ions of ammonia, carbon and oxygen.

  16. Chemistry-140 Lecture 6 Naming Compounds • Negative ions • Simple anions end in –ide, for example chloride ion. • Polyatomic anions (often oxoanions) are not systematic and must be learned (see table 3.1): • perchlorate, chlorate, chlorite, hypochlorite • hydrogen phosphate, dihydrogen phosphate • carbonate, bicarbonate • ClO4- ClO3- ClO2- ClO- HPO42- H2PO4-

  17. Chemistry-140 Lecture 6 • A convenient unit for matter that contains • a known number of particles • Definition: the amount of substance that contains as many particles as their are atoms in exactly 12 g • of the carbon-12 isotope The Mole 1 mole = 6.022136736 x 1023 Avogadro’s Number (N)

  18. Chemistry-140 Lecture 6 1 mole = 6.022136736 x 1023 Avogadro’s Number (N) The Mole • How big is this number??? • Popcorn kernels covering the continental US • $$-Dollars-$$ a national debt ($ 3.6 trillion) • computer counting at 10 million particles/second

  19. Chemistry-140 Lecture 6 • What is the Chemical Significance? • A mole of any element (or compound) always contains the same number of atoms (or molecules) • Since each type of atom has a different atomic mass, a mole of atoms of one element has a different mass from the mass of a mole of a different element • Example: 1 mole of 16O has mass = 16.0 g while • 1 mole of 19F has mass = 19.0 g The Mole

  20. Chemistry-140 Lecture 6 • The mass in grams of 1 mole of atoms of any element is the molar mass of that element • Molar mass is conventionally shown as M and expressed in grams/mole (g/mol) • For any element, the molar mass in grams is equal to the atomic mass in atomic mass units (amu). • Example: • Molar mass of Na = mass of 1 mol of Na atoms • = 22.98 g/mol • = mass of 6.022 x 1023 Na atoms Molar Mass

  21. (Moles) = grams (Grams) = moles Chemistry-140 Lecture 6 • The ability to convert from moles to mass and mass to moles is absolutely essential Mass Moles Conversion MASS MOLES CONVERSION Mass to Moles Moles to Mass molar mass 1/molar mass

  22. Chemistry-140 Lecture 6 A Question of Conversion Question: How many moles are in 454 g of silicon?

  23. Chemistry-140 Lecture 6 Answer: The molar mass of silicon is 28.09 g/mol (from the periodic table!). Convert the mass of silicon to its equivalent in moles (454 g Si) = 16.2 mol Si A Question of Conversion

  24. Chemistry-140 Lecture 6 Another Question of Conversion Question: What is the mass of 2.50 moles of lead (Pb)?

  25. Chemistry-140 Lecture 6 Answer: The molar mass of lead is 207.2 g/mol (where else but from the periodic table!). Convert the moles of lead to its equivalent mass (2.50 mol Pb) = 518 g Pb Another Question of Conversion

  26. Chemistry-140 Lecture 6 Question: A graduated cylinder contains 25.4 mL of mercury (Hg). If the density of mercury is 13.534 g/mL, how many moles of mercury are in the cylinder? How many atoms of Hg are in the cylinder? A Real Calculation

  27. x atoms/mol x g/mL x mol/g Volume, mL Atoms Mass, g Moles Avogadro’s Number molar mass density Chemistry-140 Lecture 6 Method: Requires certain conversion! A Real Calculation

  28. Chemistry-140 Lecture 6 Answer: (25.4 mL Hg) = 344 g Hg (344 g Hg) = 1.71 mol Hg (1.71 mol Hg) = 1.03 x 1024 atoms Hg A Real Calculation

  29. Chemistry-140 Lecture 7 • Chapter Highlights • define molecules • definitions of molecular formula & structural formula • understand formation of cations & anions • learn names & formula for polyatomic ions • molar mass of compounds • percent composition • empirical formula Chapter 3: Molecules & Compounds

  30. Chemistry-140 Lecture 7 • Molar Mass M, is the mass of a mole of molecules of a particluar substance = Molecular Weight • Example: • The molecular weight of PBr3 = the atomic weight of P plus 3 x the atomic weight of Br • MW (PBr3) = AW (P) + 3[AW (Br)] • = (30.97 amu) + 3(79.90 amu) • = 270.7 amu • Thus: 1 mole of PBr3 has a mass of 270.7 g Molecules, Compounds & the Mole

  31. A Mole of …………. Chemistry-140 Lecture 7 K2Cr2O7 NiCl2.6H2O CoCl2.6H2O NaCl CuSO4.5H2O

  32. Chemistry-140 Lecture 7 • Question (compare to example 3.9): • You have 23.2 g of ethanol, C2H6O. • How many moles are contained in this mass of ethanol? • How many molecules of ethanol are contained in 23.2 g? • How many atoms of carbon are contained in 23.2 g of ethanol? • What is the average mass of one molecule of ethanol?

  33. Chemistry-140 Lecture 7 Method: Conversions using molar mass & Avogadro’s number! x molecules/mol x mol/g x C atoms/molecule Mass, g Molecules Moles Number of C atoms molar mass Avogadro’s Number

  34. Chemistry-140 Lecture 7 Answer: Step 1: Calculate the molar mass of ethanol, C2H6O. 2 mol of C per mole of ethanol = (2 mol C) = 24.02 g C 6 mol of H per mole of ethanol = (6 mol H) = 6.048 g H 1 mol of O per mole of ethanol = (1 mol O) = 16.00 g O Molar mass of ethanol = (24.02 + 6.048 + 16.00) g = 46.07 g/mol

  35. 0.504 mol C2H6O Chemistry-140 Lecture 7 • Answer: • Step 2: Calculate the number of moles of C2H6O. • (23.2 g C2H6O) = • Number of moles of ethanol in 23.2 g is 0.504 mol.

  36. Chemistry-140 Lecture 7 • Answer: • Step 3: Calculate the number of carbon atoms. • (0.504 mol C2H6O) • = 3.04 x 10 23 molecules of C2H6O • 3.04 x 10 23 molecules of C2H6O • = • Number of carbon atoms in 23.2 g of ethanol is 6.07 x 10 23 . 6.07 x 10 23 atoms of C

  37. Chemistry-140 Lecture 7 • Answer: • Step 4: Calculate the mass of 1 molecule of C2H6O. • = • The mass of 1 molecule of ethanol, C2H6O, is 7.650 x 10 -23 g. 7.650 x 10 -23 g/molecule of C2H6O

  38. Chemistry-140 Lecture 7 • Composition can be given by the mass of each element relative to the total mass of the compound = Mass Percentage • Mass percentage N in NH3 = • = x 100 % = • Mass percentage H in NH3 = • = x 100 % = Percent Composition 82.27 % 17.76 %

  39. Chemistry-140 Lecture 7 • Percentage composition can be used to determine a • simplest or empirical formula • Empirical or simplest formulas show the simplest ratio of the numbers of atoms of each element in a substance. • Example, C6H6 is the molecular formula showing the numbers of C and H atoms in the molecule benzene. CH is the empirical formula showing the simplest ratio of atoms. • Therefore, to convert an empirical formula to a molecular formulawe need a molar mass! Empirical & Molecular Formulas

  40. Chemistry-140 Lecture 7 Example 3.10: Eugenol has a molar mass of 164.2 g/mol and is73.14 % C and 7.37 % H with the remainder O. What are the molecular and empirical formulas for eugenol?

  41. Chemistry-140 Lecture 7 Method: % A x mol A } x mol A y mol B AxBy % B y mol B find mole ratio ratios gives formula convert weight percentage to moles

  42. Chemistry-140 Lecture 7 Answer: Step 1: Find the number of moles of C and H in a 100 g sample of eugenol. (73.14 g C) = (7.37 g H) = 6.089 mol C 7.31 mol H

  43. Chemistry-140 Lecture 7 Answer: Step 2: Find the number of moles of O in a 100 g sample of vanillin by difference. 100.00 g = (73.14 g C + 7.37 g H) + mass of O mass of O = 100.00 g - (73.14 g C + 7.37 g H) = 19.49 g O 19.49 g O = 1.218 mol O

  44. Chemistry-140 Lecture 7 Answer: Step 3: Calculate the ratio of moles = empirical formula. = = 4.999 = = 6.00 = = 1.000 C4.999H6.00O C5H6O empirical formula

  45. Chemistry-140 Lecture 7 Answer: Step 4: Determine the molecular formula from the empirical formula and the molar mass. M(empirical formula) = [5(MC) + 6(MH) + (MO)] = [5(12.011) + 6(1.008) + (15.999] = [60 + 6 + 16] = 82 g/mol Determined molar mass of Eugenol is 164 g/mol. X 2 C5H6O C10H12O2 molecular formula empirical formula

  46. Chemistry-140 Lecture 8 • Chapter Highlights • define molecules • definitions of molecular formula & structural formula • definition of allotrope • understand formation of cations & anions • learn names & formula for polyatomic ions • molar mass of compounds • percent composition • empirical formula Chapter 3: Molecules & Compounds

  47. Sn + I2 SnxIy Chemistry-140 Lecture 8 • Empirical Formula can be determined by a • number of different experiments Determining & Using Formulas

  48. Chemistry-140 Lecture 8 Determining & Using Formulas Example 3.11: 1.056 g of tin metal and 1.947 g of solid iodine are allowed to react in 100 mL of ethylacetate. After the reaction is complete (all the iodine has reacted), 0.601 g of tin is recovered. What is the empirical formula of the product formed from the reaction between Sn & I2 in this experiment ?

  49. Chemistry-140 Lecture 8 Answer: Step 1: Calculate the mass of Sn that reacted with the I2. (original mass of Sn) - (mass of Sn recovered after the reaction) = (mass of Sn consumed in the reaction) = (1.056 - 0.601) g = 0.455 g Sn

  50. Chemistry-140 Lecture 8 Answer: Step 1: Find the number of moles of Sn and I2 used to create the sample of SnxIy. (0.455 g Sn) = (1.947 g I2) = BUT, remember there are 2 atoms of I in each molecule of I2, therefore, moles of I = 2 x (7.671 x 10-3) = 3.83 x 10-3 mol Sn 7.671 x 10-3 mol I2 1.534 x 10-2 mol I

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