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Chapter 5.1, 5.3, 5.5, 5.8, 5.9 and 5.10

Chapter 5.1, 5.3, 5.5, 5.8, 5.9 and 5.10. Homework from the book 1, 11, 35, 37, 39, 41, 43, 45, 45, 47, 67, 69, 75, 77, 79, 83, 85, 87, 89, 91, 97, 103, 107, 109, 115, 119, 121, 127, 129 and 131. H 2. H 2 O. NH 3. CH 4. Molecules and Compounds.

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Chapter 5.1, 5.3, 5.5, 5.8, 5.9 and 5.10

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  1. Chapter 5.1, 5.3, 5.5, 5.8, 5.9 and 5.10 Homework from the book 1, 11, 35, 37, 39, 41, 43, 45, 45, 47, 67, 69, 75, 77, 79, 83, 85, 87, 89, 91, 97, 103, 107, 109, 115, 119, 121, 127, 129 and 131

  2. H2 H2O NH3 CH4 Molecules and Compounds A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds A compound is a substance composed of two or more elements combined in a specific ratio and held together by chemical bonds. A molecule is formed when two or more atoms join together chemically. A compound is a molecule that contains at least two different elements. All compounds are molecules but not all molecules are compounds Which of the following is a molecule and NOT a compound? 5.1 Answer: H2

  3. There are 7 elements that occur in nature as a diatomic molecule One way to remember these elements is: Mr. BrINClHOF

  4. molecular empirical H2O Empirical and Molecular formulas A molecular formulashows the exact number of atoms of each element in the smallest unit of a substance An empirical formulashows the simplest whole-number ratio of the atoms in a substance H2O CH2O C6H12O6 O3 O N2H4 NH2

  5. Ions Atoms are neutral—meaning that the number of protons is equal to the number of electrons If an atom loses or gains electrons the atom is No longer neutral but has a charge.

  6. 11 protons 11 electrons 11 protons 10 electrons Na+ Na 17 protons 18 electrons 17 protons 17 electrons Cl- Cl Why do Ions form? Cationsform from metals because they have a low Ionization energy and will readily give up electrons to obtain the electron configuration of a noble gas + e- + energy Anionsform from non-metals because they have low Electron affinity and will readily accept electrons to obtain the electron configuration of a noble gas + energy + e- Electrons are transferred from the Cation to the Anion and the charged ions attract each other

  7. Ionic Bonds Occur between a metal and a non-metal. Ionic bonds form from the Attraction of positive and negative ions that form because of a transfer Of electrons.

  8. Lattice Energy The amount that the potential energy of the system decreases When the ions in one mole of the compound are brought from A gaseous state to the positions the ions occupy in a crystal of the Compound. This is the reason Ionic Compounds form

  9. E = k Q+Q- r lattice energy cmpd MgF2 2957 Q= +2,-1 MgO 3938 Q= +2,-2 LiF 1036 LiCl 853 Electrostatic (Lattice) Energy Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. Q+ is the charge on the cation Q- is the charge on the anion r is the distance between the ions Lattice energy (E) increases as Q increases and/or as r decreases. r F < r Cl

  10. The Octet Rule Atoms tend to gain or lose electrons until they have achieved an outer shell that contains an octet (8) of electrons Exceptions to the Octet Rule Atoms that have valence electrons in the d and f orbitals And all atoms before Carbon in the periodic table

  11. Recognizing Ionic Compounds A compound is ionic if it contains a metal from group 1 Or group 2 or one of the polyatomic ions. Binary metal Oxides and sulfides also have ionic character

  12. When Atoms Combine to make Molecules Atoms contain both positive and negative charges. When they come Together they arrange themselves so that the attractive forges of opposite Charges is greater than the repulsive forces of like charges

  13. Bond Length and Bond Energy Bond Length (Bond Distance) The distance in picometers between the nucleii of two Bound atoms in a molecule Bond Energy The strength of the bond between atoms in a molecule. The amount of energy (J) required to break the bond.

  14. Change in electron density as two hydrogen atoms approach each other.

  15. Why should two atoms share electrons? + 8e- 8e- 7e- 7e- F F F F F F F F lonepairs lonepairs single covalent bond single covalent bond lonepairs lonepairs A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Lewis structure of F2

  16. The Octet Rule Atoms tend to share electrons until they have achieved an outer shell that contains an octet (8) of electrons Exceptions to the Octet Rule Atoms that have valence electrons in the d and f orbitals And all atoms before Carbon in the periodic table

  17. single covalent bonds H H H H or H H O 2e- 2e- O 8e- O C O C O O double bonds 8e- 8e- 8e- double bonds O N N triple bond N N triple bond 8e- 8e- Lewis structure of water + + Double bond – two atoms share two pairs of electrons or Triple bond – two atoms share three pairs of electrons or

  18. Molar Mass The molar mass (MM ) of a substance is the mass in grams of one mole of the substance. The molar mass of an element is numerically equal to its atomic mass. 1 mol C = 12.01 g 1 C atom = 12.01 amu The molar mass of a compound is the sum of molar masses of the elements it contains. 1 mol H2O = 2 ×1.008 g + 16.00 g = 18.02 g 1 mol NaCl = 22.99 g + 35.45 g = 58.44 g 5.10

  19. Molecular and Formula Mass The molecular massis the mass in atomic mass units (amu) of an individual molecule. To calculate molecular mass, multiply the atomic mass for each element in a molecule by the number of atoms of that element and then total the masses Molecular mass of H2O = 2(atomic mass of H) + atomic mass of O = 2(1.008 amu) + 16.00 amu = 18.02 amu Because the atomic masses on the periodic table are average atomic masses, the result of such a determination is an average molecular mass, sometimes referred to as the molecular weight. 5.8

  20. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu Molar mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. For any molecule molecular mass (amu) = molar mass (grams/mole) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

  21. Worked Example 5.12 Calculate the molecular mass or the formula mass, as appropriate, for each of the following corresponds: (a) propane, C3H8, (b) lithium hydroxide, LiOH, and (c) barium acetate, Ba(C2H3O2)2. Strategy Determine the molecular mass (for each molecular compound) or formula mass (for each ionic compound) by summing all the atomic masses. Think About ItDouble-check that you have counted the number of atoms correctly for each compound and that you have used the proper atomic masses from the periodic table. Solution For each compound, multiply the number of atoms by the atomic mass of each element and then sum the calculated values. (a) The molecular mass of propane is 3(12.01 amu) + 8(1.008 amu) = 44.09 amu (b) The formula mass of lithium hydroxide is 6.941 amu + 16.00 amu + 1.008 amu = 23.95 amu. (c) The formula mass of barium acetate is 137.3 amu + 4(12.01 amu) + 6(1.008 amu) + 4(16.00 amu) = 255.4 amu.

  22. Interconverting Mass, Moles, and Numbers atoms or molecules

  23. 8 mol H atoms 6.022 x 1023 H atoms 1 mol C3H8O x x x = 1 mol C3H8O 1 mol H atoms 60 g C3H8O How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 72.5 g C3H8O 5.82 x 1024 atoms H

  24. Worked Example 5.14 Determine (a) the number of moles of CO2 in 10.00 g of carbon dioxide and (b) the mass of 0.905 mole of sodium chloride. Strategy Use molar mass to convert from mass to moles and to convert from moles to mass. The molar mass of carbon dioxide (CO2) is 44.01 g/mol and the molar mass of sodium chloride (NaCl) is 58.44 g/mol. Solution (a) 10.00 g CO2 × = 0.2272 mol CO2 (b) 0.905 mol NaCl × = 52.9 g NaCl 58.44 g NaCl 1 mol NaCl 1 mol CO2 44.01 g CO2 Think About It Always double-check unit cancellations in problems such as these–errors are common when molar mass is used as a conversion factor. Also make sure that the results make sense. In both cases, a mass smaller than the molar mass corresponds to less than a mole of substance.

  25. Worked Example 5.15 (a) Determine the number of water molecules and the numbers of H and O atoms in 3.26 g of water. (b) Determine the mass of 7.92×1019 carbon dioxide molecules. Strategy Use molar mass and Avogadro’s number to convert from mass to molecules, and vice versa. Use the molecular formula of water to determine the numbers of H and O atoms. Solution (a) 3.26 g H2O× × Using the molecular formula, we can determine the number of H and O atoms in 3.26 g of H2O as follows: 1.09×1023 H2O molecules × 1.09×1023 H2O molecules × = 1.09×1023 H2O molecules = 2.18×1023 H atoms 6.022×1023 H2O molecules 1 mol H2O 1 mol H2O 18.02 g H2O 2 H atoms 1 H2O molecule 1 O atom 1 H2O molecule 1 mol CO2 6.022×1023 CO2 molecules 44.01 g CO2 1 mol CO2 = 1.09×1023 H atoms Solution (b) 7.92×1019 CO2 molecules × × = 5.79×10-3 g CO2 Think About It Again, check the cancellation of units carefully and make sure that the magnitudes of your results are reasonable.

  26. Percent Composition of Compounds A list of the percent by mass of each element in a compound is known as the compound’s percent composition by mass. where n is the number of atoms of the element in a molecule or formula unit of the compound 5.9

  27. 2 x (12.01 g) 6 x (1.008 g) 1 x (16.00 g) n x molar mass of element %C = %H = %O = x 100% = 34.73% x 100% = 13.13% x 100% = 52.14% x 100% 46.07 g 46.07 g 46.07 g molar mass of compound C2H6O Percent composition of an element in a compound n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0% 3.5

  28. Percent Composition of Compounds For a molecule of H2O2:

  29. Percent Composition of Compounds We could also have used the empirical formula of hydrogen peroxide (HO) for the calculation. In this case, we could have used the empirical formula mass, the mass in amu of one empirical formula, in place of the molecular formula. The empirical formula mass of H2O2 (the mass of HO) is 17.01 amu. %H = × 100% = 5.926% %O = × 100% = 94.06% 1.008 amu H 17.01 amu H2O2 16.00 amu O 17.01 amu H2O2

  30. Worked Example 5.13 Lithium carbonate, Li2CO3, was the first “mood-stabilizing” drug approved by the FDA for the treatment of mania and manic-depressive illness, also known as bipolar disorder. Calculate the percent composition by mass of lithium carbonate. Strategy Use Equation 5.1 to determine the percent by mass contributed by each element in the compound. Think About ItMake sure that the percent composition results for a compound sum to approximately 100. (In this case, the results sum to exactly 100 percent––18.79% + 16.25% + 64.96% = 100.00%––but remember that because of rounding, the percentages may sum to very slightly more or very slightly less. Solution For each element, multiply the number of atoms by the atomic mass, divide by the formula mass, and multiply by 100 percent. %Li = ×100% = 18.79% %C = ×100% = 16.25% %O = ×100% = 64.96% 2×6.941 amu Li 73.89 amu Li2CO3 12.01 amu C 73.89 amu Li2CO3 3×16.00 amu O 73.89 amu Li2CO3

  31. Worked Example 5.16 Determine the empirical formula of a compound that is 30.45 percent nitrogen and 69.55 percent oxygen by mass. Given that the molar mass of the compound is approximately 92 g/mol, determine the molecular formula of the compound. Strategy Assume a 100-g sample so that the mass percentages of nitrogen and oxygen given in the problem statement correspond to the masses of N and O in the compound. Then, using the appropriate molar masses, convert the grams of each element to moles. Use the resulting numbers as subscripts in the empirical formula, reducing them to the lowest possible whole numbers for the final answer. To calculate the molecular formula, first divide the molar mass given in the problem statement by the empirical formula mass. Then, multiply the subscripts in the empirical formula by the resulting number to obtain the subscripts in the molecular formula. The molar masses of N and O are 14.01 and 16.00 g/mol, respectively. One hundred grams of a compound that is 30.45 percent nitrogen and 69.35 percent oxygen by contains 30.45 g N and 69.55 g O.

  32. Worked Example 5.16 (cont.) Solution 30.45 g N × 69.55 g O × The gives a formula of N2.173O4.347. Dividing both subscripts by the smaller of the two to get the smallest possible whole numbers (2.173/2.173 = 1, 4.347/2.173 ≈ 2) gives an empirical formula of NO2. Finally, dividing the approximate molar mass (92 g/mol) by the empirical formula mass [14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol] gives 92/46.01 ≈ 2. Then, multiplying both subscripts in the empirical formula by 2 gives the molecular formula, N2O4. = 2.173 mol N = 4.347 mol O 1 mol N 14.01 g N 1 mol O 16.00 g O Think About It Use the method described in Worked Example 5.13 to calculate the percent composition of the molecular formula N2O4 and verify that it is the same as that given in this problem.

  33. Empirical Formula The listing of the mass of each element present in 100 g of a compound (elemental analysis) Empirical formula: contains the smallest set of whole- number subscripts that match the elemental analysis What is the empirical formula for a compound with The following elemental analysis? 49.5% C 5.2% H 28.8%N 16.5% O

  34. Finding Empirical Formula from percentages……… • Assume you have a 100 g sample so the mass of each • element is the same as the percentage • Divide by the molar mass of each element to get moles • 49.5 g C x 1 mol C = 4.125 mol C • 12.00 g • 5.20 g H x 1 mol H = 5.2 mol H • 1.00 g • 28.8 g N x 1 mol N = 2.05 mol N • 14.00 g • 16.5 g O x 1 mol O = 1.03 mol O • 16.00 g • Divide the moles of each element by the smallest (O) to get • ratios of the atoms in the molecule • C4H5N2O ÷ 1.03 = 4 ÷ 1.03 = 5 ÷ 1.03 = 2 ÷ 1.03 = 1

  35. Finding Molecular Formula…………… Molecular formula: contains the real set of whole- number subscripts that are found in the molecule Need: Empirical Formula and Molar Mass of the compound………. • For the preceeding problem where the empirical formula was • found to beC4H5N2O with a molar mass of 291 g/ mol • Find the molar mass of the empirical formula • (97g/ mol) • Divide the Molecular Formula MM by the • empirical formula MM giving us the “multiplying factor” • 291 ÷ 97 = 3 therefore 3 is the multiplying factor • Multiply all subscripts by the multiplying factor to get the • molecular formula • C12H15N6O3

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