1 / 83

Chapter 3

Chapter 3. Chemical Compounds. 3.1 Types of Chemical Compounds and Their Formulas. Molecular Compounds. Molecular compounds are composed of molecules and contain only nonmetals . Electrons are shared AKA Covalent Compounds Represented by chemical formulas.

yovela
Download Presentation

Chapter 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 Chemical Compounds

  2. 3.1 Types of Chemical Compounds and Their Formulas

  3. Molecular Compounds • Molecular compounds are composed of molecules and contain only nonmetals. • Electrons are shared • AKA Covalent Compounds • Represented by chemical formulas

  4. The ratio of the masses of carbon and hydrogen, C:H in methane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1

  5. The ratio of the masses of carbon and hydrogen, C:H in methane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1

  6. The ratio of the masses of carbon and hydrogen, C:H in ethane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1

  7. The ratio of the masses of carbon and hydrogen, C:H in ethane is 1. 4:1 2. 1:4 3. 3:1 4. 1:3 5. 1:1

  8. Ions • When atoms lose or gain electrons, they become ions. • Cations are positive and are formed by elements on the left side of the periodic chart. • Anions are negative and are formed by elements on the right side of the periodic chart.

  9. Ionic Bonds Ionic compounds (such as NaCl) are generally formed between metals and nonmetals.

  10. A mole of solid sodium chloride, salt, contains • 22.99 g of sodium and 34.45 g of chlorine A B 2. 6.02x1023 g of sodium and 6.02x1023 g of chloride 3. 22.99 u of sodium and 34.45 u of chloride 4. 22.99 g of sodium and 34.45 g of chloride.

  11. A mole of solid sodium chloride, salt, contains • 22.99 g of sodium and 34.45 g of chlorine A B 2. 6.02x1023 g of sodium and 6.02x1023 g of chloride 3. 22.99 u of sodium and 34.45 u of chloride 4. 22.99 g of sodium and 34.45 g of chloride.

  12. A molecule of solid sodium chloride, salt, contains • 22.99 g of Na and 34.45 g of Cl A B 2. 22.99 u of Na+ and 34.45 u of Cl- 3. 22.99 g of Na and 34.45 g of Cl 4. None of the above

  13. A molecule of solid sodium chloride, salt, contains • 22.99 g of sodium and 34.45 g of chlorine A B 2. 22.99 u of sodium and 34.45 u of chloride 3. 22.99 g of sodium and 34.45 g of chloride. 4. None of the above; sodium chloride is an ionic compound, as such, it does not exist in discrete molecular units as is characteristic of covalent compounds.

  14. Would you expect the following to be ionic or molecular: • N2O • Na2O • CaCl2 • SF4 • CBr4 • FeS • P4O6 • PbF2

  15. Types of Formulas • Empirical formulas give the lowest whole-number ratio of atoms of each element in a compound. • Molecular formulas give the exact number of atoms of each element in a compound. • Structural Formulas show the order in which atoms are bonded

  16. Types of Formulas • Structural formulas show the order in which atoms are bonded. • Perspective drawings also show the three-dimensional array of atoms in a compound.

  17. Types of Formulas • Formula Unit: the smallest neutral collection of ions (most reduced, for ionic)

  18. For the ball and stick model of naphthalene to the right, the empirical formula is 1. C10H8 2. C4H5 3. C5H4 4. CH

  19. For the ball and stick model of naphthalene to the right, the empirical formula is 1. C10H8 2. C4H5 3. C5H4 4. CH

  20. For the ball and stick model of pyrimidine to the right, the molecular formula is: 1. C4N2H4 2. C2N2H2 3. C2NH2 4. (CH)4N2

  21. For the ball and stick model of pyrimidine to the right, the molecular formula is: 1. C4N2H4 2. C2N2H2 3. C2NH2 4. (CH)4N2

  22. 3.2 The Mole Concept and Chemical Compounds

  23. Formula Mass vs. Molar Mass vs. Molecular Mass • Formula Mass: the mass of a formula unit (amu) • Molecular Mass: the mass of a molecule (amu) • Molar Mass: the mass of one mole of a compound (grams) H2O: Molecule Mass 18.0153 amu (1 molecule) Molar Mass 18.0153 grams (1 mole)

  24. Molecular Formulas Diatomic molecules: H2 O2 N2 F2 Cl2 Br2 I2 Molecules: P4 (White Phosphorus) S8 (Sulfur) Distinguish between molecule and atom!

  25. Example 3-1A How many grams of MgCl2 would you need to obtain 5.0 x 10-23 Cl- ions?

  26. Example 3-2A Gold has a density of 19.32 g/cm3. A piece of gold foil is 2.50 cm on each side and 0.100mm thick. How many atoms of gold are in this piece of gold foil?

  27. Example 3-3A Halothane: C2HBrClF3 How many grams of Br are contained in 25.00 mL of halothane (d=1.871g/mL)

  28. 3.3 Composition of Chemical Compounds

  29. (number of atoms)(MM of element) x 100 % element = (MM of the compound) Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

  30. (2)(12.0 g/mol) %C = (30.0 g/mol) x 100 24.0 g/mol = 30.0 g/mol = 80.0% Percent Composition So the percentage of carbon in ethane is…

  31. Example 3-4A • What are the mass percent composition in C10H16N5P3O13?

  32. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition

  33. Establishing Formulas From % Comp • Assume 100g, %  grams • Grams  Moles • Divide by smallest moles • Get whole numbers in most reduced form Ration= MM Molecular/ MM Empirical Multiply Ratio by Empirical to get Molecular Formula

  34. Example 3-5A Sorbitol is a sweetener that has a molecular mass of 182 u and percent composition of 39.56% C 7.74% H 52.70 % O What are the empirical and molecular formulas?

  35. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

  36. Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 16.00 g 1 mol 1.01 g Calculating Empirical Formulas

  37. Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005  7 H: = 6.984  7 N: = 1.000 O: = 2.001  2 5.09 mol 0.7288 mol 0.7288 mol 0.7288 mol 5.105 mol 0.7288 mol 1.458 mol 0.7288 mol Calculating Empirical Formulas

  38. Calculating Empirical Formulas These are the subscripts for the empirical formula: C7H7NO2

  39. Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this • C is determined from the mass of CO2 produced • H is determined from the mass of H2O produced • O is determined by difference after the C and H have been determined

  40. Example The combustion of 5.00 grams of an alcohol produces 9.55 g of CO2 and 5.87 g of H2O. Find the empirical formula. Solution: First you need to find the individual masses of the elements. The general equation would look like this: CxHyOz + O2 CO2 + H2O

  41. The mass of the carbon in the CO2 all came from the alcohol so… 9.55 gCO2 X 12.01 g C 2.61 g C 44.01 gCO2 The mass of the hydrogen in H2O all came from the alcohol so… 5.87 g H2O X 2.02 g H 0.658 g H 18.02 g H2O

  42. Because the mass of the oxygen in both CO2 and H2O is derived from both the alcohol and the oxygen from combustion you need to find the mass of oxygen only in the alcohol…. By subtracting the mass of the carbon and hydrogen from the total mass of the compound. 5.00 g – (2.61 g + 0.658 g) = 1.73 g O

  43. Now you can determine the number of moles of each element in the formula….. Moles of C = 2.61 g X 1.00 mole C 0.217 mol C 12.01 g C Moles of H = 0.658 g X 1.00 mole H 0.651 mol H 1.01 g H Moles of O = 1.73 g O X 1.00 mole O 0.108 mol O 16.00 g O

  44. From the number of moles we find the mole ratios,…. C = 0.217/0.108 = 2.01 H = 0.651/ 0.108 = 6.03 O = 0.108/ 0.108 = 1.00 Thus the formula will be C2H6O or C2H5OH

  45. Practice Problem 1.540 g of an organic acid burns completely to produce 2.257 g CO2 and 0.9241 g H2O. Find the empirical formula. If the molecular mass is 90.0 grams what is the molecular formula? C 2.257 gCO2 X 12.01/44.01 = 0.6159g H 0.9241g H2O X 2.02/18.02 = 0.1036g O 1.540 –(0.6159 + 0.1036) = 0.8205g Empirical formula= CH2O Molecular formula = (CH2O)x X= molecular mass/empirical mass X= 90/30= 3 C3H6O3 C 0.6159/12.01= 0.0513 H 0.1026/ 1.01 = 0.1026 O 0.8205/16.00= 0.05128

  46. Example 3-6B • Combustion of a 1.505 g sample of thiophene, a carbon-hydrogen-sulfur compound yields 3.149 g CO2, 0.645 g H2O and 1.146 g SO2 as the only combustion products. What is the empirical formula?

  47. 3.4 Oxidation States Tell the number of electrons gained or lost when forming compounds • Handout

  48. Common Oxidation States

More Related