1 / 77

Chapter 4: Chemical Composition

Chapter 4: Chemical Composition. Questions for Consideration. How can we describe the mass composition of elements in a compound? How can we determine the number of atoms in a given mass of a material? The number of molecules? The number of formula units?

davisc
Download Presentation

Chapter 4: Chemical Composition

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 4: Chemical Composition

  2. Questions for Consideration • How can we describe the mass composition of elements in a compound? • How can we determine the number of atoms in a given mass of a material? The number of molecules? The number of formula units? • How can we use the masses of elements in a compound to determine its chemical formula? • How can we express the composition of a solution?

  3. Chapter 4 Topics: • Percent Composition • Mole Quantities • Determining Empirical and Molecular Formulas • Chemical Composition of Solutions

  4. Chapter 4 Math Toolbox: 4.1 Mole Quantities • This math toolbox gives a concise overview of the mole-quantity calculations that are introduced in more detail throughout this chapter.

  5. 4.1 Percent Composition • We have used chemical formulas to express the composition of compounds, but where do these formulas come from? • We cannot directly determine a chemical formula. However, we can measure the mass of each element in a sample of a compound by using appropriate analytical techniques. • Because the mass of each element in a compound will vary from one sample of the compound to another, we must have some expression of composition that is the same for all samples.

  6. Percent Composition • A convenient method for expressing composition is percent composition by mass. • For any element, E, in a compound, the percent composition by mass is given by the following equation:

  7. Moles, Masses, and Particles • How can we describe the composition of a compound if we know the mass of the elements in the compound? • A 3.67-g sample of the mineral chalcopyrite was determined to contain 1.27 g Cu, 1.12 g Fe, and 1.28 g S. • What is the mass percent of each element in this compound?

  8. Percent Composition • The percent composition for all the elements present in a compound must add up to 100%. Figure 4.5

  9. Composition of Chalcopyrite • Would this mass % differ for a different sample of this mineral? • If you had a 100-gram sample, what mass of copper would it contain? Figure 4.5

  10. Activity: Percent Composition • A sample of a copper compound weighs 1.63 g and contains 1.30 g of Cu. The other element in the compound is oxygen. What is the percent composition of this compound? • By difference, the sample contains 0.33 g O. Applying the formula for percent composition gives:

  11. Activity: Percent Composition • What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

  12. Activity Solution: Percent Composition • What are the percent iron and the percent sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

  13. Activity: Percent Composition • A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the percent Ca in limestone?

  14. Activity Solutions: Percent Composition • A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the percent Ca in limestone?

  15. 4.2 Mole Quantities • When working with amounts of a substance on a macroscopic scale, we cannot simply count atoms or molecules. There are too many. Instead, we use the moles of atoms, which are related by Avogadro’s number: 1 mole = 6.022 × 1023 particles • 1 mole C = 6.022 × 1023 carbon atoms • 1 mole H2S = 6.022 × 1023 H2S molecules • 1 mol Cu2O = 6.022 × 1023 Cu2O formula units

  16. The Mole • The mole unit acts as a bridge between the microscopic world and the macroscopic world. • One mole of substance contains as many basic particles (atoms, molecules, or formula units) as there are atoms in exactly 12 g of carbon-12. • One mole of a substance contains 6.022 × 1023 particles (molecules, atoms, ions, formula units, etc.) • This number is called Avogadro’s number.

  17. Moles of Various Elements and Compounds Figure 4.8

  18. Activity: Atoms in H2S • How many sulfur atoms are in 1 mol of H2S? • How many hydrogen atoms are in 1 mol of H2S? Figure 4.6

  19. Molar MassThe Mass of 1 Mole • Avogadro’s number has been defined so that the mass of 1 mol of C-12 has a mass of exactly 12 grams. • This means that the average mass of an atom of any substance in amus is the same numerical value as the mass of 1 mole of that substance in grams (molar mass). • The molar mass of carbon is 12.01 g/mol. • The molar mass of oxygen is 16.00 g/mol • The molar mass of CO2 is: 12.01 + 2(16.00) = 44.01 g/mol We’ll use 4 sig. figs

  20. Activity: Molar Mass • Complete the table.

  21. Activity Solutions: Molar Mass

  22. Molar Mass • Which contains the greatest number of atoms? • 1 mole of copper or 1 mole of gold? • 1 gram of copper or 1 gram of gold? Answer: Because 1 mole is a fixed quantity of particles (6.022 × 1023), 1 mole of copper contains the same number of atoms as 1 mole of gold. Answer: Because each copper atom has less mass than each gold atom, 1 gram of copper contains more atoms than 1 gram of gold.

  23. Converting Between Grams and Moles • Convert 10.0 grams of CO2 to moles. • Convert 0.50 mol CO2 to grams. Figure from p. 137

  24. Activity: Converting Between Grams and Moles • Convert 10.0 g O2 to moles. Figure from p. 137

  25. Grams  Moles  Particles • Once we know the number of moles of a substance, we can use Avogadro’s number (6.022×1023) to determine the number of particles in that sample of the substance. 1 mole = 6.022 × 1023 particles Figure from p. 139

  26. Activity: Number of Particles • What mass of MgCl2 contains 6.022 × 1023 Cl ions? • The molar mass of MgCl2 is 95.21 g/mol. Figure from p. 139

  27. Extra Activity: Conversions with Molar Mass • How many moles of aspartame (C14H18N2O5) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass?

  28. Extra Activity Solution: Conversions with Molar Mass • How many moles of aspartame (C14H18N2O5) are found in 40.0 mg of aspartame? How many molecules of aspartame are found in this mass? We first need to convert from mg to g: Next, we need to find the molar mass of C14H18N2O5:

  29. Extra Activity Solution: Conversions with Molar Mass • How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass? Next, convert from grams to moles: Finally, we convert from moles to molecules:

  30. Extra Activity: Conversions with Molar Mass • If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

  31. Extra Activity Solution: Conversions with Molar Mass • If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

  32. 4.3 Determining Empirical and Molecular Formulas • The formula for a substance also tells us about the composition of a compound: • A formula unit for an ionic compound tells us the ratio of ions of different elements in the compound. (MgCl2 has a 1:2 ratio of Mg2+ to Cl.) • A molecular formula tells the number of atoms of each element in a molecule and the atom ratio. (CO2 has a 1:2 ratio of C to O atoms.)

  33. Empirical and Molecular Formulas • Empirical formula • Expresses the simplest ratios of atoms in a compound • Written with the smallest whole-number subscripts • Molecular formula • Expresses the actual number of atoms in a compound • Can have the same subscripts as the empirical formula or some multiple of them

  34. Empirical Formulas • What is the same about these two compounds? Figure 4.13

  35. Empirical Formulas • The empirical formula of a substance is the ratio of atoms of different elements, in terms of the smallest whole numbers. • What are the empirical formulas of these two compounds? • What are the empirical formulas of H2O2 and H2O? Are they different compounds? Figure 4.13

  36. What is the empirical formula for copper(II) oxide? Figure 4.12

  37. Empirical or Molecular Formulas

  38. Activity: Empirical Formula Determine the empirical formulas of these substances. For which substances is the empirical formula the same as the molecular formula? Figure 4.14

  39. Activity Solutions: Empirical Formula Determine the empirical formulas of these substances. For which substances is the empirical formula the same as the molecular formula? Figure 4.14

  40. Determining Empirical Formulas from % Composition • If we know the masses of the elements in a compound, or its percent composition, we can determine its mole ratio, and therefore the compound’s empirical formula. • Consider the compound commonly called chalcopyrite. Any sample will have the following % composition: Figure 4.5

  41. Determining Empirical Formulas • Since the percent composition does not change from sample to sample, assume any size sample. The most convenient is 100 grams so % value = mass value. • Convert grams to moles for each element using its molar mass as a conversion factor. • Without changing the relative amounts, change moles to whole numbers. Do this by dividing all by the same smallest value. If all do not convert to whole numbers, multiply to get whole numbers.

  42. Determining Empirical Formulas 1. 100 grams chalcopyrite contains: • 30.5 g Fe • 34.6 g Cu • 34.9 g S 2. mol Fe = (30.5g Fe)(1mol/55.85 g) = 0.5461 mol Fe mol Cu = (34.6g Cu)(1mol/63.55 g) = 0.5444 mol Cu mol S = (34.9g S)(1mol/32.07 g) = 1.088 mol S 3. (0.5461 mol Fe)/(0. 5444) = 1.003 mol Fe (0.5444 mol Cu)/(0. 5444) = 1.000 mol Cu (1.088 mol S)/(0. 5444) = 1.999 mol S Figure 4.5 FeCuS2

  43. Steps for Determining Empirical Formula Figure from p. 143

  44. Activity: Empirical Formula from Percent Composition • A compound was determined to have the following percent composition: • 50.0% sulfur • 50.0% oxygen • What is the empirical formula for the compound? SO2

  45. Activity: Determining Empirical Formulas • Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S.

  46. Activity Solution: Determining Empirical Formulas • Determine the empirical formula for the mineral covellite, which has the percent composition 66.5% Cu and 33.5% S. First, reassign the percentages to units of grams: 66.5 g Cu and 33.5 g S. Then, convert to moles and divide both numbers by the lowest number. The whole numbers then become our subscripts. The empirical formula is therefore: CuS

  47. Activity: Determining Empirical Formulas • Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite.

  48. Activity Solutions: Determining Empirical Formulas • Shattuckite is a fairly rare copper mineral. It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite. Therefore, the empirical formula is: Cu5Si4O14H2

  49. Molecular Formulas from Empirical Formulas • Benzene and acetylene have the same empirical formulas but different molecular formulas. • How much greater in mass is benzene than acetylene? • How much greater in mass is each of these than the empirical formula? Figure 4.13

  50. Empirical and Molecular Formulas • What are the ratios of masses for the molecular and empirical formulas?

More Related