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Chapter 5

Chapter 5. Quantities in Chemistry. Setting the Stage - Bioavailability of Nitrogen. Plants need nitrogen as either ammonia or nitrate in order to use it for biosynthesis While air is 80% nitrogen, it is in the form of the relatively inert gas N 2 , hence plants cannot use it directly.

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Chapter 5

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  1. Chapter 5 Quantities in Chemistry

  2. Setting the Stage - Bioavailability of Nitrogen • Plants need nitrogen as either ammonia or nitrate in order to use it for biosynthesis • While air is 80% nitrogen, it is in the form of the relatively inert gas N2, hence plants cannot use it directly Malone and Dolter - Basic Concepts of Chemistry 9e

  3. Type and Efficacy of Nitrogen Sources • Main types of fertilizer are ammonia gas (NH3) and ammonium nitrate (NH4NO3) • 100 kg of NH3 delivers 82 kg of N • 100 kg of NH4NO3 only delivers 35 kg of N • The mass of N delivered is related to the formula of the compound and the relative masses of each element in the formula Malone and Dolter - Basic Concepts of Chemistry 9e

  4. Setting a Goal – Part AThe Measurement of Masses of Elements and Compounds • You will become proficient at working with the units of moles, mass, and numbers of atoms and molecules, and at converting between each of these Malone and Dolter - Basic Concepts of Chemistry 9e

  5. Objective for Section 5-1 • Calculate the masses of equivalent numbers of atoms of different elements Malone and Dolter - Basic Concepts of Chemistry 9e

  6. 5-1 - Relative Masses of Elements • When the masses of samples of any two elements are in the same ratio as that of their atomic masses, the samples have the same number of atoms • We can therefore use the atomic masses and the masses of samples of chemical substances to “count” the number of atoms or molecules Malone and Dolter - Basic Concepts of Chemistry 9e

  7. The Mass of an Atom • Recall that an atom has an unbelievably small mass • 12C is used as the standard and is assigned a mass of exactly 12 amu. • Other isotopes are present in natural samples (i. e. C has an overall atomic mass of 12.01) so the periodic table lists masses that are the weighed average mass of the natural sample Malone and Dolter - Basic Concepts of Chemistry 9e

  8. Relative Masses of the Elements • The amu has no practical value in a laboratory situation: 12.01 amu = 1.994  10-23 g • The best balance made can detect no less than 10-5 g, so we have to scale up the masses to something we can measure Malone and Dolter - Basic Concepts of Chemistry 9e

  9. Counting by Weighing Bolts Nuts • Hardware stores often count by weighing • If we want 175 bolts and assume that the mass of an average bolt is 10.5 g, then the mass of bolts will be: Malone and Dolter - Basic Concepts of Chemistry 9e

  10. Counting by Weighing • Nuts, being smaller, will have a lower average mass (2.25 g) • What weight of nuts will provide a nut for each bolt? Malone and Dolter - Basic Concepts of Chemistry 9e

  11. Counting by weighing • By using ratios of the average masses of bolts and nuts, or the masses of a fixed number of items, we can get equivalent numbers of bolts and nuts: Malone and Dolter - Basic Concepts of Chemistry 9e

  12. From Hardware to Atoms • We need to know the relative numbers of atoms of different elements present and the relative masses of the individual atoms • A 4He atom has a mass of 4.00 amu and a 12C atom has a mass of 12.00 amu. • If 4He and 12C are present in a 4.00:12.00 mass ratio, regardless of the units of mass, the number of atoms is the same Malone and Dolter - Basic Concepts of Chemistry 9e

  13. Objective for Section 5-2 • Define the mole and relate this unit to numbers of atoms and to the atomic masses of the elements Malone and Dolter - Basic Concepts of Chemistry 9e

  14. 5-2 The Mole and the Molar Mass of Elements • A mole is: • The number of atoms in exactly 12 g (12.00 recurring) of 12C • Avogadro’s number of atoms(6.022  1023) • The number of atoms in one atomic mass of an element expressed in grams Malone and Dolter - Basic Concepts of Chemistry 9e

  15. Avogadro Amedeo Avogadro, a pioneer in the investigation of quantitative aspects of chemistry Malone and Dolter - Basic Concepts of Chemistry 9e

  16. The Mole • One mole of an element implies • The atomic mass expressed in grams, it is different for each element • It contains Avogadro’s number (6.022 x 1023) of atoms, which is the same for all elements • A conversion factor between mass and numbers of things (allows us to count atoms by weighing) Malone and Dolter - Basic Concepts of Chemistry 9e

  17. Mass and Number of Things • For our purposes, assume that oranges are identical in mass; 12 have a mass of 2.71 kg • We can do the same with atomic mass Malone and Dolter - Basic Concepts of Chemistry 9e

  18. Moles of Elements • Here is 1 mole each of copper, iron, mercury and sulphur Malone and Dolter - Basic Concepts of Chemistry 9e

  19. Objective for Section 5-3 • Perform calculations involving masses, moles, and numbers of molecules or formula units for compounds Malone and Dolter - Basic Concepts of Chemistry 9e

  20. 5-3 The Molar Mass of Compounds • Molecular compounds • chemical formula represents a discrete molecular unit (e. g. CO2) • Ionic compounds • chemical formula represents a formula unit (the whole number ratio of cations to anions; e. g. K2SO4) Malone and Dolter - Basic Concepts of Chemistry 9e

  21. Calculation of Formula Weight • The sum of the atomic masses of all the atoms in a molecule • This is often referred to as the molecular weight • Consider CO21 C (12.01 amu) + 2 O (2  16.00 amu) = 44.01 amu • One molar mass of a compound contains Avogadro’s number of molecules Malone and Dolter - Basic Concepts of Chemistry 9e

  22. Calculation of Formula Weight • Second example, using a salt, Fe2(SO4)3 Malone and Dolter - Basic Concepts of Chemistry 9e

  23. Hydrates • Some ionic compounds can have water molecules attached within the structure • These compounds are termed hydrates and have properties distinct from the unhydrated form • The formula weight of a hydrate includes the mass of the water molecules Malone and Dolter - Basic Concepts of Chemistry 9e

  24. Hydrates • Examples • CuSO4 - [copper(II) sulfate] - a pale green solid • CuSO4•5H2O - [copper(II) sulfate pentahydrate - a dark blue solid • Often, the waters of hydration can be removed by heating Malone and Dolter - Basic Concepts of Chemistry 9e

  25. The Molar Mass of a Compound • The mass of one mole (6.022 × 1023 molecules or formula units) is referred to as the molar mass of the compound • It is the formula weight expressed in grams • For example, 44.0 g of CO2 is the molar mass of CO2 and is the mass of 6.022 × 1023 molecules of CO2 Malone and Dolter - Basic Concepts of Chemistry 9e

  26. Moles of Compounds • One mole of copper sulfate pentahydrate, sodium chloride, sodium chromate and water Malone and Dolter - Basic Concepts of Chemistry 9e

  27. Summary Chart for Part A (1) Malone and Dolter - Basic Concepts of Chemistry 9e

  28. Summary Chart for Part A (2) Malone and Dolter - Basic Concepts of Chemistry 9e

  29. Setting a Goal – Part BThe Component Elements of Compounds • You will learn about the relationship between the formula of a compound and its elemental composition Malone and Dolter - Basic Concepts of Chemistry 9e

  30. Objective for Section 5-4 • Given the formula of a compound, determine the mole, mass, and percent composition of its elements Malone and Dolter - Basic Concepts of Chemistry 9e

  31. 5-4 The Composition of Compounds Table 5-2 relates one mole of a compound (H2SO4) to all its component parts. All of these relationships can be used to construct conversion factors between the compound and its elements. Malone and Dolter - Basic Concepts of Chemistry 9e

  32. Table 5-2 Malone and Dolter - Basic Concepts of Chemistry 9e

  33. The Mole Composition of a Compound • Mole composition is the number of moles of each of the elements that make up 1 mole of the compound • CO2 – one mole of C and two moles of O • H2SO4 – one mole of S, two moles of H, and four moles of O Malone and Dolter - Basic Concepts of Chemistry 9e

  34. The Mass Composition of a Compound • Mass composition is the mass of each element in the compound • CO2 – 12.01 g of C and 32.00 g of O • H2SO4 – 2.016 g of H, 32.07 g of S, and 64.00 g of O Malone and Dolter - Basic Concepts of Chemistry 9e

  35. Percent Composition of a Compound • mass of each element per 100 mass units of compound • in 100 g of NH3, there are 82.2 g of N • therefore, the mass percentage of N is 82.2% N Malone and Dolter - Basic Concepts of Chemistry 9e

  36. CO2 • Calculation of % composition of carbon dioxide requires determining the number of grams of each element (C and O) in one mole Malone and Dolter - Basic Concepts of Chemistry 9e

  37. Objectives for Section 5-5 • Use percent or mass composition to determine the empirical formula of a compound • Given the molar mass of a compound and its empirical formula, determine its molecular formula Malone and Dolter - Basic Concepts of Chemistry 9e

  38. 5-5 Empirical and Molecular Formulas • Empirical formula - simplest whole number ratio of atoms in the compound • Procedure to find empirical formula from % composition data • Convert percent composition to an actual mass • Convert mass to moles of each element • Find the whole number ratio of the moles of different elements Malone and Dolter - Basic Concepts of Chemistry 9e

  39. Empirical Formula of Laughing Gas • Contains 63.6% N and 36.4% O • Assume 100 g of substance, so you have 63.6 g of N and 36.4 g of O • Calculation gives an empirical formula of N2O Malone and Dolter - Basic Concepts of Chemistry 9e

  40. Molecular Formula • The actual number of each atom in a formula unit • Consider acetylene and benzene • both have the empirical formula CH, but different molecular formulas: • acetylene is actually C2H2 • benzene is actually C6H6 Malone and Dolter - Basic Concepts of Chemistry 9e

  41. Molecular Formula Determination • Needs the molecular mass, which must be determined from an independent measurement (e.g. via mass spectrometry) • First determine the mass of the empirical formula Malone and Dolter - Basic Concepts of Chemistry 9e

  42. Molecular Formula Determination…contd. • Divide the empirical formula mass into the molecular mass • The resulting number (which should be a small whole number or close to it) is the number of times the empirical formula unit appears in the molecular formula Malone and Dolter - Basic Concepts of Chemistry 9e

  43. Acetylene and Benzene • The empirical formula mass for both substances is 12.0 g + 1.0 g = 13.0 g • The actual molar mass of acetylene is 26.0 g, so the empirical formula mass divides into the actual mass two times - C2H2 • Benzene’s actual molar mass is 78.0 g, so the empirical formula mass divides into the actual mass six times - C6H6 Malone and Dolter - Basic Concepts of Chemistry 9e

  44. Acetylene and Benzene: Same Empirical Formula, but DifferentMolecular Formulas These are structural formulas Malone and Dolter - Basic Concepts of Chemistry 9e

  45. Summary Chart for Part B (1) Malone and Dolter - Basic Concepts of Chemistry 9e

  46. Summary Chart for Part B (2) Malone and Dolter - Basic Concepts of Chemistry 9e

  47. Summary of Types of Formula Malone and Dolter - Basic Concepts of Chemistry 9e

  48. Worked Example Nicotine is a compound containing C, H and N only. Its molar mass is 162 g. A 1.50 g sample of nicotine is found to contain 1.11 g of C. Analysis of another sample indicates that nicotine has 8.70% by mass of H. Determine the molecular formula of nicotine. Solution. We should convert the data to masses or %, and find the mass or % of N by difference. Malone and Dolter - Basic Concepts of Chemistry 9e

  49. Worked Example Continued Malone and Dolter - Basic Concepts of Chemistry 9e

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