2. Chapter 5 Quantities in Chemistry

3. Malone and Dolter - Basic Concepts of Chemistry 9e2 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

4. Malone and Dolter - Basic Concepts of Chemistry 9e3 Type and Efficacy of Nitrogen Sources Main types of fertilizer are ammonia gas (NH 3 ) and ammonium nitrate (NH 4 NO 3 ) 100 kg of NH 3 delivers 82 kg of N 100 kg of NH 4 NO 3 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

5. Malone and Dolter - Basic Concepts of Chemistry 9e4 Setting a Goal – Part A The 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

6. Malone and Dolter - Basic Concepts of Chemistry 9e5 Objective for Section 5-1 Calculate the masses of equivalent numbers of atoms of different elements

7. Malone and Dolter - Basic Concepts of Chemistry 9e6 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

8. Malone and Dolter - Basic Concepts of Chemistry 9e7 The Mass of an Atom Recall that an atom has an unbelievably small mass 12 C 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

9. Malone and Dolter - Basic Concepts of Chemistry 9e8 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

10. Malone and Dolter - Basic Concepts of Chemistry 9e9 Counting by Weighing 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: Bolts Nuts

11. Malone and Dolter - Basic Concepts of Chemistry 9e10 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?

12. Malone and Dolter - Basic Concepts of Chemistry 9e11 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:

13. Malone and Dolter - Basic Concepts of Chemistry 9e12 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 4 He atom has a mass of 4.00 amu and a 12 C atom has a mass of 12.00 amu. If 4 He and 12 C are present in a 4.00:12.00 mass ratio, regardless of the units of mass, the number of atoms is the same

14. Malone and Dolter - Basic Concepts of Chemistry 9e13 Objective for Section 5-2 Define the mole and relate this unit to numbers of atoms and to the atomic masses of the elements

15. Malone and Dolter - Basic Concepts of Chemistry 9e14 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 12 C Avogadro’s number of atoms (6.022  10 23 ) The number of atoms in one atomic mass of an element expressed in grams

16. Malone and Dolter - Basic Concepts of Chemistry 9e15 Avogadro Amedeo Avogadro, a pioneer in the investigation of quantitative aspects of chemistry

17. Malone and Dolter - Basic Concepts of Chemistry 9e16 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 10 23 ) 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)

18. Malone and Dolter - Basic Concepts of Chemistry 9e17 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

19. Malone and Dolter - Basic Concepts of Chemistry 9e18 Moles of Elements Here is 1 mole each of copper, iron, mercury and sulphur

20. Malone and Dolter - Basic Concepts of Chemistry 9e19 Objective for Section 5-3 Perform calculations involving masses, moles, and numbers of molecules or formula units for compounds

21. Malone and Dolter - Basic Concepts of Chemistry 9e20 5-3 The Molar Mass of Compounds Molecular compounds chemical formula represents a discrete molecular unit (e. g. CO 2 ) Ionic compounds chemical formula represents a formula unit (the whole number ratio of cations to anions; e. g. K 2 SO 4 )

22. Malone and Dolter - Basic Concepts of Chemistry 9e21 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 CO 2 1 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

23. Malone and Dolter - Basic Concepts of Chemistry 9e22 Calculation of Formula Weight Second example, using a salt, Fe 2 (SO 4 ) 3

24. Malone and Dolter - Basic Concepts of Chemistry 9e23 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

25. Malone and Dolter - Basic Concepts of Chemistry 9e24 Hydrates Examples CuSO 4 - [copper(II) sulfate] - a pale green solid CuSO 4 5H 2 O - [copper(II) sulfate pentahydrate - a dark blue solid Often, the waters of hydration can be removed by heating

26. Malone and Dolter - Basic Concepts of Chemistry 9e25 The Molar Mass of a Compound The mass of one mole (6.022 × 10 23 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 CO 2 is the molar mass of CO 2 and is the mass of 6.022 × 10 23 molecules of CO 2

27. Malone and Dolter - Basic Concepts of Chemistry 9e26 Moles of Compounds One mole of copper sulfate pentahydrate, sodium chloride, sodium chromate and water

28. Malone and Dolter - Basic Concepts of Chemistry 9e27 Summary Chart for Part A (1)

29. Malone and Dolter - Basic Concepts of Chemistry 9e28 Summary Chart for Part A (2)

30. Malone and Dolter - Basic Concepts of Chemistry 9e29 Setting a Goal – Part B The Component Elements of Compounds You will learn about the relationship between the formula of a compound and its elemental composition

31. Malone and Dolter - Basic Concepts of Chemistry 9e30 Objective for Section 5-4 Given the formula of a compound, determine the mole, mass, and percent composition of its elements

32. Malone and Dolter - Basic Concepts of Chemistry 9e31 5-4 The Composition of Compounds Table 5-2 relates one mole of a compound (H 2 SO 4 ) to all its component parts. All of these relationships can be used to construct conversion factors between the compound and its elements.

33. Malone and Dolter - Basic Concepts of Chemistry 9e32 Table 5-2

34. Malone and Dolter - Basic Concepts of Chemistry 9e33 Mole composition is the number of moles of each of the elements that make up 1 mole of the compound CO 2 – one mole of C and two moles of O H 2 SO 4 – one mole of S, two moles of H, and four moles of O The Mole Composition of a Compound

35. Malone and Dolter - Basic Concepts of Chemistry 9e34 Mass composition is the mass of each element in the compound CO 2 – 12.01 g of C and 32.00 g of O H 2 SO 4 – 2.016 g of H, 32.07 g of S, and 64.00 g of O The Mass Composition of a Compound

36. Malone and Dolter - Basic Concepts of Chemistry 9e35 Percent Composition of a Compound mass of each element per 100 mass units of compound in 100 g of NH 3, there are 82.2 g of N therefore, the mass percentage of N is 82.2% N

37. Malone and Dolter - Basic Concepts of Chemistry 9e36 CO 2 Calculation of % composition of carbon dioxide requires determining the number of grams of each element (C and O) in one mole

38. Malone and Dolter - Basic Concepts of Chemistry 9e37 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

39. Malone and Dolter - Basic Concepts of Chemistry 9e38 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

40. Malone and Dolter - Basic Concepts of Chemistry 9e39 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 N 2 O

41. Malone and Dolter - Basic Concepts of Chemistry 9e40 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 C 2 H 2 benzene is actually C 6 H 6

42. Malone and Dolter - Basic Concepts of Chemistry 9e41 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

43. Malone and Dolter - Basic Concepts of Chemistry 9e42 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

44. Malone and Dolter - Basic Concepts of Chemistry 9e43 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 - C 2 H 2 Benzene’s actual molar mass is 78.0 g, so the empirical formula mass divides into the actual mass six times - C 6 H 6

45. Malone and Dolter - Basic Concepts of Chemistry 9e44 Acetylene and Benzene: Same Empirical Formula, but Different Molecular Formulas These are structural formulas

46. Malone and Dolter - Basic Concepts of Chemistry 9e45 Summary Chart for Part B (1)

47. Malone and Dolter - Basic Concepts of Chemistry 9e46 Summary Chart for Part B (2)

48. Malone and Dolter - Basic Concepts of Chemistry 9e47 Summary of Types of Formula

49. Malone and Dolter - Basic Concepts of Chemistry 9e48 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.

50. Malone and Dolter - Basic Concepts of Chemistry 9e49 Worked Example Continued