2. UNIT 2 – QUANTITATIVE CHEMISTRY (I.E. IN COMES THE MATHS!! BRING A CALCULATOR!!)

3. Scientific Notation… A (very) brief introduction

4. Stoichiometry  The study of the relationships between the quantities of reactants and products involved in chemical reactions.  Stoikheion = element  metron = measure

5. Proportions in compounds  The composition of a compound depends on two things  1. The elements they are composed of   2. The quantities of each element.

6. Law of definite proportions  A specific compound always contains the same elements in definite proportions by mass.  This is always constant even if produced other ways.

7. Relative atomic mass and isotopic abundance (p80)  Relative atomic mass = the mass of an element that would react with a fixed mass of a standard element, currently carbon-12.  One atomic mass unit (u) is described as 1/12 the mass of a carbon-12 atom.

8. Atomic mass unit  Now we know that it weighs  1 u = 1.660 540 2 x 10 -27 kg.  This is the standard used world wide

9. Isotopic abundance  Some elements have isotopes. In calculating the relative atomic mass of an element with isotopes, the relative mass and proportion of each is taken into account. For example, naturally occurring carbon consists of atoms of relative isotopic masses C-12 (98.89%) and-C 13 (1.11%). Its relative atomic mass is 12.01 u.

10. Isotopic abundance  m c = (98.89/100 x 12) + (1.11/100 x 13) =  m c = 11.8668 + 0.1443 = 12.01 u  No atom will contain 12.01 u. it is an average.

11. Example  Cl-35 = 75.53%  Cl-37 = 24.47%  (35 x.7553) + (37 x.2447) = 35.49 u  Therefore the average atomic mass is 35.49u

12.  Page 81 # 1

13. Atomic Mass and Molecular Mass  Atomic Mass  The mass of one atom of an element expressed in atomic units, u.  Molecular mass  The mass of one molecule, expressed in atomic mass units, u.

14. Example  M NH3 = 1(m N ) + 3(m H )  = 1(14.01 u) + 3(1.01 u)  = 17.04 u.  Therefore, one molecule of ammonia has a mass of 17.04 units.

15. Formula unit of mass  The mass of one formula unit of an ionic compound, expressed in atomic mass units, u

16. Example  NaCl = 1(m Na ) + 1(m Cl )  = 1 (22.99 u) + 1(35.45 u)  = 58.44 u  Therefore, one formula unit of sodium chloride has a mass of 58.44 units.

17.  Page 81 # 1-3

18. The Periodic Table and the Mass of Elements  The atomic number of a substance is written near the top of the symbol. This number tells the number of protons in the element.  The Atomic mass is usually written near the bottom. This number is given as a numerical number of the elements mass. The mass of one mole of the element measured in grams.

19. The Mole

20. It's not a spy, a machine for digging tunnels, a burrowing animal, or a spot of skin pigmentation - it's - a unit of measurement containing about 6.02 x 10 23 particles.  Chemists created the mole as their own counting unit.

21. 1 mole of atoms is 602, 214,199, 000, 000, 000, 000, 000 atoms! (This is the number of Carbon atoms there are in 12.01 g or Carbon)

22. Avogadro’s Number 6.02 x 10 23 particles / mole When you say you have a mole, it’s kind of like saying you have ‘a dozen’ of something… A dozen donuts = 12 donuts A dozen apples = 12 apples A mole of CO 2 molecules = 6.02 x 10 23 molecules A mole of Na atoms = 6.02 x 10 23 atoms

23. Just How Big is a Mole?  Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.   If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.  If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

24. Molar mass (g/mol) = M  the mass, in grams per mole, of one mole of a substance. The SI unit for a mol is g/mol  Molar mass is conventionally abbreviated with a capital M. For elements it is numerically equal to the atomic mass and is often called atomic mass units.

25. Steps to calculating molar mass 1. Write the chemical formula of the substance 2. Determine the number of atoms or ions of each element in one formula unit of the substance. 3. Us the atomic molar masses from the periodic table and the amounts in moles to determine the molar mass of the chemical 4. Write the molar mass in units of grams per mol (g/mol).

26. Example  Molar mass of copper(Cu) = mass of 1 mol of Cu atoms  M = 63.546 g/mol Cu  M = mass of 6.022 x 10 23 atoms

27. Example 2  Molar mass of glucose (C6H12O6) = mass of exactly 1 mol of glucose molecules  M = (6 x C) + (12 x H) + ( 6 x O)  M = ((6 x 12.01 (C)) + (12 x 1.01(H)) + (6x 16.00 (O)) ) / mol  M = ( (72.06 g) + (12.12 g) + (96.00g) ) / mol  M = ( 180.18 g) / mol  M = 180.18 g/mol  Therefore, 1 mol glucose has a mass of 180.18 g and contains 6.022 x 10 23 glucose molecules(C 6 H 12 O 6 )

28. Important things to note  Specify what is being measured  Ex. molar mass of oxygen atom = 16.00 g/mol  Molar mass of oxygen molecules = 32.00 g/mol  When talking about oxygen we are normally referring to the diatomic molecule  Since electrons have negligible mass compared to protons and neutrons, ions will essentially have the same mass as the atoms.

29. Moles of Elements  Why that number?  1 mole element = (atomic mass) in grams of that element 1 mole C = 12.011 g C 1 mole B = 10.81 g B 1 mole Cu = 63.55 g Cu  i.e. atoms are teeny, teeny, tiny, teeny, itty, bitty things and it is not reasonable to talk about them in their atomic weights, but if scale them up to moles, then working with balanced chemical equations etc. becomes easier.

30. Determine the mass one mole of the following in grams: a. CO 2 b. NaCl c. K 2 CrO 4 d. Fe(NO 3 ) 3

31.  PAGE 85 # 4

32. The Mole and Molar Mass  The Mole = n  Avogadro’s constant (N A ) = the number of entities in one mole = 6.02 x10 23  Eg. 1 dozen oxygen atoms = 12 oxygen atoms  1 mole oxygen atoms = 6.022 x 10 23 oxygen atoms

33. Calculations involving Molar mass  Amount in moles (n) = mass(g) Molar mass(g/mol) The Magic Triangle n = m / M m = n x M M = m / n

34. GRASP method  1 Given = What do you know?  2 Required = What do you want to know?  3 Analysis = What formula can you use?  4 Solution = Show your work  5 Paraphrase = State the answer in a sentence.

35. Example  Lets say that we are given 22 grams of graphite. How many moles do we have?

36.  Lets say we want to know how much CaCl 2 we need for a reaction that asks for 13.52 moles

37.  You are given an unknown sample that has a mass of 32.04 grams, 2.000 moles. Find M.

38. Practice: 1. Determine the number of mol in 13.7 g of CO 2. What is the mass of 2.75 mol of C 3 H 8

39. Homework  Page 86 # 5-7  Isotopic Abundance and Mole calculations Worksheet

40. Calculating Number of Entities

41. Calculating Number of Entities (N) from Mass  We are not able to count the number of atoms in reactions; we must count them indirectly by measuring out a certain mass of a substance and calculating the number of moles.  In order to estimate the number of entities in a large sample we need to take a small sample and find out the mass of a small number of entities.

42. Calculating Number of Entities and Avogadro’s number  There are times when need to know exactly how many atoms are in something. This is especially true when dealing with gases as they change in pressure with temperature.  To find the number of entities we use the formula

43.  N = nN A  Number of entities = (# of mols) x(Avogadro's #)

44. Example  Find the number of atoms in 0.004 moles of water.  Find the number of molecules in 30.0 g of Cl 2 (g)

45. Determining the number of particles in a mole

46. 6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles Note that a particle could be an atom OR a molecule! Avogadro’s Number as Conversion Factor

47. ParticlesMoleculesCompound Formula unit AtomElement

48. Equation to find number of particles! number of moles x = number of particles Example: How many particles of sucrose are in 3.50 moles? 6.02 x 10 23 particles 1 mole

49. Equation to find number of moles! number of particles x = number of moles Example: How many moles are contained in 4.50 x 10 24 atoms of zinc? 1 mole 6.02 x 10 23 particles

50. 1. Number of atoms in 0.500 mole of Al 2.Number of moles of S in 1.8 x 10 24 S atoms Learning Check

51. Molar Mass

52.  The Mass of 1 mole (in grams)  Molar mass is equal to the atomic mass (look on the periodic table) 1 mole of C atoms= 12.01 g 1 mole of Au atoms = 196.97g 1 mole of Cu atoms = 63.55 g  Always round to the hundredths place. Molar Mass

53. Mole to Mass Conversion Number of moles x number of grams = mass 1 mole Example: Calculate the mass in grams of 0.0450 moles of chromium. 0.0450 moles Cr x 52.0 g Cr = 2.34 g Cr 1 mol Cr Where will you find the number of grams per mole? The Periodic Table Look at your Periodic Table

54. Mass to Mole Conversion Mass x 1 mole = number of moles number of grams Example: How many moles of calcium are there in 525 grams of calcium? 525 g Ca x 1 mol Ca = 13.1 mol Ca 40.1 g Ca Where will you find the number of grams per mole? The Periodic Table Look at your Periodic Table

55. Mass to Atoms Conversion Two steps: 1. Use the mass to moles conversion 2. Now calculate the number of atoms using the moles to particles equation.

56. Example: How many atoms of gold are in a pure gold nugget having a mass of 25.0 g?

57. Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X 10 23 atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X 10 23 atoms Cu

58. Learning Check! How many atoms of K are present in 78.4 g of K? 78.4 g K 1 mol K 6.02 X 10 23 atoms K 39.1 g K 1 mol K = 1.20 X 10 24 atoms K

59. Empirical and Molecular Formulas 11.4

60. Percent Composition Mass of element x 100 = percent by mass Mass of compound Example: Determine the percent composition of NaHCO 3. Determine the mass of each element first! %mass element = mass of element in 1 mol compound x 100 molar mass of compound Percent Na = 22.99 g Na x 100 = 27.37% Na 84.01 g NaHCO 3 You have to complete this step for each element in the compound!!!!

61. lDivide the mass of each element by the molar mass of the compound and multiply by 100% Determine the Percent Composition from the Formula C 2 H 5 OH

62. Types of Formulas  Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present.  Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

63. ¬Convert the percentages to grams by assuming you have 100 g of the compound –Step can be skipped if given masses Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

64. ­Convert the grams to moles Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

65. ®Divide each by the smallest number of moles Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

66. ¯If any of the ratios is not a whole number, multiply all the ratios by a factor to make it a whole number –If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75 then multiply by 4 Multiply all the Ratios by 3 Because C is 1.3 Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

67. °Use the ratios as the subscripts in the empirical formula C4H6O3C4H6O3 Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

68. Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

69. uDivide the given molar mass of the compound by the molar mass of the empirical formula –Round to the nearest whole number Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3

70. uMultiply the empirical formula by the calculated factor to give the molecular formula (C 5 H 3 ) 4 = C 20 H 12 Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an empirical formula of C 5 H 3