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TOPIC III: THE MOLE

TOPIC III: THE MOLE. LECTURE SLIDES Definition and Use Avogadro’s Number Conversion Problems Empirical Formula Calculations. Kotz & Treichel, Chapter 3, Sections 3.6-3.8. From PARTICULATE (“too small to touch”) to MACROSCOPIC (amounts we can handle):. THE MOLE.

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TOPIC III: THE MOLE

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  1. TOPIC III: THE MOLE LECTURE SLIDES • Definition and Use • Avogadro’s Number • Conversion Problems • Empirical Formula Calculations Kotz & Treichel, Chapter 3, Sections 3.6-3.8

  2. From PARTICULATE (“too small to touch”) to MACROSCOPIC (amounts we can handle): THE MOLE

  3. Many different items we encounter in our daily lives come packaged in set amounts described by various “counting terms”. Let’s consider a few of them:

  4. shoes and socks and earrings come in pairs (2), • eggs come in dozens (12), • pencils are wholesaled by the gross (144), • donuts and sweet corn are often sold as “the baker’s dozen” or “the farmer’s dozen” (13), • and diet pop and beer by the 6- pack or case (24).....

  5. Chemists deal in atoms, molecules and ions, which need to be counted and measured as well. BUT: The mass of one atom of the 19F isotope is 3.156X10-23 g. The radius of a nucleus is about .001 pm and the radius of an atom about 100 pm. (1012 picometers or 1,000,000,000,000 pm = 1 m) THEREFORE......

  6. Chemists need their own unit for counting and weighingamounts of substances which come in particle size too tiny to be seen or weighed on any balance. For convenience in describing amounts of atoms, molecules, and ions , chemists have a unique unit of measure, THE MOLE

  7. The MOLE • The chemist’s counting number • Comes from the Latin meaning “whole heap or pile of” • SI base unit for measuring amount of substance • Defined as the number of atoms in exactly 12 grams of 12C, 6.022 X 1023

  8. 1 Mole always contains Avogadro’s number of particles or units: 1 Mole = Avogadro’s number of particles = 6.022136736 X 1023 particles = 602,213,673,600,000,000,000,000 particles If one used A’s number to describe macroscopic objects, one would be overwhelmed: I mole of green peas would cover the entire United States to a depth of 3 miles!

  9. Grams, amu’s, What’s the Difference? One mole is defined as the number of particles in exactly 12 g of the 12C isotope of carbon. Carbon was used as the standard for the amu scale, where the mass of one atom of 12C was defined as 12 amu. The mole answers the question: “How many atoms would you have if you took the amu scale (which describes mass of one atom) and use it as grams instead?

  10. So, How to Get a Mole: • We consult the periodic table, obtain the atomic • mass of an element in amu’s, the relative • mass of one atom..... • We weigh this amount out in grams..... • We now have one mole of atoms, A’s number, • 6.02 X 10 23 atoms, a convenient “package of • atoms”..... • We have gone into the chemist’s counting • system and can deliver not a dozen eggs • but a mole of atoms....

  11. This system works because of the relative nature of the atomic mass unit scale, in which all atoms were assigned a mass relative to 12C, the mole standard: “One mole is the number of atoms in exactly 12 g of 12C” The mole “pile or heap” of atoms for each element will weigh more or less than the mole “pile” for carbon, depending on whether the individual atoms weigh more or less than carbon.

  12. If one mole of carbon atoms weighs 12.0 g, then one mole of oxygen atoms, which weighs 1.33 times more than carbon, would be: 1.33 X 12.0 g =16.0 g = 1 mol O Since H atom is 1/12 the mass of a carbon atom, the matching pile of hydrogen atoms would be 1/12 X 12.0 g = 1.00 g = 1 mol H

  13. The “Molar Mass, M” • For any element, the molar mass, M, is the mass • in grams of a mole of atoms, “#g/mol” • M , molar mass, is NUMERICALLY equal to themass • of one atomin amu’s as given on your PT. • If, however, one weighs out the molar mass, one has • 6.022 x 10 23 atoms every time

  14. MOLES of ATOMS: the MOLAR MASS, M 1 atom 1 mole, A’s # atoms Li,6.941amu Li,6.941g Pb,207.2g Pb,207.2amu Zn,65.39g Zn,65.39amu

  15. The molar mass, “g/mol”, like density, “g/cm3”, is a convenient conversion factor: For any element: 1mole = atomic weight, grams = 6.022 X 1023 atoms Using this knowledge, the chemist can interconvert grams, moles, and atoms of any element.

  16. Suppose you weighed out 35.89 g of aluminum metal. How many moles and how many atoms of aluminum would be contained in this sample? Question:35.89 g Al= ? mol Al = ? atoms Al Relationships:1 mol Al = 26.98 g Al = 6.022 X 1023 atoms Al Setup and Solve: g ---> mol

  17. g --------> mol ---------> atoms

  18. Group Work 3.1 Suppose you weighed out15.00 g each of V and Mo metal. How many moles and how many atoms of each do you now have? 15.00 g V = ? Mol =? Atoms 15.00 g Mo= ? Mol =? Atoms

  19. What would 9.00 x 1024 atoms of mercury weigh in grams? Question:9.00 x 1024 atoms Hg= ? g Hg Relationships:1 mol Hg = 200.59 g Hg = 6.022 X 1023 atoms Hg Setup and Solve: atoms -----> mol -----> g

  20. GROUP WORK 3.2 Mercury is a liquid metal with a density of 13.534 g/cm3. If you measured out 75.0 mL of Hg into a graduated cylinder, how many atoms of Hg would be in the sample? Question:75.0 mL Hg = ? Atoms Hg Relationships: 13.534 g Hg = 1cm3 or mL Hg 200.59 g Hg = 1 mol Hg 1 mol Hg = 6.022 x 1023 atoms Hg Setup and solve:mL---> g ---> mol --->atoms

  21. Molecules, Compounds, and the Mole Let us now extend the use of molar mass, M, to include all particles chemists need to measure: not just atoms but also especially ions and molecules.... The basic principle is this: whenever you weigh out the“formula weight” of any substance or species in grams, you have A’s number of particles of that species, and the molar mass of that species...

  22. Molar Mass of Molecules The formula of any molecule describes the number of atoms making up one unit of that molecule: Br2 The diatomic bromine molecule, as bromine is found in nature: the formula tells us that 2 atoms of bromine are contained in every molecule. By extension, 2 moles of bromine atoms are contained in every 1 mole of bromine molecules. The calculation of the molar mass of molecular brominethen looks like this:

  23. The atomic weight of Br, from the PT, is 79.904 amu’s. Therefore: 2 moles of Br = 2 X 79.904 g = 159.808 g And the molar mass, M, of Br2 is 159.808 g/mol Now let’s try the molar mass of CH3CH2OH, ethyl alcohol:

  24. Molar Mass, M, CH3CH2OH M, CH3CH2OH, 46.07 g/mol

  25. Molar Mass of Ionic Compounds The formula of an ionic compound indicates the simplest ratio of ions present in any sample of the compound. It is this “formula unit” that we use for calculating the molar mass. Actually, we needn’t ask what kind of compound we are getting the M for; we simply calculate for all atoms found in the formula of any species! Cl2 (2 Cl)Fe(CN)2 (1Fe 2C 2N) (NH4)2CO3(2N 8H 1C 3O)

  26. M, CH3CH2OH, 46.07 g/mol, use in problems: Given a mass, or volume and density, solve for: a) moles of compound or individual atoms b) grams of individual atoms c) number of molecules or atoms

  27. How many moles of ethyl alcohol are contained in a sample that weighs 33.95 g? (CH3CH2OH, 46.07 g/mol). Question: 33.95 g CH3CH2OH = ? mol CH3CH2OH Relationship: 46.07 g CH3CH2OH = 1 mol Setup and Solve: ( g ---> mol)

  28. g ----------> mol

  29. How many moles of hydrogen are contained in 33.95 g CH3CH2OH? Question: 33.95 g CH3CH2OH = ? mol H Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 6 mol H Setup and Solve: ( g ---> mol CH3CH2OH --->mol H)

  30. How many grams of hydrogen are contained in 33.95 g CH3CH2OH? Question: 33.95 g CH3CH2OH = ? g H Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 6 mol H 1 mol H = 1.008 g H Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH --->mol H ----->g H)

  31. Group Work 3.3 How many grams of carbon are contained in 33.95 g CH3CH2OH? Question: 33.95 g CH3CH2OH = ? g C Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 2 mol C 1 mol C = 12.01 g C Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH --->mol C ----->g C)

  32. How many atoms of carbon are contained in 33.95 g CH3CH2OH? Question: 33.95 g CH3CH2OH = ? atoms C Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 2 mol C 1 mol C = 6.02 x 1023 atoms C Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol C -----> atoms C)

  33. GROUP WORK 3.4 What mass of ethyl alcohol, CH3CH2OH, would contain 2.06 X 1024 atoms of carbon? Question: 2.06 X 1024atoms C = ? g CH3CH2OH Relationship: 6.02 x 1023 atoms C = 1 mol C 2 mol C = 1 mol CH3CH2OH 1 mol CH3CH2OH = 46.07 g CH3CH2OH Setup and Solve: (atoms C-----> mol C ---> mol CH3CH2OH ---> g CH3CH2OH )

  34. New TOPIC: % Composition from Formula of the Compound The formula of a compound can be used to determine the mass % of each element by using the molar relationships we have learned... Consider the following: What is the % by mass of Carbon, Hydrogen and Oxygen in a sample of ethyl alcohol, CH3CH2OH?

  35. Mass % of Elements in a Compound This is done using the formula weight calculations and the approach: % by mass = total mass of one element (the part) X 100 % total mass of compound (the whole) based on one mole of the compound

  36. % Composition of Ethyl Alcohol, CH3CH2OH Molar Mass Calculation: 2 C = 2 X 12.011 = 24.022 g C 6 H = 6 X 1.008 = 6.048 g H 1 O = 1 X 15.999 = 15.999 g O 46.069 g/mol CH3CH2OH %C = 24.022 g C X 100 = 52.144% C 46.069 gCH3CH2OH %H = 6.048 g H X 100 = 13.13% H 46.069 gCH3CH2OH %O = 15.999 g O X 100 = 34.728% O 46.069 gCH3CH2OH

  37. CHECKING..... 52.144% C 13.13 % H 34.728% O 100.002 % CH3CH2OH = 100.00% (2 digits allowed after decimal)

  38. Formula from % Composition If one can go from formula to % composition, one should be able to go from % composition to the formula of the compound. This is quite true (almost): We can take % composition back to the “empirical formula”, which describes the simplest mole ratio of atoms in the formula... That is not always the same thing...

  39. Empirical Vs Molecular Formulas

  40. For “organic” or “molecular carbon containing compounds”, there exist a list of compounds which share almost every conceivable empirical formula. To determine which compound one has from analytical data, one needs the molar mass as well, as we will see... For most ionic compounds and simple molecular compounds, the empirical and molecular formulas are identical.

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