1. Welcome to the MOLE

2. What is a mole? This is not such a bad mole, but not what we need to discuss...

3. This little stinker is just plain mean and ugly...

4. and the Mole People are in the dark and clueless.....

5. We are discussing the Mole used in Chemistry: Avogadro’s Number (N A ) or 6.02214179 x 10 23 shortened to 6.022 x 10 23 This amount = 1 mole

6. The mole is a way to describe the number of something without writing a huge number. It is similar to common terms like dozen, gross or even π in geometry 1 mole of anything – atoms, molecules, cockroaches or even galaxies will number 6.022 x 10 23 We need a number like this since atoms and molecules are extremely small and so many take up such a small space

7. Key Equations – KNOW THESE! Divide by Molar Massx by N A Grams Number (mass) ofMoles ofof SubstanceSubstance Particles x by Molar Mass Divide by N A Molar Mass = Atomic Weight in Grams per Mole (g/mol)

8. Molar Mass is the mass (g) of 1 mole of an element –For example – Na is 22.989 amu which is 22.989 g, and 1 mole of Na = 22.989 g –CO 2 is 1 Carbon at 12.011 g and 2 Oxygens at 15.999 each; therefore the molar mass of CO 2 = Σ of 1 C + 2 O ≈ 44 g Mass (m) = # mols x # g # mols = m / g

9. # Mols = m / Molar Mass % Composition by Mass = m element x 100 m compound # Mols = Concentration (Molar) x Volume (L) Volume = # mols / Concentration

10. Here’s a trick – to find the needed equation, just cover up the wanted result and what is left is the equation! Mass Mols Atomic X Mols Concentration X Volume Mass

11. The Mole Concept Atomic Mass (or atomic weight) is the mass of the element in amu (μ) –It is the number under the elemental symbol –Simply make this number into grams (g) –This represents the mass (m) of 1 mole of that element; and/or the m of 6.022 x 10 23 atoms of that substance! –1 mole of any gas = 22.4 L

12. Mole (mol) – is the # of atoms, ions, molecules that is equal to N A Molar Mass (M) – this is the m in g of 1 mol of a substance (g/mol) –Example – Manganese = 54.94 μ, thus its M = 54.94 g/mol; and 54.94 g of Mn will contain 6.022 x 10 23 atoms and this is equal to 1 mol of Mn

13. Mass to Mole Calculations Remember – each element has a different amu and thus, 1 mol of each will differ in mass The Mass of a Mole –Uses the C 12 isotope as its standard –H = μ of 1; or 1/12 of 1 atom of C 12 –He has μ of 4 or 4/12 (1/3) of a C 12 atom –Remember – atomic masses use isotopes and their % abundance in nature to calculate – and the closer to a whole number – the fewer the isotopes

14. I. Molar Mass of Substance = Grams Substance 1 mol Substance Therefore: 1. Mols of A = grams of A given x 1 mol A gram A 2. Mass of B = Mols of B given x gram B 1 mol B

15. Examples: 3 mol Mn = ? Grams 3 mol Mn X 54.9 g Mn = 165 g Mn 1 mol Mn 25 g Au = ? Mols 25 g Au x 1 mol Au = 25 = 0.127 mol Au 196.97 g Au 196.97 0.127 mols Au = ? Atoms 0.127 mol Au x 6.022 x 10 23 = 7.65 x 10 22 atoms Au 1 mol Au

16. II. Moles to Mass Mols (given) x # grams = mass 1 mol Example: –2.5 mol of (C 3 H 5 ) 2 S has what mass? M = 1 mol S = 32.07 g 6 mol C = 6 x 12.01 = 72 g 10 mol H = 10 x 1 = 10 g Σ 114.07 g/mol 2.5 mol x 114.07 g = ≈ 286 g 1 mol

17. We just used the bottom left of the diagram! Divide by Molar Massx by N A Grams Number (mass) ofMoles ofof SubstanceSubstance Particles x by Molar Mass Divide by N A Molar Mass = Atomic Weight in Grams per Mole (g/mol)

18. III. Mass to Moles with Compounds Example: m of Ca(OH) 2 = 325 g (rounded off) M = ? # mols = ? M = 1 mol Ca = 40.08 g 2 mol O = 2 x 16 = 32 g 2 mol H = 2 x 1 = 2 g Σ 74.096 g/mol Given m of 325 g Ca(OH) 2 x 1 mol Ca(OH) 2 = 4.3 mol M of 74.096 g

19. IV. Mass (g) to Particles Mols x N A = # Particles Example: –m = 35.6 g of AlCl 3 What is the number of Al +3 and Cl - ions? M = Al 26.981 g/mol Cl 35.452 g/mol x 3 = 106.356 Σ 133.337 Mols Al = m given = _____ mols x N A = _____ Al ions 26.981 g/mol Mols Cl = m given = _____ mols x N A = _____ Cl ions 106.356 g/mol Continued

20. And: 35.6 g AlCl 3 = 0.267 mol 133.337 g/mol AlCl 3 0.267 mol AlCl 3 x (6.022 x 10 23 ) = 1.6 x 10 23 molecules

21. V. Percent Composition of Compounds Mass Element (m) x 100 = % by mass Mass Cmpd (M) Example: H 2 O; what percent is H and what percent is O? % H = 2 x 1 (the molar mass of H) = 2 x 100 = 11.2% (the molar mass of H 2 O) 18 Thus, all compounds equal 100%, so 100 – 11.2 = 88.8 % for O

22. What is the % of C and O in CO 2 ? g Cx 100 = 12.01 C x 100 = 27.29% total g CO 2 44.01 g CO 2 32 g O x 100 = 72.71 % 44.01 g CO 2

23. Example: H 3 PO 4 (aq) (Phosphoric Acid) % H=3 g Hx 100=3% M = 98 g % P=31 g Px 100=32% 98 g % O =64 g Ox 100=65% 98 g m Compound = H (3 x 1) + P (1 x 31) + O (4 x 16) = 98 g

24. VI. Mole Ratios Given Vitamin C (ascorbic acid) with the following percentages, determine formula: 40.92 %C 4.58 %H 54.5 %O Set up with unknown moles (n): n C = 40.92 g C / 12.01 g C=3.4 mol C n H = 4.58 g H / 1.00 g H=4.5 mol H n O = 54.5 g O / 16 g O=3.4 mol O This is the Mole Ratio

25. Set Mole Ratio values as subscripts Divide each by the lowest value: C 3.4 / 3.4 H 4.5 / 3.4 O 3.4 / 3.4 = C 1 H 1.33 O 1 The 1.33 on H needs to be Δ’d into integer –Do this by multiplying until closest to a whole number 1.33 x 2 = 2.66 1.33 x 3 = 3.99 which can be rounded off to 4 –So the “magic” number is 3 – must multiply all subscripts by 3 Result is C 1 x 3 H 1.33 x 3 O 1 x 3 or C 3 H 4 O 3 !

26. What is the molecular formula that has 92.2% C and 7.8%H. The molar mass is 52.1. First – assume a 100 g sample of the substance The element’s percentages are assumed to be masses (g) Determine the moles of elements in compound: 92.2 g C x 1 mol C=7.68 mol C 12.01 g C 7.8 g H x 1 mol H=7.72 mol H 1.01 g H Divide all mols by the lowest value 7.68 C = 1 mol C7.72 H = 1.01 mol H 7.68 Continued

27. The Empirical Formula is C 1 H 1 (this is the basic form) To get the Molecular Formula: –Molar mass of the Empirical Formula is 12.01 + 1.01 = 13.02 g/mol –Molar mass of unknown is 52.1 g/mol So: Molar Mass Compound = Whole # to multiply Molar Mass Emp. Formula subscripts to get molecular formula 52.1 g/mol = 4 13.02 g/molthus, (CH) x 4 = C 4 H 4

28. Once again, a molecular formula calculation : Given 38.7% C, 9.7% H, and 51.6% O with a molecular formula mass of 62.0 g. What is the true molecular formula? First - find the empirical formula (this is CH 3 O) Find the formula mass:C 1 x 12 = 12 H 3 x 1.01 = 3.03 O 1 x 16 = 16.0 Σ 31.0

29. Divide the molecular mass by the empirical formula mass: 62 g (given) / 31 g (mass of emp. form.) = 2 Multiply each subscript by (n) or 2 in this example... Thus, the molecular formula: (CH 3 O)(n)  (CH 3 O)(2)  C 2 H 6 O 2

30. VII. Hydrates Hydrates are substances that include H 2 O in their formulas, but are not wet! –Hydration – adding H 2 O –Dehydration – removing it –Anhydrous – no H 2 O present

31. Methane Hydrate is found on the ocean’s floor –The methane will burn – but the water in it keeps the skin from burning!

32. The methane molecule (CH 4 ) is in a cage of water molecules There is 1 mole CH 4 per 5.75 mols H 2 O It is found at depths of 300 meters or more There is an estimated 1,300 trillion cubic feet of methane hydrate in the oceans However – the problem is that methane is one of the major greenhouse gases which contributes to global warming – so more study on retrieval and use is needed

33. Example: Barium Chloride Hydrate –Mass = 5 g. How many H 2 O per molecule? –BaCl 2 ____H 2 O The sample is heated and the result is 4.26 g anhydrous BaCl 2 –The difference between the 5 g hydrate and the 4.26 anhydrate is.74 g H 2 O So.........

34. A common hydrate is... MgSo 4 7H 2 O Magnesium Sulfate Heptahydrate

35. 4.26 g BaCl 2 =0.0205 mols BaCl 2 M = 208.23 g/mol 0.74 g H 2 O=0.041 mol H 2 18.02 g/mol # H 2 O = x = mols H 2 O = 0.041 = 2 mols cmpd 0.0205 Thus: BaCl 2 2H 2 O or barium chloride dihydrate

36. Summary