2. 1 The Mole 6.02 X 10 23 Everybody knows Avogadro’s Number

3. 2 STOICHIOMETRYSTOICHIOMETRY - the study of the quantitative aspects of chemical reactions.

4. 3 The Mole A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000 6.02 X 10 23 (in scientific notation) Amedeo Avogadro (1776 – 1856)This number is named in honor of Amedeo Avogadro (1776 – 1856),

5. 4 The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 10 23 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X 10 23 cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X 10 23 atoms Note that the NUMBER is always the same, but the MASS is very different! Mole is abbreviated mol: n = m/MM gmol -1

6. 5 = 6.02 x 10 23 C atoms = 6.02 x 10 23 H 2 O molecules = 6.02 x 10 23 NaCl “molecules” 2 x 6.02 x 10 23 ions A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles 1 mole C 1 mole H 2 O 1 mole NaCl

7. 6 Mole Ratio: http://www.files.chem.vt.edu/RVGS/ACT/n otes/The_Mole.htmlhttp://www.files.chem.vt.edu/RVGS/ACT/n otes/The_Mole.html

8. 7 n= m/MM the unit for MM is gmol -1

9. 8 Learning Check

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11. 10 The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of Cu atoms =63.5 g Molar Mass

12. 11 Other Names Related to Molar Mass Molecular Mass/Molecular Weight: If you have a single molecule, mass is measured in amu’s instead of grams. But, the molecular mass/weight is the same numerical value as 1 mole of molecules. Only the units are different. (This is the beauty of Avogadro’s Number!) Formula Mass/Formula Weight: Same goes for compounds. But again, the numerical value is the same. Only the units are different. THE POINT: You may hear all of these terms which mean the SAME NUMBER… just different units

13. 12 Find the molar mass (usually we round to the tenths place) Learning Check! A.1 mole of Br atoms B.1 mole of Sn atoms =79.9 g/mole = 118.7 g/mole

14. 13 Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl 2 = 111.1 g/mol 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl 2 1 mole of N 2 O 4 = 92.0 g/mol Molar Mass of Molecules and Compounds

15. 14 A.Molar Mass of K 2 O = ? Grams/mole B. Molar Mass of antacid Al(OH) 3 = ? Grams/mole Learning Check!

16. 15 Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass. Learning Check

17. 16 molar mass Grams Moles Calculations with Molar Mass

18. 17 Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al ? g Al Converting Moles and Grams

19. 18 Answer = 81.0 g Al

20. 19 The artificial sweetener aspartame (Nutra-Sweet) formula C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? Learning Check!

21. 20 Atoms/Molecules and Grams Since 6.02 X 10 23 particles = 1 mole AND 1 mole = molar mass (grams) You can convert atoms/molecules to moles and then moles to grams! (Two step process) You can’t go directly from atoms to grams!!!! You MUST go thru MOLES. That’s like asking 2 dozen cookies weigh how many ounces if 1 cookie weighs 4 oz? You have to convert to dozen first!

22. 21 molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

23. 22 Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? = 3.4 X 10 23 atoms Cu

24. 23 Learning Check! How many atoms of K are present in 78.4 g of K?

25. 24 Learning Check! What is the mass (in grams) of 1 molecule of glucose (C 6 H 12 O 6 )?

26. 25 Learning Check! How many atoms of O are present in 78.1 g of oxygen?

27. 26 Volume of Gases Avogadro’s Law http://www.chemistry.co.nz/avogadro.htm

28. 27 2 liters of CO gas react with 1 liter of O 2 gas to yield 2 liters of CO 2 gas. 2 CO (g) + 1 O 2(g)  2 CO 2(g) HOW CAN THIS BE POSSIBLE? Avogadro's Law : EQUAL VOLUMES OF DIFFERENT GASES CONTAIN EQUAL NUMBERS OF MOLECULES WHEN MEASURED AT THE SAME TERMPERATURE AND PRESSURE.

29. 28 One mole of any gas occupies 22.41L at 0 o C and 1 atm(stp) and contains the same number of particles - atoms - molecules. 22.41L is know as molar volume. Molar Volume = 22.4 dm3 mol at s.t.p. And 24.0 dm3 mol at r.t.p.

30. 29 1. What volume of carbon dioxide, measured at s.t.p. will be liberated from 5 g of calcium carbonate, by the action of an excess of dilute hydrochloric acid? 2. 120 cm3 of hydrogen gas is released at S.T.P. when a small amount of Magnesium is placed in excess dilute hydrochloric acid. What mass of Magnesium was added.

31. 30 Problems 1. Calculate the molar volume of each of the following: a. 46.00 grams of nitrogen gas b. 200.0 grams of carbon monoxide c. 3.01 x 1023 molecules of ethane (C2H6)

32. 31

33. 32 If the amount of gas in a container is increased, the volume increases. If the amount of gas in a container is decreased, the volume decreases. V1/n1 = V2/n2

34. 33 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? Answer: V 1 n 2 = V 2 n 1 (5.00 L) (1.80 mol) = (x) (0.965 mol)

35. 34 What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG (monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C Percent Composition

36. 35 Chemical Formulas of Compounds (HONORS only) Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions).Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). NO 2 2 atoms of O for every 1 atom of N NO 2 2 atoms of O for every 1 atom of N 1 mole of NO 2 : 2 moles of O atoms to every 1 mole of N atoms If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

37. 36 Types of Formulas (HONORS only) Empirical FormulaEmpirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula Molecular FormulaMolecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

38. 37 To obtain an Empirical Formula (HONORS only) 1.Determine the mass in grams of each element present, if necessary. 2.Calculate the number of moles of each element. 3.Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4.If whole numbers are not obtained * in step 3), multiply through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

39. 38 A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole 14.01 g/mole moles of O = 5.34 g = 0.334 moles of O 16.00 g/mole 16.00 g/mole Formula: Formula: (HONORS only)

40. 39 Calculation of the Molecular Formula (HONORS only) A compound has an empirical formula of NO 2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

41. 40 Empirical Formula from % Composition (HONORS only) A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance?

42. 41 IB Problems

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44. 43 Percent Composition Experiment http://classes.mhcc.edu/enh/ch151_mr/pdf151/PercentKClO3.pdf A student heated 100g of HgO according to 2 HgO => 2Hg + O 2 and the remaining residue weighed 91.5g. Find the percent error in the residue determination. Given: mass of crucible=15g

45. 44 http://www.savitapall.com/matter/labs/Lab-Percentage%20Composition-MgO.pdf http://www2.ucdsb.on.ca/tiss/stretton/chem3/Lab_1_Percent_COmposition.html http://www.chemistry.ucsc.edu/teaching/roland/Chem1M/procedures/1M_02_Em piricalproc.pdf