1. Chapter 10: Sec 1 and 2 The Mole

2. Vocabulary 2 Formula Mass - Atomic Mass (A) - Mole (mol) - Avagadro’s - Number (N) Molar Mass ( M ) - The sum of the atomic masses of all the elements in a compound The mass of an atom expressed relative to the mass assigned to carbon-12 The number of atoms of an element relative to the number of atoms in exactly 12g of carbon-12 The number of particles (atoms, molecules or formula units) in a mol (6.022 x 10 23 particles/mol) The mass in grams (g) of one mole of a substance

3. The Mole 3 Mole (mol) - The number of atoms of an element relative to the number of atoms in exactly 12g of carbon-12 Avagadro’s - Number (N) The number of particles (atoms, molecules or formula units) in a mol (6.022 x 10 23 particles/mol) So what’s the relationship? It’s a relationship between numbers Example: Describe the following?

4. The Mole 4 So… 12Eggs=1 Dozen Eggs 12Donuts=1 Dozen Donuts 12Baseballs=1 Dozen Baseballs The Relationship? 12Eggs 1 Dozen Eggs 12Baseballs 1 Dozen Baseballs 12Donuts 1 Dozen Donuts So how does this relate to moles?

5. The Mole 5 12Eggs=1 Dozen Eggs As… 1 moleC atoms Like… 12Eggs 1 Dozen Eggs 6.022 x 10 23 C atoms 1 moleC atoms 6.022 x 10 23 C atoms =

6. Calculating Particles 6 Atoms - Molecules - Formula Units - Individual elements (ex. Na, Zn, Si) Diatomic elements or molecular substances (ex. O, N, H, Group 7A or CH 4, NH 3 ) ionic substances (ex. NaCl, KBr, LiF) ** All three (3) are calculated using Avagadro’s Number ** 1 mol = 6.022 x 10 23 atoms (single element) 6.022 x 10 23 molecules (covalent molecules) 6.022 x 10 23 formula units (ionic compounds)

7. Rules to follow when solving word Problems! 7 1.Write down WHAT you know. Given in the problem. 2.Write down WHAT you want to know. What the problem asks for. 3.Determine HOW to get there. Conversion factors (TOOLS) you will use. Practice solving word Problems using the following conversion factor

8. PRACTICE USING THE TOOLS 8 Use the cards (TOOLS) to solve the following problems. “A” represents any particle (atom, molecule, formula unit) Align the cards so that the units cancel. Example: How many mols of A are in 3.4 particles of A. Follow your rules… 1 mol A 6.022 x 10 23 particles A 1 mol A (P.T.) g of A 1 mol A 3.4 particles of A “Given” = mols of A 1. What do I know? 2. What do I want? 3. How do I get there?(Tools)

9. 0.36 mol Al 6.022 x 10 23 atoms Al 1 mol Al x = 2.1679 x 10 23 atoms Al - - - - - Example13 mol of sodium carbonate (Na 2 CO 3 ) was used to neutralize an acid before throwing it away. How many formula (fu) units were used. 13 mol Na 2 CO 3 6.022 x 10 23 fu Na 2 CO 3 1 mol Na 2 CO 3 x = 7.8286 x 10 24 fu Na 2 CO 3 - - - - - - - - - ExampleIt takes 50 mols of propane (C 3 H 8 ) to cook a steak on a gas barbeque. How many molecules are burned. 50 mol C 3 H 8 6.022 x 10 23 molecules C 3 H 8 1 mol C 3 H 8 x = 3.011 x 10 25 molecues C 3 H 8 - - - Example A piece of foil contains 0.36 mol Al. Calculate the number of atoms in the piece of foil. = 2.2 x 10 23 atoms Al = 7.8 x 10 24 fu Na 2 CO 3 = 3 x 10 25 molecues C 3 H 8

10. 10 Section 2: Mass and the Mole Which one weighs more? Why? Same number different masses 1 mole = 6.022 x 10 23 atoms (single element) But… atoms are not all the same size. That is why… 1 mol of carbon atoms = 12 g C (Periodic Table) 1 mol of iron atoms = 55.85 g Fe (Periodic Table) Same number…different mass Same concept in chemistry…

11. Calculating Moles and Mass 11 1 mol of element = atomic mass of element 1 mol of compound = formula or molar mass of compound Example1 mol of Fe = 55.847 g 1 mol of Na =22.990 g 1 mol of sodium chloride (NaCl) =58.443 g (Add Them Together!) 1 mol of carbonic acid (H 2 CO 3 ) = 62.024 g

12. 12 Example Calculate the formula mass for ammonia (NH 3 ) Step 1 Write down the different elements and their atomic mass N 14.007g H 1.008g Step 2 Count how many of each element is present and multiply the quantity by the atomic mass N 14.007g x 1 H 1.008g x 3 = 14.007 g = 3.024 g Step 3 Add all the totals together Formula mass = 14.007 + 3.024 = 17.031 g

13. PRACTICE USING THE TOOLS 13 Example: Determine the mass in grams of 3.57 mol Al Follow your rules… 1 mol Al 6.022 x 10 23 particles Al 1 mol Al 26.98 g of Al 1 mol Al 3.57 mols of Al “Given” = g of Al 1. What do I know? 2. What do I want? 3. How do I get there?(Tools) 9.63

14. 14 ExampleHow many mols of oxygen are in 52 g of oxygen 52 g O 2 31.998 g O 2 1 mol O 2 x = 1.625 mol O 2 - - - ExampleHow many mols of calcium are in 37 g of calcium 37 g Ca 40.08 g Ca 1 mol Ca x = 0.9232 mol Ca - - ExampleHow many mols of sodium chloride are in 100 g NaCl 100 g NaCl 58.443 g NaCl 1 mol NaCl x = 1.711 mol NaCl - - - - - = 2 mol NaCl = 1.6 mol O 2 = 0.92 mol Ca

15. Solving the problems Using the Tools 15 1 st …Follow your rules. 1. What do I know? 2. What do I want? 3. How do I get there?(Tools) You have 2 tools in your tool box! 1 mol Al 6.022 x 10 23 particles Al 1 mol Al Avogadro’s # 1 mol Al 107.87g of Ag 107.87g of Al 1 mol Al Molar Mass

16. PRACTICE USING THE TOOLS 16 Example: Determine the number of moles 25.5 grams Ag. Follow your rules… 1 mol Al 6.022 x 10 23 particles Al 1 mol Al 107.87g of Ag 107.87g of Al 1 mol Al 25.5 g of Ag “Given” = mols of Ag 1. What do I know? 2. What do I want? 3. How do I get there?(Tools) 0.236

17. 17 ExampleHow many grams of hydrogen are in 1.67 mol of hydrogen 1.67 mol H 2 1 mol H 2 2.016 g H 2 x = 3.367 g H 2 - - - - - - - ExampleHow many grams of carbon are in 0.85 mol of carbon 0.85 mol C 1 mol C 12.011 g C x = 10.209 g C - - ExampleHow many grams of ammonia are in 2.3 x 10 5 mol ammonia 2.3 x 10 5 mol NH 3 1 mol NH 3 17.031 g NH 3 x = 3.917 x 10 6 g NH 3 - - - - - - - - = 3.37 g H 2 = 10. g C = 3.9 x 10 6 g NH 3

18. Converting Mass and atoms 18 Mass (g)Moles (mol)Particles Example: How many atoms are in 55.2 g Li Golden Rule…GET TO MOLES!!!! Then you can do what ever you want! 18 1 mol Li 6.022 x 10 23 particles Li 1 mol Li 6.94 g of Li 1 mol Li 55.2 g of Li “Given” = mols of Li 7.95 = atoms of Li4.79 x 10 24

19. 19 ExampleIn a small firework, 3.467 x 10 22 atoms Sr are used to make the firework burn red. How many grams of strontium are in the firework. 3.467 x 10 22 atoms Sr 6.022 x 10 23 atoms Sr x= 5.044 g Sr - - - - - - - x 87.620 g Sr 1 mol Sr - - - - - - - ExampleAn extra strength aspirin tablet contains 0.500g of aspirin (C 9 H 8 O 4 ). How many molecules of aspirin are in one tablet 0.500 g C 9 H 8 O 4 180.16171 g C 9 H 8 O 4 x = 1.671 x 10 21 molecules C 9 H 8 O 4 - - - - - - - - - - - - x 6.022 x 10 23 molecules C 9 H 8 O 4 1 mol C 9 H 8 O 4 - - - - - - - = 1.67 x 10 21 molecules C 9 H 8 O 4

20. Section 3: Molar Mass and Compounds 20 HH O Water (H 2 O) ----- 1.00794 g1.00794 g ---- ----- 15.999 g Formula Mass = Sum of atomic mass for each element Formula Mass = 1.008 + 1.008 + 15.999 = 18.015g

21. 21 Section 3: Moles of Compounds

22. 22 Example Calculate the formula mass of acetic acid (HC 2 H 3 O 2 ) Step 1 H 1.008 g C 12.011 g O 15.999 g Step 2 H 1.008 g x 4 = 4.032 g C 12.011 g x 2 = 24.022 g O 15.999 g x 2 = 31.998 g Step 2 H 1.008 g x 4 C 12.011 g x 2 O 15.999 g x 2 Step 3 Formula mass = 4.032 + 24.022 + 31.998 = 60.052 g Example Calculate the formula mass of carbonic acid (H 2 CO 3 ) Step 1 H 1.008 g C 12.011 g O 15.999 g Step 2 H 1.008 g x 2 = 2.016 g C 12.011 g x 1 = 12.011 g O 15.999 g x 3 = 47.997 g Step 2 H 1.008 g x 2 C 12.011 g x 1 O 15.999 g x 3 Step 3 Formula mass = 2.016 + 12.011 + 47.997 = 62.024 g

23. PRACTICE USING THE TOOLS 23 Use the cards (TOOLS) to solve the following problems. “A” represents any particle (atom, molecule, formula unit) Align the cards so that the units cancel. Example: How many mols of A are in 5.0 g of A. Follow your rules… 1 mol A 6.022 x 10 23 particles A 1 mol A (P.T.) g of A 1 mol A 5 g of A “Given ” = mols of A 1. What do I know? 2. What do I want? 3. How do I get theres?(Tools)

24. Section 3: Moles of Compounds Recall that a chemical formula indicate the numbers and types of atoms contained in one unit of the compound Example Freon – CCl 2 F 2 1- Carbon 2 – chlorine 2 - fluorine This ratio also represents a mole ration. For example… 1 mol CCl 2 F 2 = 2 mol of Cl atoms How many moles of fluorine atoms are in 5.50 moles of freon (CCl 2 F 2 ) 5.50 mol CCl 2 F 2 1 mol CCl 2 F 2 2 mol F atoms x - - - - - = 11.0 mol F atoms

25. Calculating Volume 25 1 mol of gas @ STP = 22.4 L ExampleA chemical reaction produces 0.82 mol of oxygen gas. What volume of O 2 was produced 0.82 mol O 2 1 mol O 2 22.4 L O 2 x = 18.368 L O 2 - - - - - STP = Standard Temperature and Pressure T = 25 o C = 298 K (Room Temperature) P = 1 atm = 18 L O 2

26. 26 ExampleHow many liters of oxygen gas are in 3.250 mol of oxygen gas 3.250 mol O 2 1 mol O 2 22.4 L O 2 x = 72.800 L O 2 - - - - - ExampleHow many liters of nitrogen gas are in 100 g N 2 100 g N 2 28.014 g N 2 1 mol N 2 x = 79.960 L N 2 - - - - - ExampleAn average size room has a volume of around 4000L of air at STP. How many moles of air are in a average size room. 4000 L air 22.4 L air 1 mol air x = 178.571 mol air - - - x 1 mol N 2 22.4 L N 2 - - - - - = 80 L N 2 = 200 mol air = 72.80 L O 2

27. Vocabulary 27 Percent Composition - Tells you the percentage of the mass made up by each element in the compound Empirical Formula - Gives the simplest whole number ratio of the atoms of the element Molecular Formula - Gives actual number of atoms of each element in the molecular compound

28. Calculating Percent Composition 28 % of element = mass of element x 100 formula mass compound ExampleWhat is the percent composition of hydrogen and oxygen in water? % of H = 2.01588g x 100 18.01488g = 11.2% % of O = 100 – 11.2 = 88.8% Formula mass = 2(1.00794g) + 15.999g = 18.01488g Chemical Formula = H 2 O

29. 29 ExampleFind the percent composition of a compound that contains 2.30g Na, 1.60g O and 0.100g H. Total mass = 2.30g + 1.60g + 0.100g = 4.00g % of Na = 2.30g x 100 4.00g = 57.5 % % of O = 1.60g x 100 4.00g = 40.0 % % of H = 0.100g x 100 4.00g = 2.5 %

30. 30 ExampleA sample of an unknown compound with a mass of 0.562g has the following percent compositions: 13.0% C, 2.2% H and 84.8% F. What is the mass of each element? Formula mass = 0.562 % of C = mass C x 100 0.562g = 13.0 % % of H = mass H x 100 0.562g = 2.2 % % of F = mass F x 100 0.562g = 84.8 % Mass C = 0.0731g Mass H = 0.0124g Mass F = 0.477g

31. Calculating Empirical Formula 31 ExampleA compound was analyzed and found to contain 13.5g Ca, 10.8g O, and 0.675g H. What is the empirical formula for the compound? Step 1 Convert the mass of each element into moles and box the smallest mole value 13.5 g Cax 40.08 g Ca 1 mol Ca - - = 0.337 mol Ca 10.8 g Ox 15.9994 g O 1 mol O - - - = 0.675 mol O 0.675 g Hx 1.00797 g H 1 mol H - - - = 0.670 mol H

32. 32 Step 2 Divide each mole value by the smallest mole value and round answers to the nearest whole number 0.337 0.337 mol Ca = 1 mol Ca 0.337 0.675 mol O = 2 mol O 0.337 0.670 mol H = 1.98 = 2 mol H Step 3 Express mole ratio of each element and empirical formula using mole values Mole ratio1 Ca : 2 O : 2 H Empirical FormulaCaO 2 H 2 = Ca(OH) 2

33. 33 Step 4 Check answer by comparing initial percent composition to final percent composition. Total mass = 13.5g + 10.8g + 0.675g = 24.975g % of Ca = 13.5g x 100 24.975g = 54.1 % % of O = 10.8g x 100 24.975g = 43.2 % % of H = 0.675g x 100 24.975g = 2.7 % Initial: Formula mass = 40.08g + 2(15.999) + 2(1.00794g) = 74.094g % of Ca = 40.08g x 100 74.094g = 54.1 % % of O = 31.998g x 100 74.094g = 43.2 % % of H = 2.0159g x 100 74.094g = 2.7 % Final:

34. 34 ExampleDetermine the empirical formula for a compound containing 2.128g Cl and 1.203 g Ca 2.128 g Clx 35.453 g Cl 1 mol Cl - - = 0.060 mol Cl 1.203 g Cax 40.078 g Ca 1 mol Ca - - - = 0.030 mol Ca Step 1 Step 2 0.030 0.060 mol Cl = 2 mol Cl 0.030 0.030 mol Ca = 1 mol Ca Mole ratio1 Ca : 2 Cl Empirical FormulaCaCl 2 Step 3

35. 35 Formula mass = 2.128g + 1.203g = 3.331g % of Cl = 2.128g x 100 3.331g = 63.9 % % of Ca = 1.203g x 100 3.331g = 36.1 % Initial: Formula mass = 2(35.453) + 40.078g = 110.986g % of Cl = 70.906g x 100 110.986g = 63.9 % % of Ca = 40.078g x 100 110.986g = 36.1 % Final: Step 4

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37. Calculating Molecular Formula 37 Example  -carotene, compound in carrots, can be broken down to form vitamin A. The empirical formula for  -carotene is C 5 H 7. The molar mass of  -carotene is 536 g/mol. What is the molecular formula for  -carotene Step 1 Calculate the empirical formula Empirical formula = C 5 H 7 Step 2 Calculate empirical formula mass and divide that value into the molar mass. Empirical Formula Mass = formula mass of empirical formula Empirical Formula Mass = 5(12.01125) + 7(1.00797) = 67.111g molar mass empirical mass 536g 67.111g = 7.987 = 8 =

38. 38 Step 3 Multiply the empirical formula by the ratio of molar mass to empirical formula Empirical Formula C 5 H 7 = C 5 x 8 H 7 x 8 Molecular Formula C 40 H 56 = C 40 H 56

39. 39 ExampleFind the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Step 1 4.90 g Nx 14.007 g N 1 mol N - - = 0.350 mol N 11.2 g Ox 15.999 g O 1 mol O - - - = 0.700 mol O 0.350 0.350 mol N = 1 mol N 0.350 0.700 mol O = 2 mol O Mole ratio1 N : 2 OEmpirical FormulaNO 2

40. 40 Step 2 Empirical Formula Mass = 14.007 + 2(15.999) = 46.005g molar mass empirical mass 92.0g 46.005g = = 2 Step 3Empirical Formula NO 2 = N 1 x 2 O 2 x 2 = N 2 O 4 Molecular Formula N 2 O 4

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42. Calculation Relationships 42 Moles Mass Particles (atoms/molecules/ formula unit) Volume of gases at STP Molar volume = 22.4 L/mol Avagadro’s number 6.022 x 10 23 particles/mol Molar mass / Formula mass or atomic mass