1. 1 Chapter 10 The Mole Chemical Quantities

2. 2 How do you measure how much in Chemistry? How do you measure how much in Chemistry? We count chemical pieces in We count chemical pieces inMOLES Measuring Moles But….our balances measure in grams

3. 3 Moles n 1 mole is 6.02 x 10 23 particles. (think of it like a very large dozen) Ex: 1 doz = ? bricks? pencils ? eggs? Atoms ? eggs? Atoms Ex: 1 mol = ? bricks? pencils ? eggs? atoms ? eggs? atoms n 6.02 x 10 23 is called Avagadro’s number.

4. 4 1 H 1.0079 17 Cl 35.45 92 U 238.0 Moles

5. 5 Measuring Moles n Remember relative atomic mass? amu n the atoms have mass relative to each other. n Therefore if we increase the “size” of the samples (say to 6.02 x 10 23 or 1 mole), this relative “relationship” still exists!! n Therefore the atomic mass on the periodic table is also the mass of 1 mole of those atoms in grams.

6. 6 1 H 1.0079 17 Cl 35.45 92 U 238.0 g g g Atomic Mass or Gram Atomic Mass or Gram Atomic Mass Moles: (Measuring Moles)

7. 7 0.5 mole of C = _____ atoms How many moles are there in 3.01 x 10 23 atoms of zinc ? Moles to Atoms / Atoms to Moles Moles to Atoms / Atoms to Moles 3.01 x 10 23 atoms 6.02 x 10 23 atoms 1 mol = 0.5 mol 0.5 mol 1 mol 6.02 x 10 23 atoms = 3.01 x 10 23 atoms

8. 8 0.335 mole of Pb = _____ atoms How many moles are there in 1.65 x 10 24 atoms of tin ? Moles to Atoms / Atoms to Moles Moles to Atoms / Atoms to Moles 1.65 x 10 24 atoms 6.02 x 10 23 atoms 1 mol = 2.74 mol 0.335 mol 1 mol 6.02 x 10 23 atoms = 2.02 x 10 23 atoms

9. 9 Atomic Mass or Gram Atomic Mass Atomic Mass or Gram Atomic Mass n The mass of 1 mole of an element in grams. 1 mole of C = 55.9 g 55.9 g 1 mole of Fe = 1 mole of tin = 12.0 g 118.7 g l

10. 10 Atomic Mass or Gram Atomic Mass Atomic Mass or Gram Atomic Mass n Remember we can “count things” by determining the mass. 1 mole of oxygen 1 mole of oxygen 24.3 g of Mg = 24.3 g of Mg = 1 mole of Mg 1 mole of Mg 1 mole of Cl 16.0 g of oxygen = 35.5 g of Cl =

11. 11 n How many moles of magnesium in 14.7 g of Mg? How much would 2.34 moles of carbon “weigh”? Moles to Mass (g) / Mass to Moles (g) Moles to Mass (g) / Mass to Moles (g) 2.34 mol12.0 g = 28.1 g 14.7 g1 mol 24.3 g = 0.605 mol 1 mol

12. 12 How many atoms of lithium in 1.00 g of Li? n How much would 3.45 x 10 22 atoms of U “weigh”? Moles; Atoms; Grams Moles; Atoms; Grams 3.45 x 10 22 atoms 6.02 x 10 23 atoms 238.0 g = 13.6 g 1.0 g 6.9 g 6.02 x 10 23 atoms = 8.72 x 10 22 atoms

13. 13 What about compounds? n in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms. n To find the mass of one mole of a compound. –determine the moles of the elements they have. –Find out how much they would weigh. –add them up.

14. 14 Compounds: Formula Mass n The number of grams of 1 mole of a compound. n Ex: Water H 2 O H =2 x1.0g= 2.0g O =1 x16.0g= 16.0g 18.0g 18.0 g/mol

15. 15 Compounds: Molar Mass The generic term for the mass of one mole. The generic term for the mass of one mole. The same as: molecular mass / gram molecular mass formula mass / gram formula mass formula mass / gram formula mass atomic mass / gram atomic mass atomic mass / gram atomic mass

16. 16 Compounds: Gram Molecular Mass CH 4 C =1 x12.0g= 1 2.0g H =4 x 1.0g= 4.0g 16.0 g 16.0 g/mol What is the mass of one mole of CH 4 ?

17. 17 Compounds: Gram Formula Mass Al 2 (SO 4 ) 3 Al =2 x27.0g= 54.0 g S =3 x 32.1g= 96.3 g 342.3 g 342.3 g/mol What is the mass of one mole of Al 2 (SO 4 ) 3 ? O = 12 x 16.0g= 192.0 g 342 g

18. 18 Using Molar Mass* in problems Given the Mass of a compound Find the moles of the compound. Given the Moles of a compound Find the mass of the compound. * molecular mass / gram molecular mass formula mass / gram formula mass formula mass / gram formula mass atomic mass / gram atomic mass atomic mass / gram atomic mass

19. 19 5.69 g1.0 mol g n How many moles in 5.69 g of NaOH? n 2 steps. 1. Find the Molar mass of the compound. 2. Then Convert using dimensional analysis technique. technique. ? Using Molar Mass*

20. 20 1 st Step: NaOH Na =1 x23.0g= 23.0g H =1 x 1.0g= 1.0g 40.0 g 40.0 g/mol What is the molar mass of NaOH? O =1 x16.0g = 16.0g

21. 21 5.69 g1.0 mol = 0.142 mol g n How many moles in 5.69 g of NaOH? 40.0 2 nd Step:

22. 22 1.58 mol 1.0 mol g n What is the mass of 1.58 moles of Cu(ClO 3 ) 2 ? n Remember - 2 steps. 1. Find the Molar mass of the compound. 2. Then Convert using dimensional analysis technique. technique. ? Using Molar Mass*

23. 23 1 st Step: Cu(ClO 3 ) 2 Cu =1 x63.5g= 63.5 g O = 6 x 16.0g= 96.0 g 230.5 g 230.5 g/mol n What is the mass of 1.58 moles of Cu(ClO 3 ) 2 ? Cl =2 x35.5g = 71.0 g

24. 24 1.58 mol 1.0 mol = 364 g What is the mass of 1.58 moles of Cu(ClO 3 ) 2 ? 230.5 g 2 nd Step: = 364.19 g

25. 25 n Many of the chemicals we deal with are gases. n Need to know how many moles of gas we have. n Two things effect the volume of a gas n Temperature and pressure n Must compare at the same temp. and pressure. These are n Standard Conditions Gases and the Mole

26. 26 Standard Temperature and Pressure n Avagadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles. n Standard Temperature and Pressure –abbreviated STP 0ºC and 1 atm pressure 0ºC and 1 atm pressure At STP 1 mole of gas occupies molar volume = 22.4 L** molar volume = 22.4 L**. **at this time : all problems at STP, therefore always use 22.4 L always use 22.4 L

27. 27 1 H 2 1.0079 17 Cl 2 35.45 7 N 2 14.007 g g g Molar Volume Volume Moles of Gases: (Molar Volume) @ STP @ STP

28. 28 1.58 mol 1.0 mol L n What is the volume of 1.58 moles of Nitrogen (N 2 ) gas at STP ? 22.4 Using Molar Volume = 35.4 L = 35.392

29. 29 8.85 L L g n What is the mass of 8.85 liters of Methane (CH 4 ) gas at STP ? 22.4 Using Molar Volume = 6.32 g = 6.3214 C = H = 1 x 4 x 12.0g 1.0g = 12.0 g = 4.0 g 16.0 g 16.0

30. 30 The 7 Diatomic Molecules HONClBrIF H 2 O 2 N 2 Cl 2 Br 2 I 2 F 2 n The molar mass of these molecules will be multiplied by 2 when they’re by themselves. Examples n Oxygen gas = O 2 n 32.0 grams n What’s the mass of 1 mole of chlorine gas? n Remember: chlorine gas = Cl 2 !! Cl 2 = 2 x 35.5g = 71.0 grams

31. 31 Percent Composition n Like all percents Part Part whole whole n Find the mass of each component, –Then: divide by the total mass. n and multiply by 100% x 100 %

32. 32 1 st Step: Fe =2 x55.8g= 111.6g 159.6 g O = 3 x16.0g = 48.0g Percent Composition What is the % composition of of Fe 2 O 3 ? 2 nd Step: Fe 2 O 3 111.6g Fe 159.6 g O 48.0g 159.6 g X 100% Fe = 69.9 %O = 30.1 %

33. 33 1 st Step: Ag 29.0 g 33.3 g S 4.30 g Percent Composition Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S. 2 nd Step: 29.0g Ag 33.3 g S 4.30g 33.3 g X 100% Ag = 87.1 %S = 12.9 %

34. 34 The Empirical Formula n The lowest whole number ratio of elements in a compound. –CH 2 –C2H4–C2H4–C2H4–C2H4 –C3H6–C3H6–C3H6–C3H6 –H2O–H2O–H2O–H2O empirical formula empirical formula no CH 2

35. 35 Empirical Formula From percentage to formula From mass to formula Determine the Molar Ratio. It reflects the ratio of atoms in a molecule.

36. 36 Calculating Empirical Formula (from percent composition) A compound is found to consist of 27.3% carbon and 72.7% oxygen. Determine its empirical formula. CO 72.7%27.3% = 100% 72.7g27.3g = 100g 27.3g72.7g 12.0g/mol16.0g/mol Need a Molar ratio: therefore Convert to moles 2.28 mol 4.54 mol 2.28 mol 1 = 1.991 2 CO 2

37. 37 Calculating Empirical Formula (from mass) A compound is found to consist of 29.0 g of silver and 4.30 g of sulfur. Determine its empirical formula. AgS 4.30g29.0g 29.0g4.30g 107.9g/mol32.1g/mol Need a Molar ratio: therefore Convert to moles 0.269 mol 0.134 mol 2 = 2.00746 1 Ag 2 S 0.134 mol

38. 38 Molecular Formula n The actual whole number ratio of elements in a compound. –CH 2 –C2H4–C2H4–C2H4–C2H4 –C3H6–C3H6–C3H6–C3H6 –H2O–H2O–H2O–H2O empirical formula empirical formula molecular formula (Molecular formulas are not reduced)

39. 39 Calculating Molecular Formulas A compound with a gram formula mass of 120.0 g is found to consist of 40.0% carbon and 53.4% oxygen and 6.60% hydrogen. Determine its molecular formula. CO 53.4%40.0% = 100% = 100g 12.0g/mol 12.0g/mol16.0g/mol 3.34 mol 6.60 mol 3.33 mol 1 = 1.982 2 COH 2 H 6.60% 40.0g53.4g6.60g 1. First find the empirical formula 1.0g/mol 3.33 mol 1

40. 40 COH 2 2. Next find the mass of the empirical formula C O H 1 x 2 x 1 x12.0g 16.0g 1.0g =12.0g =16.0g = 2.0g 30.0g 3. Now divide the gfm by the efm gram formula mass of 120.0g 120.0g 30.0g = 4.0 4. Now multiple the empirical formula by 4 (COH 2 )x4 C4O4H8C4O4H8Calculating Molecular Formulas (cont’) (cont’)

41. 41 Example n A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 99.0 g. What is its molecular formula?