1. Agenda 2/12/2015 5 A Day Slip Quiz The Mole is a number – and so much more! Practice Problems from 11.1 Study Guide – self-check and grade Hydrated crystals eg. CuSO 4 ·5H 2 O Empirical and Molecular formulas Work to do at home

2. 5A Day Slip quiz Write the correct formulas (including states) for the following substances – use your Naming flow chart – 3 mins then turn in – I will grade today. 1.Hydrochloric acid 2.Nitric acid 3.Phosphoric acid 4.Sulfuric acid 5.Hydrogen chloride (gas)

3. Section 11.1 Measuring Matter Counting Particles Practice problems 1.Determine the number of atoms in 2.50 mol Zn 2.50 mol x 6.02 x 10 23 atoms = 1 mol = 15.05 x 10 23 atoms = 1.51 x 10 24 atoms

4. 2. Given 3.25 mol AgNO 3 determine the number of formula units (because it’s an ionic compound, there are no molecules, remember). 3.25 mol x 6.02 x 10 23 formula units = 1 mol 19.565 x 10 23 formula units = 1.96 x 10 24 formula units

5. 3. Calculate the number of molecules in 11.5 mol H 2 O. 11.5 mol x 6.02 x 10 23 molecules = 1 mol 69.23 x 10 23 molecules = 6.92 x 10 24 molecules

6. 4. How many moles contain each of the following? a)5.75 x 10 24 atoms Al x 1mol 6.02 x 10 23 atoms = 57.5 x 1mol = 9.53488 mol 6.02 = 9.53 mol

7. 4. How many moles contain each of the following? b)3.75 x 10 24 molecules CO 2 x 1 mol. 6.02 x 10 23 molecules = 37.5 x 1mol = 6.2292 mol 6.02 = 6.23 mol

8. 4. How many moles contain each of the following? c)3.58 x 10 23 formula units ZnCl 2 x 1 mol. 6.02 x 10 23 formula units = 3.58 x 1mol = 0.59468 mol 6.02 = 0.595 mol

9. 4. How many moles contain each of the following? d)2.50 x 10 20 atoms Fe x 1 mol. 6.02 x 10 23 atoms = 2.50 x 1mol = 0.41528 x 10 -3 mol 6.02 x 10 3 = 4.15 x 10 -4 mol

10. Chapter 11. Study Guide for Content Mastery The Mole Section 11.1 Measuring Matter Smallest pair (2) 5 dozen (12) gross (12 x 12 = 144) 200 ream (500 sheets paper)

11. Section 11.1 Measuring Matter 200 ream (500 sheets paper) 6 000 000 000 (6 x 10 9) 0.5 mol ( 3.01 x 10 23 ) 1 mol (6.02 x 10 23 ) Largestfour moles (2.40 x 10 24 )

12. Section 11.1 Measuring Matter 11. 1.20 mol Cu x 6.02 x 10 23 Cu atoms 1 mol Cu = 7.22 x 10 23 Cu atoms 12. 9.25 x 10 23 molecules CH 4 x 1 mol CH 4 6.02 x 10 23 molecules CH 4 = 1.54 x 10 -1 mol CH 4

13. Section 11.1 Measuring Matter 13. 1.54 x 10 26 atoms Xe x 1 mol Xe _ 6.02 x 10 23 atoms Xe = 2.56 x 10 2 mol Xe 14. 3.01 mol F 2 x 6.02 x 10 23 molecules F 2 1 mol F 2 = 1.81 x 10 24 molecules F 2 15 poi nts

14. Section 11.2 Mass and the Mole 1. The isotope hydrogen-1 is the standard used for the relative scale of atomic masses. Carbon-12 2. The mass of an aotm of helium-4 is 4 amu. 3.The mass of a mole of hydrogen atoms is 1.00 x 10 23 amu 1.01g 4. The mass in grams of one mole of any pure substance is called its molar mass. Units g/mol VIP lear n

15. Section 11.2 Mass and the Mole 5. The atomic masses recorded on the periodic table are weighted averages of the masses of all the naturally occurring isotopes of each element. 6. The molar mass of any element is numerically equal to its atomic mass in grams. 7. The molar mass unit is mol/g g/mol

16. Section 11.2 Mass and the Mole 8. If the measured mass of an element is numerically equal to its molar mass, then you have indirectly counted 6.02 x 10 23 atoms of the element in the measurement. (One mole of the element.)

17. Section 11.2 Mass and the Mole 9. Number of moles in 23.0g of zinc 23g x 1 mole zinc mass 1 mole zinc 23g x 1mole zincb. 65.4g Zn

18. Section 11.2 Mass and the Mole 10. Find the mass of 5.0 x 10 20 zinc atoms 5.0 x 10 20 atoms Zn x 1 mol Zn 6.02 x 10 23 atoms then x 64.5g Znd then a. 1mol Zn

19. Section 11.2 Mass and the Mole 11. Find the mass of 2.00 moles of zinc. 2.00 moles Zn x 64.5 g Zn 1 mole Zn a.

20. Section 11.2 Mass and the Mole 12. Find the number of atoms in 7.40 g of zinc. 7.40g Zn x 1 mole Zn x 6.02 x 10 23 atoms 64.5g Zn 1moleZn b then c.

21. Section 11.2 Mass and the Mole 13. Find the number of moles that contain 4.24 x 10 23 of zinc atoms. 4.24 x 10 23 Zn atoms x 1 mole Zn 6.02x10 23 atoms Zn d.

22. Section 11.2 Mass and the Mole 14. Find the number of atoms in 3.25 moles of zinc. 3.25 moles Zn x 6.02 x 10 23 atoms 1moleZn c. 15 points for the page

23. Section 11.3 Moles of Compounds 1.A molecule of methane contains 1 carbon atom, 4 hydrogen atoms. CH 4 ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

24. Section 11.3 Moles of Compounds 2. A molecule of trichloromethane contains 1 carbon atom, 1 hydrogen atom, and 3 chlorine atoms. CHCl 3 ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

25. Section 11.3 Moles of Compounds 3. How many moles of each element are in a mole of methane? CH 4 1 mole of carbon, 4 moles of hydrogen ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

26. Section 11.3 Moles of Compounds 4. How many moles of each element are in a mole of trichloromethane? CHCl 3 1 mole of carbon, 1 mole of hydrogen, 3 moles of chlorine ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

27. Section 11.3 Moles of Compounds 5. The number of carbon atoms in one mole of methane is 6.02 x 10 23. ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

28. Section 11.3 Moles of Compounds 6. The number of chlorine atoms in one mole of trichloromethane is 3(6.02 x 10 23 ) = 1.81 x 10 24. ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

29. Section 11.3 Moles of Compounds 7. The molar mass of methane is 12.01g/mol + 4(1.01)g/mol = 16.05g/mol ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

30. Section 11.3 Moles of Compounds 8. Chloromethane (CH 3 Cl) has a molar mass of 50.49g/mol. (12.01 + 3(1.01) + 35.45) ElementMolar mass (g/mol) hydrogen1.01 carbon12.01 chlorine35.45

31. Section 11.3 Moles of Compounds 8. Chloromethane (CH 3 Cl) has a molar mass of 50.49g/mol. (12.01 + 3(1.01) + 35.45) Number of molecules of CH 3 Cl in 101 grams of the substance? 101g x 1 mol x 6.02 x 10 23 molecules = 50.49g 1mol = 1.20 x 10 24 molecules 10 points for the page

32. Molar Mass So if 1 mole of Carbon-12 atoms is exactly 12 gram, 1 mole of any element on the periodic table will be its average atomic mass in grams. Molar mass (or mass of one mole of atoms) of any element is numerically equal to its atomic mass and has the units grams per mole (g/mol)

33. If the atomic mass value in grams is the mass of one mole of an element, then the sum of the average atomic masses in grams for all the atoms in a compound will be the mass of one mole of the compound. Example H 2 O Atomic mass H 1 amu Molar mass H 1g/mol Therefore Molar mass 2 H 2g/mol Molar mass O 16g/mol For mass of one mole of an element, look up atomic mass, then add grams as the unit

34. If the atomic mass value in grams is the mass of one mole of an element, then the sum of the average atomic masses in grams for all the atoms in a compound will be the mass of one mole of the compound. Example H 2 O Atomic mass H 1 amu Molar mass H 1g/mol Therefore Molar mass 2 H 2g/mol Molar mass O 16g/mol Molar mass H 2 O (2 + 16)g/mol= 18g/mol (or 1 mole of H 2 O has a mass of 18g) Look up average atomic mass for each element Convert to molar mass units Add everything up Law of conservation of mass

35. Stuck at home? Try signing up for a free account at http://education-portal.com/ Type in “the Mole” and there is a clearly explained video going over this again.

36. Empirical and Molecular Formulas Percent composition Mass of elementx 100 = percent by mass Mass of compound The percent by mass of each element in a compound is called the percent composition of the compound.

37. Empirical and Molecular Formulas Empirical formula Formula with the smallest whole number ratio of the elements. Empirical formula may be same as molecular formula like water H 2 O Or it may not be: hydrogen peroxide emprical formula is OH and molecular formula H 2 O 2

38. Calculating Empirical formula From masses or percent compositions Compound is 40% S 60% O So if we had 100g of compound would be 40g S and 60g O

39. Calculating Empirical formula From masses or percent compositions Compound is 40% S 60% O So if we had 100g of compound would be 40g S and 60g O Convert to moles 40g S x 1 mole = 40g S x 1 mole mass 1 mole S 32g

40. Calculating Empirical formula From masses or percent compositions Compound is 40% S 60% O So if we had 100g of compound would be 40g S and 60g O Convert to moles 40g S x 1 mole = 40g S x 1 mole = 1.25 mol S mass 1 mole S 32g

41. Calculating Empirical formula 40g S and 60g O Convert to moles 40g S x 1 mole = 40g S x 1 mole = 1.25 mol S mass 1 mole S 32g Convert to moles 60g O x 1 mole = 60g O x 1 mole = 3.75 mol O mass 1 mole O 16g

42. Calculating Empirical formula 40g S and 60g O Now convert to simplest ratio: 1.25 mol S = 3.75 mol O = 1.25

43. Calculating Empirical formula 40g S and 60g O Now convert to simplest ratio: 1.25 mol S = 1 mol S 3.75 mol O = 3 mol O 1.25 Simplest whole number mole ratio is of S to O is 1:3 Empirical formula SO 3

44. Calculating Empirical formula 40g S and 60g O Now convert to simplest ratio: 1.25 mol S = 1 mol S 3.75 mol O = 3 mol O 1.25 Simplest whole number mole ratio is of S to O is 1:3 Empirical formula SO 3 In cases where the calculated mole values are not whole numbers, multiply by smallest factor that will make them whole numbers. (See Example Problem 11-11. Page 332)

45. Calculating Molecular formula Start from calculated Empirical formula And Molar mass (found by experimentation). Experimentally determined molar mass = n Mass of empirical formula Then n(empirical formula) = molecular formula

46. Work to do at home Written work due on: 18 Feb, Wednesday Read textbook and complete Study Guide page 64 and 65 (not qu. 10) Section 11.4 Empirical and Molecular Formulas Read textbook and complete new page 66, Section 11.5 The formula for a hydrate Mole as a number of particles practice problems (10) And Grams and the Mole – Practice problems (14) www.chemistrymc.comwww.chemistrymc.com Chapter 11 test – if you are looking for a challenge / thinking of doing AP Chem. (Post to Google Classroom)