1. Chapter 10 “Chemical Quantities” Y ou will need a calculator for this chapter!

2. Section 10.1 p. 287 The Mole: A Measurement of Matter

3. How do we measure items? How do we measure items?  You can measure mass,  volume,  or count pieces  We measure mass in grams  We measure volume in liters  We count pieces in MOLES

4. Other Ways to Measure Amount  Pair: 1 pair of socks = 2 socks  Dozen: 1 dozen donuts = 12 donuts  Gross: 1 gross of pencils = 144 pencils (12 dozen)  Ream: 1 ream of paper = 500 sheets of paper Guided Practice Problem p. 289

5. Practice Problem #2 pg. 289 Assume 2.0 kg of apples is 1 dozen and that each apple has 8 seeds. How many apple seeds are in 14 kg of apples? (work INDEPENDENTLY to solve)

6. What is the mole? Not this kind of mole!

7. Moles (abbreviated mol ) Moles (abbreviated mol)  Derived from German word molekül (molecule)  SI measurement of an amount  1 mole = 6.02 x 10 23 of representative particles, or…..  # of carbon atoms in exactly 12 g of Carbon-12 isotope Avogadro’s number  Called Avogadro’s number

8. What are Representative Particles? (Table 10.1 p. 290)  The smallest pieces of a substance: 1)molecular cmpd - molecule 2)ionic cmpd - formula unit (made of ions) 3)element: is the Remember the 7 diatomic elements? (made of molecules) BrINClHOF H2H2 N2N2 O2O2 F2F2 Cl 2 I2I2 Br 2 atom Guided Practice Problem #3 p. 291

9. Mole VideoMole Video 3:49 Mole Video

10. Quick Quiz How big is a mole? If everyone in the world got a mole of pennies, how much $ would every person have? If you stacked a mole of paper how many times would it go from the Earth to the moon? How long would it take for every person in the world to eat through a mole of marshmellows? 6.02 x 10 23 1 trillion bucks $1,000,000,000,000 80 billion times 80,000,000,000 40,000,000 years w/o a bathroom break!

11. Consider these questions: How many oxygen atoms in the following? CaCO 3 Al 2 (SO 4 ) 3 How many ions in the following? CaCl 2 NaOH Al 2 (SO 4 ) 3 3 atoms of oxygen 12 (3 x 4) atoms of oxygen 3 total ions (1 Ca 2+ ion and 2 Cl 1- ions) 2 total ions (1 Na 1+ ion and 1 OH 1- ion) 5 total ions (2 Al 3+ + 3 SO 4 2- ions)

12. Practice problems Practice problems

13. The Mass of a Mole of an Element  Atomic mass of element (mass of 1 atom) expressed in amu -atomic masses - relative masses based on mass of C-12 (12.0 amu) -1 amu is 1/12 mass of C-12 atom

14. Molar Mass…. in grams  = mass of 1 mol of element in grams (periodic table)  12.01 grams C has same # particles as 1.01 g H & 55.85 g Fe  12.01 g C = 1 mol C  1.01 g H = 1 mol H  55.85 g Fe = 1 mol Fe All contain 6.02 x 10 23 atoms

15. Molar Mass Practice Problems Molar Mass Practice Problems

16. What about compounds?  1 mol of H 2 O molecules has 2 mol of H atoms & 1 mol of O atoms (think of a compound as a molar ratio)  To find mass of 1 mol of a cmpd: o determine # moles of elements present o Multiply # times their mass (from periodic table) o add up for total mass

17. Calculating Molar Mass Calculate molar mass of magnesium carbonate, MgCO 3. 24.3 g + 12.0 g + 3 x (16.00 g) = 84.3 g So, 84.3 g = molar mass for MgCO 3

18. Section 10.2 p. 297 Mole-Mass and Mole- Volume Relationships

19. Molar Mass  Molar mass - generic term for mass of 1 mol of any substance (expressed in grams/mol)  Same as: 1) Gram Molecular Mass (for molecules) 2) Gram Formula Mass (ionic compounds) 3) Gram Atomic Mass (for elements) o molar mass is more broad term than these other specific masses

20. Examples  Calculate the molar mass of: Na 2 S N 2 O 4 C Ca(NO 3 ) 2 C 6 H 12 O 6 (NH 4 ) 3 PO 4 = 78.05 g/mol = 92.02 g/mol = 12.01 g/mol = 164.10 g/mol = 180.12 g/mol = 149.12 g/mol

21. Molar Mass is…  # of g in 1 mol of atoms, formula units, or molecules  Make conversion factors from these - To change btwn g of cmpd and mol of cmpd

22. Using the Mole Roadmap  How many moles is 5.69 g of NaOH? 0.142 mol NaOH

23. The Mole-Volume Relationship  gases - hard to determine mass  how many moles of gas?  2 things affect gas V:  a) Temp & b) Pressure  compare all gases at = temp & pressure

24. Standard Temperature and Pressure  0ºC & 1 atm pressure - abbreviated “STP” any  At STP, 1 mol of any gas has V of 22.4 L - Called molar volume  1 mol of any gas at STP = 22.4 L

25. Practice Examples Practice Examples

26. Mole Day Celebrated on October 23 rd from 6:02 am until 6:02 pm (6:02 on 10-23)

27. Density of a gas  D = m / V (density = mass/volume) - for gas units are: g / L  find density of a gas at STP if formula known  You need: 1) mass and 2) volume  Assume 1 mol, so mass is molar mass (from periodic table)  At STP, V = 22.4 L

28. Practice Examples (D=m/V) Practice Examples (D=m/V)

29. Another way:  If given density  If given density, find molar mass of gas  Assume 1 mol at STP, so V = 22.4 L modify: D = m/V to show:  “m” will be mass of 1 mol, given 22.4 L  What is molar mass of a gas with density of 1.964 g/L?  How about a density of 2.86 g/L? = 44.0 g/mol 64.0 g/mol m = D x V

30. Summary all equal: all equal: a) 1 mole b) molar mass (in grams/mol) c) 6.02 x 10 23 representative particles (atoms, molecules, or formula units) d) 22.4 L of gas at STP make conversion factors from these 4 values (p.303)

31. Notice all conversions must go through the MOLE! Copy this conversion map into your notes!

32. Section 10.3 p. 305 Percent Composition and Chemical Formulas

33.  All percent problems: part whole 1)Find mass of each element, 2)Divide by total mass of cmpd; & x 100 x 100 % = percent %mass of element = mass of element x 100% mass of cmpd

34. % composition from mass  Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S. 29.0 g Ag 33.3 g total X 100 = 87.1 % Ag 4.30 g S 33.3 g total X 100 = 12.9 % S Total = 100 %

35. % comp from the chemical formula  If we know formula, assume you have 1 mole,  Subscripts used to calculate mass of each element in 1 mole of cmpd  sum of masses is molar mass

36. % Composition Examples % Composition Examples

37. % composition as conversion factor conversion factor  We can also use % as conversion factor to calculate # grams of element in cmpd  Calculate % C in  Calculate % C in C 3 H 8  What is mass of C in 82.0 g sample of propane (C 3 H 8 ) 67.1 g C

38. % Composition % Composition 4:15

39. What is an Empirical Formula? ratioLike ingredients for recipe – double recipe, you double each ingredient, but ratio of ingredients stays same Empirical formula: lowest whole number ratio of atoms in cmpd

40. Calculating Empirical  Find lowest whole number ratio C 6 H 12 O 6 CH 4 N atoms  A formula is not just ratio of atoms, it is also ratio of moles moleculeatom of C 2 atoms of O  1 molecule of CO 2 = 1 atom of C and 2 atoms of O  1 mol of CO 2 = 1 mol C and 2 mol O = CH 2 O = this is already the lowest ratio.

41. Calculating Empirical  get a ratio from % composition 1)Assume you have a 100 g sample percentagegrams75.1%75.1 grams - the percentage become grams ( 75.1% = 75.1 grams ) gramsmoles 2)Convert grams to moles. moles 3)Find lowest whole number ratio by dividing each # of moles by smallest value

42. Example  Calculate empirical formula of cmpd composed of 38.67 % C, 16.22 % H, and 45.11 %N.  Assume 100 g sample, so  38.67 g C x 1mol C = 3.22 mole C 12.0 g C  16.22 g H x 1mol H = 16.22 mole H 1.0 g H  45.11 g N x 1mol N = 3.22 mole N 14.0 g N Now divide each value by the smallest value CH 5 N

43. Example  The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N  The ratio is 16.22 mol H = 5 mol H 3.22 mol N 1 mol N = C 1 H 5 N 1 which is = CH 5 N

44. Practice Problem 36 p. 310 Practice Problem 36 p. 310

45. What is a Molecular Formula? Molecular formula: true # of atoms of each element in formula of cmpd molecular cmpds only Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Therefore, empirical formula = CH (the lowest whole number ratio)

46. Formulas (continued) ionic compounds ionic compounds ALWAYS empirical (cannot be reduced). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

47. Formulas (continued) molecular compounds Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11 (Correct formula) (Lowest whole number ratio)

48. Empirical to molecular  Since empirical formula is lowest ratio, the actual molecule weighs more Molar mass Empirical formula mass whole # to increase each coefficient in empirical formula =

49. Empirical to molecular practice problem Empirical to molecular practice problem

50. Empirical and Molecular Formulas Empirical and Molecular Formulas 3:29