1. Chapter 10 “Chemical Quantities” Yes, you will need a calculator for this chapter!

2. Section 10.1 Da Mole: A Measurement of Matter

3. How do we measure items?  Measure mass in.  Measure volume in.  Measure amount in.

4. What is the mole? We’re not talking about this kind of mole!

5. Moles (mol)  It is.  Defined as the number of carbon atoms in exactly 12 grams of carbon-12.  of representative particles.  6.02 x 10 23 is called: number.  1 mole = 6.02 x 10 23 = Avogadro’s number

6. Similar Words for amount Dozen:  1 dozen muffins = muffins  2 dozen puppies = puppies  36 m&m stuffed donuts = dozen donuts Mole :  1 mol muffins = muffins  2 mol puppies = puppies  2.4 x 10 24 m&m stuffed donuts = mol donuts

7. What are Representative Particles?  The smallest pieces of a substance:  based on what we’re looking at 1)For a molecular compound: it is the. 2)For an ionic compound: it is the formula unit (ex: NaCl, MgS). 3)For an element: it is the. 1 mol of CO 2, 1 mol of NaCl, and 1 mol of H equal 6.02 x 10 23 of that thing.

8. Measuring Moles  The mass on the PT is also the mass (g/mol) – mass (in grams) of 1 mole of that atom  =

9. Nitrogen Aluminum Zinc Find the Molar Mass (g/mol) of the following:

10. What about compound mass? In 1 mole of H 2 O, there are moles of H atoms and mole of O atoms  To find the mass of a compound: – Determine of each element present – the number times their mass (from the periodic table) – them up for the

11. Calculating Compound Mass Calculate the mass of magnesium carbonate, MgCO 3.

12. Practice Problem: What is the mass of one mole of CH 4 ? 1 mole of C = 12.01 g/mol 4 mole of H x 1.01 g = 4.04g/mol 1 mole CH 4 =

13. 10.1 Review 1.1 mol = particles 2.2.5 mol MgCl 2 = ? particles MgCl 2 = 3.7.2 x 10 24 Fe particles = ? mol Fe =

14. 4. Molar mass of BeF 2 = 5. Molar mass of C 6 H 12 O 6 = Be = 9.0 g/mol F = 19.0 g/mol C = 12.0 g/mol H = 1.0 g/mol O= 16.0 g/mol

15. Section 10.2 Mole-Mass and Mole-Volume Relationships

16. Since Molar Mass is…  Number of in 1 mole.  grams per mole (g/mol)  Use to make _ factors from these. - Use molar mass to convert to or of a substance.

17. For example How many moles is 5.69 g of NaOH? = How many grams in.53 mol of NaOH? =

18. Practice Problems:  How much would 2.34 moles of carbon weigh?  How many moles of magnesium is 24.31 g of Mg?

19. The Mole-Volume Relationship  Under different circumstances, can.  Two things effect the volume of a gas: a) Temperature and b) **We need to compare all gases at the same temperature and pressure.

20. Standard Temperature and Pressure abbreviated (273K) and pressure At STP 1 mole of gas occupies 22.4 L= molar volume of any gas at STP

21. Practice Problems:  What is the volume of 4.59 mole of CO 2 gas at STP?  How many moles is 5.67 L of O 2 at STP?

22. What is the volume of 8.8 g of CH 4 gas at STP? =

23. Summary: These four items are all : a) 1 mole b) molar mass (in grams/mol) c) particles (atoms, molecules, or formula units) d) of a gas at **Thus, we can make factors from them.

24. Avog. # Practice problems:  How many molecules of CO 2 are in 4.6 moles of CO 2 ?  How many moles of water is in 5.87 x 10 22 molecules?  How many atoms of carbon are in 1.230 moles of Carbon?  How many moles is 7.78 x 10 24 formula units of MgCl 2 ?

25. Mixed Practice Problems:  How many atoms of lithium is 1.0 g of Li?  How much would 3.45 x 10 22 atoms of U weigh?  What is the volume of 10.0 g of CH 4 gas at STP?

26. Section 10.3 Percent Composition and Chemical Formulas

27. Percentage Composition Percentage by of element in a compound

28. Calculating Percent Composition of a Compound  Like all percent problems: part whole 1)Find mass of (from the periodic table) 2)Next, divide by total mass of ; then _ x 100 % = _

29. Percentage Composition Find the % composition of Cu 2 S. %Cu = %S =

30. %H 2 O = Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? Percentage Composition

31. Empirical Formula (EF) whole number of atoms in a compound C2H6C2H6C2H6C2H6 simplify subscripts

32. Empirical Formula (EF)  Just find whole number ratio C 6 H 12 O 6 CH 4 N  Formula is not just ratio of atoms, it is also ratio of.  In 1 mole of CO 2 there is mole of carbon and moles of oxygen.  In one molecule of CO 2 there is 1 atom of C and 2 atoms of O. = already lowest ratio.

33. Formulas (continued) Formulas for compounds are ALWAYS empirical (the lowest whole number ratio = be reduced). *Remember, we simplify beforehand Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

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

35. Formulas Empirical Formula (EF) = lowest whole number ratio of elements in a compound. Molecular Formula (MF) = the actual ratio of elements in a compound. The two can be the same. CH 2 = C 2 H 4 = C 3 H 6 =H 2 O =

36. Calculating EF We can get a ratio from the percent composition. 1.) Assume you have 100 g. -the percentage becomes grams (75.1% = 75.1 grams) 2.) Convert grams to moles. 3.) Find lowest whole number ratio by dividing by the smallest # of moles. **4.) If not a whole #, use a multiplier.

37. Example Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so: 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.06 mole H 1.01 gH 45.11 g N x 1mol N = 3.220 mole N 14.01 gN Now divide each value by the smallest value

38. Example 3.220 mol C = 1 mol C 3.220 16.06 mol H = 5 mol H 3.220 3.220 mol N = 1 mol N 3.220 EF = C 1 H 5 N 1 = CH 5 N

39. Empirical Formula (EF) Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N 74.1 g 1 mol 16.00 g = 4.63 mol O 1.85 mol = 1 N = 2.5 O

40. Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5N2O5N2O5

41. EF Practice Find the empirical formula for a sample of 43.64% P and 56.36% O.

42. Molecular Formula (MF) “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6C2H6C2H6 empiricalformula molecularformula ?

43. Empirical to Molecular Since the empirical formula is the lowest ratio, the actual molecule would weigh more.  By a whole number multiple. **Divide the actual molar mass by the empirical formula mass.

44. Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass is 28.1 g/mol? 28.1 g/mol 14.03 g/mol = 2 empirical mass = 14.03 g/mol (CH 2 ) 2  C 2 H 4

45. Practice Problem: Caffeine (EF= C 4 H 5 N 2 O) has a molecular mass of 194 g. What is its molecular formula?