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

2. 2 Section 10.1 The Mole: A Measurement of Matter n n OBJECTIVES: – –Describe methods of measuring the amount of something. – –Define Avogadro’s number as it relates to a mole of a substance. – –Distinguish between the atomic mass of an element and its molar mass. – –Describe how the mass of a mole of a compound is calculated.

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

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

5. 5 Moles (is abbreviated: mol)   It is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon-12.   1 mole = 6.02 x 10 23 of the representative particles.   Treat it like a very large dozen   6.02 x 10 23 is called: Avogadro’s number.

6. 6 Similar Words for an amount   Pair: 1 pair of shoelaces = 2 shoelaces   Dozen: 1 dozen oranges = 12 oranges   Gross: 1 gross of pencils = 144 pencils   Ream: 1 ream of paper = 500 sheets of paper

7. 7 What are Representative Particles?   The smallest pieces of a substance: 1) 1)For a molecular compound: it is the molecule. 2) 2)For an ionic compound: it is the formula unit (made of ions). 3) 3)For an element: it is the atom. » »Remember the 7 diatomic elements? (made of molecules)

8. 8 Types of questions n n How many oxygen atoms in the following? CaCO 3 Al 2 (SO 4 ) 3 n n 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)

9. 9 Measuring Moles   Remember relative atomic mass? - The amu was one twelfth the mass of a carbon-12 atom.   Since the mole is the number of atoms in 12 grams of carbon-12,   the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

10. 10 Gram Atomic Mass (gam)   Equals the mass of 1 mole of an element in grams (from periodic table)   12.01 grams of C has the same number of pieces as 1.01 grams of H and 55.85 grams of Fe.   We can write this as:   12.01 g C = 1 mole C molar mass (this is also the molar mass) (**Remember we can count things by weighing them.)

11. 11 n nitrogen n aluminum n zinc 14.01 g 26.98 g 65.39 g Find the Gram Atomic Mass (gam) of the following:

12. 12 What about compounds? In 1 mole of H 2 O molecules there are 2 moles of H atoms and 1 mole of O atoms (think of a compound as a molar ratio)   To find the mass of one mole of a compound: – –Determine the number of moles of the elements present – –Multiply the number times their mass (from the periodic table) – –Add them up for the total mass

13. 13 Calculating Gram Formula Mass (gfm) Calculate the formula mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3 x (16.00 g)= 84.32 g Thus, 84.32 grams is the formula mass for MgCO 3.

14. 14 GFM Practice Problem: n What is the mass of one mole of CH 4 ? 1 mole of C = 12.01 g 4 mole of H x 1.01 g = 4.04g 1 mole CH 4 = 12.01 + 4.04 = 16.05g

15. 15 Section 10.2 Mole-Mass and Mole-Volume Relationships n n OBJECTIVES: – –Describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass. – –Identify the volume of a quantity of gas at STP.

16. 16 Since Molar Mass is…   The number of grams in 1 mole of atoms, ions, or molecules   We can make conversion factors from these. - To change between grams of a compound and moles of a compound.

17. 17 For example n How many moles is 5.69 g of NaOH?

18. 18 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles for NaOH l 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

19. 19 Practice Problems:   How much would 2.34 moles of carbon weigh?   How many moles of magnesium is 24.31 g of Mg? 28.1 grams C 1.000 mol Mg

20. 20 The Mole-Volume Relationship   Many of the chemicals we deal with are: gases.   Two things effect the volume of a gas: a) Temperature and b) Pressure **We need to compare all gases at the same temperature and pressure.

21. 21 Standard Temperature and Pressure n abbreviated STP n 0ºC (273K) and 1.0 atm pressure n At STP 1 mole of gas occupies 22.4 L= molar volume 1 mole = 22.4 L of any gas at STP

22. 22 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?   What is the volume of 8.8 g of CH 4 gas at STP? = 103 L CO 2 = 0.253 mol O 2

23. 23 Summary: n These four items are all equal: a) 1 mole b) molar mass (in grams/mol) c) 6.02 x 10 23 particles c) 6.02 x 10 23 particles (atoms, molecules, or formula units) d) 22.4 L of a gas at STP **Thus, we can make conversion factors from them.

24. 24 Avog. # Practice problems:   How many molecules of CO 2 are in 4.6 moles of CO 2 ?   How many moles of water is 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 ? 2.8 x 10 24 m.c. CO 2 0.0975 mol H 2 O (or 9.75 x 10 -2 ) 7.405 x 10 23 atoms C 12.9 moles MgCl 2

25. 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? 8.7 x 10 22 atoms Li 13.6 g U 14.0 L CH 4

26. 26 Section 10.3 Percent Composition and Chemical Formulas n n OBJECTIVES: – –Describe how to calculate the percent by mass of an element in a compound. – –Interpret an empirical formula. – –Distinguish between empirical and molecular formulas.

27. 27 Percentage Composition n the percentage by mass of each element in a compound

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

29. 29  100 = Percentage Composition %Cu = 127.10 g Cu 159.17 g Cu 2 S  100 = %S = 32.07 g S 159.17 g Cu 2 S 79.85% Cu 20.15% S n Find the % composition of Cu 2 S.

30. 30  100 = %H 2 O = 36.04 g 147.02 g 24.51% H 2 O n Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? Percentage Composition

31. 31 Empirical Formula (EF) C2H6C2H6C2H6C2H6 CH 3 reduce subscripts n Lowest whole number ratio of atoms in a compound

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

33. 33 Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

34. 34 Formulas (continued) 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)

35. 35 Calculating EF n 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.

36. 36Example n Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. n Assume 100 g so: n 38.67 g C x 1mol C = 3.220 mole C 12.01 gC n 16.22 g H x 1mol H = 16.06 mole H 1.01 gH n 45.11 g N x 1mol N = 3.220 mole N 14.01 gN Now divide each value by the smallest value

37. 37 Example n 3.220 mol C = 1 mol C 3.220 n 16.06 mol H = 5 mol H 3.220 n 3.220 mol N = 1 mol N 3.220 EF = C 1 H 5 N 1 = CH 5 N

38. 38 Empirical Formula (EF) n 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

39. 39 Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers  multiply by 2 N2O5N2O5N2O5N2O5

40. 40 EF Practice Find the empirical formula for a sample of 43.64% P and 56.36% O.

41. 41 Molecular Formula (MF) n “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6C2H6C2H6 empiricalformula molecularformula ?

42. 42 Empirical to Molecular n Since the empirical formula is the lowest ratio, the actual molecule would weigh more.  By a whole number multiple. n **Divide the actual molar mass by the empirical formula mass.

43. 43 Molecular Formula n 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

44. 44 Practice Problem: n Caffeine (EF= C 4 H 5 N 2 O) has a molecular mass of 194 g. What is its molecular formula?

45. 45 Formulas n EF = lowest whole number ratio of elements in a compound. n The molecular formula (MF) = the actual ratio of elements in a compound. n The two can be the same. n CH 2 is an empirical formula n C 2 H 4 is a molecular formula n C 3 H 6 is a molecular formula n H 2 O is both empirical & molecular

46. 46 Density of a gas (10.2 revisited)   D = m / V (density = mass/volume) - for a gas the units will be: g / L   We can determine the density of any gas at STP if we know its formula.   To find the density we need: 1) mass and 2) volume.   If you assume you have 1 mole, then the mass is the molar mass (from periodic table)   And, at STP the volume is 22.4 L.

47. 47 Practice Problems (D=m/V)   Find the density of CO 2 at STP. D = 44.01g/22.4L = 1.96 g/L   Find the density of CH 4 at STP. D = 16.05g/22.4L = 0.717g/L

48. 48 Note page 313 – Gas Chromatography used for chemical analysis

49. 49