1. 1 Chapter 10 Chemical Quantities Get ready for some serious finger exercise!

2. 2 10.1 The Mole: 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?   You can measure mass,   or volume,   or you can count pieces.   We measure mass in grams.   We measure volume in liters (or cm 3 ).   We count matter in MOLES.

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

5. 5 Mole (abbreviated mol)   A mole is simply an amount (like a dozen)   It is defined as the number of carbon atoms in exactly 12 g of carbon-12 ( 12 C).   That amount (1 mole) = 6.022 x 10 23 of the representative particles   6.022 x 10 23 = Avogadro’s number.

6. 6 Similar Words for Amounts   Pair: 1 pair of shoes = 2 shoes   Dozen: 1 dozen roses = 12 roses   Ream: 1 ream of paper = 500 sheets of paper

7. 7 Representative Particles?   Representative particles are the smallest “units” of a substance that define it 1) 1)For a molecular compounds: it is the molecule. 2) 2)For an ionic compounds: it is the formula unit (lowest whole-number ratio of the ions). 3) 3)For an element: it is the atom. » »Remember the 7 diatomic elements? (made of molecules) » »Br 2 I 2 N 2 Cl 2 H 2 O 2 F 2

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 Parts of a Whole Mole n Molecules are made of groups of atoms n Ionic compounds are made of ions n The subscripts in the formulas of compounds give the numbers of each type of element –CO 2 has 2 O and 1 C atom per molecule »So one mole of CO 2 has one mole of C atoms and 2 moles of O atoms –CaCl 2 has 2 Cl − and 1 Ca + ion per formula unit »So one mole of CaCl 2 has 2 moles of Cl − and 1 mole Ca + ions n Tricycle analogy –One tricycle has 1 seat, 2 pedals and 3 wheels –So the “formula” would be StPe 2 Wh 3 –How many of each “element” in a dozen tricycles? How many in one mole of tricycles?

10. 10 Practice problems   How many molecules in 4.56 moles of CO 2 ? How many atoms?   How many moles of water is 5.87 x 10 22 molecules?   How many atoms are in 1.23 moles of C 6 H 12 O 6 ?   How many moles is 7.78 x 10 24 formula units of MgCl 2 ?

11. 11 Measuring Moles   Remember the mole is based on the number of C atoms in 12 grams of carbon-12.   1 AMU in atoms = 1 g in moles   This proportion is true for all elements in the periodic table.

12. 12 Gram Atomic Mass   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 1.008 grams of H and 55.85 grams of iron.   We can write this as: 12.01 g C = 1 mole C (this is the molar mass) ( = )   We can count things by weighing them! 12.01 g C1 mole C 1 mole C12.01 g C

13. 13 Practice Problems   What is the mass in grams of 2.34 moles of carbon?   How many moles of magnesium is 24.31 g of Mg?   How many atoms of lithium is 1.00 g of Li?   How much would 3.45 x 10 22 atoms of U weigh?

14. 14 What about compounds?   in 1 mole of H 2 O molecules there are two 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

15. 15 Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO 3. 24.3 g + 12.0 g + 3 x (16.0 g) = 84.3 g the formula mass (or molar mass) for MgCO 3 is 84.3 g/mol

16. 16 More Practice n 1. What is the molar mass of sucrose (C 12 H 22 O 11 )? n 2. What is the molar mass of each of the following compounds? a)phosphorus pentachloride (PCl 5 ) b)uranium hexafluoride (UF 6 ) n 3. Calculate the molar mass of each of the following ionic compounds: a)KMnO 4 b)Ca 3 (PO 4 ) 2 n 4. How many moles is 3.52 x 10 24 molecules of water? n 5. How many atoms of zinc are in 0.60 mol of zinc? n 6. What is the mass of 1.00 mol of oxygen (O 2 )?

17. 17 10.2 Mole-Mass and Mole-Vol 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 molar quantity of gas at STP

18. 18 Molar Mass   Molar mass is the generic term for the mass of one mole of any substance (expressed in grams/mol)   The same as:   1) Gram Molecular Mass (for molecules) 2) Gram Formula Mass (ionic compounds) 3) Gram Atomic Mass (for elements) – – molar mass is a broad term that encompasses all these other specific masses

19. 19 Examples   Calculate the molar mass of the following and tell what the units are for each: 1. 1. Na 2 S 2. 2. N 2 O 4 3. 3. C 4. 4. Ca(NO 3 ) 2 5. 5. C 6 H 12 O 6 6. 6. (NH 4 ) 3 PO 4

20. 20 Molar Mass = Conversion Factor   Molar mass is the number of grams in 1.0 mole of atoms, ions, or molecules   We can make conversion factors from molar masses to change: - -from grams of a substance to moles - -OR, from moles to grams

21. 21 For Example n n How many moles is 5.69 g of Ca(OH) 2 ? – –Want to convert grams to moles, so need a conversion factor with grams in denominator (to cancel grams) and moles in numerator to change the units to moles. 5.69 g 5.69 g Ca(OH) 2 ? mole ? mole Ca(OH) 2 ? g ? g Ca(OH) 2

22. 22 For Example n n Molar masses are equivalent values of moles ↔ grams of a substance l 1mole Ca = 40.1 g x 1 = 40.1 g l 1 mol O = 16.0 g x 2 = 32.0 g l 1 mole H = 1.0 g x 2 = 2.0 g 1 mole Ca(OH) 2 = 74.1 g 1 mol Ca(OH) 2 74.1 g Ca(OH) 2 74.1 g Ca(OH) 2 1 mol Ca(OH) 2 = 2 O and 2 H atoms per formula unit of Ca(OH) 2

23. 23 For Example Ca(OH) 2 n How many moles is 5.69 g of Ca(OH) 2 ? –Looking to convert grams to moles, so need a conversion factor with grams in denominator (to cancel grams) and moles in numerator to change the units to moles. 5.69 g Ca(OH) 2 1 mol Ca(OH) 2 74.1 g Ca(OH) 2

24. 24 For Example Ca(OH) 2 n How many moles is 5.69 g of Ca(OH) 2 ? –Looking to convert grams to moles, so need a conversion factor with grams in denominator (to cancel grams) and moles in numerator to change the units to moles. 5.69 g Ca(OH) 2 1 mol Ca(OH) 2 74.1 g Ca(OH) 2 = 0.077 mol Ca(OH) 2

25. 25 The Mole-Volume Relationship   Many chemicals we’ll deal with are gases - difficult to measure their masses   But, we will still need to know how many moles of gas we have   we must know how many particles (atoms, ion, molecules) are taking part in chemical reactions   Easy to measure the volume of a gas, but 2 things effect volumes of molar amounts of gases: a) Temperature and b) Pressure   If we compare all gases at the same T and P they all occupy the approximately same volume

26. 26 Standard Temperature & Pressure (STP)  =  = 0ºC (273 K) and 1 atm pressure   At STP, 1 mole of any gas occupies a volume of 22.4 L - Called the molar volume   This equality can also be a conversion factor   1 mole of any gas at STP = 22.4 L 1 mole gas 22.4 L 22.4 L 1 mole gas

27. 27 Practice Examples   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?

28. 28 Density of a gas   Density = mass / volume   for a gas the units are usually g / L   We can determine the density of any gas at STP if we know its formula.   Use formula to calculate molar mass (= grams in 1 mole of gas)   Divide molar mass by molar volume (all gases are 22.4 L / mole) Mass of 1 mole in grams Volume of 1 mole in L = grams in 1 L = density!

29. 29 Practice Examples (D=m/V)   Find the density of CO 2 at STP   Find the density of CH 4 at STP

30. 30 Another way:   If given the density, we can find molar mass of a gas.   Again, “pretend” you have 1 mole at STP, so V = 22.4 L. D = m/V so…   “m” = mass of 1 mole m = D x V

31. 31 Using Density to Find Molar Mass   What is the molar mass of a gas with a density of 1.964 g/L at STP?   How about a density of 2.86 g/L at STP?

32. 32 Summary n n These four items are all equal a) 1 mole b) molar mass (in grams) c) 6.02 x 10 23 representative particles (atoms, molecules, or formula units) d) 22.4 L of gas at STP Thus, we can make conversion factors from these 4 values!

33. 33 Practice Problems 1. What is the molar mass of each of the following compounds? a. C 6 H 12 O 6 b. NaHCO 3 c. C 7 H 12 d. KNH 4 SO 4 2. Calculate the mass in grams of each of the following: a. 8.0 mol lead oxide (PbO) b. 0.75 mol hydrogen sulfide (H 2 S) c. 1.50 x 10 -2 mol oxygen (O 2 ) d. 2.30 mol ethylene glycol (C 2 H 6 O 2 ) 3. How many grams are in 1.73 mol of dinitrogen pentoxide (N 2 O 5 )? 4. How many grams are in 0.66 mol of calcium phosphate [Ca 3 (PO 4 ) 2 ]? 5. Calculate the number of moles in each of the following: a. 0.50 g sodium bromide (NaBr) b. 13.5 g magnesium nitrate [Mg(NO 3 ) 2 ] c. 0.0010 g chloromethane (CH 3 Cl) d. 1.02 g MgCl 2

34. 34 More Practice Problems 6. A chemist plans to use 435.0 grams of ammonium nitrate (NH 4 NO 3 ) in a reaction. How many moles of the compound is this? 7. A solution is to be prepared in a laboratory. The solution requires 0.0465 mol of quinine (C 20 H 24 N 2 O 2 ). What mass, in grams, should the laboratory technician obtain in order to make the solution? 8. What is the volume at STP of 2.66 mol of methane (CH 4 ) gas? 9. How many moles is 135 L of ammonia (NH 3 ) gas at STP? 10. What is the density of carbon dioxide (CO 2 ) gas at STP? 11. What is the molar mass of ethene (C 2 H 4 ) if its density at STP is 1.25 g/L?

35. 35 10.3: % Composition, 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

36. 36 Calculating Percent Composition   Like all percent problems: part whole 1) 1)Find the mass of each of the components (the elements), 2) 2)Next, divide by the total mass of the compound; then x 100 x 100 % = percent

37. 37 Example   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 %

38. 38 Using Formulae to Calculate % Mass   assume you have 1 mole of the compound…   If we know the formula we know the mass of the elements and the whole compound (from the periodic table!).

39. 39 Practice Problems   Calculate the percent composition of C2H4C2H4 Aluminum carbonate   Sample Problem 10.10, p.307   We can also use the percent composition as a conversion factor (see p. 308)

40. 40 n Read the explanation at left. n Use that information to help you complete the practice problems below.

41. 41 Empirical and Molecular Formulas Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) Therefore, the empirical formula = (the lowest whole number ratio) Therefore, the empirical formula = CH (the lowest whole number ratio) Empirical formula: the lowest whole number ratio of atoms/ions in a compound Molecular formula: the true number of atoms of each element in the molecules of a molecular compound.

42. 42 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

43. 43 Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 8 H 18 Empirical: H2OH2O CH 2 OC4H9C4H9 Water (Lowest whole number ratio) GlucoseOctane

44. 44 Determining Empirical Formulas   Just find the lowest whole number ratio C 6 H 12 O 6 CH 2 Cl 2   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 = already the lowest ratio

45. 45 Calculating Empirical Formulas Can be calculated from the percent composition because the sum of the parts always = 100% 1) 1)Assume you have a 100 g sample - the percentage become grams (75.1% = 75.1 grams) 2) 2)Convert grams to moles. 3) 3)Find lowest whole number ratio by dividing each number of moles by the smallest value.

46. 46 Example   Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. 1) 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 2) Now divide each value by the smallest value

47. 47 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 = CH 5 N

48. 48 Practice Problem   Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

49. 49 Empirical to Molecular Formula   An empirical formula is the lowest ratio of atoms in the compound, the actual molecule may have more atoms.   By a whole number multiple.   For this type of problem, you are usually given the data to determine an empirical formula, then given a molar mass of the compound.   Divide the compound’s actual molar mass by the empirical formula mass   yielding a whole number to increase each coefficient in the empirical formula   Caffeine has a molar mass of 194 g. what is its molecular formula?

50. 50 Empirical to Molecular Formula   Caffeine has a molar mass of 194 g. what is its molecular formula? n n Recall the empirical formula for caffeine from the previous example problem: C 4 H 5 N 2 O Find empirical formula mass C = 4 x 12 g = 48 g/mol H = 5 x 1 g/mol = 5 g/mol N = 2 x 14 g/mol = 28 g/mol O = 1 x 16 g/mol = 16 g/mol Molar mass C 4 H 5 N 2 O= 97 g/mol Molecular mass = 194 g EmpForm mass = 97 g = 2

51. 51 Video Homework Remember, every time you watch a video for chemistry, you are expected to take notes and complete all practice problems pointed out in the lecture. You are also expected to THINK. From now on, you are also expected to bring to class one intelligent question to ask me about the lecture.