2. Unit 5 Counting Atoms 1, 2, 3, 4, ….. 10, 11, ……… 1000, 1001, 1002…………

3. Counting Suppose I asked you to count all the peanuts in this bag? But what if I knew that 100 peanuts had a mass of 5 grams – can you think of another way to estimate the number of peanuts in the sample?

4. Connecting Mass to Number of Particles One of the greatest challenges early chemists faced was trying to find a way to connect the mass of a substance to the number of particles in the sample. Recall it was determined that particles combined in fixed ratios by weight.

5. Connecting Mass to Number of Particles This led Dalton to the “atomic model” of matter Example: The mass ratio of oxygen to hydrogen in water is 8:1 This does not tell us how many atoms of each element are involved It could tell us this if we knew the relative mass of each kind of atom

6. Relative Mass To assign relative masses to elements it is necessary to know that the samples being compared have the same number of particles

7. Relative Mass Which has more balls in it: a bucket of golf balls or an identical bucket of baseballs? If this is true in the macroscopic world, why wouldn’t it be true in the sub- microscopic one?

8. Relative Mass Was iron more dense than aluminum because iron had more particles per given volume than aluminum or because iron’s individual particles were more massive than aluminum’s? Could it be some combination of both?

9. Relative Mass The truth is, based on the experiments we conducted earlier in the year, we couldn’t say which was true. Dalton did not know what was true during his time either. Since the mass of individual atoms could not be determined, a system of atomic masses had to be determined by comparison.

10. Relative Mass To determine a system of masses by comparison, one element would have to be chosen as the basis of comparison for all others Dalton chose hydrogen and assigned it a mass of 1.

11. Relative Mass To find the mass of another element like oxygen: Compare the masses of equal number of oxygen and hydrogen atoms OR Find the combining masses of oxygen and hydrogen in water

12. Avogadro’s Hypothesis Remember..Avogadro assumed: Equal volumes of gases have equal numbers of molecules These molecules can be split during chemical reactions That molecules of elemental gases could contain more than a single atom

13. Avogadro’s Hypothesis If we accept Avogadro’s Hypothesis, we can compare the mass of various gases and deduce the relative mass of the molecules To do this, we pick a weighable amount of the lightest element (how about 1.0 g?) then use mass ratios to assign atomic masses to the other elements

14. Implications If two volumes of hydrogen combine with one volume of oxygen gas, it is reasonable to assume that two molecules of hydrogen are reacting with each molecule of oxygen

15. U5 worksheet 1 Back in Unit 1 we found that iron was more dense than aluminum. Two possible models arose to account for this difference: 1.The masses of Al and Fe atoms are about the same, but there are more atoms of Fe than atoms of Al in each cm 3 sample. 2.One cm 3 samples of Fe and Al contain about the same number of atoms, but the Fe atoms are more massive. 3.Another possibility – that both the size and the mass of the atoms of these two elements were different – also came up. At the time, we did not have enough evidence to make a decision about these possible models.

16. U5W1 cont’d While the reason for density variation between particle types is difficult to determine for liquids and solids, a conclusion can be reached more easily for gases due to the fact that particles in a gas are widely spaced. This means that particle size does not have an effect on the volume that a given number of gaseous particles occupy.

17. Relative mass activity vials containing several different items pretend that the items in the vials are enlarged particles of various gases. Since the vials are the same size, we will assume that each contains the same number of particles. Your job is to determine the relative mass of each kind of item in the vial, without opening the vial to examine single items.

18. Comparison of Counting Methods P vs. n Lab to Avogadro’s Hypothesis Our Previous Way of Counting Particles Avogadro’s Way of Counting Particles Assumptions Equal volumes of air in an open syringe contain equal numbers of particles (based on density – equal vol – equal mass, then equal # of particles) Equal volumes of any gas contain equal numbers of gas molecules. Comparing masses of gas with the same volume (therefore, count) gives us relative masses of substances. Unit Definition Volume of 1 syringe-ful at constant density = 1 ‘puff’’ Actual count unknown; must be measured indirectly using volume 2 grams of hydrogen gas = 1 ‘mole’ of gas molecules (mole = ‘lump’) Actual count unknown; must be measured indirectly using volume and mass.

19. Relative Mass From Compounds Measurements of the density of the gaseous phase of many of the elements would be difficult, if not impossible. However, we are going to see that chemists could use another tool – the percent composition of compounds to determine molar masses. Many substances combine with oxygen to form a type of compound called an oxide. such combinations often occur in multiple proportions.

20. U5W1 cont’d 12.49 g 75.01 g 87.51 g 348.8 g 1250. g 1349 g

21. Relative masses Assuming that the elements combine in a 1:1 ratio in these compounds, we now have relative masses for these elements. One could conclude that each atom of oxygen is times as heavy as an atom of hydrogen. However, you have seen evidence that two atoms of hydrogen combine with one atom of oxygen. This makes the relative mass of oxygen twice as great as this value.

22. Molar mass determinations We can then set up simple proportions to determine the molar mass of the other elements 16.0 g oxygen x 75.01 g C=12.0 g C 100 g O Therefore the molar mass of carbon would be 12.0 g

23. U5W1 cont’d

24. The Mole The word chosen to represent the standard weighable amount of stuff, the mole, comes from the Latin “mole cula” or little lump

25. Mole A mole of any element is defined as: the amount of the element that contains as many atoms as there are in exactly 12 g of the carbon-12 isotope.

26. Avogadro’s Number A mole was found experimentally to be equal to 6.022 X 10 23 atoms of C-12, which is called Avogadro’s Number (N A ). 12 g Carbon-12 = 1 mol Carbon-12 atoms = 6.022 X 10 23 Carbon-12 atoms

27. The MOLE (mol) The molar mass of any substance is the mass of one mole of that substance. Molar mass is numerically equal to the atomic mass of the substance. A mole of any substance contains Avogadro’s number of units of that substance (6.022 X 10 23 units).

28. Mass 2 ways to measure mass: Atomic Mass Units (amu) The atomic mass of an element is the mass (in amus) as shown on the periodic table for an atom. It is a average relative mass. C = 12.01 amu

29. Average Atomic Mass of Elements ELEMENTAVERAGE ATOMIC MASS (amu) AVERAGE ATOMIC MASS, rounded (amu) Hydrogen1.007941.01 Carbon12.011112.01 Oxygen15.999416.00 Chlorine35.45335.45 Iron55.84755.85

30. Formula Mass Formula Mass: the sum of the atomic masses of all atoms in one compound. Water (H 2 O) has 2 Hydrogen atoms and 1 Oxygen atom. Formula Mass of H 2 O is H 2 X 1.01 amu = 2.02 amu O 1 X 16.00 amu = 16.00 amu 18.02 amu

31. Your turn…..Formula Mass What is the formula mass of methane, CH 4 ? Of NaCl? Of ammonia, NH 3 ? Of glucose, C 6 H 12 O 6 ?

32. Molar Mass Molar Mass (grams per mole) The mass of 1 mol of a compound. SO 3 = 1S + 3O = 32.10 g + 3(16.00 g) = 80.10 g CaI 2 = 1 Ca + 2 I = 40.10 g + 2(126.90 g) = 293.00 g

33. Molar mass and formula mass 1 atom of Carbon weighs 12.01 amus (formula mass) 1 mole of Carbon weights 12.01 grams (molar mass) 1 molecule of H 2 0 weighs 18.02 amus (formula mass) 1 mole of H 2 0 weighs 18.02 grams (molar mass) Same numerically but different units!!!!!

34. What’s in a Mole? Atoms – one mole of an element contains 6.022 X 10 23 atoms of the element. Molecules – one mole of a molecular (covalent) compound contains 6.022 X 10 23 molecules. Gizmos – one mole of gizmos contains 6.022 X 10 23 gizmos. Anything – one mole of anything contains 6.022 X 10 23 anythings! N A (6.022 X 10 23 ) is very practical for counting small particles, especially things like atoms, ions and molecules.

35. EXAMPLES How many pens in 1 mole of pens? How many atoms in 63.546g of Cu? How many atoms in 6.3546g of Cu?

36. EXAMPLES How many molecules in 1 mole of sugar (C 6 H 12 O 6 )? How many molecules in 10 moles of sugar? How many carbon atoms in 1 mole of sugar? How many oxygen atoms?

37. Solving ‘Mole Problems” MOLES MASS Molar mass

38. Mole Conversions by Factor Label Method Mass and Moles Use the molar mass of the substance. 1 Mole = n grams of the substance, so have 2 conversion factors: 1 moleorn grams n grams 1 mole

39. Mole Conversions by Factor Label How many moles in 75.0 g of iron? Example A: How many moles in 75.0 g of iron? How many grams in 0.250 mol Na? Example B: How many grams in 0.250 mol Na? 75.0g Fe X 1 mol Fe 55.85g Fe = 1.34 mol Fe 0.250 mol Na X 22.99g Na 1 mol Na = 5.75g Na

40. Solving ‘Mole Problems” MOLES PARTICLES Number of Particles in 1 mole (6.02 X 10 23 )

41. Mole Conversions by Factor Label Particles and Moles Use Avogadro’s Number (6.022 X 10 23 ) of particles. (6.022 X 10 23 )particles  1 mol 1 mole (6.022 X 10 23 )particles Again, set up Factor Labels to cancel units!

42. Mole Conversions by Factor Label Method = 0.697 mol CO 2 How many atoms in 0.25 mol Na? Example 1: How many atoms in 0.25 mol Na? 0.250 mol Na X 6.022 X10 23 atoms Na mol Na 1 mol Na = 1.51 X 10 23 atoms Na Example D: How many moles in 4.20 X 10 23 molecules of CO 2 ? 4.20 X 10 23 Molecules CO 2 X 1 mol CO 2 6.022 X 10 23 molecules CO 2

43. Mole is the bridge… “How much” “How many” Mass Number of particles MOLEBridge

44. Summary: Solving ‘Mole Problems” using the Mole Map MOLES MASS Molar mass PARTICLES Number of Particles in 1 mole (6.02 X 10 23 )

45. Summary of Mole Conversions by Factor Label Method Mass and Moles Use the molar mass of the substance. Particles and Moles Use Avogadro’s Number (6.02 X 10 23 ) of particles. Set up Factor Labels to cancel units! DON’T GET LAZY! Include labels to ensure that ALL units cancel correctly. Multi-step conversions are easily done if you are careful with the labels!

46. The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100%. Ex: CH 4 C 1 x 12.00g=12.00 g H 4 x 1.01g= 4.04 g total 16.04 g %C = 12.00g/16.04g X 100 = 74.8% %H = 4.04g/16.04g x 100 = 25.2% Percentage Composition

47. Example 2 2.45 g aluminum oxide decomposes into 1.30 g aluminum & 1.15 g oxygen. What is the percentage composition? %O = 1.15g O X 100% = 46.9% O 2.45g Al Oxide note: Oxygen (O, not O 2 ) %Al = 1.30g Al X 100% = 53.1% Al 2.45g Al Oxide As a check, note that 46.9% + 53.1% = 100.0%. Percentage Composition

48. Determining Empirical Formulas Empirical Formula The formula that gives the simplest whole number ratio of the atoms of the elements in the formula. Example 1 What is the empirical formula of a compound containing 2.644g of gold and 0.476g of chlorine? 0.476g Cl X 1 mol Cl = 0.0134 mol Cl 35.45g Cl 2.644g of Au X 1 mol Au = 0.0134 mol Au 196.97g Au Empirical formula = Au 0.0134 Cl 0.0134 or AuCl

49. Determining Empirical Formulas Example 2 What is the empirical formula of a compound with 5.75 g Na, 3.50 g N & 12.00 g O? First, find the mole amounts. 5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na 3.50 g N X (1 mol N/14.01g N) = 0.250 mol N 12.00g O X (1 mol O/16.00g O) = 0.750 mol O Empirical formula = Na 0.250 N 0.250 O 0.750 Divide each mole quantity by the smallest to get whole numbers. (0.250 in this case) Empirical formula = NaNO 3

50. Determining Molecular Formulas Molecular Formula The formula that gives the actual number of atoms of each element in a molecular compound. Example Hydrogen peroxide has a molar mass of 34.00 g/mol and a chemical composition of 5.90% H & 94.1% O. What is its molecular formula?

51. Determining Molecular Formulas 1.First, find the empirical formula, assuming the percents are mass. 5.90gH X (1 mol H/1.01g H) = 5.84 mol H 94.1g O X (1 mol O/16.00g O) = 5.88 mol O Empirical formula = H 5.84 O 5.88 or HO. 2.Find the molar mass of the empirical formula = 17.01g/mol 3. Divide by molar mass of the empirical formula 34.00/17.01 = 2 4. Multiply the empirical formula subscripts by this number: = 2(HO) or H 2 O 2

52. The story so far…. From Avogadro’s Hypothesis we are able to count molecules by weighing macroscopic samples. For gases at the same temperature and pressure we can deduce the following: 1. From combining volumes we can determine the ratio in which molecules react. 2. From masses of these gases we can determine the relative mass of individual molecules.

53. The story so far…. From these results it is possible to determine: the molar masses of the elements; using these masses and formulas of compounds, one can determine molar masses of compounds. One can also determine empirical formulas and molecular formulas These tools allow one to relate “how much stuff” to “how many particles”.