3.1-3.3 Macroscopic versus Microscopic

1. Chemical symbols Chemical equations moles Balancing chemical equations 3.1-3.3Macroscopic versus Microscopic I II Balanced rxn is lowest ratio III IV

2. I. What chemical symbols mean big, thick wall macroscopic microscopic phase Bond angles Bond lengths temperature density Dipole moments Specific heat pressure

3. II. Chemical Equations: Symbols glucose Microscopic viewpoint: 1 molecule of glucose reacts with 6 molecules of oxygen to form 6 molecules of carbon dioxide and six molecules of water. Macroscopic viewpoint: Solid glucose reacts with gaseous oxygen to form gaseous carbon dioxide and gaseous water. The mass of all the glucose and oxygen is the same as the mass of all the carbon dioxide and water Clearly, a chemical reaction has microscopic scale information (molecule names and ratios), and macroscopic scale information (phases), but the majority of the important information (for us) is on the microscopic scale.

4. II. Chemical Equations: Conservation of mass Recall from chapter two that the law of mass conservation is intimately related to chemical equations. Since chemical equations contain information both macroscopic and microscopic, let’s see how the law of mass conservation plays on both levels. Microscopic Macroscopic Every atom of every element in the reactants must appear in the products. Atoms cannot be created or destroyed. The total mass of the reactants must be exactly the same as the total mass of the products. Mass cannot be created or destroyed. This is the origin of balancing reactions! link

5. III. Moles: the mathematical bridge While you already know the microscopic meaning of a chemical equation, historically they didn’t. Their microscopic information was severely limited. • They knew: • Under some conditions, 12 g carbon reacts with 16 g oxygen (making 28 g of product). • Under different conditions, 12 g carbon reacts with 32 g oxygen (making 44 g of product). • And thousands of similar reactions So they designed a relative scale (amu) and set one carbon atom at 12 amu and one oxygen atom at 16 amu. This bridges the macro/micro divide. But how many carbon atoms in one gram? What’s the relationship between amu and grams?

6. III. Moles: the mathematical bridge On one level, we don’t care the exact number of amu per gram. We can’t really count all the atoms and molecules anyway, so a relative scale is good enough—the number of amu in one gram is real, even if we don’t know it. It still bridges the microscopic and macroscopic divide. therefore, whatever number of atoms of carbon there are in 12 g of carbon, there are exactly that many atoms of oxygen in 16 g of oxygen. Because: • The mass (in grams) equal to the atomic weight (in amu) is called a molar mass. • The collection of atoms in one molar mass is called a mole. • The exact number of atoms in one mole is Avogadro’s number.

7. III. Moles: the mathematical bridge Because of the law of conservation of mass, we can simply add up the weights of the individual atoms to come up with the weight of the individual molecule (or one formula unit, for ionic compounds). And the same reasoning we used to create the molar weights of the elements works to create the molar weights of compounds as well. Since we will often use molar mass to convert between microscopic and macroscopic viewpoints, we will usually represent molar mass with units of grams per mole, Microscopic units Macroscopic units link

8. II. & III. Chemical reactions & moles A new view of the chemical equation can now be given: The “mole” unit is very much like the “dozen” unit, except, umm, a whole lot bigger. Remember that the coefficients are only ratios. If you double the moles of hydrogen and nitrogen, then you double the moles of ammonia. This will be VERY important in stoichiometry.

9. IV. Balancing chemical reactions The new, “bridging” view of the chemical reaction demands correct chemical coefficients. We need to know how to make them. This is called “balancing the reaction”.

10. IV. Balancing chemical reactions

11. IV. Balancing chemical reactions: table of chemical lovin’ Its usually easiest to start with elements other than C, H, and O. But they’re already balanced, so let’s start with C.

12. IV. Balancing chemical reactions: table of chemical lovin’ We could try to fix the O, but it doesn’t look like a whole number coefficient helps. Let’s fix hydrogen instead, and hope the oxygen problem goes away.

13. IV. Balancing chemical reactions: table of chemical lovin’ Now we have to face oxygen. We need to increase oxygen on the reactant side, where it appears in two places. In both cases the coefficient will affect two other elements. But in the first compound, the other two elements only appear together on the product side. Whereas in the second compound, the other two elements are in different products; that sounds more complicated.

14. IV. Balancing chemical reactions: table of chemical lovin’ We’re almost done. We just need to fix the KCl and we’re through.

15. IV. Balancing chemical reactions: table of chemical lovin’ Boo ya!

16. Two Hearts of Stoichiometry 3.4 – 3.6Stoichiometry II. I. Outputs Inputs II.

17. I. Stoichiometry Defined Stoichiometry is the mathematics of chemicals and chemical reactions. One “heart” of stoichiometry is the repeated bridging of the microscopic/macroscopic divide. We have only one mathematical concept that achieves this: the mole. Thus, a large part of stoichiometry is the manipulation of moles. The second “heart” of stoichiometry involves reactions only. Because a balanced reaction is a statement of ratios of the substances involved, we often develop microscopic knowledge of one component of the reaction and leverage it to find out something about a different component of the reaction. If in doubt, convert to moles.

18. I. Converting to/from moles We have two “facts” about moles: Avogadro’s number and molar masses. One relates to numbers of particles and the other relates to masses. Converting from numbers to moles: Converting from moles to numbers:

19. I. Converting to/from moles

20. I. Mole relationships within a compound • In one mole of H2O, there are one mole of oxygen atoms (one O atom per molecule) • In one mole of H2O, there are two moles of hydrogen atoms (two H atoms per molecule) For example:

21. I. Moles and reactions But remember, the coefficients represent ratios. So then if, instead of 3 moles of hydrogen, I actually have 19.4 moles of hydrogen. Then the reaction would consume 6.47 moles of nitrogen because:

22. I. Quick list • Massmoles • Molesnumbers • Numbersmass • Numbersmoles • Moles Amoles B (in balanced rxn) Therefore we can also do any of these in combination. To go from mass to numbers, convert mass to moles and then moles to numbers. But notice, pretty much any combination you devise has to go through moles. Moles are the linchpin to stoichiometry. If in doubt, convert something to moles!

23. II. Given the balanced reaction: 2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l) How many grams of NaOH are needed to react with 25.0 g of Cl2? How many molecules of NaOCl are produced from the above reaction?

24. Reaction Yields The reaction yields you calculate from the balanced reaction are called theoretical yields. It is the amount you are supposed to get. It is not usually the amount you actually get. You are clumsy (admit it). And there are usually other reactions going on that use up your reactants making other things. The actual yield is not something you get from theory. It is an experimental fact and can only be learned from doing calculations with numbers from the experiment. Often written as a percent Normal stoichiometry calculation

25. Limiting Reagents (I)molecular picture

26. Limiting Reagents (II) Balanced rxn To determine which reagent is the limiting reagent, divide the moles of each reagent by its stoichiometric coefficient. I call this the limited reagent lovin’ index (LRLI). Whichever LRLI is smallest is the limiting reagent. 2 C8H18 + 27 O2 16 CO2 + 18 H2O 3 moles 16 moles 3 moles octane / 2 = 1.5 16 moles oxygen / 27 = 0.59 smallest • Once you have the limiting reagent, do two things: • Base all stoichiometric calculations off of your limiting reagent. It determines your theoretical yields, etc. • Throw away all the LRLI. You won’t use it again.

27. Moles B Moles A Stoichiometry Possibilities % yield Volume A grams B grams A atoms/molecules A atoms/molecules A

28. % yield Volume A Actual yield density grams B grams A MW MW Balanced rxn Moles A Moles B Avog. # Avog. # atoms/molecules A atoms/molecules A Limiting Reagent

29. S Solutions • Solutions are homogeneous mixtures (think salt dissolved in water). • Mixtures do not have molecular weights, so “converting to moles” is different. • Molarity (M) measures the moles of solute (the salt) per liter of solvent (the water) • Molarity is a lot like a density. Instead of mass per volume its moles per volume.

30. % yield Volume A Actual yield grams B grams A MW MW Liters sol’n Molarity B Balanced rxn molarity Moles A Moles B Liters sol’n Avog. # Avog. # atoms/molecules A atoms/molecules A S Solutions (II) • Molarity serves the same stoich fxn as molar weights. That is, it is another way to convert to moles.

31. S How many mL of 0.300 M sulfuric acid is to be added to 50.0 mL of 0.250 M barium hydroxide solution to complete the following balanced reaction: • Convert to moles • Moles B  Moles A • Moles A  volume A =0.042 L = 42 mL

32. S Solutions (III) • While the solute is primarily what we use in solution stoichiometry, the amount of solvent matters too. • Increasing the amount of solvent is called “diluting” the solution. • The mathematics of dilution is quite simple: “i” means inital “f” means final molarity volume

33. What volume of 18.0 M H2SO4 is required to prepare 250.0 mL of 0.500 M aqueous H2SO4? Re-solve equation for the unknown and substitute in the values:

34. Describe how you would prepare 2.10 L of a 0.360 M solution of glacial acetic acid starting with concentrated sulfuric acid that is 12.5 M. After extracting the 60.5 mL of glacial acetic acid, you would then dilute until you have 2.10 L, in other words you would add approximately 2.04 L.

35. Titration • Lab technique—details don’t matter for us • Just stoichiometry—nothing new for us % yield Volume A Actual yield grams B grams A MW MW Liters sol’n Molarity B Balanced rxn molarity Moles A Moles B Liters sol’n Avog. # Avog. # atoms/molecules A atoms/molecules A

36. Titration example • What is the molarity of a sulfuric acid solution if a 25.0 mL sample is titrated to equivalence with 50.0 mL of 0.150 M sodium hydroxide solution? • H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) Titration-speak for “reacted with” Convert to molarity (check it out) Convert NaOH to moles Moles NaOH to moles sulf. acid

37. The reaction between sulfuric acid and potassium hydroxide can be described by the equation Matt will write on a side board. A 25.00 mL sample of sulfuric acid requires 42.61 mL of of a 0.1500 M potassium hydroxide solution to titrate to its end point. Calculate the concentration of the sulfuric acid solution.

38. Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage. Percent Composition

39. Percent Composition example Saccharin: C7H5NO3S FW = 183.18

40. Percent Composition Backwards • Start with percents, and work back to mole ratios • Percents are independent of mass, so choose a convenient mass (100 g is a good choice) A compound has the following mass percents: 27.931% Fe, 24.057% S, and 48.012% O. What is its formula? Assume 100 g of total mass, then percents of each element turn into masses of each element Now convert to moles to get the mole ratio The lowest whole number ratios emerging from these three numbers are: 2 Fe, 3 S, and 12 O. The formula from this is Fe2S3O12, which can be written more informatively as Fe2(SO4)3.

41. Mass Spectrometer

42. Mass Spectrograph

43. Formulas • The percent composition technique just given finds the lowest whole number ratio of atoms in the formula. This formula is called the empirical formula. • The following compounds have the same empirical formula (CH): acetylene, benzene. They have different molecular formulas. • Mass spectrometry can find a molar weight (but gives no info on what elements are in the compound) So percent composition and mass spectrometry combined can find the molecular formula A compound with 92.3% C and 7.74% H was subjected to mass spectrometry and found to have a molecular weight of 26.038 amu. What is its molecular formula? • Use percent comp. to get moles of each. • Construct empirical formula • Use mass spec. number to get molecular formula. This is a 1:1 ratio. The empirical formula CH has a molecular mass of 13.019 amu. 26.038amu/13.019amu = 2

45. The compound most responsible for the odor of garlic has the empirical formula C3H5S2. What is its percent composition? Mass spectrographic analysis shows its molar weight to be 146 g/mol. What is its molecular formula?

46. Combustion Analysis • Combustion analysis is one of the most common methods for determining empirical formulas. • A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram. • It is particularly useful for analysis of hydrocarbons. (C, H, O) How do you get oxygen? Subtract the mass of carbon and hydrogen from the sample mass. The rest of the mass is oxygen. The CO2 and water will weigh more together than the sample, because some of the oxygen came from the air, adding to the mass!

47. Combustion Analysis Example Ascorbic acid is known to contain only C, H, and O. A 6.49 mg sample was subjected to combustion analysis which produced 9.74 mg CO2 and 2.64 mg H2O. What is the empirical formula of ascorbic acid? • Get the masses of carbon and hydrogen • Get the mass of oxygen • Get the moles of each component • Construct an empirical formula. Putz around to find the ratio. In this case the C:H:O ratio is: This gives an empirical formula of C3H4O3.

48. Combustion analysis of an unknown compound containing carbon and hydrogen produced 4.544 g of carbon dioxide and 2.322 g of water. What is the empirical formula of the compound? C2H5