1. + Mole and Stoichiometry HW: Read CH 6 and CH 9

2. + Measuring Matter A mole is an amount of a substance Similar to: 12 in a dozen, 100 pennies in a dollar, etc. A mole of any substance represents 6.02 x 10 23 particles of that substance 6.02 x 10 23 = Avogadro’s Number 602000000000000000000000 6.02 x 10 23 X atoms OR1 mol X 1 mol X 6.02 x 10 23 X atoms NOTE: the words atoms/molecules/particles are all the same

3. + IMPORTANT HINT!!! When you do calculations, you cancel the unwanted units (divide by unwanted units)

4. + Problem 1: If you had 1 mole of donuts, how many donuts would you have? 6.02 x 10 23 Problem 2: In 4.0 moles of glucose (C 6 H 12 O 6 ), how many molecules/particles are there? 4.0 mol glucose x 6.02 x 10 23 = 2.4 x 10 24

5. + Molar Mass (formula mass) Must know how to determine molar mass (formula mass) Molar mass measures the mass of one mole of any element or compound The molar mass can be calculated using the atomic masses of each element (1 amu = 1 gram) from the periodic table 1 bromine atom = 79.90 amu Molar mass can be used to convert grams of a substance into moles or vice versa

6. + Calculating the Molar Mass Molar Mass: add up all the amu values (make sure you look at subscripts and coefficients) What is the molar mass of 1 mole of H 2 O H: 2 x 1.01 = 2.02 O: 1 x 16.00 = 16.00 1 mole H 2 O = 18.02 g What is the molar mass of 1 mole of Cu(OH) 2 Cu: 1 x 63.55 = 63.55 O: 2 x 16.00 = 32.00 H: 2 x 1.01 = 2.02 1 mole Cu(OH) 2 = 97.57 g

7. + YOU TRY! Find the molar mass of the following: A) H 3 PO 4 B) N 2 O 5 (don’t forget to make sure neutral!) C) copper (II) nitrate D) potassium hydroxide

8. + Calculations with mass and grams (amu) Mass of sample x conversion factor = number of moles in sample Calculate the mass of a sample that contains 23 moles of nitrogen (given: 23 moles N, unknown: N mass) 1 mol N= 14.01 grams 23 mole N x 14.01 grams= 322.23 grams = 1 mol N320g Calculate the number of oxygen moles present in a sample that has a mass of 288 grams. 1 O mol = 16.00 grams 288 grams x 1 mol O= 18 O moles 16 grams

9. + Calculations with moles and particles particles x conversion factor = number of moles Chromium is a metal that is added to steel to improve its resistance to corrosion. Calculate the number of moles in a sample of chromium containing 5.00 x 10 20 atoms. 5.00 x 10 20 atoms Cr x 1 mol = 8.31 x 10 -4 mol Cr 6.02x10 23 atoms Cr

10. + Calculations with grams to particles grams x conversion factor = moles; moles x conversion factor = particles Calculate the number of salt molecules present in a sample of 63.2 grams of salt 63.2 g NaClx1mol NaCl= 1.0815 mol NaCl 58.44 g NaCl 1.0815 mol NaClx 6.02 x 10 23 molecules= 6.5103 x 10 23 1 mol NaCl= 6.51 x 10 23 molecules

11. + More molar mass problems YOU TRY! A) How many moles are in 25 grams of water? B) How many grams are in 3.2 moles of C 12 H 22 O 11 ?

12. + More molar mass problems YOU TRY! C) How many molecules are in 2.32 moles of HNO 3 ? D) How many moles are in 3.60 x 10 24 molecules of Li 2 O? E) Calculate the number of iron (Fe) atoms present in a sample that has a mass of 4021.2 grams. (two steps!)

13. + Percent by Mass Composition When given a sample of a compound, you can determine how much of each element is present Percent composition is the percent by mass of each element in a compound The following formula is used to calculate percent composition: % Mass of Element = (grams of element/grams of compound) x 100%

14. + Percent Composition Example: Calculate the percentage of carbon in ethane, C 2 H 6 grams of carbon = 2 x 12 = 24 grams grams of compound = 24 + (6 x 1) = 30 grams (24 /30) x 100 = 80% Example: Find the percentage composition of a compound that contains 2.7369 g of chlorine, 0.4116 g of oxygen, and 0.7971 g of phosphorus in a 3.9460 g sample of the compound Cl (2.7369 g/3.9460 g) x 100 = 69.4% O (0.4116 g/3.9460 g) x 100 = 10.4% P (0.7971 g/3.9460 g) x 100 = 20.2%

15. + Percent Composition Example: A sample of an unknown compound with mass of 2.876 g has the following composition: 66.07% carbon, 6.71% hydrogen, 4.06% nitrogen, 23.16% oxygen. What is the mass of each element? Equation: (% Mass x grams compound) *you can use other methods 100% C (66.07x2.876 g)/100 = 1.90g H (6.71x2.876 g)/100 =.193g N (4.06x2.876 g)/100 =.117g O (23.16x2.876 g)/100 =.66g

16. + YOU TRY! A) Calculate the percent by mass composition of dimethylether, CH 3 OCH 3 B) What is the percent composition in each element of Iron (III) oxide?

17. + Empirical Formula The empirical formula gives the simplest whole number ratio of the elements in a compound Example: empirical formula for hydrogen peroxide, H 2 O 2, is HO. The ratio is 1:1 To determine: 1) convert mass of each element given into moles using molar mass (from periodic table) 2) divide each mole value by the smallest # of moles 3) round the mole ratio of the elements to the nearest whole # (NOTE: if something rounds to.5 then you have to multiply everything by 2)

18. + Empirical Formula Example: A compound was found to contain 7.30 g Na, 5.08 g S, and 7.62 g O. Determine the empirical formula.  7.30 g Na x 1 mol Na =.327 mol Na5.08 g S x 1 mol S =.158 mol S 23.0 g Na 32.1 g S  7.62 g O x 1 mol O =.476 mol O 16.0 g O SMALLEST # MOLES =.158  Na.327/.158 = 2.1S.158/.158 = 1.0  O.476/.158 = 3.0  Answer: Na 2 SO 3 (Also accept: Na 2 O 3 S)

19. + Empirical Formula Example: Determine the empirical formula of a compound containing 20.23% aluminum and 79.77% chlorine  Al 20.23%: 20.23 g Al x 1 mol Al =.750 26.98 g Al  Cl 79.77%: 79.77 g Cl x 1 mol Cl = 2.25 35.45 g Cl SMALLEST # MOLES =.750  Al.750/.750 = 1.0  Cl 2.25/.750 = 3.0  Answer: AlCl 3

20. + Empirical Formula YOU TRY! A hydrocarbon (a compound containing only hydrogen & carbon) is found to be 7.690% H and 92.31% C by mass. Calculate its empirical formula. A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 g chromium, and 72.67 g oxygen. What is the compound’s empirical formula?

21. + Molecular Formula The actual number of atoms in a compound A whole-number multiple of the empirical formula CH 2 O; C 6 H 12 O 6 ; C 12 H 24 O 12 First find the empirical formula, then divide the molar mass of the entire compound by the mass of the empirical formula to determine the multiple

22. + Molecular Formula Example: Ribose has a molar mass of 150 g/mol and a chemical composition of 40.0% carbon, 6.67% hydrogen, and 53.3% oxygen. What is the molecular formula for ribose?  C: 40.0 gC x 1 mol C = 3.33 mol C H: 6.67 gH x 1 mol H = 6.60 mol H 12.0 g C 1.01 g H  O: 53.3 gO x 1 mol O = 3.33 mol O 16.0 g O Empirical Formula= C 1 H 2 O 1 empirical formula mass 30.0 g/mol (12+1+1+16)  molar mass = 150 g/mol  molecular formula = 150 / 30 = 5 x empirical C 5 H 10 O 5

23. + Molecular Formula YOU TRY! A) A hydrocarbon (a compound containing only hydrogen & carbon) is found to be 7.690% H and 92.31% C by mass. The molar mass is 78.12 g/mol. Calculate its molecular formula (use the previous empirical formula of CH).

24. + Balancing Chemical Equations Balancing equations is done to satisfy the Law of Conservation of Mass to make the # of atoms the same on both sides of the equation  Add coefficients but leave the subscripts alone Step by Step Instructions: 1. Draw a vertical line down from below the arrow in the equation. 2. On EACH side of the line, list all of the elements that appear. Do not repeat elements on the same side. 3. For EACH side of the line, count the number of atoms for each element. 4. Add coefficients in front of formulas so that the number of atoms on each side matches. 5. Once all element counts match, make sure all coefficients are reduced to their lowest whole-number ratio.

25. + Balancing equations This process in rather unscientific and is essentially a process of trial and error TIP #1: If an element appears in only one compound on each side of the equation, try balancing that first. TIP #2: If one of the reactants or products appears as the free element, try balancing that last. TIP #3: Remember that you can only change the coefficient. NOTE: coefficients should be whole numbers

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27. + YOU TRY! Balance the following: A) Fe + Br 2  FeBr 3 B) C 4 H 8 + O 2  CO 2 + H 2 O C) Pb(NO 3 ) 2  PbO + NO 2 + O 2

28. + A Typical Plan for Solving Stoichiometry Problems between different compounds (There is a basic pattern to all stoichiometry problems, with variations depending on what information is given and what questions must be answered.) A. You must start with a balanced equation. B. Convert the units of any starting substances into moles. Since the molar ratio is your stoichiometric link between substances (the coefficients from the balanced equation) and they are in moles, you must work the problem in moles. C. Reread the problem to determine the information that you need to calculate. Use the molar ratios of the relevant substances to convert from a known substance to a desired substance that you need to answer the question. D. Note that the molar ratio is set up with the known substance on the bottom (so it will cancel out) and with the desired substance on the top. E. If necessary, convert any answers back into grams.

29. + Mole Ratios Balanced equation: C 3 H 8 +5O 2  3CO 2 + 4H 2 O is really 1 mol C 3 H 8 +5 mol O 2 yields 3 mol CO 2 + 4 mol H 2 O What number of moles of CO 2 will be produced by the decomposition of 3.5 mol of C 3 H 8 ? Unknown = moles of CO 2 Given = 3.5 moles C 3 H 8 Also, 1 mol C 3 H 8 = 3 mol CO 2 3.5 mol C 3 H 8 x 3 mol CO 2 = 10.5 mol CO 2 = 11 mol 1 mol C 3 H 8

30. + Another example Solid iron reacts with oxygen to produce iron(III) oxide. If you wanted to produce 2.5 moles of iron(III) oxide, what mass of oxygen must react? 4Fe + 3O 2  2Fe 2 O 3 Unknown = Grams of O 2 Given = 2.5 Moles Fe 2 O 3 2.5 moles Fe 2 O 3 3 mol O 2 32 grams O 2 2 mol Fe 2 O 3 1 mol O 2 Answer: 120 grams O 2 (2.5 x 3 x 32) / 2

31. + Mole Ratios YOU TRY! CH 4 + 2O 2  CO 2 + 2H 2 O A) Calculate the mass of O 2 required to produce 2.23 g of carbon dioxide. B) Calculate the mass of water produced when 34.0 g of CH 4 is burned.

32. + Limiting Reactants The reactant that runs out first and thus limits the amounts of products that can form is called the limiting reactant (or limiting reagent)

33. + Limiting Reactant problem-solving plan For limiting reactant problems, the problem will give you information about two reactants A. Start with a balanced equation. B. First determine which reactant will run out (which reactant is the limiting reactant) 1) You must change all starting masses into moles. 2) The trick to determine which reactant limits is to divide the moles of each reactant by the coefficient (from the balanced equation) associated with that reactant. The number that comes out the smallest indicates which reactant is the limiting one. The limiting reactant is the one that you must base all your other calculations on because it is the substance that limits how much of everything else can be made or is needed. C. The other reactant (if there’s only two) will be the excess reactant, and some of it will be left over. **Use molar ratios to convert the moles of the known limiting substance to the moles of other compounds (use coefficients from balanced equation).

34. + Limiting Reactant problem-solving plan For limiting reactant problems, the problem will give you information about two reactants A. Start with a balanced equation. B. First determine which reactant will run out (which reactant is the limiting reactant) 1) You must change all starting masses into moles. 2) The trick to determine which reactant limits is to divide the moles of each reactant by the coefficient (from the balanced equation) associated with that reactant. The number that comes out the smallest indicates which reactant is the limiting one. The limiting reactant is the one that you must base all your other calculations on because it is the substance that limits how much of everything else can be made or is needed. C. The other reactant (if there’s only two) will be the excess reactant, and some of it will be left over. **Use molar ratios to convert the moles of the known limiting substance to the moles of other compounds (use coefficients from balanced equation).

35. + Limiting Reactant problem-solving plan For limiting reactant problems, the problem will give you information about two reactants A. Start with a balanced equation. B. First determine which reactant will run out (which reactant is the limiting reactant) 1) You must change all starting masses into moles. 2) The trick to determine which reactant limits is to divide the moles of each reactant by the coefficient (from the balanced equation) associated with that reactant. The number that comes out the smallest indicates which reactant is the limiting one. The limiting reactant is the one that you must base all your other calculations on because it is the substance that limits how much of everything else can be made or is needed. C. The other reactant (if there’s only two) will be the excess reactant, and some of it will be left over. **Use molar ratios to convert the moles of the known limiting substance to the moles of other compounds (use coefficients from balanced equation).

36. + Limiting Reactant problem-solving plan For limiting reactant problems, the problem will give you information about two reactants A. Start with a balanced equation. B. First determine which reactant will run out (which reactant is the limiting reactant) 1) You must change all starting masses into moles. 2) The trick to determine which reactant limits is to divide the moles of each reactant by the coefficient (from the balanced equation) associated with that reactant. The number that comes out the smallest indicates which reactant is the limiting one. The limiting reactant is the one that you must base all your other calculations on because it is the substance that limits how much of everything else can be made or is needed. C. The other reactant (if there’s only two) will be the excess reactant, and some of it will be left over. **Use molar ratios to convert the moles of the known limiting substance to the moles of other compounds (use coefficients from balanced equation).

37. + Limiting Reactants Fe + CuSO 4  FeSO 4 + Cu If 120. g of Fe are reacted with 200. g of CuSO 4, identify the limiting reagent. Which reagent is in excess? 120 g Fe x 1 mol Fe= 2.15 mol Fe / 1 = 2.15 (excess) 55.85 g Fe 200 g CuSO 4 x 1 mol CuSO 4 = 1.25 mol CuSO 4 /1 = 1.25 (limiting) 159.62 g CuSO 4

38. + Limiting Reactant If you are asked to determine the amount of excess reactant leftover: A. Knowing which reactant limits and which is excess, use the limiting reactant to determine the moles of the excess reactant that is actually needed to do the reaction. Subtract. B. Convert these moles into grams.

39. + Limiting Reactants Fe + CuSO 4  FeSO 4 + Cu If 120. g of Fe are reacted with 200. g of CuSO 4, identify the limiting reagent. Which reagent is in excess? Calculate the mass of Copper formed. How much of the excess reagent is left over at the end of the reaction? 120 g Fe x 1 mol Fe= 2.15 mol Fe (excess) 55.85 g Fe 200 g CuSO 4 x 1 mol CuSO 4 = 1.25 mol CuSO 4 (limiting) 159.62 g CuSO 4 1.25 mol CuSO 4 x 1 mol Cu x 63.55 g Cu = 79.4 g Cu 1 mol CuSO 4 1 mol Cu Out of 2.15 mol Fe, only 1.25 mol is used so 0.9 mol Fe in excess 0.9 mol Fe58.85 g Fe = 53.0 grams in excess 1 mol Fe

40. + Limiting Reactants YOU TRY! (Ouch! There is LOTS to remember) Which is the limiting reactant? Calculate the mass of lithium nitride formed from 56.0 g of N 2 gas and 56.0 g of Li in the unbalanced reaction: Li + N 2  Li 3 N

41. + Percent Yield Percent yield = actual yield x 100% theoretical yield In previous example 93.7 g Li 3 N should have formed. If this reaction actually gave 55.2 g Li 3 N, what is the percent yield of Li 3 N? 55.2 g Li 3 N x 100 = 58.9% 93.7 g Li 3 N

42. + Percent Yield YOU TRY! Good Luck! TiCl 4 + O 2  TiO 2 2Cl 2 Suppose that 6.71 x 10 3 g of TiCl 4 is reacted with 2.45 x 10 3 g of O 2. Calculate the maximum mass of TiO 2 that can form. If the percent yield of TiO 2 is 75%, what mass is actually formed?