1. AP CHEMISTRY CHAPTER 3 STOICHIOMETRY

2. Average atomic mass- weighted average based on isotopic composition To calculate : % Isotope A (mass of A) + % Isotope B (mass of B) + …= avg. atomic mass

3. What is the average atomic mass in grams of lithium if 7.42% exists as 6 Li (6.015 g/mol) and 92.58% exists as 7 Li (7.016 g/mol)? (0.0742)(6.015) + (0.9258)(7.016) = ? 6.94 g/mol Don’t forget to change percents to decimals! Don’t divide by the number of different isotopes! Don’t round until the end. Your answer must be between the masses of the individual isotopes. Check!

4. A mass spectrometer is an instrument used to determine the relative masses of atoms by the deflection of their ions in a magnetic field. One use is to determine the isotopic abundance of a sample of an element. Mass Spectrometer Samples are vaporized and ionized. The ions are then separated by mass and the data is graphed.

5. Figure 3.2 Neon Gas These show the Mass spec data for samples of neon. Using this data, what would the approximate average atomic mass of neon be?

6. Figure 3.3 Mass Spectrum of Natural Copper

7. What element results in this mass spec data?

8. Mole -the # of C atoms in 12g of pure carbon-12 Avogadro’s Number = 6.022 × 10 23 The mass of one mole of an element is equal to its atomic mass in grams.

9. Cody found a gold nugget that had a mass of 1.250 oz. How many moles was this? How many atoms? (1 lb = 16 oz, 453.59g = 1 lb) 1.250 oz Au 1 lb 453.59g 1 mol Au 16 oz 1 lb 196.97g Au = 0.1799 moles 0.1799 moles Au 6.022 × 10 23 atoms 1 mol Au = 1.083 × 10 23 atoms

10. Molar Mass = mass in grams of one mole of a substance

11. Calculate the molecular mass of cisplantin, Pt(NH 3 ) 2 Cl 2. Pt 1 × 195.08 = 195.08 N 2 × 14.01 = 28.02 H 6 × 1.01 = 6.06 Cl 2 × 35.45 = 70.90 300.06 g

12. How many grams are 3.25 moles of cisplantin? 3.25 mol cisplantin 300.06g 1 mol cisplantin = 975g

13. Percent Composition -“mass percent” or “percent by mass” Total mass of element × 100 = % comp of element Total mass of compound

14. Find the percent composition of all elements in cisplantin, Pt(NH 3 ) 2 Cl 2. Pt 195.08 × 100 = 65.01% 300.06 N 28.02 × 100 = 9.34% 300.06 H 6.06 × 100 = 2.02% 300.06 Cl 70.90 × 100 = 23.63% 300.06

15. Determining the Formula of a Compound Empirical formula- simplest whole number ratio of the various types of atoms in a compound Molecular formula - the exact formula (empirical formula) x

16. A sample of a compound contains 11.66g of iron and 5.01g of oxygen. What is the empirical formula of this compound? 11.66g Fe 1 mol Fe = 0.2088 mol Fe 55.85g Fe 5.01g O 1 mol O = 0.3131 mol O 16.00g O Fe:O 0.2088 : 0.3131 1:1.5 or 2:3 0.2088 0.2088 Fe 2 O 3

17. What is the empirical formula of hydrazine, which contains 87.5% N and 12.5%H? 87.5gN 1 mol N = 6.246 mol N 14.01g N 12.5g H 1 mol H = 12.38 mol H 1.01 g H N:H = 6.246 : 12.38 1 : 1.98 = 1 : 2 6.246 6.246 NH 2

18. Combustion Analysis Additional O 2 is added to burn the sample of a compound containing C, H, and sometimes O. All of the H 2 O is absorbed in the first chamber and all of the CO 2 produced is absorbed in the second chamber. Increases in the masses of the two chambers are used to determine the mass of CO 2 and H 2 O produced.

19. Suppose you isolate an acid from clover leaves and know that it contains only the elements C, H, and O. Heating 0.513g of the acid in oxygen produces 0.501g of CO 2 and 0.103g of H 2 O. What is the empirical formula of the acid? Given that another experiment has shown that the molar mass of the acid is 90.04g/mol, what is its molecular formula? C + O 2  CO 2 0.501g CO 2 1 mol CO 2 1 mol C 12.01gC 44.01g CO 2 1 mol CO 2 1 mol C = 0.1367 g C in compound

20. 2H 2 + O 2  2H 2 O 0.103g H 2 O 1 mol H 2 O 2 mol H 1.01g H 18.02g H 2 O 1 mol H 2 O 1 mol H = 0.01155 g H 0.513g cmpd  (0.137 + 0.012)= 0.364g O 0.364g O 1 mol O = 0.02275 mol O 16.00g O 0.137g C 1 mol C = 0.01141 mol C 12.01g C 0.01155g H 1 mol H = 0.01144 mol H 1.01 g H

21. C:H:O 0.01141: 0.01144 : 0.02275 0.01141: 0.01144 : 0.02275 0.01141 0.01141 0.01141 1: 1: 1.98 1:1:2 CHO 2 efm = 12.01 + 1.01 + 32.00 = 45.02 90.04/45.02 = 2 C2H2O4C2H2O4

22. Chemical Equations Reactants  Products Yields A balanced equation must have the same # of atoms of each element on each side. Symbols representing physical states: (s) (l) (g) (aq)

23. Balancing Chemical Equations We balance equations by adding coefficients, never by changing formulas. Most equations can be balanced by inspection. Some redox reactions require a different method.

24. Ex. Al(s) + Cl 2 (g)  AlCl 3 (s) 2Al(s) + 3 Cl 2 (g)  2AlCl 3 (s) N 2 O 5 (s) + H 2 O(l)  HNO 3 (l) N 2 O 5 (s) + H 2 O(l)  2 HNO 3 (l) Pb(NO 3 ) 2 (aq) + NaCl(aq)  PbCl 2 (s) + NaNO 3 (aq) Pb(NO 3 ) 2 (aq) + 2 NaCl(aq)  PbCl 2 (s) + 2 NaNO 3 (aq)

25. 2PH 3 (g) + 4O 2 (g)  3H 2 O(g) + P 2 O 5 (s) Phosphine, PH 3 (g) is combusted in air to form gaseous water and solid diphosphorus pentoxide.

26. 2NH 3 (g) + 2Na(l)  H 2 (g) + 2NaNH 2 (s) When ammonia gas is passed over hot liquid sodium metal, hydrogen is released and sodium amide, NaNH 2, is formed as a solid product.

27. Stoichiometric Calculations Mole ratio = moles required/moles given

28. What mass of NH 3 is formed when 5.38g of Li 3 N reacts with water according to the equation: Li 3 N(s) + 3H 2 O  3LiOH(s) + NH 3 (g)? 5.38g Li 3 N 1 mol Li 3 N 1 mol NH 3 17.04g NH 3 34.83g Li 3 N 1 mol Li 3 N 1 mol NH 3 = 2.63g NH 3

29. Limiting reagent- reagent that limits or determines the amount of product that can be formed Animation If you are given amounts of two or more reactants in a stoichiometry problem and asked to determine how much product forms, the easiest thing to do is to work a problem with each reactant and take the smaller of the answers.Animation

30. You work in a sandwich shop. You MUST make sandwiches according to the following formula: You use two pieces of bread, 4 pieces of meat, 2 pieces of cheese, 3 pieces of bacon and 2 ounces of special sauce. If you have 20 pieces of bread, 32 pieces of meat, 20 pieces of cheese, 21 pieces of bacon and one quart of special sauce, how many sandwiches can you make?

31. X + 2Y  XY 2 X Y XY 2 Before reaction After reaction Assume that the reaction goes to completion. Draw the resulting particles in the right-hand box. What is the limiting reagent?

32. X + 2Y  XY 2 X Y XY 2 Before reaction After reaction Assume that the reaction goes to completion. Draw the resulting particles in the right-hand box. What is the limiting reagent?

33. How many moles of Fe(OH) 3 (s) can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O and 3.0 mol O 2 to react? 2Fe 2 S 3 (s) + 6H 2 O(l) + 3O 2 (g)  4Fe(OH) 3 (s) +6S(s) 1.0 mol 2.0 mol 3.0 mol limiting 2.0 mol H 2 O 4 mol Fe(OH) 3 6 mol H 2 O = 1.3 mol Fe(OH) 3

34. If 17.0g of NH 3 (g) were reacted with 32.0g of oxygen in the following reaction, how many grams of NO(g) would be formed? 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(l) 17.0g 32.0g Xg 17.0g NH 3 1 mol NH 3 4 mol NO 30.01 g NO = 29.9g NO 17.04g NH 3 4 mol NH 3 1 mol NO 32.0 g O 2 1 mol O 2 4 mol NO 30.01 g NO = 24.0g NO 32.00g O 2 5 mol O 2 1 mol NO 24.0g is the smaller yield, so O 2 is limiting and 24.0 g is the correct answer.

35. Theoretical yield- amount of product that should form according to stoichiometric calculations Actual yield- experimental yield Percent yield = actual yield × 100 Theoretical yield

36. In the reaction of 1.00 mol of CH 4 with an excess of Cl 2, 83.5g of CCl 4 is obtained. What is the theoretical yield, actual yield and % yield? Actual yield = 83.5g CCl 4 CH 4 + 2Cl 2  CCl 4 + 2H 2 1.00 mol CH 4 1 mol CCl 4 153.81g CCl 4 1 mol CH 4 1 mol CCl 4 = 154g CCl 4 (theoretical yield) 83.5 × 100 = 54.2% yield 154